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Gauge Transformations

  1. May 30, 2016 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. May 30, 2016 #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]?
     
  4. May 30, 2016 #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
     
  5. May 30, 2016 #4
    So for example for the euler lagrange I have gif.gif
     
  6. May 30, 2016 #5
    correction gif.gif
     
  7. May 30, 2016 #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]
     
  8. May 30, 2016 #7
    Oh wow, this is painful how much I overlooked that. Thanks!
     
  9. May 30, 2016 #8
    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?
     
  10. May 30, 2016 #9

    stevendaryl

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    Yes, if [itex]G[/itex] is only a function of [itex]q[/itex] and [itex]t[/itex].
     
  11. May 30, 2016 #10
    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?
     
  12. May 30, 2016 #11

    stevendaryl

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    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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