Register to reply 
EulerLagrange equation derivation 
Share this thread: 
#1
Jun2611, 07:14 PM

P: 84

I'm trying to understand the derivation of the EulerLagrange 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. 


#2
Jun2711, 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*D_{x}f+g*D_{y}f where D_{x} and D_{y} are the partial derivatives in the xdirection and ydirection respectively. So just let y be x dot, and f be your Lagrangian, and you get the result. 


#3
Jun2711, 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.



#4
Jun2711, 11:50 AM

P: 969

EulerLagrange equation derivation
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_{x}f(x,y+g)+g*D_{y}f(x,y) which you get just from the 1variable definition of the derivative. Now the key is that applying the one variable definition of the derivative again: h*D_{x}f(x,y+g)=h*D_{x}f(x,y)+g*h*D_{x}D_{y}f(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_{x}f(x,y+g)+g*D_{y}f(x,y)=h*D_{x}f(x,y)+g*D_{y}f(x,y) 


#5
Jun2711, 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.



Register to reply 
Related Discussions  
Derivation of first integral EulerLagrange equation  Advanced Physics Homework  3  
A question about the derivation of the EulerLagrange equation  Calculus  3  
Help with derivation of eulerlagrange equations  Calculus  1  
Derivation of EulerLagrange Equation  Classical Physics  3 