# Gauge Transformations

1. May 30, 2016

### dynamicskillingme

1. The problem statement, all variables and given/known data
For a gauge function G(t,q) where ,
does or have any alternative form or can they be expressed in any other way?

2. Relevant equations

3. The attempt at a solution

2. May 30, 2016

### stevendaryl

Staff Emeritus
Could you give a little more context to your question? I assume you're talking about electromagnetism? Or are you talking about quantum field theory? What is $L$ and $L'$?

3. May 30, 2016

### dynamicskillingme

Its a mechanical system, but just working on proofs and theory so

I'm working on proving how different quantities change, such as the euler lagrange equations, the generalised momenta, mechanical energy and power

4. May 30, 2016

### dynamicskillingme

So for example for the euler lagrange I have

5. May 30, 2016

### dynamicskillingme

correction

6. May 30, 2016

### stevendaryl

Staff Emeritus
Oh, so you're just talking about Lagrangian mechanics.

If $G$ is a function of $q$ and $t$, then you have:

$\dot{G} = \frac{\partial G}{\partial q} \dot{q} + \frac{\partial G}{\partial t}$

So in that case, $\frac{\partial \dot{G}}{\partial \dot{q}} = \frac{\partial G}{\partial q}$

7. May 30, 2016

### dynamicskillingme

Oh wow, this is painful how much I overlooked that. Thanks!

8. May 30, 2016

### dynamicskillingme

A quick question about the same function, would be a true statement?

9. May 30, 2016

### stevendaryl

Staff Emeritus
Yes, if $G$ is only a function of $q$ and $t$.

10. May 30, 2016

### dynamicskillingme

G is a function of q and t, (G(t,q) to be exact). Could you explain why you can change the order of the derivatives in this case?

11. May 30, 2016

### stevendaryl

Staff Emeritus
If you have a function $X(t,q)$of $q$ and $t$, then $\dot{X} = \frac{d}{dt} X = (\frac{\partial}{\partial t} + \frac{\partial}{\partial q} \frac{dq}{dt}) X = (\frac{\partial}{\partial t} + \frac{\partial}{\partial q} \dot{q}) X$

So $\dot{G} = \frac{\partial G}{\partial t} + \dot{q} \frac{\partial G}{\partial q}$. Therefore, $\frac{\partial \dot{G}}{\partial q} = \frac{\partial^2 G}{\partial t \partial q} + \dot{q} \frac{\partial^2 G}{\partial q^2}$.

Similarly, $\frac{d}{dt} \frac{\partial G}{\partial q} = (\frac{\partial}{\partial t} + \frac{\partial}{\partial q} \dot{q}) \frac{\partial G}{\partial q} = \frac{\partial^2 G}{\partial t \partial q} + \frac{\partial^2 G}{\partial q^2} \dot{q}$