The Dancing 无理 Masters

Magic Square: Lo Shu Metric!

2011-12-11T21:29:00.001-08:00

Let us indulge in the topic of Magic Squares today – simply put, it is a square grid into which whole numbers are placed. The whole numbers themselves need not follow any order but they must be distinct, and placed in such a way that each horizontal row, vertical column and diagonal add up to the same number.

Let us illustrate by means of an example:

A 3 x 3 'Lo Shu' Square

Observe:

1st Row: 8 + 1 + 6 = 15

2nd Row: 3 + 5 + 7 = 15

3rd Row: 4 + 9 + 2 = 15

1st Column: 8 + 5 + 2 = 15

2nd Column: 1 + 5 + 9 = 15

3rd Column: 6 + 7 + 2 = 15

1st Diagonal: 8 + 5 + 2 = 15

2nd Diagonal: 6 + 7 + 2 = 15

This magic square is known as the 'Lo Shu' square and was known in China around 3000 BC. Legend says that it was first seen on the back of a turtle emerging from the Lo river - the natives then took it as a sign from the gods that they would not be freed of pestilence unless they increased their offerings.

But really, is there a systematic way of generating such magic squares? Indeed there is! There are several methods, of which the likes of trial and error I shall not mention, haha. Suppose we go by a logical algebraic method:

1) For a 3 x 3 square, there will be 9 numbers, whose sum is given by:

S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Since each row must add up to the same number, dividing this by 3 means that each row/column/diagonal must add up to 15.

2) Let us now consider the central cell and label it x:

The Blank Square

We know that each row, column and diagonal must add up to 15, and so if we add up the middle row, middle column and both diagonals, we have them adding up to 60:

15 + 15 + 15 + 15 = 60

Now, pause for a moment - if you take the sum of the middle row, middle column and both the diagonals, that's equivalent to taking S + 3x = 60 because you're adding all the numbers from 1 through 9 once, and then adding them to 3x.

Since S = 45, then surely:

3x = 15 and therefore x = 5

And of course, we can carry on ad infinitum, ad nausea, via a system of simultaneous equations that can be solved for every single element in the grid we see above.

But surely, surely there must be an easier method!

Well, guess what? There is one! Simon de la Loubere, the French ambassador to the King of Siam in the late 17th century, wrote down an algorithm for constructing magic squares that have an odd number of rows and columns:

1) Place a '1' in the middle of the first row.

2) Go up one cell and then one cell to the right (order doesn't matter) to place successive numbers.

3) If the cell is blocked, the successive number should be placed beneath the current number.

Let's illustrate this awesome rule:

The 'Loubere' Algorithm

And there you have it:

A 3 x 3 'Lo Shu' Square

Using this algorithm, you can easily create any magic square that has an odd number of rows and columns. :)

Pendulum Wave

2011-12-10T06:24:00.000-08:00

It's been about 2 years since I last posted anything, so here goes:

Suppose for a minute there you believe me that the angular frequency of a pendulum is given by:

Where L is the length of the pendulum and g is the gravitational field strength. Well, none of that matters really, suffice to say that the longer the pendulum, the faster the oscillation for a given gravitational field.

In the video below, you’ll see that the pendulum bobs with a longer string length swing with a lower frequency than those with shorter string lengths, in accordance with the equation above.

What do you notice? Notice how the pendulum bobs start to swing in a synchronized fashion and then gradually start to lose their synchronization. This generates the nice pattern we see. Can we apply any quantitative explanation to this nice pattern of synchronization we see? Yes we can!

Now, suppose you now believe me (again) that the pendulum swings in such a way that when you follow the horizontal component of its displacement from equilibrium, it traces out a sinusoidal curve with respect to time as follows (I’ve left out all axes for simplicity):

Now imagine you have many different pendulums, all of different lengths – each pendulum bob differs from its preceding and successive pendulum bob by just a small change in its length. Then you you would expect the sinusoidal curves traced out by each bob to be slightly different due to different frequencies, as such:

The sinusoidal curves are all of different frequencies, though differing only slightly from one another – occasionally, one might expect the pendulum bobs to swing in phase and synchronize with one another, but most of the time, the pendulum bobs swing out of phase and are asynchronous due to their different frequencies.

So what does your eye see when you superpose all of the pendulum motions together? Well, essentially, there will be a period of time when the pendulum bobs all seem to be swinging in sync and then because of their different frequencies, they slowly start to swing out of sync, and then slowly again, they swing in sync. So if you add up all of the sinusoidal curves for a representative interpretation, you obtain this:

This is known as a wavepacket – the amplitude (how high the wave is) represents how coherent, or how in sync the individual pendulums are with one another. In the central portion, the amplitude is high, meaning the pendulums are almost swinging in sync. As time progresses, the pendulums swing out of sync, and you see the height of the wave packet decreases.

The motion is however, periodic, so you see the same patterns repeating over and over again, though with decreased amplitudes due to loss in kinetic energy.

That’s essentially it, but I haven’t made any additional effort to correctly quantify the patterns that we see. So sorry!

Ass'N Tool (Pun Intended!)

2009-09-07T11:13:00.000-07:00

Well, so we all know (or do we? Haha!) that the famous SN2 reaction proceeds via a 'backside' attack.

Yep, you got that right - when the nucleophile attacks an alkyl halide, it hits it right on its gluteus maximus (actually, I see it from another angle, but never mind).

And of course the age-old question: why? Why in particular, the 'backside'? Well, everything will be made clear with Molecular Orbital theory, so let us consider what the LUMO (Lowest Unoccupied Molecular Orbital) of a typical alkyl halide (in this case, chloromethane) looks like:

Notice the huge lobe on the 'backside' of the carbon atom? That's where the nucleophile attacks if if wants to bond to the carbon atom - by donating electrons into this antibonding orbital, the carbon-chlorine sigma bond effectively breaks, and a SN2 reaction ensues.

Ah well. I probably haven't explained in detail enough. But hey, I'm tired. :)

Recollection

2009-09-01T16:51:00.000-07:00

I know I've asked this before, but it keeps coming back to me, again and again:

Does a proton possess orbitals?

If you say I'm silly, then consider the thousands of Chemists worldwide who speak of the hydrogen ion (H+) as having orbitals.

Isn't a hydrogen ion a proton?

But isn't a proton a free particle by itself?

Quantum mechanics is just plain weird, period. :)

Conquering the FORTress

2009-08-25T16:33:00.000-07:00

Mr Yeo has successfully written his quadratic equation solver program, so now all you people out there can use it for their homework. :)

And that's the first step to conquering Fortran! :)

Unfortunately, Yahoo Geocities is closing its server this October, so I shan't bother to post it up; will probably do so when I've found another host. :)

So many smilies. Yay! :)

Water Me Down!

2009-08-24T13:00:00.001-07:00

You know, I’ve always wanted to construct a Walsh Diagram for a simple molecule? So what exactly is a Walsh diagram? Well, let me illustrate using water as an example; one should know that water has an equilibrium geometry that is bent, with a H-O-H angle of 104.5 degrees as such:

And of course, the other limiting (non-equilibrium) geometry would be a linear geometry with a H-O-H angle of 180 degrees as follows:

So now the question is, why is the former geometry adopted by water molecule? Phrasing this question another way: is it such that the energy of the water molecule is lower when it has a H-O-H bond angle of 104.5 degrees?

Well, the easiest way to find out whether this is true or not (not so much why it is true!) is to plot a graph of the energy of the water molecule against the H-O-H bond angle.

Now usually you wouldn’t be able to calculate the total electronic energy of the water molecule by hand, but since I’m bundled up with this molecular calculation suite called Gaussian03 (yes, I know the latest version is G09!), I’ve performed the calculations using my trusty laptop.

So here’s the graph, or like how they call it, the Walsh Diagram for the water molecule (click to enlarge):

This graph actually shows that the 104.5 degrees H-O-H bond angle gives rise to a lower total electronic energy of water than the linear geometry! So be convinced! That the Singaporean education system isn't wrong! :)

So the next step comes in explaining why that's the case.

Another time. :p

And all calculations were clumsily performed with a minimal STO-3G basis set using the Hartree Fock level of theory, which merely took 5 seconds or so for each single point energy calculation.

Yay! Time to sleep. :)

Don't Point So Much!

2009-08-15T10:01:00.001-07:00

I'm undertaking this module on group theory, and I noticed something that I never really thought about because I just took things for granted. Now, given that:

And given that the first two results are definitely correct, what can one say about the third?

And well well, here are two molecules where students usually assign the wrong point group to:

If you'd like to try, then don't play cheat and figure it out from scratch! :)

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Because I am the lazy person that I am (I'm too tired, :p), the answers are as follows without explanation or elaboration:

Satisfaction

2009-08-01T06:52:00.000-07:00

I've been feeling rather dry lately, but I'd just like to say:

Science is nothing more than intellectual satisfaction to me. It represents not the truth, neither is it any philosophy; it is simply a set of tools, principles and logic obtained from empirical evidence that appeals to the human mind in such a way that I am excited by it.

That being said, truth is not discovered, but revealed only to those who seek it. :)

So continue to seek Him YYK. :)

Druggie Druggie Addict!

2009-06-22T03:23:00.000-07:00

There's a hidden birthday message, can you find it?

So here's the case; as is the case with the USA, Singaporean soldiers are not allowed to take illegal drugs (besides the usual smoke I guess, which is perfectly legal). The question that I pose to you now, is:

If you'd like to do a survey to find out how many Singaporean soldiers consume illegal drugs, how would you do it?

You must keep in mind that if they do confess or are caught answering "Yes", then they will face the death penalty or some other kind of huge fine. The nice guy you are, you'd want to come up with a way to prevent this, and yet come up with some sort of estimate of figures.

Knowing this blog, it's definitely a statistical trick that you'll have to employ, but what? Heh.

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Well, the thing is, you've got to let them confess in secret, but yet you have to know what they confessed somehow. But you need to ensure their anonymity and safety as well!

As such, leaving any high-technology mind reading devices out, we are left with no choice but to play Jedi mind tricks.

Yeah right. We just use what we learnt in high school - that's right, simple statistics, for a good estimation.

We come up with three types of cards, as shown (of course their backs will be all of the same colour):

Each type of card is associated with one type of question - these questions need not be written on the card, but may be posed to the soldiers verbally. Now, to ensure that the soldiers draw out a card at random, so the chance of drawing a red, blue or green card is 1/3 for all, we have say, a good number of cards (say 72 cards) laid down like this:

So now, the chance or probability that any soldier draws the green, blue or red card, is exactly 1/3, even if he or she has a propensity or tendency to draw from the corners or from the middle of the deck. Make sure that the cards are laid in the order as shown for the probability to be true.

Let's give it a go; let's say for example, that out of 12000 soldiers who were surveyed, 5600 of them answered "Yes". Assuming all soldiers are sane, and that they completely understand English and are well, disciplined enough to not want to play with the system, they'll be truthful.

Therefore, all "Yes" replies must mean that the soldiers either chose the red or blue card.

Since 12000 is a huge enough number, and this is a fair test, we should be confident enough to say that on average, we expect 4000 soldiers to say "Yes" to the question of whether "Is this card red?"

If that is the case, then we expect that on average, at least 1600 soldiers do take some kind of illegal drug.

On the other hand, we also expect 4000 soldiers to say "Yes" to the question of "Do you take drugs?"

So what does all these mean? It means that on average, we expect at best, 1600 soldiers to be taking drugs, but at worst, 4000 soldiers to be taking drugs.

It'd be good to perform another test on another group first, to obtain the standard deviation! Haha.

So... what do you think? Aren't statisticians rather useful? Heh.

Representations

2009-05-30T08:54:00.000-07:00

Well, for those who have no knowledge of matrices, here's a challenge I have for you; given that the rules of 2x2 matrix mutliplication are summarised as such:

Can you find me a 2x2 matrix, such that when it's multiplied by itself, yields:

There's about 10000000000000000000000000000000000 such matrices by the way. So if you can find 1, and you think you're smart, think again. Haha.

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Well, even though this post has been here for the longest time, it seems that no one has ventured beyond perfunctory reading, so I guess I'll just carry on.

One possible matrix would be:

And another would be:

There's actually a pattern: the number in the upper left and lower right positions are the same! But of course, you'd need the right combination. Heh. Can you find any more? There's probably a million more of these, haha.

Stay Still!

2009-01-20T19:26:00.000-08:00

I think I’ll go into the idea of stationary states today; what exactly is a stationary state? Well, it’s essentially a state of a system, where the energy of the state remains constant over time. Defined more rigorously, it’s a state where the expectation values of observables remain constant over time.

For instance, let us take a look at the particle-in-a-box wavefunctions (as functions of time and position):

So let’s consider the probability density function of finding the particle within the box:

Notice that the phase factors (exponential time factors) cancel out when the complex conjugation is taken and multiplied! That is, the probability density function isn’t a function of time! It’s solely a function of position and thus, doesn’t vary with time.

How about the expectation value of the position? Well let’s take a look:

Hey! It turns out that the expectation value of the position doesn’t depend on time as well! So, it turns out that for all states of the system that are eigenstates (that is, if the system exists only as an eigenstate), the state is a stationary state!

So what isn’t a stationary state? Well, let’s look at linear combinations of eigenstates; we know that any linear combination (properly normalized of course) of eigenstates will still result in an arbitrary state that is a solution of the Schrödinger equation, so let’s try a positive linear superposition of the ground state and first excited state (I’ve normalized it for all of you already):

Let us now evaluate the probability density function (click on it to enlarge it):

Notice that now the probability density function is a function of time as well! The time phase factors no longer cancel out! It turns out that for any linear combination of eigenstates, the state will no longer be a stationary state – and therefore observables like its energy, momentum and even position will not have time-constant expectation values.

If you’ve heard physics professors go, “It’s all because of the cross terms!” this is what it means. Haha.

Postulato Negativito!

2009-01-20T18:19:00.000-08:00

So I was looking, and I was just a little bit amused:

"So... there are two types of fundamental postulates in science. The first type is the self-evident kind, like how heat flows from hot to cold regions. The second type is one that is more subtle, and needs to be unfolded in a series of arguments before you get the point of it all."

And why am I laughing?

"Unfortunately, in Quantum Mechanics, the fundamental postulates all belong to the second type."

Lol!

Time Evolution

2009-01-19T17:21:00.000-08:00

So if you look at it this way, then the time evolution of the wave function of a system is no more than an evolution that is governed by the energy of the system:

Then only separating out the time-dependent wavefunction:

It would appear that the time evolution of a system lies only in the changing of the phase of the system!

Strange. Weird. I still don't get what this means, haha.

Double Pendulum Revisited

2009-01-18T04:35:00.000-08:00

Remember this post: http://wulidancing.blogspot.com/2008/07/double-pendulum.html?

Well, it turns out that you don't really need any high-level mathematics for this question, and I'm just wondering how come I didn't figure it out at that time, haha. So here's the whole set up again:

It'd be useful to see that the equilibrium position should have zero gravitational potential energy:

The trick to solving this question is to lift each pendulum up one after the other; so let's give the first pendulum a push and see what happens to the potential energy:

Notice that both pendulum bobs are raised by the same length, which explains the coefficient of '2'. Now let's give the lower pendulum another push and let's see what happens:

Notice that we've just added another term to the potential energy, and this time the coefficient is '1' because only one bob is raised.

Quite easy right? Haha.

