# Space-time symmetry (Langrangian Mechanics)

1. Dec 22, 2009

### vertices

When deriving the conserved quantity in the case of space-time symmetry, a line in my notes goes from:

$$\int{dt.(1+\epsilon\dot{\xi}).L[q(t+\epsilon\xi)+{\delta}q(t+\epsilon\xi)]} - \int{dt.L[q(t)+{\delta}q(t)]}$$

where L is the Lagrangian and $$\xi$$ is a function of time and both integrals are over the same time interval, to:

$$\int{\dot{\xi}L+\xi\frac{dL}{dt}+O(\epsilon^{2})}$$

I can't see how these two lines equal one another.

How does the $$O(\epsilon^{2})$$ come about?

Thanks.

2. Dec 22, 2009

### diazona

That would be the result of a Taylor expansion in $\epsilon$. Also I think you're missing a factor in the last expression, and that it should be
$$\int \mathrm{d}t\,\epsilon\biggl(\dot{\xi}L + \xi\frac{\mathrm{d}L}{\mathrm{d}t}\biggr) + O(\epsilon^2)$$

3. Dec 22, 2009

### vertices

thanks diazona, much appreciated:

I tried Taylor expansion as you suggested (and I did get the right answer) but I am not sure whether I have done so mathematically correctly or not. Can you check whether or not I am on the right lines...

We have $$\int{dt.(1+\epsilon\dot{\xi(t)}).L[q(t+\epsilon\xi(t))+{\delta}q(t+\epsilon\xi(t))]} - \int{dt.L[q(t)+{\delta}q(t)]}$$

start with $$L[q(t+\epsilon\xi(t))]$$

if we first Taylor expand the bit in parenthesis (ie. $$q(t+\epsilon\xi(t))$$) we get

$$q(t+\epsilon\xi(t))=q(t)+\frac{{\partial}q(t)}{{\partial}t}\epsilon\xi(t)+...$$

therefore$$L[q(t+\epsilon\xi(t))=L[q(t)+\frac{{\partial}q(t)}{{\partial}t}\epsilon\xi(t)+...]$$

This is equal to:

$$L(q)+\frac{{\partial}L}{{\partial}q}\frac{{\partial}q(t)}{{\partial}t}\epsilon\xi(t)+...$$

but

$$\frac{{\partial}L}{{\partial}q}\frac{{\partial}q(t)}{{\partial}t} = \frac{{\partial}L}{{\partial}t}$$ (CANCELLING the 'dq's)

So:

$$L[q(t+\epsilon\xi(t))]=L(q)+\frac{{\partial}L}{{\partial}t}\epsilon\xi(t)+smaller terms]$$

Thanks.