Question about derivation of Euler-Lagrange eq.

[Agrasin]

I've attached the part from Landau & Lifschitz Mechanics where I got confused.

"The necessary condition for S(action) to have a minimum (extremum) is that these terms (called the first variation, or simply the variation, of the integral) should be zero. "

Why is this a necessary condition? If you could point me towards a definition AND explain intuitively that'd be great.

(Background: I've learned calculus and Taylor expansions-- I think this is related. But, visually, I don't see why a variation in the function at the minimum should be zero. Imagine a paraboloid. The bottom is the minimum. If you go a tiny bit in any direction (a variation) then the first-order change in the function is NOT zero. Maybe it's just that I don't understand what first-order change means.)

 

[DEvens]

It's just saying that if it is a minimum then it has to be at the bottom of a bowl, so to speak. Just as the ordinary derivative of a function has to be zero at the function's minimum, so to does the action have to be unchanged by small variations in the path near the minimum of the action.

 

[Agrasin]

"Just as the ordinary derivative of a function has to be zero at the function's minimum, so to does the action have to be unchanged by small variations in the path near the minimum of the action."

Okay, please explain this analogy. I agree that the derivative of a function is zero at a minimum. But f(x+h) for any small variation h is NOT equal to f(x) necessarily. In fact, at a minimum, f(x+h) > f(x).

 

[haisydinh]

For simplicity, let's just deal with the 2-dimensional case. It's true that at the minimum, whichever way you go, you will always 'go up the hill' (so to speak). Hence, it's obvious that $f(x-h) >f(x)$; and $f(x+h) >f(x)$ (for any small variation $h$). So visually, the graph decreases as it goes from $x-h$ to $x$, and it increases as it goes from $x$ to $x+h$. In calculus, when a function decreases, it'll have a negative slope, which means that its derivative is negative. Similarly, when a function increases, it'll have a positive slope, which means that its derivative is positive. And since the derivative changes sign (from negative to positive) at the minimum point as the function goes from $x-h$ to $x + h$, the derivative must be zero at this minimum point. All of this is the basic idea behind the first derivative test in calculus. So I would suggest you to refresh your mind a bit on calculus 101 because that will probably clear up all your confusion.

Once you understand what I've said above, then you can easily apply it to the case of minimizing the action in 3-dimensional coordinate. The idea is just the same!

 

[Agrasin]

So "the first order variation in S (action) has to be zero for S to be at a minimum" is just a fancy way of saying "the derivative of S is zero at a minimum"?

 

[haisydinh]

So "the first order variation in S (action) has to be zero for S to be at a minimum" is just a fancy way of saying "the derivative of S is zero at a minimum"?

Well, I think you got it. You can re-write the quote by just replacing the word 'variation' with the word 'derivative' (because they mean basically the same thing as far as differentiation is concerned). In other words: "the first-order [derivative] in S (action) has to be zero for S to be at a minimum". It is totally not a fancy way of saying things, because it describes exactly what a minimum is.

 

[brainpushups]

It may help to read a little about functional derivatives: http://www.physics.byu.edu/faculty/berrondo/wt752/functional%20derivative.pdf