Lagrangian in Action

2009-01-17T05:45:00.001-08:00

As a follow up to my previous post, let's see some Lagrangian Mechanics in action, with the use of a very easy example - the simple pendulum! So let us consider the following set-up:

For your convenience, I've worked out the various quantities to take note already, and I've labelled all of them on the above diagram. Now, with all of these observables in place, we can now write down the Lagrangian immediately:

Notice that I'm no longer using the x coordinate! This explains why the Lagrangian is so versatile, because it can be expressed in terms of any general coordinate, and still work! So given this, let us work out the Euler-Lagrange Equation:

And we see that if we equate the two derivatives, we obtain the equation of motion:

Notice that we didn't even have to go about resolving out the various forces, like tension or weight or whatever! It's really easy with Lagrangian Mechanics!

I believe your teacher might have told you to "only make small oscillations when setting up the pendulum", and the rationale for that being that "small oscillations result in simple harmonic motion." But how?

Easy, let us consider small angular displacements, and our equation of motion automatically reduces to:

Voila! Notice that the second derivative with respect to time of the angular displacement (i.e. angular acceleration) is now directly proportional to the angular displacement! This is the very definition of simple harmonic motion!

Haha. Easy right? Lagrangian Mechanics really simplifies a lot of things. :)

1, 2, 3... Action!

2009-01-16T21:38:00.000-08:00

Last night I was getting down to understanding the Lagrangian as it is used in Classical Mechanics, which is something I’ve been putting off for quite a while now; but taking a closer look at it, it doesn’t seem as tough as I thought it’d be! Well well, so how does one start?

Well, a proper derivation proved too tedious for me to want to type it out (I tried deriving it for my friend Shaun, but I gave up halfway because it’s really too long and I was lazy, haha!) so here’s a really general derivation, which needs some prior knowledge of calculus of variations.

But I think all of you are smart. So I’m going to go ahead with this general derivation. Haha.

Alright, let’s start with the following equation, known as the Euler-Lagrange Equation:

As above, we have an equation that is pretty much obvious: it’s simply a differential equation that involves a function of three variables x, y and y’ (the derivative of y with respect to x).

Now, for any function F(x, y, y’) that fulfills this condition, we can immediately say that over the interval as shown below:

I won’t be explaining why this is the case, because it took me quite some pages to type out and I don’t think I’m in the mood to type everything out for this blog post, haha. So please do accept this for the time being!

So what does this have to do with the Lagrangian function in Classical Mechanics? Oh yeah, what is the Lagrangian?

Well, it’s simply this function right here:

Where T is the total kinetic energy and V is the total potential energy of the system. Notice also that:

So we now rewrite the Lagrangian as:

Now, let us consider the following derivatives of the Lagrangian:

Now for all conservative central potentials, which we are all familiar with, we know that:

That is, the negative of the gradient of the potential energy is the force acting on the system at that position (JC students might remember this from basic Gravitational Theory). So let’s keep this in mind first! Now onto another derivative of the Lagrangian:

Of course, we know that:

Which is acceleration! Let us now put everything together:

Now let us compare the Lagrangian differential equation with the Euler-Lagrange equation:

Hey! Doesn’t this mean that the Lagrangian should then imply that the following integral is a minimum:

This integral is known as the action; notice that Newton’s Second Law is actually an Euler-Lagrange Equation when it is written in the form of the Lagrangian, which in turn means that the above integral, the action, is to be a minimum.

This came to be known as what is now called the Principle of Least Action.

And that’s about as watered-down as I can make it for all of you, haha.

Separate Them!

2009-01-11T06:53:00.000-08:00

I would like to highlight in this post, a nice mathematical technique known as separation of variables, which reduces a more complicated partial differential equation into a less complex ordinary differential equation. To illustrate this technique, I shall be using the time-dependent Schrodinger wave equation in 1 dimension, which is as follows:

This equation tells us that the time evolution of the wavefunction depends on how the Hamiltonian operator acts on the wavefunction. That is, the time evolution depends in some manner, on the energy of the system.

With that, we shall now consider the Hamiltonian operator acting on the wavefunction:

However, in most cases, the potential is independent of time, say like how an electron always orbits around a fixed centre of charge; therefore, we indicate this by re-writing:

Notice now that the potential function only depends on the position coordinate, and not time. Now, we can write out the full time-dependent wave equation as:

In quantum mechanics, it turns out that the overall time and position dependent wavefunction can be factored out into a product of separate time wavefunctions and position wavefunctions. That is, we can now write:

This assumption (generally valid when the potential is not a function of time) leads to the technique known as separation of variables, where you effectively factor out one variable from a function. And if we make the substitution, we see that:

Dividing throughout by φ(x)τ(t):

Alright! Now on the left we have a function depending solely on time, and on the right a function depending solely on position! Notice that we have separated the variables! So what’s so good about this you say? Consider this:

‘If I vary x only, then only the right hand side of this equation should change, since the left hand side doesn’t depend on time. However, if the left hand side doesn’t change and remains constant, so should the right hand side! This means that both sides are actually equal to a constant!’

With this revelation, can you fashion a guess as to what this constant might be?

If you said energy, you are absolutely right! We can now equate both sides to the energy of the system:

Also notice that the equations are no longer partial differential equations, but ordinary differential equations! We’ve made life simpler! Hurrah!

Let us look at the time-dependent wavefunction:

Voila! The time-evolution depends on the energy of the system as shown! Now, you might be wondering about the position-dependent wavefunction, so let me rearrange it as shown:

Hark! Isn’t this the usual time-independent Schrodinger wave equation that we always see? Haha.

QM!

2009-01-02T10:39:00.000-08:00

After a long break, and with Quantum Chemistry looming in the near future of the coming semester, let us go into a short (and I do mean ‘short’) discourse on the structure of Quantum Mechanics.

So, what is the principal difference between the so-called Classical Physics and Quantum Physics? Is it… that quantities are quantized and no longer continuous? Well actually no, even in the classical realm things are quantized, just that in such small portions that they seem continuous to the human senses! So what is the main difference that makes quantum mechanical treatment ‘quantum’?

First off, you should be acquainted with the term ‘observable’, which in general refers to any dynamical quantity that can be observed in real life; that is, properties such as momentum, position, weight, mass, energy etc. By the word ‘observe’, we mean to say that the property belongs to a system, and the system possesses that property or quantity and what we measure depends on the state of the system.

In all Physics, we are concerned with two types of quantities: ‘parameters’ and observables. So what is a ‘parameter’? Time, for instance, is a parameter – you can measure time no doubt, but time doesn’t belong to a system. A system evolves with time, but it doesn’t possess time per se, so as to speak. With that, time is a parameter, and not an observable. That is to say, the measure of time, doesn’t depend on the state of any system, and evolves very much by itself. In that case, time isn’t an observable!

Alright, so with the definition of an observable laid out, we can now state the fundamental difference:

‘In classical physics, observables are represented by functions
but in quantum physics, observables are represented by operators.’

So what is this difference? Now… a function by itself, means something. For instance, if I tell you the position of say, my Chemistry professor depends on time as such:

I’m pretty sure you’d be tracing out a parabola in your head. Or maybe not. But the point is, in classical physics, the observable takes on a pretty much, well, for lack of a better word, ‘observable’ form. But if I tell you now, that say, the momentum of my Chemistry professor depends on an operator as such:

Then, what exactly does this mean? The derivative isn’t operating on anything at all! How can I say that the momentum is equal to this derivative – the derivative of what? So it seems like, hey, quantum observables aren’t that easy to observe, huh?

Therein lies the first magic of quantum physics that all science students should be aware of: without any measurement, you can’t say anything about any system. Which makes sense: if you don’t allow the operator to ‘operate’ (i.e. measure) on a system, then no information can be obtained, because an operator needs to operate on a system before it can extract any information!

So that’s why in quantum physics, there needs to be something called the ‘wavefunction’ – essentially, this wavefunction contains all information about a system. If you want to find out something about this system, easy! Just use the appropriate operator on the wavefunction, and it’ll give you a numerical value about just the thing you want to know about the system.

I guess you should be aware of the Hamiltonian; the Hamiltonian is the ‘total energy operator’, meaning, when you apply the Hamiltonian to a wavefunction, you obtain the total energy of the system. And of course, we see this most often in the much-celebrated Schrodinger wave equation:

That is, if I wish to determine the total energy of a system, I just apply the Hamiltonian to the wavefunction of the system, and bam! I get the energy ‘E’!

Alright, I’m tired, and it’s 2:37 am in the morning, and I think I’ll be taking some of these thoughts to bed. Haha.

Salen This!

2008-10-09T09:18:00.000-07:00

You know the perfect gift for someone who likes shiny, yellow crystals? Well, watch this space and I'll teach you how! :p

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Enter salicylaldehyde, commonly referred to as salen, which is a very common ligand in Organometallic Chemistry:

And in its pure state, it manifests itself in the form of flaky, shiny yellow crystals! It's really intriguing, because one usually thinks of crystals as inorganic compounds, consisting of metal ions and their salts. Yet this is but an organic compound, and whenceforth the colour?

Only time will give the answer!

Now You Oxy-It, Now You Don't!

2008-10-04T02:33:00.000-07:00

Alright, after a failed attempt at the retrosynthesis of haemoglobin, let's just marvel at nature's molecule:

Freaky. :p

Carboshift!

2008-09-16T20:36:00.000-07:00

I remember reading in this article, where there was a study made on carbocations in solution; apparently they managed to stabilise the formation of carbocations in superacidic solution, and therefore observed the characteristics of carbocations, in particular their fluxional character. And they found that when they prepared the carbocation that had the structure:

After NMR analysis, they found that they had two other carbocations:

Can you fashion a guess? Haha.

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And if you still haven't figured it out yet, it's due to carbocationic rearrangements as such:

Quite cool huh? Haha.

Circular Argument I

2008-09-14T08:09:00.001-07:00

Here's something that caught my attention when I was tutoring one of my students in 'E' Math at the Secondary School level: the angle subtended by a chord of a circle at the centre of the circle is twice that subtended by the same chord at the circumference.

Perhaps a diagram might make things much easier:

The chord is the line AB, and the angle subtended at the centre is the angle AOB, and the angle subtended at the circumference is the angle ACB, as indicated in Greek above, haha. So this property of a circle says that angle AOB is twice that of angle ACB. But why?

Easy enough, I shall use the method that Shaun used during our lecture, which is a rather neat and easy proof; so kudos to Shaun! First of all, you divide the triangle ABC into half, down the centre as shown below with a line CD, and then notice I've coloured an isoceles triangle in green:

And then you should notice that this line should bisect the angles AOB and ACB, and we can conclude first that:

And from this, the following statements should therefore make sense:

Quite a neat proof huh? Shaun came up with it all by himself, pro!

Speeding Towards Light

2008-09-11T03:02:00.000-07:00

So, someone tagged that she wants to know why light always travels at the speed of light huh? Haha. Well here’s some information to get you started:

This is a typical wavefunction, which is a function of both position (x) and time (t). If you’re a good JC student in Singapore taking Mathematics at the H2 level, then you should be able to figure out that this function travels in the direction of positive x (if you can’t tell you can come ask me, heh) and it also travels in this direction with increasing time.

Now, we see that there’s an amplitude (A), and there’s an angular frequency (ω), and there’s a wavenumber (k). These quantities should, hopefully, be familiar to you, or at least I hope you’ve heard of them! Haha. These are all A Level knowledge, so hopefully you have!

I’d like to say that although what I’m about to say applies to all waves, not all waves have such a nice looking function like the one above! I just chose the easiest one of all so that it’s easy to digest what I’m going to say.

Going on, I’d like you to see that:

What I’ve done here is to take the second partial derivative, meaning I differentiate the wavefunction with respect to x only, twice, each time keeping t constant.

Aright, now, let’s try to take the partial derivative again, now with respect to t:

Cool! To sum it all up so that you follow me better:

It should be obvious to you then, that:

Does anyone want to fashion a guess what this means? Heh. Well, for waves we know that (from basic JC level Physics):

And therefore, we can say that:

But hey! We learnt in Secondary School that the speed v of any wave is related to the frequency f and wavelength λ by:

And therefore:

And thus our previous equation shows that:

That is, if we differentiate the wavefunction with respect to position twice, we obtain the derivative of the wavefunction with respect to time divided by the speed of the wave squared! What an interesting inherent symmetry! And this holds for all waves!

I’d really like to go on, but well, I don’t think most of you are acquainted with electromagnetic waves, are you? For electromagnetic waves, the equation turns out such that:

If you don’t know, ε is the permittivity of vacuum, and μ is the permeability of vacuum. Interesting enough, we note that:

Since all quantities on the left are constants, then we conclude that the speed of light (or rather, electromagnetic waves) must also be a constant regardless of anything, and therefore we now have:

Well well, I hope this is good enough! Because really, I can’t explain why light has a constant speed unless one deals with more advanced electromagnetic theory and wave mechanics. But haha, this is good enough I think, for all of you to chew upon! :p

Double Angle Madness III

2008-09-04T08:30:00.000-07:00

And I might as well show the last identity, right? Haha, and so here it is:

And that's all for now. :)

Double Angle Madness II

2008-09-04T08:25:00.000-07:00

And continuing with the streak, I might as well show how to prove a very commonly used identity; for convenience, I've reused the diagram from the previous post:

Trigonometry isn't that hard right? :)

Double Angle Madness

2008-09-04T08:14:00.000-07:00

Well, I gave one of my tuition students this identity to prove:

sin 2x = 2 sinx cosx

And is it really that hard? Well, actually all it needs is a single diagram and two lines of working, as I have illustrated below:

Cheers! :)

Newton's Legacy

2008-08-30T20:28:00.000-07:00

Here’s something for you to think about – Newton’s Second Law is often quoted as the net external force on an object is the product of the object’s mass and its acceleration. Well I wouldn’t say that’s wrong, but that’s only truly correct in a classical sense. That is, the following equation only holds for non-relativistic speeds:

The more accurate way to quote Newton’s Second Law, which holds for all situations, is this: the net external force on an object is the time rate of change of momentum of the object itself. And we express this mathematically as:

And I’ll illustrate the generality of this expression by using this to derive a relativistic expression for the acceleration of a body. Now, to start off, let us recall the relativistic momentum expression:

And therefore the rate of change of momentum of a body must be given by a direct differentiation with respect to time:

And if you simplify this, it becomes:

And we recognize two things, that is:

And therefore:

Notice that Newton’s Second Law no longer just involves the product of the mass and acceleration – this means that the expression F = ma is not a generally valid expression! And hence my conviction that Newton’s Second Law should be reformulated in terms of rate of momentum change instead.

Now, let us rearrange what we have above:

So let’s say you want to accelerate a body to the speed of light (c ms^-1), notice that as the speed v tends towards c, the acceleration tends to zero for a finite force:

So how do you accelerate something to a speed of c given that the acceleration fades away to zero when you near the speed of light? Easy, you use an infinite force! But is there such a thing as an infinite force? No! However, is there another way to go about it? Well, yes! Notice that as the mass goes towards zero:

Therefore, if the mass of a body is zero, then it is possible to accelerate the body all the way to the speed of light!

Ask yourself, what is the mass of a photon? :p It’s simply zero, which is expected! Otherwise it can’t go at the speed of light! :)

God's Thoughts

2008-08-30T07:44:00.000-07:00

"I want to know how God created this world. I am not interested in this or that phenomenon. I want to know His thoughts, the rest are details." - Albert Einstein, Jewish Physicist

And this is what one of the greatest minds who ever lived on this Earth once said - and I find it so encouraging that even a giant like Albert Einstein was after God's heart and mind, and how his Creation came into place.

