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Euler-Lagrange equation derivation

by silmaril89
Tags: action, derivation, euler-lagrange
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silmaril89
#1
Jun26-11, 07:14 PM
P: 84
I'm trying to understand the derivation of the Euler-Lagrange equation from the classical action. This has been my main source so far. The issue I'm having is proving the following equivalence:

[tex]
\int_{t_1}^{t_2} [L(x_{true} + \varepsilon, \dot{x}_{true} + \dot{\varepsilon},t) - L(x_{true}, \dot{x}_{true},t)] \mathrm{d}t = \int_{t_1}^{t_2} (\varepsilon \frac{\partial L}{\partial x} + \dot{\varepsilon} \frac{\partial L}{\partial \dot{x}}) \mathrm{d}t
[/tex]

I understand the idea behind their equivalence intuitively, The derivative of a function is the change in that function, and I see how on the left side there is a representation of a small change in the lagrangian, but I'm having a hard time proving this to myself mathematically and I'd like some help.

I understand all the other steps shown in the derivation.

Thanks to anyone that responds.
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RedX
#2
Jun27-11, 02:02 AM
P: 969
You should ignore the integral, and just focus on showing that the things under the integrals are equal. If you have a function f(x), then by definition of the derivative:

[f(x+h)-f(x)]/h=f'(x)

when h goes to zero.

Therefore f(x+h)-f(x)=h*f'(x)

This is for one variable, and the generalization to two variables is:

f(x+h,y+g)-f(x,y)=h*Dxf+g*Dyf

where Dx and Dy are the partial derivatives in the x-direction and y-direction respectively.

So just let y be x dot, and f be your Lagrangian, and you get the result.
silmaril89
#3
Jun27-11, 09:22 AM
P: 84
Ok, that makes sense. I understand how it works with just one independent variable, but I did not realize the generalization to two variables. Thanks for that.

RedX
#4
Jun27-11, 11:50 AM
P: 969
Euler-Lagrange equation derivation

Quote Quote by silmaril89 View Post
Ok, that makes sense. I understand how it works with just one independent variable, but I didn't not realize the generalization to two variables. Thanks for that.
O, I didn't know that you know how it works with one variable. In that case:

f(x+h,y+g)-f(x,y)=[f(x+h,y+g)-f(x,y+g)]+[f(x,y+g)-f(x,y)]=
h*Dxf(x,y+g)+g*Dyf(x,y)

which you get just from the 1-variable definition of the derivative.

Now the key is that applying the one variable definition of the derivative again:

h*Dxf(x,y+g)=h*Dxf(x,y)+g*h*DxDyf(x,y)

and if g and h are really small, just keep the first term.

So f(x+h,y+g)-f(x,y)=h*Dxf(x,y+g)+g*Dyf(x,y)=h*Dxf(x,y)+g*Dyf(x,y)
silmaril89
#5
Jun27-11, 08:42 PM
P: 84
Thanks for clearing that up for me. That's a clever trick to apply the one variable definition again. To be honest, it's bit unsettling to me that we disregard a term based on g and h being very small. However, I do realize in the derivation on the wikipedia page I linked that that term would be very small.


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