The Principle of Stationary Action -Symmetry Lectures 2
The principle of stationary action is a very mathematical principle - it states that there is a quantity, known as the action, which is always minimised in Nature. An example is the 'action' for throwing a ball - every possible path it takes has an associated action, and the one it takes is the one where the action is lowest. Strictly speaking the principle is not actually that it is always the minimum, but simply a stationary or extreme point - so it could also be a minimum, although this is less likely in real life.
Sound abstract? Of course it is, but it's also powerful, because you've left the world of forces and vectors in the past. Whatever happens will simply be the thing that happens to minimise the action. This particular lecture was pretty much a straight derivation, and frankly, I don't understand the subject well enough to provide some super interesting exposition to go around the maths - there's a reason this post was preceded by a description of Taylor series and Integration by Parts!
So, here's the plan for showing that the Principle of Least Action is equivalent to Newtonian mechanics:
1) Select the path of minimum action
2) Create another path with a small deviation
3) Show that the difference to first order is zero. (Thus proving that the first path was a minimum)
4) Show that F=ma
Now, this seems perhaps far-fetched. We have done nothing to justify that the last step will come out at all, but we expect it to if the Principle of Stationary Action is identical to Newtonian mechanics.
Actually, the very first thing to do is that I haven't told you how to calculate the action yet! The action is defined (for reasons that may or may not become clear) as the path integral of the difference between kinetic and potential energies.
That is, A = int ( T - U)dt , where A is the action, T is kinetic energy and U is potential energy. A path integral is just a fancy way of adding up the value (T-U) at every point along the path, and is the same as a definite integral evaluated at the start and end of that path. So, without much further ado, here's the mathsy bit for which I apologise that Wix doesn't support LaTeX yet:
We let the path x(t) represent the true path of a particle. It's velocity is given by x'(t) and it's potential energy by U(x). A similar path has a small error term e(t), position y(t)=x(t)+e(t) and velocity y'(t). To first order, we expect the actions of both to be zero if and only if the path x(t) is a minimum.
A = int [(m/2) x'(t)^2 - U(x)]dt --> 0 = int[(m/2)y'(t) - U(y)]dt - int[(m/2)x'(t)^2 - U(x)]dt
Let's take a look at the term involving y:
A(y) = Int [ (m/2)( x'(t) + e'(t) )^2 - U(x+e) ] dt
where we have simply written that as y=x+e, y'=x'+e'. This is valid and is easier to recognise if you use the dy/dx notation which is unfortunately very difficult to format (suffice to say that you have a fraction and you split into two parts). Expanding the square you get:
A(y) = Int [ (m/2)( x'(t)^2 + 2x'(t)e'(t) )+ e'(t)^2 - U(x+e) ] dt
The next step is to take advantage of the fact that e is small: firstly we can remove the terms in e^2 or higher:
A(y) = Int [ (m/2)( x'(t)^2 + 2x'(t)e'(t)) - U(x+e) ] dt
Then we can taylor expand U(x+e) = U(x) + edU/dx :
A(y) = Int [ (m/2) (x'(t)^2 + 2x'(t)e'(t)) - U(x) - eU'(x) ] dt
Now, with a little algebra to expand the bracket in there and move some stuff about we get:
A(y) = Int [ (m/2)x'(t)^2 - U(x) + mx'(t)e'(t) - eU'(x) ] dt
A(x) = Int [ (m/2)x'(t)^2 - U(x) ] dt
A(y)-A(x) = 0 = Int [ m x'(t)e'(t) - eU'(x) ] dt
Did you follow all of that? The basic goal was to rearrange A(y) to look as similar as possible to A(x), substituting the relationship y=x+e wherever possible. Then, subtracting becomes pretty easy. I'll also add that we are getting close to having F=ma, because U'(x) is one way of writing a force (forces can always be written as the gradient of a potential in this way. If you know the formulas for gravitational potential energy and gravitational force, you can check this for yourself).
Now we integrate by parts:
Int [ m x'(t)e'(t) - eU'(x) ] dt = ?
Integrals can be split up, so we're going to work out the first part first:
Int [m x'(t)e'(t)] dt = m Int[x'(t)e'(t)] dt
Let u = x' and v' = e' , then Int[uv'] = uv - Int[u'v]
I.e. Int[x'e'] dt= x'e - Int[x''e] dt
Here's the thing about integration by parts: the integrals can be definite integrals, but the (uv) bit has to be evaluated at both ends of the integral. In this case, since the error in y=0 at both ends, x'e actually evaluates to 0 as well. This gives us an equation for the change in the action:
dA = 0 = Int [ m x'(t)e'(t) - eU'(x) ] dt
0 = Int [m (-x''(t)e) - U'(x)e] dt
Factoring out -e:
0 = -e Int [ m x''(t) + U'(x) ] dt
Since the error is not zero except at the endpoints, we are allowed to cancel it, as clearly it isn't causing the integral to be zero.
0 = Int [mx''(t) + U'(x)] dt
We still have an integral though. However, the part being integrated (the integrand) must be zero because of the following reason: imagine it is not. Then the only way for the integral to be zero is for the path to cancel itself out somewhere else. But that means that for some parts of the journey, the integral was not equal to zero, but either a positive or negative value for the cancellation to happen, and we have just shown that all legitimate paths must have this integral equal to zero.
Therefore, we write:
mx''(t) + U'(x) = 0
-U'(x) = mx''(t)
What do these mean? Well the first is the negative derivative of potential energy, which is a force F. The second is mass times the second derivative of position, that's an acceleration a.
Therefore,
F=ma.
A lot of effort, but hugely satisfying to come to that final conclusion.
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Further reading: Classical Mechanics: The Theoretical Minimum, by Leonard Sussking and George Hrabovski. They have an alternative derivation, lots of information on Lagrangian and Hamiltonian mechanics and is an introduction to the more advanced areas of classical mechancs that include such things as Stationary Action.
A similar derivation to this one is in Vol II of the Feynman Lectures, which is available for free online. That has some commentary at the end about the importance of minimum principles in general, and the entire chapter stands seperate from the rest of the 3 Volume set. Other than that, try searching Analytical mechanics or Lagrangian mechanics (or 'Calculus of Variations').