I'd like to emphasise that all Science around us, all Mathematics, is only but an attempt to describe and understand this world, which is based on the laws that God engineered for all of us. Seeing how this intricate clockwork has come together, His handiwork is truly in the firmaments. :)

*Note: Albert Einstein wasn't a Christian even though he pursued the thoughts of God.

At The Speed of Light

2008-08-27T17:21:00.000-07:00

It's now 8:00 am in the morning, and I just had this flash of inspiration whilst in the toilet; not too glamourous a situation for a brainwave, but oh well. Hopefully you all can understand what I'm about to type out.

Many years ago, Einstein as a young lad was thinking to himself: "What would happen or what would I see if I travelled at the speed of light? Would I see light waves or photons freeze in their tracks because I'm moving just as fast as them and thus the relative velocity between them and me is zero?"

And many years later, Einstein was convinced that no matter how fast one travels at, even at the speed of light, one will still see light travel at the speed of light.

Why?

Well, to first understand the situation, you must first understand the fundamental problem in Einstein's gedanken, that is, can you even see a stationary light wave?

The answer is a resounding and definite NO! Light, being composed of photons, are massless particles. From Einstein's mass energy equation, we therefore know that the total energy of a photon must be composed of its momentum-energy, that is, its kinetic energy, because it has no mass at all.

Now, to view a photon that is stationary, is then to view a photon without its kinetic energy - by denying a photon its kinetic energy, one essentialy annihillates that photon from sight. By travelling up to the speed of light, and to insist that one can still see light, then the light waves that one sees while travelling at the speed of light, must still be moving, and can't be at zero velocity.

You all got that? Haha.

And I guess I'll stop here for a while - I'll come back to explain further (in another post) why you need to be massless to move at the speed of light. :)

Derivator!

2008-08-20T23:56:00.000-07:00

Look at this second-order equation:

How would you solve it? I mean, there’s always the conventional method of assuming a solution that has an exponential form, but did you know that you can treat this like a quadratic equation?

The method is known as the solution of second order differential equations via factorization, which isn’t really an original method thought out by me, but well, it helps to explain why the solution is of an exponential form. First of all, we factorize out the y term:

Then of course, we make a simplification; that is we allow:

And therefore we must insist that:

Of course, let us factorize this further:

In which case we must therefore have:

And of course if we solve for these two equations, we have:

And of course, the general solution being the sum of the two particular solutions we obtained earlier, from the principle of superposition for linear equations:

And if you only want a real solution, then:

And this explains why we always assume the solution is of an exponential form; simply because if you do solve it from first principles, it always is! :)

Resonance Imaging

2008-08-19T17:54:00.001-07:00

So, some of you think Resonance Theory is outdated huh? Well, try this question! Do you think you can tell me which carbon atoms would most likely bear the positive charge in this cation without Resonance Theory:

I've drawn out the resonance hybrid (i.e. the actual molecular structure) of this ion. I'll give you a hint: there are four carbon atoms that could bear the positive charge. :)

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It seems like no one has given a go at this, but oh very well; if you can mentally draw the set of resonance structures in your head, this should really be problem at all:

And that's that! Haha.

Summation Woes

2008-08-17T05:07:00.000-07:00

Well, here's something that both JX and Luoning can understand and appreciate; it's basically something I discussed with Shaun over MSN, and it's an alternating series, taking the form of:

Notice that we can group the terms like this and obtain a sum for an odd number of terms:

Also, we can group the terms like this and obtain a sum for an even number of terms:

The strange thing is, notice that if you consider an even number of terms, the sum goes to positive infinity, and if you consider an odd number of terms, the sum goes to negative infinity! This is what we call an oscillating series, which happens to be diverging as well.

I’ve included a graph for your reference, to illustrate the oscillating and diverging nature of this series:

Notice the diverging nature of the sum to n terms, and consequently, there can be no sum to infinity, simply because the sum to n terms depends on the very number of terms, and thus on the last term. A sum to infinity where the number of terms and the last term is not defined can therefore produce no well defined result.

Nuclear Worries

2008-08-13T19:53:00.001-07:00

Nuclear worries should be of a small magnitude right? But it sure ain't. Sigh.

You know, one of the most interesting things about Nuclear Physics is that there’s no fixed boundaries that define the study of this field of Physics. What do I mean? Well, take Electrodynamics for instance: you have Maxwell’s Four Equations, that clearly set a basis from which all other theories of Electrodynamics are derived. For Classical Mechanics, there’s Newton’s Laws of Motions, and for Classical Dynamics, there’s the Lagrangian. So clearly, for these fields, there’s a fixed set of rules to go by.

But for Nuclear Physics, we base it on Quantum Mechanics (for the behavior of nuclei), Electrodynamics (for the charge distributions), Relativity (taking reduced mass and binding energies into account) and basic Mechanics and Dynamics (collisions etc.). So there’s really no limit to what can be done in this field.

Which explains why I’m actually finding myself doing so much for the Physics module I’m doing this semester, haha. So what does my Professor want me to figure out? Well, for now it’s Electrodynamics!

So my Professor goes and says this:

“The potential energy within a homogenously charged sphere of charge Q and radius R due to an interaction with another charge q within its interior at a distance r from its centre is given as:

And well, you all should know this already.”

I should? Well I sure don’t! So I’m going to try and derive it now instead of taking it for granted, haha.

Well so how do we start off? Easy, let us consider what we’re looking at in terms of a pictorial representation:

So what we have here is just a big blue sphere of charge Q, of radius R, and the small charge is placed within its interior, at a distance r away from the centre. So how do we go about solving this question? Well it’s easy – we assume the charge is homogenously distributed within the sphere, meaning each portion of the sphere bears the same amount of charge (i.e. uniform charge density ρ).

In that case, the charge q experiences the effective amount of charge that is contained within the bound radius of r instead of the whole sphere’s radius of R, marked out in pink:

In which case we can then write the electric force on the charge q as:

But we all know that the effective charge contained within the radius r is actually:

But we do know that the effective volume V of the pink region is:

And of course, the charge density is given by dividing the total charge Q by the total volume of the sphere:

Putting everything together:

The next part is to recognize that the Coulombic force is related to the potential energy as such:

And therefore conclude that:

So integrating this expression:

Now, to determine the constant k, we note first that when r = R, that is, where the small charge is on the surface of the big spherical charge, the potential energy is simply:

And therefore we must insist that:

Putting everything together:

Voila!

Arithmetic This!

2008-08-09T02:50:00.000-07:00

There’s a special name for this type of sequence:

Notice that each term differs from its preceding and succeeding term by a constant difference of 3 – this type of series can be represented as such:

Where a is the first term, and d is the common difference. Notice that also that the n-th term is given by a + (n – 1)d. This is known as an Arithmetic Progression, or an Arithmetic Sum.

Now, can we find a sum to n terms? That means, can we find a general formula for the following summation of the first term to the last term:

Well, of course we can! Let us see how; first, let us consider the sum to n terms on its own:

We started off by writing the sum from the first term a – but we could have done so by starting from the last term A as well, and thus the sum to n terms written backwards from the last term
A is simply:

Now let’s put them side by side and add them up and let’s see if you notice something:

What do you notice? Well, notice that the ‘d’s and the ‘-d’s all cancel one another out, leaving us with:

The n is there because we have n terms in each sum, and adding up two sums should give us n of A and n of a. Now, what is the last term A in terms of the first term a? We’ve already said that we want the sum to n terms, and thus the last term A must be the n-th term, given by:

In which case we make a substitution:

And therefore the sum to n terms follows nicely:

Sweet eh? :)

Sum To Infinity, and Beyond!

2008-08-09T02:15:00.000-07:00

Well, what exactly is a Geometric Progression? To put it very simply, it is a sum or series that has the following pattern:

Of course, you may then ask what is the sum to infinite terms, or the sum to infinity for this series, and we can then represent it as such:

The previous post dealt with the problem of 0.11111… as a recurring decimal using the idea of a Geometric Progression, and I’m about to show you exactly how right now.

First of all, let us consider the sum to n terms (this is often called the partial sum of the first n terms):

It must then be agreeable and logical that if I multiply this by the common ratio r, I obtain:

Taking the difference of the two, I see that:

And rearranging, I immediately see that:

Now what is the sum to infinity then? Easy, it can be evaluated by considering the limit as n tends towards infinity:

Notice that this limit can only exist, if the common ratio r has a magnitude smaller than one, such that it decreases to effectively zero for huge values of n:

With this in mind, the sum to infinity is simply:

So applying this to the previous problem, we see that:

And therefore there is no actual need for any algebraic manipulation to solve for recurring numbers once you have grasped this theory. :)

Simple Algebra, or Not?

2008-08-09T01:48:00.000-07:00

Well, I won't be doing much explaining in this post, so if you don't understand maybe I'll post another explanation for the second part; Luoning's tag has directed me to a very nice solution for determining fractional expressions for recurring decimals - it's a method taught to me by Mrs Pauline Kan way back in JC1, and I always turn to it for some mathematical fun when dealing with young Secondary School kids during tuition.

The question goes like this, given a recurring decimal like the one below, can you determine a fraction that equals it:

Well, the trick is to recognise that you can make use of the commutative properties of algebraic manipulation as follows:

And then by equating the two equations we must concur that:

And therefore:

But of course, for those budding Mathematicians out there, you may have seen this as a simple Geometric Progression sum to infinity, and therefore the general formula follows directly for such a sum:

I'm taking for granted that all of you know what Geometric Progressions are, haha.

For The Record

2008-08-06T11:04:00.000-07:00

Well, here's just something Dr. Phil Chan, a really really really interesting Physics Professor, mentioned during my course in my first Semester at NUS:

Do you understand why? Haha. I probably won't be revealing this even if you don't get it, because Physicists truly are number one! :p

How Many Ways?

2008-08-06T00:54:00.000-07:00

So, given this picture of four rooms interlinked by pink gates or doors, do you think you can find a route through the doors such that you pass through every door only once? Or do you think there's no such route?

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Well, Luoning nailed down the essence of my explanation; basically, it's got to do with the number of entrances to each room you see in the picture above. To facilitate the explanation, let me redraw the network of rooms above as such:

Notice I've represented each room by a black dot, which I shall now refer to as a node; so there's nodes A, B, C and D, each representing the rooms. Now, notice each node has lines connecting it to other nodes - these lines represent the doors or entrances leading from one room to another room.

Looking at this diagram, you'll find that it's much more easier to figure out possible routes.

So what did Luoning say again? Ah yes, the even or odd number of entrances! Indeed! Ask yourself what it means to be able to trace out a path where you don't re-use an entrance: this means that you must be able to arrive at the room the same number of times as you leave the room.

For you to arrive at the room the same number of times as you leave the room, one obvious condition is needed: you need each room to have an even number of entrances. Try it!

Of course, there is another condition: you could also have two rooms having an odd number of entrances but of course you need these rooms to be directly linked to one another as well. The direct linkage of these two rooms then means that the two rooms can be visualised as one big room and therefore reduces the number of effective entrances by one, turning an odd number of entrances into an even number of entrances!

Well I didn't think of this myself, but I did think of how to put forward this explanation myself! So give me some credit, won't ya? Haha.

So for this problem, of course no single route can be found! :)

[Edit: Hmm, this post isn't that well explained - I'll come back to this once I've figured out Graph Theory for myself!]

Double Pendulum

2008-07-26T06:40:00.001-07:00

The double pendulum is always an interesting question because it's a special physical system composed of well, surprise surprise, two pendulums! Haha. Anyway, here's a question Funman asked me, so I might as well share it:

Well, just click on it to enlarge it!

What say I master Lagrangian Mechanics before I attempt this? :p

Complex Power

2008-07-21T08:19:00.000-07:00

Well, here's something I asked one of my students, Cassandra, and let's see how many people actually manage to figure it out by, say, the end of the week:

Have a go at it!

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Some of you might be thinking this problem might be adequately solved if one considers Euler’s exponential form of complex numbers:

But here’s another problem; we know that:

In fact, the number i is obtained each time we make a full revolution in the imaginary phase space, such that:

Does that mean now that:

Go figure!

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Well, the answer actually is very simple: you just have to define the allowed values of the argument that define the imaginary number i, such that it can be placed to the power of something. :)

Oh yeah, this answer was contributed by Kenneth Tay Jingyi!

Expansion Woes

2008-07-18T00:04:00.000-07:00

My my, have you ever worried about how things expand when heated? I once explained this to one of my friends in Physics in my first year of University, and looking back now, it’s rather interesting. The question he asked me was:

“Eh, if you got this ring right, when you heat it and it expands… what will it look like?”

Oh yes, he asked me with reference to this diagram:

And I wonder... Mmhmm...

A Light Burden

2008-07-17T20:26:00.000-07:00

Did you know that radiation carries with it a kind of pressure known as radiation pressure? That means that when I shine a torch on you, it's supposed to be pushing you as it gets absorbed or scattered off of you. Let's try and determine the force exerted on the Earth by the light rays from the Sun, and you'll see what a staggering figure this is!

Assuming that the Sun is a perfect sphere, of radius R = 695 500 km and that it is a perfect blackbody with emissivity e = 1.00, and a roughly constant surface temperature of T = 5 778 K, we can use the Stefan-Boltzmann Law to determine the power of radiation emitted by the sun:

Where σ is the Stefan Boltzmann constant and A is the surface area of the Sun. And yes I've conveniently assumed the Earth is of so a low temperature with respect to the Sun such that it hardly radiates back any radiation back to the Sun.

Let us denote the distance of the Earth from the sun to be d (which is an enormous d = 149 600 000 km) and its radius as r = 6371 km. With these figures, we can then determine the intensity of solar light received at the position of our Earth, which is the initial power emitted by the Sun divided by the surface area of a sphere of radius d since the light from the Sun is isotropic in all directions:

Since all light rays coming from such a far off source will be parallel to one another, we can take the cross sectional area of the Earth to be the area that intercepts the light rays, which can be taken to be:

The power intercepted by the Earth is therefore:

Now, we know that for electromagnetic radiation, pc = E, where p is the equivalent momentum of a quantum of energy, c is the speed of light in a vacuum, and E is the energy of radiation considered. This simply means that the power is related to the momentum as such:

Don't forget, we know that the force exerted by the radiation on Earth is equivalent to the rate of change of momentum of the radiation photons (of course, we're assuming all radiation is absorbed, which is an approximation!), and thus the force exerted on Earth by the Sun's light is:

And this force is about 10^8 N! Which turns out to be an astronomical number! :) So why do you think the Earth is not being pushed away by the Sun? Easy, there's still the force of gravity, which when worked out by Newton's Law of Gravitation:

And this amounts to about 10^22 N! Do you see the difference in the two forces? The gravitational force far outweighs the force exerted by light, and thus continues to aid us in our journey through intergalactic space! :)

Infinity, Or Not?

2008-07-16T22:00:00.000-07:00

As always, click the above comic strip to enlarge it; you'll notice that there's a really peculiar question which involves an infinite 2 dimensional array of resistors. Haha. Anyone game for this one? :p It's solvable!

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Well, I was thinking of Kirchoff's Laws as a possible way of solving this, so let me just direct you through my methodology first.

Oh yes, a simple application of Kirchoff's Laws won't work in this question - I'll explain why later.

