Euler-Lagrange equation derivation

silmaril89

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.

RedX

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*D_xf+g*D_yf

where D_x and D_y 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

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

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*D_xf(x,y+g)+g*D_yf(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*D_xf(x,y+g)=h*D_xf(x,y)+g*h*D_xD_yf(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*D_xf(x,y+g)+g*D_yf(x,y)=h*D_xf(x,y)+g*D_yf(x,y)

silmaril89

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.