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Proving that a Noether charge is constant

  1. Oct 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey guys, so I gota prove that the charge

    [itex]Q=\int d^{3}xJ^{0}(\vec{x},t)[/itex]

    is constant in time, that [itex]\dot{Q}=0[/itex]
    2. Relevant equations

    [itex]J^{\mu}=i[\phi^{\dagger}(\partial^{\mu}\phi)-(\partial^{\mu}\phi^{\dagger})\phi][/itex]

    3. The attempt at a solution

    So first what I did was find [itex]J^{0}=i[\phi^{\dagger}\dot{\phi}-\dot{\phi^{\dagger}}\phi][/itex]

    Then plug this into Q and differentiate it w.r.t. time, which gives us:

    [itex]\dot{Q}=i\int d^{3}x(\phi^{\dagger}\ddot{\phi}-\ddot{\phi^{\dagger}}\phi)[/itex]

    And erm, provided I've done it all right (which I probably havent lol!) i dont know how to show that this is 0?

    Thanks in advance guys
     
  2. jcsd
  3. Oct 22, 2014 #2

    Orodruin

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    What is the equation of motion for ##\phi##?
     
  4. Oct 22, 2014 #3
    Hmm, I get something like [itex]2\Box\phi+m^{2}\phi^{\dagger}=0[/itex]...but I dont know how to use this?

    EDIT - actually that's wrong lol you cant have phi and phi^dagger mixing! omg I need to find out what Im doing wrong
     
  5. Oct 22, 2014 #4
    Okay so im sure the equation of motion for phi is

    [itex](\Box+m^{2})\phi=0[/itex]
     
  6. Oct 22, 2014 #5

    Orodruin

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    So what does this tell you about ##\partial_\mu J^\mu##?
     
  7. Oct 22, 2014 #6
    i know that [itex]\partial_{\mu}J^{\mu}=0[/itex] but I dont see how this helps...i know im stupid lol -.-
     
  8. Oct 22, 2014 #7

    Orodruin

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    There are several ways of using it leading to the conservation of charge (assuming that the current vanishes at infinity). Let us start with the straight forward one: What does ##\partial_\mu J^\mu =0## tell you in terms of components? Can you relate this to something that you can find in an expression for ##\dot Q##?
     
  9. Oct 22, 2014 #8
    right, so [itex]\partial_{\mu}J^{\mu}=\partial_{0}J^{0}+\partial_{i}J^{i}[/itex] where i goes from 1 to 3 for the 3 spatial components, and the 0th component is time. The problem is that Q dot is just a time derivative...so how do I proceed?
     
  10. Oct 22, 2014 #9
    wait...if [itex]\partial_{\mu}J^{\mu}=0[/itex] does that mean that each term in the sum expansion is 0? so that [itex]\partial_{0}J^{0}=0[/itex]?
     
  11. Oct 22, 2014 #10

    Orodruin

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    So you have ##\partial_0 J^0 = - \partial_i J^i = - \vec\nabla \cdot \vec J## and your integral is over space. To paraphrase Monty Python: "Hint, hint, nudge, nudge, know what I mean?" ;)
     
  12. Oct 22, 2014 #11
    Hmm its gona be something obvious I just cant see it :S
     
  13. Oct 22, 2014 #12
    Hey is it true that [itex]\phi\partial_{0}(\partial^{0}\phi^{\dagger})=\phi^{\dagger}\partial_{0}(\partial^{0}\phi)[/itex]? If so then I think I got it lol!
     
  14. Oct 22, 2014 #13

    Orodruin

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    Yes, this is one of those things you will bang your head into the desk when you realise. ;)
    I really find it hard to tell you more without giving the answer away ...

    But let me drop a hint on the other approach before I go to bed: You can rewrite the charge ##Q## as
    $$
    Q(t) = \int_S d^3x J^0(\vec x, t) = \int_S d^3x\, n_\mu J^\mu,
    $$
    where ##S## is space at time ##t## and ##n## is the time-like normal vector to ##S## (i.e., (1,0,0,0) in your coordinate system). What is then ##Q(t) - Q(0)##?

    Edit: Just one more thing which is basically just writing down what you know already ... How can you rewrite the following?
    $$
    \int_V (\vec\nabla \cdot \vec A(\vec x)) d^3x
    $$
     
  15. Oct 22, 2014 #14
    aww I think I like the other approach lol XD You've helped me a bunch man I owe you one! I think I'll get it eventually :P

    Thanks again man sleep well!!
     
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