You start off with a single point in this infinite array and reconstruct the infinite array from scratch - let us assume we have this point (0, 0) that is the origin. We pump in 1A of current into this origin, and thus from this origin, we have four current vectors as such, coming out as shown below.

It should make sense to you that the current is divided equally into four, because all four paths are of equal resistance. We're taking one black line to be 1/4 A of current. Now, we then extend the network as such:

Notice that now each blue line is 1/12 A of current, because at each junction there are 3 paths of equal resistance. We can go on and extend the network and you'll see this:

Each red line is now worth 1/12 A of current as well! Surprise, surprise! This is because you've two '1/12 A' currents going into one junction, and then leaving into two junctions of equal resistance. It should then make sense that the two paths of red carry 1/12 A each.

So let's view our small portion in the infinite array of resistors:

Notice that in order to get from point O to point A, you go from one black to one blue to one red, and you experience a potential drop of:

V = (1/4)(1.0) + (1/12)(1.0) + (1/12)(1.0) = 0.417 V

And this method will only get us the potential drop between these two points, and nothing else.

Why? You might be thinking that as in the cube resistor problem in the previous post, we can simply divide the potential drop by the current to obtain the resistance between these two points. But this problem doesn't allow for that, because the initial current was 1 A, and the final current we dealt with at point A isn't 1 A but 1/12 A. This means we can no longer divide the potential drop by the initial current to obtain the equivalent resistance.

The previous problem has the starting current and end current being the same value, since by Kirchoff's First Law we know that all current going into a junction must equal all current going out of the junction if there is to be no buildup of charge.

The actual mathematics required for this question is beyond me, and it involves Fourier analysis. If you're still interested however, you can have a look at the solution at www.geocities.com/frooha/grid/node2.html.

I guess this post is another fiasco! :(

The Cube of Resistance

2008-07-16T21:41:00.000-07:00

Sounds rather mystical! Haha. Well, this question is a really popular question, and it appears in this Physics textbook by Resnick and Halliday and well, in many other textbooks and publications. But I first heard of this problem from my SCONE senior Shaun Lin, who in turn heard it from his classmate Guoliang, so I shall credit this post to Guoliang (a.k.a. Chewie!).

Envision a cube of resistors as below, where there are twelve resistors, all of the same resistance value:

Please find the resistance between points A and H (Oh yes, I took the above figure from this PDF file at www.radioelectronicschool.net, because I couldn't really be bothered to draw so much anymore, it's killing my eyes and finger dexterity, haha.)

There's more than one way of doing it, but personally, I prefer one method over the other. So... I shall just leave this here while I go for lunch. You can go figure!

2nd Round

2008-07-16T10:09:00.001-07:00

This is the second question of the 1st IPhO - try it for kicks! (Click on it to view the enlarged picture - same for all other pictures in this post!)

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I actually remember doing this in Physics ‘S’ paper! Haha. So let’s give it a try, using my well, noob method. Let’s consider a simple system at the start first:

Obviously the resistance here is just the simple sum of two resistors in series:

Let’s continue to add two more resistors; notice that I’ve renamed one of the resistors x (you’ll know why in a second!):

This is a little trickier; the resistance here must take into account the parallel circuitry, which I’ve circled out in red. If you can’t see this, well, let me redraw it in a more accessible manner:

And of course, if you still remember the formula for parallel resistors, this new set-up has a fairly easy to calculate expression for its resistance:

And then let’s do one more addition of a branch to see how everything adds up, to understand how I’m going to simplify the situation:

Well, without redrawing it, I’d expect you to say that the set-up now consists of two parallel set-ups. For those who can’t see yet, I’ve tried to point out using circles again. Look above at the blue circle; I can represent this as a resistor x, and then the diagram is simplified into:

And what do you notice? Haha, we’ve already obtained the resistance for this, which was determined earlier to be:

So what’s x? Well, let’s find out; referring to the blue circle, I see that x’s resistance is simply equivalent to a set of parallel resistors:

Okay, so that you’ve roughly got my trick of simplifying, let’s do a long set so that you’ll be seeing what I’m seeing:

Let’s do the red circle first, and let’s call this new resistor a:

Now, treating the red circle as a resistor a, we can simplify the diagram into:

Alright, so now let’s do the blue circle and let’s call this new resistor b:

So now we can simplify again:

And now let’s do the green circle and let’s call this resistor c:

And now, let’s simplify again:

This couldn’t be simpler, and let’s do the orange circle this time and call it a resistor d, and find its resistance:

With this, we can then simplify our diagram as:

And the overall resistance is therefore simply:

Now, if you haven’t seen or noticed the pattern yet, let me just write out everything in full:

What a monster! You may want to view it in full size, haha. Notice that there is a repeating unit! And if the series goes on and on indefinitely, let me just re-write it in a more digestible way:

Can you spot the repeating unit yet? Haha. If you haven’t spotted the recurring unit yet, it’s simply this monster right here:

But what is this creature? Notice that the overall resistance R is just the repeating unit:

So how do we solve this thing? Well, algebra gives us a good method:

And then now we manipulate this expression into a quadratic equation, as such, to obtain R:

Which can then be solved using the standard quadratic equation (notice I’ve rejected the negative answer since no resistance is negative):

And as some of you might have already noticed, the Golden Ratio (phi) appears in the answer:

Such that:

Well, what is the Golden Ratio? Let’s leave that for another entry, because I’m absolutely tired of typing out equations and expressions for the day. I’ll give a hint though; recall the Fibonacci sequence as:

If you look closely, each succeeding term is obtained via addition of its two preceding terms. Well, if you take any term and divide it by its preceding term, you’ll obtain a number close to the Golden Ratio (phi). Let’s say we take 89 and 55:

And this ratio gets increasingly closer to the Golden Ratio as one continues on into the series.

And wow! What a lengthy post! And it's rather strange isn't it? Then adding more and more resistors forces your resistance into a fixed, finite value instead of an infinite resistance, and the fact that the Golden Ratio pops up in such an instance! Simply amazing! :p

1st IPhO Ever!

2008-07-16T01:34:00.000-07:00

I'm very sure everyone has at least heard of the International Physics Olympiad, an annual event that has its origins in the year 1967, which targets brilliant high school students who have an aptitude as well as an attitude for Physics.

Unfortunately, yours truly wasn't one of those brilliant students. :( Haha.

But in any case, I've gotten my hands on the first IPho paper, which was held in 1967, Warsaw, Poland, and here's the first question:

If you can't see the picture clearly, please click on it for a full-sized image.

But anyway, the first few olympiads weren't too tough, in my opinion - how do I know? Well the fact that I can at least do them shows that they're not as hard as the recent ones! However, the recent ones tend to be more guided.

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Neglecting air resistance, we can consider our system, comprising of the ball and the bullet (of course, with the inclusion of the Earth), to be not acted on by external forces – in such a system, we can then apply the Principle of Conservation of Momentum, which states that:

The total momentum in an isolated system remains constant before and after any event.

With this statement, we can then say that:

Initial Momentum = Final Momentum

What is the initial momentum? Well, at the start the bullet only has a horizontal velocity component and the ball is stationary, so we say that the initial momentum vector is towards the right, and is solely possessed by the ball. Specifically, we are comparing the total momentum along the horizontal axis. The final momentum must then be compared along the same horizontal axis, and thus we now have to consider the horizontal velocities of both the ball and bullet.

So let’s start by calculating the initial momentum of the bullet:

Calculating the final momentum of the system is a little more tricky and involved; first of all, let’s focus only on the horizontal component of the ball’s velocity, and notice that the horizontal velocity is in no way affected by gravity, such that it remains at a fixed magnitude throughout the descent of the ball.

Notice also that the horizontal distance through which the ball falls is determined by the time of descent as well as the horizontal velocity; and we are given the horizontal distance, and thus we need only determine the time of descent to figure out the horizontal velocity. So we proceed:

Where h is the height given in the question, u is the initial vertical speed (which is zero) and g is acceleration due to gravity. And therefore, we have:

With this, the horizontal velocity of the ball is then:

And now, with this piece of information, we can then determine the final momentum of the system:

Where v is the final horizontal velocity component of the bullet. Of course, by equating this to the initial momentum, we can determine v:

Now, we figure out the time of descent of the bullet, using the same equation of motion:

And well, it should’ve occurred to you that the bullet should reach the ground the same time as the ball does actually, since there is no air resistance, and all objects dropped vertically will reach the ground at the same time. So, we say that the horizontal distance, x, travelled by the bullet after collision is given by:

Amazing, the bullet travels 103 metres! A staggering distance as compared to the more massive ball (20 m).

Now, to find the amount of kinetic energy lost as heat, we simply find the difference between the total final kinetic energy and initial kinetic energy:

That simply means that 1158 J of energy was converted from bulk kinetic motion of the bullet into heat transferred to ball, and thus the fraction lost is then:

Gosh, that’s a mighty sum! :p

Another Youtube Conundrum

2008-07-14T18:29:00.001-07:00

Here’s another interesting video of a false proof I found on Youtube, which shows a rather interesting argument of why 1 = 2. See if you can spot the mistake (the mistake comes in rather early!):

I'll probably post the answer when I get home from work later, so you can have a go at it while there's time. Heh.

------------------------------------------

Well, the answer is easy: you can never divide by zero, because to divide by zero really means multiplying both sides of the equation by infinity, and everyone knows that infinity isn’t a number, it’s just an indication of the tendency of a number towards huge value.

For instance, look at the following treatment:

2 > 1

Hence: 2(∞) > ∞

Does it makes sense? Well, of course not! There’s no such thing as something that is twice as infinite as another! Infinity isn’t a number, but just a limit.

Harkening back to the question, notice that we allowed b = a, which means that (b – a) = 0. And thus this step is wrong:

Reminds me that grace is God’s way of loving us infinitely as much as an infinite God can already possibly love.

Youtube Conundrum

2008-07-14T09:24:00.001-07:00

Well well, look what I found on Youtube:

It's a false proof by the way: can you spot the mistake? :p

----------------------------------------------------------------

If you haven't figured it out, here's the explanation. The mistake lies in the very first line, where we have:

Now it makes sense that the right hand side involves a square root of (-1) squared - now a square root of something that is squared, immediately yields back itself, as such:

And so:

So how can you even write this first line down? It is a fallacy and not even an identity to start with! In fact, this is one of the problems Secondary School kids and JC kids struggle with. Do you know that what goes into the square root sign is important?

If you don't know what goes into the square root sign, then we can say both positive and negative answers are ok; but if you know that what goes inside is positive for sure, or negative for sure, then obviously the answer must be positive, or negative, and not both! I've shown this below for you to see:

So now, can we say that this is true:

Of course not! The left hand side has (-1) as its ingredients, and it is made explicit, and the right hand side has (1) as its ingredient, and it is made explicit as well. So how can you say that they're equal!

Don't get conned kids! :p

Sir, You're Making Things Too Hard!

2008-07-14T09:14:00.000-07:00

Sometimes I wish someone would help to tell my Physics professors that they make things too complicated. Haha. I was tinkering with my equation editor when I realised that:

Have a go at it. :)

Harmony

2008-07-14T07:07:00.000-07:00

I’m going to analyse a little bit of the nature of the harmonic mean, but first, let us consider where in nature does this type of averaging take place. In order to do so, let us consider a very simple PSLE question:

“I want to go to the beach from my house, and I travel there at an average speed of 30 km/hr. Going home, I travel back to my house at an average speed of 20 km/hr. What then is my overall average speed?”

Interesting enough, 95% of everyone I’ve asked (to be fair, I asked them to give me an intuitive answer without working) gives the answer 25 km/hr, the average of 30 and 20. But remember, this is the arithmetic mean – is it truly the answer? Well let’s do some algebra to find out:

Doesn’t the last statement look familiar? Let’s contrast what we have with the harmonic mean:

Yes! You have just used the harmonic mean to calculate the overall average speed! And the normal arithmetic mean will not work in this case.

Well apart from the fact that algebra obviously must work, is there a physical reason as to why we used the harmonic mean? There is! Recall I mentioned that the harmonic mean is a special mean that weighs the magnitude of the two numbers we’re averaging? Haha, yes, but how exactly does it weigh? I’ll give you an example; let’s say I have the following harmonic mean:

And I say

And of course, what does this mean? It means that we can then neglect the presence of the second x value in the denominator:

Therefore we see the harmonic mean actually weighs out both numbers and gives a number that sort of “favours” (technically favour isn’t the word but never mind) the smaller of the two numbers!

With this concept in mind, we go on to note that harmonic means typically occur in time-based problems. The above PSLE question was one good example. Another example is the concept of reduced mass in a diatomic molecule, where we say the reduced mass is given by the harmonic mean:

So what happens if say, the second mass (i.e. one of the atoms in the diatomic molecule) is super huge and massive? Then we can neglect the first mass in the denominator and say that:

So what does this physically mean? It means that most of the time, mass 1 is doing the vibrating of the molecule – this makes sense, because lighter atoms move faster, and thus mass 1 should be doing most of the vibrating of the diatomic molecule.

In relation to the PSLE question, the harmonic mean favours the smaller number, and thus gives us a value of 24 km/hr, which is closer to 20 km/hr. Notice that the person travelled to the beach at 30 km/hr and then back at 20 km/hr – so I ask you, which speed is used for a longer time? That’s right, the 20 km/hr speed is being used for a longer time! And this is the crux! The harmonic mean actually deals with the number that has the greatest time concentration (time concentration isn't exactly the right term here, but oh well)!

For instance, in a diatomic molecule, the smaller mass is moving most of the time, and thus the harmonic mean favours it. In the PSLE question, you spent more time moving at 20 km/hr, so the harmonic mean favours it. In another physical situation where you have rotating masses, the reduced mass favours the smaller one, because the smaller mass is moving most of the time, and the larger one is effectively stationary.

Therefore, the harmonic mean actually weighs the two quantities in accordance with how much time has been associated with each of the quantities. Oh well, this has been a weak, incomplete and unmathematical justification of the true nature of the harmonic mean, but it’ll do – I’m super tired from work, heh.

How Mean Are They?

2008-07-14T06:05:00.000-07:00

Recall that in an earlier post, I pointed out that there were four different kinds of means that commonly pop up in physical situations, and to recap your memory, they are:

The Arithmetic Mean

The Geometric Mean

The Harmonic Mean

The Power Mean

Oh and before anything, I shall just assume both numbers are positive, so it simplifies my working. :)

And of course, I’ll now endeavour to show the relative magnitudes of the above means I’ve stated, and I’ll start with the power mean (I’ll consider a specific one, the quadratic mean) and geometric mean first, since these two are easiest to compare:

Another easy pair is the arithmetic mean and the geometric mean:

The next easy pair should then be the arithmetic mean and the quadratic power mean:

Notice that I’ve used the same inequality over and over again, namely:

Is that amazing, or what? Let’s now recap:

So this leaves us with the harmonic mean, which can be a rather nasty affair, but with some tricks again, we’ll see:

There you have it: by proving that the geometric mean is greater than the harmonic mean, we no longer have to prove the other pairs of means, because the geometric mean was the least in our previous analysis. And therefore, we can now say:

So why have I determined the order of the magnitudes of the different means? Well, it's to show you that there is no such thing as "one mean is more mean than the other"; their magnitudes can still differ considerably from one another! Therefore, we can't really say one mean is better than the other, but rather, we have to look at the situation, and what the situation calls for, before we choose the appropriate mean to use.

