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Gauge Transformations

  • #1

Homework Statement


For a gauge function G(t,q) where
gif.gif
,
does
gif.gif
or
gif.gif
have any alternative form or can they be expressed in any other way?

Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #2
stevendaryl
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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 [itex]L[/itex] and [itex]L'[/itex]?
 
  • #3
Its a mechanical system, but just working on proofs and theory so
gif.gif


I'm working on proving how different quantities change, such as the euler lagrange equations, the generalised momenta, mechanical energy and power
 
  • #4
So for example for the euler lagrange I have
gif.gif
 
  • #5
correction
gif.gif
 
  • #6
stevendaryl
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Oh, so you're just talking about Lagrangian mechanics.

If [itex]G[/itex] is a function of [itex]q[/itex] and [itex]t[/itex], then you have:

[itex]\dot{G} = \frac{\partial G}{\partial q} \dot{q} + \frac{\partial G}{\partial t}[/itex]

So in that case, [itex]\frac{\partial \dot{G}}{\partial \dot{q}} = \frac{\partial G}{\partial q}[/itex]
 
  • #7
Oh wow, this is painful how much I overlooked that. Thanks!
 
  • #8
Oh, so you're just talking about Lagrangian mechanics.

If [itex]G[/itex] is a function of [itex]q[/itex] and [itex]t[/itex], then you have:

[itex]\dot{G} = \frac{\partial G}{\partial q} \dot{q} + \frac{\partial G}{\partial t}[/itex]

So in that case, [itex]\frac{\partial \dot{G}}{\partial \dot{q}} = \frac{\partial G}{\partial q}[/itex]
A quick question about the same function, would
l%20G%7D%7B%5Cpartial%20q%7D%20%3D%20%5Cfrac%7B%5Cpartial%20%5Cdot%7BG%7D%7D%7B%5Cpartial%20q%7D.gif
be a true statement?
 
  • #9
stevendaryl
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A quick question about the same function, would
l%20G%7D%7B%5Cpartial%20q%7D%20%3D%20%5Cfrac%7B%5Cpartial%20%5Cdot%7BG%7D%7D%7B%5Cpartial%20q%7D.gif
be a true statement?
Yes, if [itex]G[/itex] is only a function of [itex]q[/itex] and [itex]t[/itex].
 
  • #10
Yes, if [itex]G[/itex] is only a function of [itex]q[/itex] and [itex]t[/itex].
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
stevendaryl
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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?
If you have a function [itex]X(t,q)[/itex]of [itex]q[/itex] and [itex]t[/itex], then [itex]\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[/itex]

So [itex]\dot{G} = \frac{\partial G}{\partial t} + \dot{q} \frac{\partial G}{\partial q}[/itex]. Therefore, [itex]\frac{\partial \dot{G}}{\partial q} = \frac{\partial^2 G}{\partial t \partial q} + \dot{q} \frac{\partial^2 G}{\partial q^2}[/itex].

Similarly, [itex]\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}[/itex]
 

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