Maths Primer 1 - Symmetry Lectures 2
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Lecture 2 introduced the concept of Lagrangian mechanics, which is an alternative formulation to Newton's mechanics. The rest of the lectures didn't require a good understanding of this to be honest, but it was covered due to how well it matched the purposes of the lectures, which was to give an introduction to how modern physics is done and what it looks like. And modern physics is done by finding a Lagrangian which matches your problem, so we have to learn about Lagrangians.
Since the lecture was necessarily maths heavy, here is a quick overview of essential ideas. I don't have time to draw the necessary annotations right now, but they might be added later.
</intro>
Lagrnagian mechanics is founded upon the 'principle of least action', this suggests that we need a good understanding of minimisation and the best place to start is actually with geometric optics, which Fermat showed can be derived from a 'principle of least time'. There are three textbook results of this:
1) Light travels in straight lines. In Euclidean geometry straight lines are the shortest distances, and assuming a constant speed of light (or uniform refractive index) this means light should take the shoortest path. An amusing fact is that in certain liquids with smoothly varying refractive indexes, light can be seen to 'bounce' due to minimising time rather than distance, but for light rays in air or vacuum, they're pretty much straight.
Proof: Draw a three points A, B, and a midpoint M. A light ray travelling from A to B is shorter than one travelling A to M to B unless M is collinear with A and B. This can be seen by the fact that a triangle is formed and by the triangle inequality, the sum of two sides is always longer than the other side. (The triangle inequality can itself be proved using varous methods, one somewhat overpowered one is just to use the cosine rule.)
2) Light obeys the law of reflection: angle of incidence = angle of reflection. This can be geometrically proven in a problem known as Heron's problem. Or, you can use calculus, name the distances x and (l-x), use some geometry and you get the result that sin(incidence)=sin(reflection). This then simplifies to the familiar result.
3) Refaction (Snell's law): n_1sin(a_1) = n_2sin(a_2) where a_1 is angle of incidence, a_2 is angle of refraction. This drops out of the same calculus that you use for the reflection problem, but because of the differing refractive indexes, you can no longer get rid of the sines easily so they stay in the final result.
The calculus proofs of the laws of reflection and refraction are under the Proof! tab. As you can see, minimising functions requires setting thed derivative to zero. That's because minima are stationary points where the function changes direction, and so it's tangent briefly becomes horizontal before the function starts to increase again. If you think about it, a minima is exactly the location where a function stops decreasing and starts to increase again - that exact point is when the tangent to the curve becomes horizontal and the derivative is therefore zero. (Maxima are the same, but upside down. Actually, most laws don't state whether these things are maxima or minima, but they are often minima in physical reality.)
Another way of thinking about minima is using Taylor's formula. This is really where I wish I could write in LaTeX, because his formula involves multiple derivatives, factorials and summations. The basic idea with a Taylor formula is to rewrite a function at a certain point as a power series like this:
F(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + ...
The n'th term in the series will be a_nx^n where the a_i's in our formula are some constants that we can tweak to make the series come out with the correct values. Actually, the most general power series would use expressions like a_i(x-k)^n where k is some constant that's the same for all terms.
The constants aren't hard to find - a_0 = F(0), because then all the ther terms are zero. You can then differentiate and substitute in zero again to find a_1=F'(0). Again to find a_2=F''(0)/2. In general a_n=Fn(0)/(n!) where Fn represents the n'th derivative of F wrt x. We can use all of this to write an expression that looks hideous, especially without proper formatting!
It goes like this: F(x) = (Sum over n, from zero to infinity) { Fn(a)(x-a)^n / (n!)}
When you set say a=0, we say you are 'taking a Taylor expansion around 0'. Here's a useful result from Taylor expansions, that we'll want to use a lot: take a function F(x) at a point x. Increment x by a small amount h, so that you have a new value F(x+h). Now, Taylor expand about x, to get:
F(x+h) = (Sum over n, from zero to infinity) { Fn(x)(x+h-x)^n / (n!)}
F(x+h) = (Sum over n, from zero to infinity) { Fn(x)(h)^n / (n!)}
F(x+h) = F(x) + F'(x)h + F''(x)h^2/2 + F'''(x)h^3/6 + ...
This allows us to see the increase in F(x) - usually the dominant terms is the F'(x)h term. All other terms have higher powers of h which is already a very small number. But at a stationary point, F'(x)=0, so that term drops out entirely. This tells us a fact about stationary points that we already knew: at a stationary point, the function is roughly flat, that is it doesn't increase if you vary x by a little bit. Of course, it does increase, but we say that it's constant to first order in h, because a linear approximation would show no increase (a linear approximation can be thought of as moving along the tangent line, which obviously doesn't increase the value of the function at a stationary point, because it's horizontal!)
Taylor expansion are incredibly useful, and show up all over the place in physics. Often you might want to find an increase, so you'd do something similar to what we did just now. Other times, you need an approximation to a function and a Taylor expansion is easy to evaluate, then cut out higher order terms. And, as seen above, they help to understand more rigorously what we mean by maxima and minima.
The last bit of maths used in the lecture was integration by parts. The particular integral wasn't too hard, but I'll introduce some notation: let u and v be functions of x and u', v' be their derivatives. Then by the product rule: (uv)' = u'v + uv'. We can now rearrange a little to get:
uv' = (uv)' - u'v
Integrating, gives us Int(uv') = uv - Int(u'v). How did we do that? Well, when we integrate we try to guess a function that differentiates to the given function - but we already know what differentiates to (uv)'! That's just uv. The other two are obvious because we didn't actually do anything, but it does make some integrals much easier to evaluate.