Another conclusion can be drawn if you look at their mathematical structure: if you look at the geometric mean, one may think it's not giving equal weightage to both numbers; but what if you think about the powers? Multiplying two numbers together means you're adding their powers together, and taking the square root halves their powers added, and thus the geometric mean effectively gives you the mean of the powers of the two numbers.

How about the harmonic mean? It's interesting, because the product of the two numbers is weighed by the sum of the two numbers - it is as if this mean weighs the two numbers according to how big they are, or how small they are! I'll speak more on this in a later post.

And all of these things are nice to chew upon. Heh. :)

Quadratic Magic!

2008-07-13T19:49:00.000-07:00

I'm working at MOE now, and I'm drowned out by the boredom here! But after some reflection, I find the method of differences a rather interesting tool to help me derive some basic ‘A’-Levels identities and general formulae, such as a general expression for the sum as shown below:

Well, I guess I’d call this the quadratic series, and it occurs rather frequently in applications, and thus it’d pay to know how to derive a general formula for this sum. We start off by writing down the general term expression to work with first:

Of course, you might wonder why I wrote down the general term as a cubic term, but be patient! Next we find the difference as such:

Ah hah! Notice that by using a cubic term, we effectively forced out the quadratic term (together with a linear and constant term)! Looks really easy and neat so far! Now, the next step, as always, is to sum up the above difference to n terms:

Now, we’ve gotten ourselves a nice equation; to make use of this equation, we notice that we can simplify the two summation terms at the top of the above expression as such:

Voila! It becomes a sum of a quadratic equation! So if we rewrite this properly, separating individual terms into their sums, we have:

On the left hand side, we notice that the last two terms are particularly easy to reduce into simpler expressions, as such (well, you should know this if you’ve been through ‘A’-Levels!):

So if we now make a simple substitution, we obtain:

And yes, a simple, neat and beautiful proof for the quadratic series! :) Heh.

Meanie!

2008-07-13T08:51:00.000-07:00

You know, there’s the usual arithmetic mean (pops up nearly everywhere!):

And of course, there’s the geometric mean (appears in the calculation of the end point pH for a titration):

The harmonic mean sometimes pops up in physical solutions (appears in the calculation of reduced mass and average speeds):

And finally, there is the power mean (appears in the calculation of root mean square speed in the kinetic theory of gases):

I'll deal more with these different types of means in later posts, so keep a lookout for them!

Method of Differences

2008-07-13T08:28:00.000-07:00

[Edit: Oh my goodness, I took a look at Professor Adrian Yeo's derivation, and it's nearly the same! Or rather, equivalent! I feel good! Haha.]

I was doing some ‘A’-Levels tuition when I chanced upon a method to determine the general formula for many types of series. In fact, I'm about to derive the sum to n terms for the following series:

To start, let’s try this method out on the harmonic series:

And of course we can simplify this into:

So how do we start? Well, personally, I like to start with the Method of Differences, which is a very useful methodology taught now in ‘A’-Levels (but wasn’t taught in my time!), but often under-rated method. So let’s find the difference between:

This being the case, we then carry out a summation to n terms:

But do think about it:

And we already know that:

And therefore we write:

With this, we see that:

And immediately, we can substitute this back into equation (a):

Now the ingenuity of this method is that it allows us to determine the sum to n terms of the following series:

And voila, the sum to infinite terms can be seen to be a finite sum:

What an excellent and neat proof! :)

A Proof For 0!

2008-07-12T23:46:00.000-07:00

I still remember:

YYK: "Mrs Kan, why is 0! = 1? Shouldn't it be zero?"
Mrs Pauline Kan: "Well, it's just by definition!"

So here's how you can actually show that 0! = 1; first of all, the '!' function is pronounced as factorial, and not just an exclaimation mark. So what does this function do? Well, easy, if you have a positive integer n, then n! simply means to do the following:

n! = n(n-1)(n-2)(n-3)...3(2)(1)

Well, so this means that n! is equal to the product of all positive integers preceding n. But you see, there's another way to define n!, and notice that:

n! = n(n-1)!

And then you must agree that:

1! = 1(0!)

But we all know that 1! = 1, and therefore we must agree that 0! = 1. :)

Cards N' Such

2008-07-12T23:07:00.001-07:00

I'm reading through random articles and here's an interesting one from the Department of Mathematics and Statistics at Stanford University:

"The way that a magic trick works can be just as amazing as the trick itself. My favorite way of illustrating this is to talk about shuffling cards. In this article, I will try to explain how there is a direct connection between shuffling cards and the Riemann Hypothesis — one of the Clay Mathematics Institute’s Millennium Prize Problems. Let us begin with perfect shuffles. Magicians and gamblers can take an ordinary deck of cards, cut it exactly in half, and shuffle the two halves together so that they alternate perfectly as in figure one, which shows a perfect shuffle of an eight-card deck.

If the shuffle is repeated eight times with a fifty-two card deck, the deck returns to its original order. This is one reason that perfect shuffles interest magicians." - Persi Diaconis, Professor of Statistics and Mathematics, Stanford University

Different Dots!

2008-07-12T22:41:00.000-07:00

I wonder if people really know what differentiation is? What does this monster of a “dee-wai-dee-axe” mean to everyone? To be frank, there are so many sides of differentiation that I can’t put my finger down on an exact meaning that fits the nature of differentiation. Differentiation is simply one of the most beautiful tools in mathematics that allows you to obtain wonderful relations.

Differentiation, to Richard Feynman, was a game; he allowed for the convention:

And then differentiation was reduced to nothing but a mere game of placing the dots:

Differentiation has never been easier! Dot placing has been the most interesting way to do differentiation:

And it’s all about playing with dots. Sometimes one dot, sometimes two! But it’s all about using dots efficiently. I’ve just illustrated an extended version of the product rule in differentiation. I very much like to call it the ‘dot’ rule.

How about the quotient rule? Well, that’s easy! Looky here:

And if I put it altogether, we have:

Mod Sum Does It All - Part I

2008-07-09T17:55:00.000-07:00

I modified a mathematical trick that is a favourite amongst some, well, I like to call them Mathemagicians! And it goes like this:

First, pick out a three digit number, any one will do, that has its digits in decreasing order. Now remember this number, don’t tell me what you have with you.

Second, reverse the order of that number. Then subtract this new number from the original number you thought of. Having done this, remember also this new number that you’ve obtained via subtraction.

Finally, reverse this new number and add to itself.

Well, you have obtained a very, very, very special number! And because it’s so special, I know what it is already, because it’s simply 1089!

And of course this isn’t the trick, because now I’m going to up the difficulty level of this trick and impress you some more; I’m going to give you a range of numbers to choose from, and I’ll say this before hand: ‘the larger the number you choose, the harder it’ll be for me’. So here goes:

54, 1089, 3060, 42012, 80001, 103410, 500031

Now, just choose any one of the above numbers (don’t worry, they aren’t that special actually, I could give you one whole list of other numbers!). Have you chosen one yet? Well if you have, I’m going to up the difficulty one more time!

I want you to choose any number you want, any number with any number of digits, and I want you to multiply your chosen number by that new number you just chose. To make it really hard for my, I implore you to multiply it by say, a 3 digit number.

Now, if you’re trying this out, you’ve to tell me one thing, and I’ll tell you one thing in exchange:

If you tell me all the digits of your final number, save for one, I’ll tell you the last digit you left out.

I’m going to up the difficulty for myself again: you needn’t tell me the digits of your number in order and you can jumble them up. Still, I can tell you the last digit which you didn’t tell me.

Try it! And then you can always come ask me and challenge me, or just tag on the tagboard on the right. :)

An Equation for Bacon Sandwiches

2008-07-06T09:10:00.001-07:00

Well, well, what do we have here:

This is none other than what Britons refer to as a bacon butty! To me it's just a bacon sandwich, but it is apparently so important, that the Britons have come up with an equation just to determine how much force is required to produce how much crunch for every bite of the sandwich! Believe it or not, the research fellows at Leeds University spent more than 1000 hours testing (with their mouths and guts I presume!) dozens of such sandwiches!

And their brainchild? Haha, it amounts to no more than a mere equation:

To quote, "0.4 newtons should be applied to crunch the sandwich, creating 0.5 decibels of noise. The formula uses these values: N = force in newtons; fb is the function of the bacon type; fc is the function of the condiment or filling effect; Ts is the serving temperature; tc is cooking time; ta is the time taken to insert the condiment or filling; cm is the cooking method and C represents the breaking strain in newtons of uncooked bacon."

Well, don't take my word for it! View the original article here at The New York Times: http://www.nytimes.com/2007/04/11/world/europe/11bacon.html?_r=2&ref=world&oref=slogin&oref=slogin

Clapping for Kepler! - Part II

2008-07-05T08:50:00.000-07:00

I've been thinking of a way to explain Kepler's 2nd Law, and well, Newton provided two methods; in his Principle of Natural Philosophy, he outlined a geometrical method which involves highly accurate sketches.

Well too bad for you guys, I'm not about to go drawing things on my computer, because the angles to work out are simply too tedious! So kudos to Isaac Newton for working it out by hand.

But hey, he did come up with another easier method, which we all know now as Calculus - however, his work on his three laws of motion never once included the mechanics of calculus in his proofs.

So... how?

Like that lor.

No lah, of course not - let's just go through the main idea of using Calculus in proving Kepler's 2nd Law. Now, let us say we have an area A that is swept out by a planet moving around the Sun. If we can prove that the rate of change of area swept out (i.e. dA/dt) is zero, then we can then prove that the area swept out doesn't change with time!

Hurhur. Sounds easy.

Well, strictly speaking the area we should be considering should really be the area below, which shows the area swept out by Earth with the Sun at one focus:

But now we're considering the infinitesimal area that the planet sweeps out, and thus we are allowed to consider dA instead - that is, we consider the area swept out in a very small amount of time, such that the area swept out can be represented adequately by a triangle as such:

Well, I've removed the Earths for clarity's sake; and you may not yet be convinced as to why this small area can be a triangle - it is clearly a segment, is it not? Well, let me enlarge the diagram for you:

Are you now convinced that the segment is effectively reduced to a simple isoceles triangle when infinitesimally small areas are considered? It's as if the radius vector's length didn't change much! That's why we consider dA instead of A itself, and this allows us to work with nice approximations that hold. With this in mind, let us determine a formula for dA and I've enlarged the triangle OAB (yes, enlarged again!) for your viewing pleasure:

Notice that I've obliterated the Sun and the trajectory as well for clarity (Physicists tend to do that, haha!). Now,

Clapping for Kepler! - Part I

2008-07-05T08:29:00.000-07:00

Have you heard of Kepler's three laws of planetary motion? Do you even know who Johannes Kepler is? Well, he was the great astronomer and mathematician who worked on information on Tycho Brahe (another interesting astronomer!), and realised that planets don't move in perfect circular paths around the Sun!

Well, you might say that's rather commonsense - but try saying it about 250 years ago, when no one knew of Newton's Laws of Motion, and when everyone thought that force was required to sustain motion! Clearly then, it would've been an astronomical task (no pun intended!) to derive and deduce the path taken by the celestial planets.

Let's dive in immediately, because I simply cannot stand dwelling on simple historical facts anymore; to get into the heat of action, let us have Kepler's 2nd Law in a nice statement:

Equal areas are swept out by the radius vector connecting the planet to the Sun in equal times.

Well, I'll explain that in another post, but harkening back to what I'm really interested in for this entry, let us assume for a second that there is no gravitational force or whatsoever in this Universe. Would Kepler's 2nd Law still hold? Well, very much! And I shall endeavour to prove this to you now.

First of all, what does a force do to the state of motion? Easy, it changes either the direction and/or the magnitude of the velocity of the body in motion. Next, where exactly and in what direction does gravity act if you're travelling around the Sun? Easy also: gravity acts also in a straight line towards the centre of the Sun and acts at your centre of mass. Basically this means that the gravitational force is a radial force, a force that acts towards the centre of the mass providing the gravitational field.

So, let us now assume that the Sun is infinitely massive with respect to the Earth or any other object (and it is rather much so!), which is justified by its very much larger mass - if so, then we can neglect the Sun's motion and say that it is effectively in rest in our mathematical treatment. Now, let us have a moving rocket - this rocket moves up, in a straight line, at constant velocity of 10 m/s. And we have three positions outlined below, A, B and C, all in equal intervals of 5 s. So let's refer to the picture below that I've so kindly (ahem!) drawn out for your ease of interpretation of the physical situation at hand:

Now, I've mentioned already this is a thought experiment to see what happens to Kepler's 2nd Law if there were no gravity, and this explains why the rocket has no force on it and continues moving in a straight line with constant speed.

So, the two triangles to compare now, are obviously triangles OAB and OBC; notice that both have the same height h, and because the speed of the rocket is the same, and equal time intervals have passed, we must agree that the rocket travels the same distance in the same amount of time, and therefore AB = BC.

What then is the formula giving the area of triangles? Well, if you've forgotten, it's:

Area of a triangle = 1/2(base)(height)

Since OAB and OBC have the same height h, and the same base (AB = BC), we conclude that they must have the same area! And this must then mean that Kepler's 2nd Law holds true even in the absence of gravity! That equal areas are swept out in equal times. :)

Mathematical Synthesis

2008-07-04T19:23:00.000-07:00

I know all of us know what long division is, but do you know what synthetic division is? Well it's a very fast method of calculation taught to me by my Additional Mathematics teacher in Secondary Three, Mr Michael Doyle. However, there are some limitations and downsides to using this method. I shall illustrate using an example:

Let's say I'd like to simplify the following fraction using long division first:

Notice that I haven't done it step by step because I'm assuming you, the reader, knows enough of long division already. Well notice the important things: we have a polynomial in x, being divided by a divisor D(x), which results in a quotient Q(x) and a remainder R(x).

We now introduce what we call the remainder theorem, that is, for any polynomial, we can write it as such:

If the remainder R(x) = 0, then we say that the divisor D(x) is a factor of the polynomial. So clearly for this particular polynomial we have it written as:

Now, that being done, we naturally have to ask ourselves: is there an easier method? Well, there is! And this method is known as synthetic division. We shall use the previous polynomial to illustrate. This method requires us to write down the following first:

Notice that I've written down the coefficients of the polynomial - these coefficients will represent our polynomial of interest. Now, to the left of these coefficients is the number -3, which you should recognise as representing the divisor (x + 3). We put -3 to represent (x + 3) because earlier we said that:

Now clearly when x = -3, notice that the divisor term (x + 3) goes to zero, and we're only left with the remainder term (x - 6). So to divide by (x + 3), a really simple method is to substitute in the value that makes (x + 3) = 0. Now, there is a basic limitation: clearly then the remainder is obtained no longer as a polynomial, but as a specific number. So this explains why long division is preferred in some instances.

No matter the case, we move on the next step, as shown below:

And of course, the next step:

And the last step:

Our final presentation is therefore:

Voila! Do you see how synthetic division has worked out the same exact answer as long division? Try it yourself today!

On The Dimensional God

2008-07-03T01:45:00.000-07:00

In trying to determine whether it is essential to have an intelligent Creator or not, I have written, rewritten and investigated many possible reasons, and none are so persuasive to the human intellect as to those which I shall mention and discuss here. To discuss at length the exact mathematics involved would be a rather lofty task, one which I dare not undertake in view of the numerous other more qualified experts in this world can verify, but nevertheless, I’ll persist in providing a more technical view because I think through it one can see the validity of my claims and judge for yourself whether it makes sense to have an intelligent creator or not.

I however, admit that all of these evidence merely point towards an engineer of precision who placed the universe into its intricate clockwork, and it is in no way sufficient to prove and convince anyone that it is the Christian God who single-handedly created this world. I leave it up to faith and conscientious work on the part of the individual to realize for himself that the God of the Christian Bible, is the God that created this world.

There are main intellectual barriers that simply dismiss that the universe itself could have ever come into being and existence solely due to chance alone:

1) The rejection of evolution based on statistics and thermodynamics.

2) The Big Bang model that necessitates a Creator.

3) The use of dimensional analogy to explain how the nature of a Creator allows for his role.

4) The precise intricacy of the universe that can only result from Creation.

Perhaps it is a little mundane for me to harp on the same piece of evidence over and over again, in which case I've done so with several friends over the course of this week itself. That being said, let me offer you my views of retrodimensional analysis, which makes use of dimensional analogy to show how Jesus himself can be fully God, and yet fully man at the same time. And that brings the emphasis of this entry to point number 3.

You got that right (wow, what a sudden change in tone!), Jesus was fully man and God, but how do we even come to understand such a fact? Surely it can be justified and rationalised? Of course it can, and we do so first by doing away with certain disputes.

The first dispute is that Jesus died, and God does not ever die, and therefore Jesus can't be God, for he wouldn't have died. Now, in saying this, you have just proven yourself wrong, because Jesus didn't die - He rose again on the third day (well, technically speaking, it wasn't a full three days actually) and therefore He is said to have overcome death itself. Surely that is a feat that only God himself can achieve?

The second dispute is, if Jesus really were God, then how was it possible that He was manifested in the image of man, and yet not lose his deity? That is, I'm asking, how is it possible that both man and God can both be two sides of the same coin.

Alright, I'm really making a big claim here, so if you know I'm wrong, please do correct me; for the scriptures we have read clearly say that "So God created man in his own image, in the image of God he created him; male and female he created them." - Genesis 1:27.

In His image, get that settled in your head first; and I would take that to mean that man is but an image of God, and therefore it is possible that in each and every one of us, we see the slightest hint of God.

So scripture says so, does science say otherwise? Well, this question has to be dealt with by geometry, a branch in mathematics, but to be more specific, hypergeometry. Let me explain further the basics of dimensional analogy before we go on; dimensional analogy is simply the study of the relationships between a dimension of order n and another dimension of order n-1. That is, the study of how higher and lower dimensions relate to one another.

Sounds rather abstract at this point doesn't it? Well let's give you an example: we all know all 3-dimensional objects are encased by 2-dimensional surfaces (for instance, a cube has six surfaces) and all 2-dimensional surfaces are encased by 1-dimensional lines (like how a square is enclosed by four lines). Of course, a single 1-dimensional point would be enclosed by empty space (zero dimensions). The intellectual difficulty in visualising four dimensions can then be partially reduced by making the logical leap that a four dimensional object is enclosed by 3-dimensional volumes.

Well, that doesn't make things all that better, but at least we've now one extra piece of information to use to analyse things of a higher dimension. So let's give you another piece of information to help you bolster your thoughts; 3 dimensional beings like us, naturally live in a 3 dimensional world (we have 3D space and time), while 2 dimensional things will exist in a 3 dimensional space (like how a piece of paper must exist in a 3D space, because stacking two 2D pieces of paper automatically generates the existence of depth, introducing 3D qualities and thus 2D objects, in order to exist independently of its own geometry, must be in a 3D world). So of course, a 1D object must be living in a 2D world, and in a 1D world, only things of zero dimensions can exist (that is to say, nothing really exists in a 1D world except for dots).

May I pause for a second to say that God was wise to create us in a 4D world? :) For life without time is simply a life without existence, for it is time that gives meaning to many things, and it is time that builds up many things in life.

Now, I shall give you one more piece of information for your understanding; when you have a three dimensional object and you cast light say, directly overhead, you'll obtain a 2D shadow. Of course, a 2D object like a sheet of paper, if you cast light along its edge, you'll obtain a 1D shadow in the form of a line. And if you have a 1D object like a string, and cast light head on, you'll obtain a point like shadow, that is, a shadow of effectively no dimensions.

So have you caught on to what I'm saying yet?

If you haven't made the logical jump, it's alright, let me illustrate. For instance, let us assume we have a 4D being - wouldn't this 4D being first of all live in a world of higher dimensions as we've just said? So this world would then be effectively 5D, and of course I'm assuming this higher being is a 4D being, but it could be of any dimensions higher anyway. Now, when a shadow of this 4D being is being cast, wouldn't it be a 3D shadow then, which we realise to be an object that can be visualised in real space and time.

Now let's make the bold hypothesis that this 4D being is God - then the 5D realm could be heaven (notice I'm being very conservative with my language here for this is just a mere hypothesis). Now that explains why we can't physic ally see God nor heaven, for they are of higher dimensions!

Now, you're now probably getting what I'm talking about - earlier, I was hinting at God being able to manifest himself as Jesus to us, visible to the naked eye, because a shadow of God would be something that can be materialised as an object in real space, or at least in our world. But to say Jesus is a shadow of God completely misses the picture, for a shadow is not the object that casted it, but a mere aspect of the object, and shadows can be distorted, and they aren't the complete replica of the object, for the sense of depth, length and breadth can be distorted using shadows.

I know I know, you're probably saying now 'but you just talked about shadows, so why not?' Well I'm saying now that the example I gave isn't really that adequate, for it doesn't tie in with the true nature of who Jesus really was. Let me now borrow an idea from Edwin Abbott, the author of Flatland (a truly exciting and mind blowing book written way back in the 1900s!).

Well, imagine you have a 2D world (well, a piece of paper!), in which 1D men and women live (oh those poor souls! I'm talking about stick men basically); imagine one day, there's this 3D person that comes by - let's say he's a cube, in quite the literal sense. So if this cube were to sit on the 2D world (ouch!), what would the 1D people see? Well, the 1D people can only perceive as much as the 2D world will allow, and what does the 2D world allow? The 2D world allows only one surface of the cube to be on the paper, meaning all the stick men will see, is a square.

That's it! Jesus, is simply God, and the above analogy helps us to understand that. Let me elaborate a little further - the square that the 1D people see, is not a square at all, but is simply something that the 1D people are forced to see because they cannot see further. So is it right to say that the 2D square is a shadow of the 3D cube? Well, no! And a rather resounding no at that!

Why? Because the square is part of the cube, and the cube is part of the square, but of course we do see that the square must eventually return to be part of the greater cube, for it is in the realm of higher dimensions that the cube can really be in all its glory of having 3 full dimensions.

So isn't that what Jesus is saying? He says: "You heard me say, 'I am going away and I am coming back to you.' If you loved me, you would be glad that I am going to the Father, for the Father is greater than I." - John 14:28

Clearly, Jesus is saying, if you love Him, then you must understand through death, he will return to where he truly belongs. If you read the original Greek, it doesn't mean Jesus is going back to his father, but that Jesus will be glorified in going back. (Tells you how important Greek is huh?)

So clearly, the square isn't separate from the cube, for the square is just what they see of the cube, and the square was never apart from the cube - the square retains the quality of the cube just as much as it appears as a square to the eyes of the stick men.

And just as this has made clear, that is how God shows Himself as Jesus. Now, I'm making a very bold claim in saying this, and it is by no means, scientifically justifiable - but I am making an assertion that an engineer once made in a VCF (Varsity Christian Fellowship) meeting, just that I'm now piling up more information, now that I've acquainted myself more with the topic of dimensional analysis.

So why must God be a 4D or higher dimensional being? Well, let me put it this way: if you believe in a Creator, what attributes would he have? For starters, he must exist outside of our dimensions, for only one who is of higher dimensions can effectively create something of lower dimensions. It completely makes sense. 2D beings, with no sense of height, could never visualise buildings, for buildings require the vision of height and depth. But they could most certainly feel it, because they would know that something that has height cannot be simply passed through. Similarly, 3D beings can't visualise time - it is simply something to be felt.

Only the Creator of higher dimensions could effect the Big Bang (now that is to be included in a later discussion), for the Big Bang was something that happen in spacetime fabric.

Now, how does God see into your heart? This can be explain by retrodimensional analysis as well. For instance, if you're a three dimensional being, you see all sides of a 2D object at once. If you're a 2D being, you see the entire 1D object (all sides doesn't make sense here). If you're a 1D being, well then, haha, there's nothing much left for you to see.

But the point is, a being of higher dimensions, sees all aspects of objects of lower dimensions. Now, the higher the order of your dimensional existence, the more you can see of and into objects of lower dimensions. That being said, I shall leave the rest to your intellectual wit. :p

This concludes all I have to say on the dimensional nature of God - but please note that God is more than just plain dimensions!

Psychic!

2008-05-03T09:14:00.000-07:00

Have you ever thought about it? A proton is just a proton, and it has no orbitals or whatsoever to speak of, because it has no electrons, and it is simply a lone existence. However, when we talk about a hydrogen ion, we speak of it possessing orbitals into which electrons are added when a hydrogen ion accepts them.

But, isn't a hydrogen ion just simply, a proton? Why then do we speak of it as if orbitals exist?

The matter, clearly then, is that orbitals themselves simply don't exist at all! And that is why one can speak of them in hydrogen ions, and not speak of them for a proton. For an orbital is merely a probability space of finding an electron, and not an actual dimensional entity that an electron can occupy.

And that is how a proton itself can possess all orbitals at once, and yet, possess no orbitals at the same time.

Wow. Deep, sweet!

So... who tells the electron where to go? And how does it know where to go when there's no one telling it?

Haha, that's what interactions are for!

Enthalpical Feat

2008-02-23T22:11:00.001-08:00

Here's a very good question:

We all know that the enthalpy of vapourization varies with temperature, that is for a phase transition of the form A(l) to A(g), the enthalpy change ΔH will generally change with temperature. Therefore, the question in mind is: what do you think the value ΔH for vapourization is when the temperature tends towards infinity?

A good Physical chemist should have a good gut feeling! :p

Entropical Catharsis

2008-02-09T02:46:00.000-08:00

If you haven't noticed, I'm sticking to a very wise mantra by posting short tidbits these few days:

"A man of few words... is a wise man!"

Very true indeed - I often find myself the butt of all jokes whenever I open my mouth too many times.

In any case, I was reading an excerpt from a Statistical Mechanics book when it occurred to me just how much I've changed in the past four or five years:

*Back then in Physics Olympiad training tutorials with Mr Daniel Khor*

YYK: ".... alright, remember this Yong Kiat, entropy change is good if it's positive! Because that's the correct answer, can get you marks!"

*Now after 4,5 years...*

YYK: "Oh man, entropy goes up again. We're screwed. This sucks. We're all going to die."

Haha, heat death of the Universe never did sound appealling to me, just that I was always too caught up in getting the right answer.

Absolute Zero Ain't Zero-Point

2008-02-08T18:17:00.001-08:00

I was just wondering:

"Even when the temperature reaches 0 K, atoms in a solid still manage to vibrate with a zero-point energy..."

It's as if even when all of the energy has been sapped from particles, the particles are still resilient, showing us that there is some forbidden life-energy that cannot be touched no matter how low the temperature goes. It's amazing isn't it? It's as if everything around us has some fundamental life-force!

Reminds me of a verse from the bible, but too bad I can't remember it. :p

Water (II): Resilience

2008-02-06T19:57:00.000-08:00

I just checked out Mount Everest's height at Wikipedia, and it's at a staggering height of 8,840 m! You know what that means?

That means I, weighing 95 kg, would need approximately 8840(95)(9.81) = = 8238438 J of energy to scale that mountain fully! What a heck of a mountain!

Now just look at this: let's say I have 62.5 moles of water - that isn't very much, just about 1 litre of pure water. Assuming it is at room temperature, and assuming that I need very hot water at 373.15 K to make good hot chocolate for my family, the energy I need is given by a simple calculation:

I've done the above calculation assuming that the boiling process was done at constant volume inside a simple flask, and I used the latent heat of vapourization and specific heat capacity of water, as above. I mean, you probably won't boil away all the water right?

What in the world? I climb so high up, and all the energy I expended, can only be used to boil 3 litres of water? This is ridiculous!

Indeed. Heh. :p

Granted of course, I've neglected the inefficiency of the human body, the heat loss from the body due to the cold, the sapping of mental strength, the time loss due to sleep, as well as the energy loss because of additional weight (i.e. equipment). So the above calculation isn't really anything actually. :p

Pre-Eminent Molecule: Water (I)

2008-02-05T20:11:00.000-08:00

God uses water in the bible, in very powerful ways - He used water to flood the Earth in the Noah's Ark saga; water is used for baptism, even to baptise Jesus; Jesus turned water to wine; Jesus and Peter both walked on water itself; Moses knocked his staff on a rock and water came out, springing forth life, that quenched the dying Israelites' thirst and the list is endless! The bible makes so many referenes to the humble water molecule, and it's a really incredible molecule!

I bet you the only reason why it's such an unsung hero in our lives is because it seems to have an ubiquitous existence, and that's why we tend to take the precious water we have for granted. I feel that Singaporeans have themselves taken the potable water flowing out of their taps for granted, myself inclusive.

Therefore, inspired by my friend, Shaun Lin Darong, I shall now start posting a series of posts on the water molecule, and the first post is a simple one, to whet all of your appetites: have you ever wondered why the water molecule can only accept a single proton when it has two lone pairs of electrons?

What I'm trying to say is, why can't the following reaction occur:

After all, water has two lone pairs of electrons right? So why not? A seasoned Chemistry student would simply retort: Of course it can't! The first protonation results in a positive charge on the water molecule, which naturally repels the addition of another positively charged proton!

However, another could simply argue - then if I just increase the concentration of acid in water, eventually this will result in both lone pairs of electrons on the oxygen being protonated right?

And the argument could go on endlessly in fact, and we'd have no idea who's right! The problem with such arguments is that all of them are purely qualitative explanations, and there is no rigour to back them up at all. So how should we go about solving this problem?

Easy, we fall back to first principles: we do a simple molecular orbital calculation! All will come to light when we analyse the water molecule from its very fundamental electronic structure, and here goes:

Water is a simple triatomic bent molecule, with a bond angle of 109.5 degrees - upon closer examination, we see that its point group is C2v, such that we say it has C2v symmetry:

I've included the symmetry operations above for reference, and here is the associated character table for C2v molecules:

So let's start assigning each irreducible representation in the table to each orbital in the three atoms of water! Let's however, not forget one postulate in molecular orbital theory, that in order to form bonds, orbitals must go into overlap effectively, both in a spatial and energetical sense. However, this means that the only orbitals in effective overlap will be the valence orbitals, since the inner core orbitals are buried deep within the atoms. Therefore in our analysis, we shall only consider the valence orbitals of both oxygen and hydrogen atoms.

If so, let us then consider the symmetry labels of the oxygen valence atomic orbitals first, where we concern ourselves with the 2s and three 2p orbitals. Taking a look at the 2s atomic orbital of the central oxygen atom, we see that:

Indeed, it has a symmetry label of a1, corresponding to the totally symmetric irreducible representation of A1, since all operations result in the same 2s orbital. If we take a look at the 2pz orbital, we find that it also has the a1 symmetry label:

The 2px and 2py orbitals however, have symmetry labels of b1 and b2 respectively, because they correspond to the irreducible representations of B1 and B2 respectively (please verify this yourself by looking at the diagrams below and performing the symmetry operations yourself!):

If so, then we can summarise everything in a single table, on the oxygen atomic orbitals:

It is now time to take a look at the hydrogen atomic orbitals, namely the 1s atomic orbitals; however, we require not the 1s atomic orbitals but rather, a symmetry-adapted linear combination of the 2 1s atomic orbitals, which means that we could either have an additive linear combination, or a subtractive linear combination (note that this is only the case for simple additions of two atomic orbitals, since these are the only two combinations that can result). The SALCs are illustrated below:

I coloured them differently to show whether they're in phase or not, and leaving the verification to the reader, these two SALCs have symmetry labels of a1 (additive linear combination) and b2 (subtractive linear combination), and thus we summarise this in a table again:

So let's plot the molecular orbital energy level diagram:

One now easily sees that the b1 molecular orbital is non-bonding, indicating a lone pair of electrons centred on the oxygen atom a la 2px fashion. Where is the other lone pair of electrons? Easy, it's the lowest a1 molecular orbital, which is also non-bonding (notice it isn't lowered very much in energy), which has the the form of a 2s atomic orbital centred on the oxygen as well.

But what does this mean? Notice that the b1 molecular orbital has an energy of -13.6 eV, which is just nice for the bonding to a proton, whose energy is also at -13.6 eV! Howevr, the lowest a1 molecular orbital is simply too low in energy (i.e. too unreactive) to even bond with a proton, and therefore we expect only one proton to be able to bind, for there is only one reactive binding site.

Wow. This post has been heavy on graphics. :p

Final Fantasy Elixir: Ethers

2008-02-05T18:37:00.000-08:00

I was thinking, what better way to revise for Organic Chemistry, than to share it with others on this blog? Plus, all my peers in Chemistry will be reading this blog, so I guess it's the best way to learn Chemistry!

Alright, here goes!

Let's start off with the synthesis of ethers; we've learnt that alcohols undergo an intramolecular dehydration in order for it to become alkenes, but did you know that alcohols can also undergo an intermolecular dehydration to become ethers? From my point of view, this is the easiest way to partition your way of thinking in order to better understand the two competing reactions.

Of course, there are other ways, heh. But let us just keep this point in view for the moment: only primary alcohols may actually undergo such a dehydration reaction to produce ethers.

The fundamental difference between alkene-dehydration and ether-dehydration is that while the former is a simple elimination reaction, the latter is a simple nucleophillic substitution reaction, in particular a SN2 reaction. The nucleophile in this reaction is the alcohol itself, and the substrate is a protonated alcohol, an alkyloxonium ion.

Recall also that elimination reactions are favoured at higher temperatures, and this explains why the alkene forms when the reaction is carried out at 180 degrees, and the ether forms when the reaction is carried out at 140 degrees.

As shown above, the first step is a simple acid-base reaction, where the lone pair of the alcoholic oxygen functions as a base that forms a bond to a proton from the concentrated sulphuric acid. This grants the oxygen a positive charge that polarises and weakens all bonds connected to oxygen. The alcohol, being in excess as the solvent and reactant, functions as a nucleophile (now that the hydroxy group has been converted to a better leaving group) and attacks the hydroxy carbon in a SN2 mechanism (a.k.a back-side attack!) in a single step to produce the protonated ether (you can call it a dialkyloxonium ion as well!). The last deprotonation step is then carried out by other alcohol molecules.

However, this method of synthesis is only good for industrial synthesis of large amounts of ether, and has some very obvious limitations. For instance, unsymmetrical ethers are almost impossible to synthesise with a very precise degree of control! Look at the following example:

Indeed, if two different alcohols are used, then three possible ethers can be formed! Such lack of control is definitely not a desirable quality of synthesis reactions. Moreover, let us consider another important factor - by virtue of the fact that this is a simple SN2 reaction, it means that secondary and tertiary alcohols are unlikely substrates, because using them leads to the hindering of the formation of the transition state, and therefore, elimination is favoured instead, resulting in alkenes as the major product.

A much better synthesis route is known as the Williamson-Synthesis, a rather simple trick but oh-so-effective! Just take a look:

The leaving group L can simply be any good leaving group, such as a tosylate or even a halide group. Notice that it is simply another SN2 reaction, but one uses a much stronger nucleophile, and one uses a much better leaving group in the substrate! It's just playing around with mechanisms, really! This is a much better method for producing asymmetric ethers, because there is no ambiguity to the structure of the ether that may be formed!

However, if this is still a SN2 reaction, then the usual limitations hold for this reaction scheme as well. What I mean to say is, if the substrate is a secondary or tertiary halide, then elimination is bound to occur significantly. Therefore, one must use proper conditions such as the choice of an aprotic polar solvent, lower temperatures etc. to favour the SN2 reaction. In which case, if you want to form an asymmetric ether that has a bulky alkyl group R1 on one side and a relatively unhindered alkyl group R2 on the other side, one should use a R2 alkyl halide and a R1 base for the synthesis to favour a SN2 reaction.

There is also another class of ethers that are very special, which we term as epoxides, which is simply a short-form nomenclature for epi-oxide molecules, a three-membered ether ring, which is also known as an oxirane, as follows:

A very easy method to synthesise such epoxides is to use an alkene and a peroxycarboxylic acid, where the acid actually forms a cyclic transition state with the pi-bond of the alkene, and the mechanism is a single concerted step where all bonds break and reform at once:

We can therefore think of the acid as being a very good electrophile, and the alkene being the nucleophile in this reaction!

Now, because there is a three-membered ring, such epoxides are rendered highly susceptible to nucleophillic substitution due to the ring strain (i.e. it is favourable to break the ring that is highly strained), and this can be done in two ways, via an acid-catalysis or a base-catalysis.

The acid catalysis mechanism involves the protonation of the epoxide oxygen on the lone pair of electrons, resulting in a better leaving group, and also polarises the bonds connected to oxygen. The nucleophile in this case, is water - notice that this type of catalysis makes the hydrolysis easier because a better leaving group and a more reactive carbon centre is formed.

Notice that the reactive substrate is a protonated epoxide, meaning that the intermediate has a positive charge. Similarly, this means that the transition state involves a species with a positive charge that needs to be dispersed effectively. Indeed! One notices that if an asymmetric epoxide is being used in this reaction, the two oxygen-carbon bonds are of different length - in fact, the oxygen forms a weaker bond to the carbon that is more able to retain a positive charge (i.e. the more highly alkyl substituted carbon atom).

In fact, in the transition state, it is this bond that breaks, because then the carbon is more able to disperse its developing positive charge as the nucleophile attacks. Therefore acid catalysis usually results in the more hindered carbon being attacked (but not exclusively though!).

The base catalysis mechanism doesn't invoke the use of a better leaving group - it focusses much more on providing a high concentration of a very strong nucleophile that does the attack (in this case, a hydroxide ion). Notice that the leaving group is highly basic (an alkyloxonium ion) that immediately protonates itself in water to stabilise itself.

However, this attack is more of a nucleophillic nature, and therefore we see that it is a simple SN2 intramolecular reaction (sort of!). As such, the nucleophillic prefers to attack the more unhindered carbon atom.

And well, I guess I've run out of things to say. :p

WTH - What The Helmholtz!

2008-02-01T12:40:00.000-08:00

Yes, one of the many great names embedded into Thermodynamics, and one often wonders what one has to do to engrave his name into the annals of history. Well, let us take a look at what Gibbs and Helmoholtz formulated, which van't Hoff eventually utilized to derive his very own van't Hoff equation. We're going into a very useful relation in Thermodynamics, which is known as the Gibbs-Helmholtz Relation, which is simply:

The interesting thing about this equation is that if we know the enthalpy of the system, we can then know the temperature dependence of the Gibbs free energy, which is very useful in many situations. For now, let us focus on the derivation of this relation. From first principles, we do a product rule differentiation, where we obtain:

Now, using what we've all learnt in Physical Chemistry class, we have the following two relations:

If we do a simple substitution, we then obtain:

There you have it! A simple derivation of the relation, which holds for all systems, because we have made no mention of the system at all, and thus this must be a very general equation! (Am I the only one who's so excited about this result?) But you may ask: why would I need this result? And how would I use it? Good question!

Far too often one teaches the Gibbs-Helmholtz relation without any application, and I came up with a simple example to illustrate how to use this relation properly. Let us recall the definition of enthalpy:

For a monoatomic ideal gas, we must also agree that:

And if we do a substitution into the Gibbs-Helmholtz relation, we obtain:

And therefore the result becomes:

We can now integrate this expression with respect to temperature, and we actually obtain a very neat expression:

If we graph this equation out, namely on a plot of G versus T, we obtain:

Interesting! Notice that there is a maximum point! To find this maximum point, we take the derivative of this graph (i.e. gradient function) and set it to zero:

We find that we end up with:

And solving for the temperature, we have:

But what does this entire analysis mean to us? Let us fall back to a more qualitative physical explanation: the Gibbs energy of an ideal gas is zero at absolute zero, because both its enthalpy and entropy are zero at this temperature. However, instead of the Gibbs energy decreasing when temperature is raised, the Gibbs energy increases initially! This is counter-intuitive if one thinks of the Gibbs energy as the amount of work available - shouldn't this decrease when the temperature is raised because at higher temperatures there is more entropy?

This just means that at lower temperatures, the enthalpy content must be larger than the entropy content! Which makes sense! Because of a lower entropy, more work can actually be harnessed from a system. But what's the drawback? Your system can't really do much work in the first place!

In such a case the two laws of Thermodynamics tells men that they are not real masters, for the 1st law says that you can only break even, but the 2nd law says that you can't even win!

This situation also allows us to determine the absolute entropy of an ideal gas! Recall that the slope of the graph was calculated previously, but this slope of the graph is the derivative of the Gibbs function with respect to temperature, which in effect, is the negative of the entropy! Therefore we see that:

Since the term in parentheses must be negative when T is lesser than 1, and entropy is never negative, we must insist that c must be a negative term!

Perhaps one might wish to work out a more exact relation for a Dieterici gas? Or perhaps a van der Waals gas?

Heh. Try it!

Volume Up! Volume Down!

2008-01-30T00:42:00.000-08:00

My friend and classmate, Aaron Thong (currently a teacher in ACJC teaching Chemistry!) asked a very interesting question which he thought of after a student asked him a question, that is:

For a reaction of the form NaOH (aq) + HCl (aq), does the final volume of the system increase, decrease, or stay the same, on the provision that there is no loss of mass from the system?

For those taking Physical Chemistry this semester, this is a typical problem that makes use of partial molar quantities (if you haven't heard of it yet, how can you be a University Chemistry student? Haha.), and well, have a go at it!

Kudos to Aaron!

Canon Not In 'D'

2008-01-27T08:01:00.000-08:00

Yup, I'm talking about a canonical ensemble here; my Physical Chemistry professor, Dr. Fan Wai Yip, completely missed out this essential concept in my course last semester, which is essential to the concept of partition functions, statistical weight and the entropy of systems.

So what exactly is a canonical ensemble? Let us consider what it means by the word ensemble first:

Above we have a collection of atoms - any simple collection of microscopic objects can be referred to as an ensemble. Typically, we consider an ensemble to be a macroscopic system. So now, what is a canonical ensemble? Well, look at this:

Well, a canonical ensemble is simply a collection of ensembles as above! It is simply partitioning a huge, gigantic macroscopic system into smaller subsystems, each one still a macrosystem by itself. What this means is that each subsystem by itself, still obeys the statistical laws of statistical mechanics perfectly well, and therefore, by calculating the properties of one subsystem, one can simply multiply that property by the number of subsystems (if it is an extensive quantity) to obtain the overall property or quantity.

And the reverse is true: if you can calculate a quantity for the entire canonical ensemble, then it follows that you can divide by the number of such ensembles to obtain the quantity for the small subsystem by itself.

We shall now use this idea to derive a very general expression for the entropy of a system in terms of the probabilities of a macrosystem being in a certain energy state r. Let us consider a very general case where we have a huge macrosystem that can be partitioned into n identical subsystems. Each subsystem will then be in thermal contact with one another, interacting weakly, and each system will also have the same probabilities of being in certain energy states.

Let us then assume that each subsystem can exist in energy states 1, 2, 3, ..., r... and we allow the associated probability of each subsystem being found such an energy state be p(r). Then for n such subsystems, the number of systems to be found in the energy state r is simply:

With this, we then proceed to calculate the number of ways all of the subsystems can be partitioned within the huge system. What I mean to say is that we now proceed to determine the number of possible ways to distribute all of the subsystems into the various energy states 1, 2, 3, ..., r ... and we do this by determining the statistical weight of the entire canonical ensemble:

The corresponding entropy of the entire ensemble is then:And if we use Stirling's approximation we obtain: We then substitute in the very first relation we stated at the start of this derivation:

And the next step is to simply divide the entropy of this huge macrosystem by the number of subsystems in the ensemble, so that we obtain:

And there you have it, the entropy of a single macroscopic system entirely in terms of probabilities!

Repeat After Me: Schottky, Not Schotty Defect!

2008-01-26T06:41:00.001-08:00

What a lovely Saturday night for reminiscing; ever since learning about point defects in my Inorganic Chemistry module, I've never gotten the pronunciation right before. Can you believe it?

Schottky Defect

Amazing. I keep pronouncing it as 'Schotty' by insisting on the absence of a 'k'; I should really give the guy some credit! Haha.

In any case, this Schottky Defect is a prime example of what we call a point defect; basically defects occur in all solids, and this can be easily proven via a consideration of thermodynamic parameters. Instead of the usual ΔG = ΔH - TΔS treatment that we usually apply in Chemistry, let us consider a more top-down approach that is much more complete, albeit still lacking in details.

Now, at absolute zero, we know that according to the Third Law of Thermodynamics the atoms of a perfect crystalline solid are arranged perfectly in a regular crystal lattice, hence giving rise to zero entropy and zero disorder (we neglect any residual entropy in this discussion for simplification purposes!). Now, as the temperature increases, there will be a corresponding increase in thermal agitation that tends to produce defects in the crystalline structure - defects are basically misalignments or irregularities in an otherwise regular array of atoms in the solid.

When the temperature increases, the atoms go into a more frenzied state of vibration about their lattice positions, and sometimes, the atoms can actually be displaced from their lattice site, and migrate to the surface of the solid. This results in a vacant lattice site, which is referred to as a Schottky defect. The diagram below shows a perfect ordering in two dimensions, and a Schottky defect occuring:

And that's basically it, a Schottky defect that forms due to disorganizing thermal motion. Now what we wish to do now is to determine how the number of Schottky defects varies with the absolute temperature for a crystal that is in thermal equilibrium at a temperature T. Well, let us put into place certain assumptions:

1) The energy associated with a Schottky defect is ε: in other words, we say that the zero of energy is an atom within the lattice, and the energy of an atom on the surface of the solid with respect to an inner atom is ε, which is what is required to produce a defect.

2) For N atoms, let there be n defects, such that the total energy is nε: by saying this, we are saying that there are relatively few defects as compared to the total number of atoms (n is much lesser than N). As such, as according to the diagram below, all defects are well spaced away from one another such that each defect is surrounded by a regular array of atoms:

3) We have assumed that n is much lesser than N: in general, this is true at temperatures below 500 K, because the energy of a Schottky defect is roughly of the order of 1 eV, while the thermal energy at 300 K is around 1/40 eV. As such, we say that very few defects form at normal to moderate temperatures.

With this three assumptions, we can now say that:

Total energy of system, E(n) = nε

The next step is to then determine the statistical weight (or thermodynamic probability) of a typical macrostate of the solid crystal. Let us go by the above assumptions, so let's consider a crystal composed of N atoms, with n defects as our macrostate, and the number of microstates (i.e. number of ways to obtain such a configuration) is simply the number of ways one can select n lattice sites from N lattice sites:

The corresponding entropy term is then given by:

Now you might say: isn't this expression too easy? Using simple statistics for the statistical weight? And then for the entropy? Well, you're right, we have made some additional assumptions:

1) We have neglected surface effects: basically, there is an additional entropy term that we neglected, which corresponds to the number of ways you can arrange the atoms on the surface of the solid. But the reason for this is simple: for a typical amount of solid, say one mole, we have 10^23 atoms present. Since the number of atoms on the surface is proportional to (10^23)^2/3, we have roughly 10^16 sites on the surface of the solid. Comparing this number (10^16) to the actual number of inner atoms (~10^23), we say that surface effects can be suitably neglected under normal conditions.

2) We have neglected vibrational effects: earlier, we mentioned that the atoms do vibrate about their equilibrium positions, and this in fact contributes to the entropy as well at higher temperatures.

However, having neglected these two effects doesn't mean we're calculating the wrong entropy - it just means that we're calculating an incomplete entropy term. The key point here is that surface effects, defects and vibrations each have their own entropy and energy term - we can then calculate their energies separately (same goes for their entropy) and then sum them up later on, and it is still correct. The basis for such a treatment is that each component is a subsystem and they all interact weakly. Because they interact weakly, they are essentially independent of one another, and thus we can separate out their thermodynamic variables.

This is reminiscent of having an isolated system being split into two or more subsystems, where each subsystem has its own characteristic temperature; it is these temperatures (and other variables) that must be equal to one another when thermal equilibrium ensues within the solid.

Now that the problems are out of the way, let us consider a mathematical simplification by application of Stirling's rule, which states that for large values of x:

In which case if we apply it to the entropy of the system, we now have:

I've obtained the derivative of S with respect to n in anticipation that we'll need it later. Now, we'll make use of the basic definition of temperature in terms of entropy so that:

The right hand side is simply an extension of the middle term where I've used the chain rule; now that we've obtained the partial derivative of S with respect to n, let us obtain the other derivative:

Putting everything together we obtain the final expression:

Let us keep in mind that n <<>

And voila! This expression has the form of the much-celebrated Boltzmann distribution! An amazing thing, don't you think?

Just ruminate over it. :p

Certain Uncertainty

2008-01-25T05:34:00.000-08:00

Ever wondered how the Heisenburg Uncertainty Principle works? I mean, after so many textbooks and explanations, have you ever wondered what the following statement means:While the first term (ΔE) refers to an uncertainty in energy of the system, the second term (Δt) refers to the natural lifetime of the system in that state. And how would you use it?

Well here’s a classic example that is not often used in textbooks – consider the ground state of a molecule, and given that the ground state is the lowest energy state of the molecule, then the ground state must correspondingly be the most stable energy state of the molecule. That is the most basic postulate in statistical mechanics, where the lowest energy state is the most stable state.

Now, if we are given a molecule in the ground state, and provided that there is no perturbation (in the form of energy given to it via radiation, heat, etc.), then we can be very sure that the molecule will always stay in the ground state. We therefore say that the molecule stays in the ground state for an indefinitely long time unless energy is given to it to excite it.

Now, this would just mean that the natural lifetime of the ground state is:

Yes, in other words, we say that the lifetime is infinitely long! So what does the Uncertainty Principle tell us? It tells us that:

Indeed, we conclude that the uncertainty in the energy of the ground state must be zero! That is, the ground state is exactly known with a well defined energy! Our first conclusion must therefore be: the ground state of any molecule must therefore be a state of definite energy.

With this first concluding statement in mind, let us consider a typical quantum mechanical transition between two energy levels:

Now, we know that the ground state has a definite energy (G), as explained earlier, but what about the excited state (E)? Does it possess also a definite energy? Now, let us consider a fact, an experimental fact: excited molecules always decay back to the ground state, either via thermal collisions or re-radiation of electromagnetic radiation, giving rise to emission spectra lines.

This means that we must insist that the molecules don’t have an infinite life time, because if they did, then the molecules once excited, would never decay. Once again, we turn to the use of the Uncertainty Principle, where we see that:

This time round, the uncertainty in energy isn’t zero, but of a certain magnitude, depending on the lifetime of the excited state. Now, we should more properly depict the transition as:

Interesting! The transition is no longer as properly defined as it was earlier! This explains why spectroscopic graphs don’t possess infinitely sharp peaks, because this natural line broadening phenomenon takes place.

However, this equation has very often been misused, where the student very frequently mistakes the term ΔE for the energy gap between the excited state and the ground state, and simply substitutes this into the Uncertainty Principle to calculate the lifetime – notice that this isn’t correct, because the term ΔE more accurately refers to the blurring of the excited state, rather than the energy of transition itself.

Nevertheless, we can still say that as the energy of transition increases, the excited state is higher in energy, and thus less stable, which still translates to a smaller lifetime – it is for this reason that students fail to make the proper distinction between energy of transition, and energy of state.

Particulate This!

2008-01-25T02:30:00.000-08:00

Everyone's heard of Feynman Diagrams (or maybe not! :p) but what exactly are these? To begin understanding such a pictorial diagram of quantum processes, one first needs an example, which is aptly shown below:

I haven't quite drawn the diagram in the conventional way (without the axes being labelled) because I want the meaning of the diagram to be explicit. In this diagram, the bottom left arrow depicts an electron moving to the right - notice that it moves upwards (in time) meaning it is travelling forward in time, and it moves right (in position) meaning it is travelling to the right in real life.

The bottom right arrow represents a positron (the anti-electron) - in this convention, all anti-particles are drawn with a reversed arrow, so while the electron has an arrow that points forward in time, the positron has an arrow that points backwards in time. You might ask: why so? Simply put, an antiparticle is a particle travelling forward in time! We'll have more on that later; for the time being, suffice it to say that this is the convention we're going to adopt.

You'll notice two dotted lines at the top, which corresponds to two identical photons - you might notice that there are no arrows at all. Want to fashion a guess? It's because a photon is its own anti-particle, and hence no arrow is needed to distinguish between a photon and an anti-photon at all. Hey wait a minute! Are you saying that photons can actually be travelling forward and backwards in time? Perhaps, for light itself defines the very limits of time travel, and maybe only a photon can experience true time travel without being altered itself.

So how do I read this diagram? Easy, if we use the conventional wisdom of our real world, we say that an electron and a positron travel towards one another, collide, annihilate, and their mass energy is converted into electromagnetic radiation in the form of two photons that travel away from one another.

But in modern Physics, there is an alternative viewpoint: an electron moves from the left to the right, and at one moment in time, emits two photons. The emitting of these two photons causes the electron to change its momentum in time, and causes it to move backwards in time, becoming a positron that goes on towards the right, but backwards in time.

Well, that's all for this post - till when I'm feeling awake again!

Thermodynamics and Mechanics: Complementarity

2008-01-23T01:36:00.001-08:00

Sometimes separate fields in Physics come together for an unified application in certain problems. Don't believe me? Just take a look at the following thought experiment:

We have two identical metal spheres (consider them perfectly spherical!) with radius r, mass m, in the presence of a gravitational field on Earth g - but one of them hangs on a thread of negligible thickness and size, while the other lies motionless on the ground (assume that the ground is rigid and doesn't absorb any heat from anything including the ball). Now, the ball has a heat capacity c, which we can assume to be constant throughout this experiment (of course it won't be, but why torture ourselves here?)

The experimental procedure is as follows: we use a flame to impart E joules of energy to both balls and simply wait.

The question I have for you is: What is the final temperature of both balls? Will they be the same? If yes why? If not, why? [Hint: Of course the final temperature won't be the same!]

Here's another 2 questions for the really psychedelic hardcore Physics lover: Can you work out for me the difference in temperature? Can you also work out the difference in size? [Assume that the ball has a constant expansivity of a]

Working this problem out just demonstrates how intricately thermodynamics is linked to mechanics; indeed, the First Law of Thermodynamics itself embodies the essence of one of the important principles in mechanics, the Principle of Conservation of Energy. :)

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Well, it's been days, and no one has replied or commented or whatsoever, so I guess it's time to go on and reveal more of the answer to fulfill my own needs and complete the post. This question is a very simple one, as long as one bothers to draw out the initial state and final state of the system - notice that I'm making use of one of the basic concepts in Thermodynamics, the fact that initial and final states determine the change in any state function, and in this case, the temperature of the ball.

Consider the two expansion processes:

What do you notice? I'll leave this post hanging here for another few days before someone comes along to comment on what happens. :p

Time Based Entropy

2008-01-22T19:49:00.000-08:00

Call me silly, but I'm just very persistent in Thermodynamics, and hence the need to justify the point of my last post: namely, that entropy is just as much a function of time as it is a function of heat flow, for heat flow is also a function of time, for without time, how can heat flow?

However, let us first consider the importance of entropy - what exactly is this monster called entropy? Well, as all elementary definitions put it, entropy is the amount of disorder within a system. So if you have a cup of hot water, and cup of cold water, naturally the hot water will have its molecules sloshing around (alright, not quite literally) with kinetic motion, far more than the cold miserable molecules of the cold water would. As such, by virtue of its internal molecular motion, we say that the hot water has more entropy of motion because of a higher temperature.

Yeah yeah, so it's disorderliness, so what? Sure enough, that's not all. Let us then dive into the Second Law of Thermodynamics again, in its simplest form: the entropy of the universe must always increase in an irreversible process. Ah, but why must entropy increase? This is always a difficult concept to explain, and thus let us have a thought experiment:

If I have a container with a partition separating hot and cold water, I say that the hot water is a region of higher entropy, and the cold water a region of lower entropy - but is this really the state of maximum entropy? No! Why? Because there exists a very obvious sense of order! Because the container can be exactly divided into an ordered region and a disordered region, and therefore, there is an intrinsic order associated with the system!

So how would I increase the entropy further? Easy, break that orderliness! So we break the partition and allow the regions to mix, producing lukewarm water, and thereby increase the entropy of the system to a maximum. Now, ask yourself - this is an irreversible process, is it not? You will never see a lukewarm glass of water separating itself into hot and cold regions spontaneously by itself! No way man! And the direction of increasing entropy tells us how this works.

Now, let us consider this again: why is entropy so important? Think about it: the hot water and cold water have internal energy U - this internal energy can be used to do work, because the hot water has thermal energy that can be used to generate electricity by perhaps, driving a thermocouple.

Now consider the hot and cold water mixing - it still has internal energy U right? But yet the amount of work it can do is lesser! When you use warm water to drive a thermocouple, not so much energy is produced as work because the temperature isn't that high.

Weird! The amount of internal energy available is the same, but yet the amount of work that can be done is different! And this is explained because of entropy. As entropy increases, what this means is that the energy contained within a system is more spread out.

Ask yourself again: what kind of energy is useful? Why, of course, it must be energy that is able to flow from one region to another, energy that can flow! If energy can't be transported or unable to flow, then we simply can't tap it or harness it! Imagine if the chemical energy from your food couldn't be moved from the food itself into your cells for you to utilize! You'd be unable to do anything with the food you ate!

Entropy causes a spreading of the energy into an equilibrium state, such that there is an even mix of energy (and mass) everywhere within the system, such that in such an even distribution of energy, energy can't move anymore! Or rather, if energy moves in one direction, an equal amount will move in the opposite direction that compensates such movement, and thus there is no net movement of energy observable. Work is the net movement of energy, a loose definition, and if energy can't even move, work can never be done. Of course work has a more rigorous definition, but oh well, it's sufficient for this point.

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Well, here comes the main point of this post, to prove (in a very non-rigorous manner) that entropy is also a function of time. Please bear with me as I plough you through some essential basics before that. Let us first take a look at how the change in entropy (dS) is mathematically defined in basic Thermodynamics:

Simple enough, a change in entropy (dS) is caused by a reversible flow of heat (dQrev) in or out of the system, divided by the temperature of the system (T). If the flow of heat is into the system, then dQrev is postive, otherwise it's negative. The reason why it's defined like this can't be explained using any simple ideas, but suffice it to say (the mathematical derivation is very complex and time consuming) that heat is the flow of energy brought about by molecular motion, which therefore increases the disorderliness of the system. As such, we use the flow of heat as a measure of the change in entropy of the system.

The temperature is present in the equation because obviously for a very high temperature a small flow of heat wouldn't cause that much significant a change in the entropy of the system. Thus in Physics one would say that entropy change is a change weighted by the temperature of the system.

Let us now first prove an important theorem in Thermodynamics, the Clausius Inequality, which actually shows that no matter what, as long as you have a cycle, the entropy of the universe must definitely increase! Interesting right, a proof! Now let's first start from the First Law of Thermodynamics, which states that:

This means the change in internal energy of a system is equal to the heat flow in/out of the system and the work done on/by the system. Notice that this two quantities can be either reversible (rev) or not. We then make the distinction between reversible work done on the system and irreversible work done on the system:

That is, reversible work done on the system is lesser or equal to the irreversible work done on the system. The equality holds only when the work done is reversible, in which case a subscript rev is added to dW. And if we do some mathematical arrangement, we see that:

Which must lead us to the conclusion that:
And therefore we have:

Which tells us that if we divide throughout by the absolute temperature of the system we obtain:

And recognising that the quantity on the left is simply the change in entropy, we write:

And we proceed to determine the total change in entropy when we have a cycle, by integrating over the cycle, which is indicated with a small circle on the integral to indicate a closed integral:

And recall by definition that a cycle is a process that brings a system from state A to some state, and then back to state A. If entropy is a state function, then the system being at state A, will possess a fixed entropy, and thus the change in entropy of the system must be zero since the final and initial entropy of the system is the same:

Which concludes the derivation of the Clausius Inequality:

So what exactly does this inequality mean? Well, you must notice that if the entropy change of the system is zero, then we must agree that change in entropy of universe = change in entropy of surroundings, am I right? Now look at the equation above: it says that the heat flow is always negative in a cycle, and therefore the heat that flows must flow out of the system into the surroundings.

Wait a minute, doesn't the heat flowing into the surroundings mean that the surroundings' entropy change must be positive? Hey that means that the entropy change of the universe increases right?

Correct! In any cycle, we end up increasing the entropy of the universe - we can't go against this principle. This means that everytime you turn on and use the engine in your car, you're killing the universe. :p

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As for time-based entropy, let us consider the change in entropy again:

Then using the chain rule in basic calculus, we have:

Which then rearranges into:
We have recognised that dQrev/dt simply refers to power, and we explicitly refer to power as a function of time by putting the bracket Prev(t). And from the previously worked out example, we have:

Which allows us to say:

So that we can conclude that:
In other words, the entropy changes with time because heat must flow as a function of time! If there exists no time for heat to flow, there can be no change in entropy. This last equation took for granted that the power is always causing heat flow into the system, which is not necessarily true, but it's just a special case anyway. :p

Therefore, by knowing the mathematical form of P(t), we can simply integrate within the limits, and obtain entropy as a function of time. Which is actually easily done if we consider heat transmission via radiation. The law of heat transmission by radiation is summed up in the Stefan-Boltzmann Law of Radiation, where:

And there you have it, you actually can express the entropy of a system as a function of time (notice in the above I omitted the constant of integration because I'm lazy, :p).