1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conservation of Noether current

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey guys. So I gota prove that the currents given by

    [itex]M^{\mu;\nu\rho}=x^{\nu}T^{\mu\rho}-x^{\rho}T^{\mu\nu}[/itex]

    is conserved. That is:

    [itex]\partial_{\mu}M^{\mu;\nu\rho}=0.[/itex]
    2. Relevant equations
    Not given in the question but I'm pretty sure that

    [itex]T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial^{\nu}\phi-\mathcal{L}g^{\mu\nu}[/itex]

    And we're considering a real Klein-Gordon theory, so we have

    [itex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi)-\frac{m}{2}\phi^{2}[/itex]

    3. The attempt at a solution

    So here's what I've done so far:

    [itex]T^{\mu\rho}=(\partial^{\mu}\phi)(\partial^{\rho}\phi)-\mathcal{L}g^{\mu\rho}[/itex]
    [itex]T^{\mu\rho}=(\partial^{\mu}\phi)(\partial_{\mu}\phi)g^{\mu\rho}-\mathcal{L}g^{\mu\rho}[/itex]
    [itex]T^{\mu\rho}=\left[ (\partial^{\mu}\phi)(\partial_{\mu}\phi)-\mathcal{L}\right]g^{\mu\rho} [/itex]
    [itex]T^{\mu\rho}=\left[ \frac{1}{2}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+\frac{m}{2}\phi^{2}\right]g^{\mu\rho} [/itex]

    Doing the same thing to [itex]T^{\mu\nu}[/itex] gives

    [itex]T^{\mu\nu}=\left[ \frac{1}{2}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+\frac{m}{2}\phi^{2}\right]g^{\mu\nu} [/itex]

    Now putting it together gives

    [itex]M^{\mu;\nu\rho}=( \frac{1}{2}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+\frac{m}{2}\phi^{2}))(x^{\nu}g^{\mu\rho}-x^{\rho}g^{\mu\nu})[/itex]

    Now I have to hit this with [itex]\partial_{\mu}[/itex]. So i get:


    [itex]\partial_{\mu}M^{\mu;\nu\rho}=( \frac{1}{2}\partial_{\mu}(\partial^{\mu}\phi)(\partial_{\mu}\phi)+m\phi(\partial_{\mu}\phi))(x^{\nu}g^{\mu\rho}-x^{\rho}g^{\mu\nu})[/itex]

    And I'm stuck on what to do next. Don't know how to deal with [itex]\partial_{\mu}(\partial^{\mu}\phi)(\partial_{\mu}\phi)[/itex]
     
  2. jcsd
  3. Nov 12, 2014 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You have the same problem as in the other thread of using the same summation index for two different sums.

    That aside, have you proven that ##T^{\mu\nu}## is a conserved current? If you have the problem is trivial. If not, I suggest starting by that.

    The alternative is deriving the Nother current associated with Lorentz transformations (rotations and boosts).
     
  4. Nov 12, 2014 #3
    Proving the currents is the previous question which I am also stuck on lol! I'm basically asked to derive that expression for M using Noether's theorem (I cant get to the end for some reason) and then I have to prove that the charge is 0.

    I dont see which index I can change in my equation though...can you help me out please?
     
  5. Nov 12, 2014 #4
    • Message originally posted in new thread.
    Hey guys,

    Consider an infinitesimal Lorentz transformation of coordinates given by [itex]\delta x^{\mu}=\alpha^{\mu}_{\nu}x^{\nu}.[/itex] What would be the corresponding transformation to the field [itex]\delta \phi[/itex]?

    I'm also working with the real Klein-Gordon theory so i know that [itex]\phi(x)=\phi'(x').[/itex] Also I know:

    [itex]\delta\phi(x)=\phi'(x')-\phi(x)-(\delta_{\mu}\phi)\delta x^{\mu}[/itex]

    So we get [itex]\delta\phi=-(\partial_{\mu}\phi)\alpha^{\mu}_{\nu}x^{\nu}[/itex]

    Is this right? I mean I have another relation [itex] \phi'(x)=\phi(\Lambda^{-1}x)[/itex] but no idea how to use this, or even if I need it to find [itex]\delta\phi[/itex]?
     
  6. Nov 12, 2014 #5

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you see an index appearing three times in a term, you can tell there is something wrong.
    Whenever you introduce a new index that is summed over, you must use a letter that is not already present. So your second equation above is incorrect.
     
  7. Nov 12, 2014 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Can you post your working for ##T^{\mu\nu}##? Are you using Noether's theorem or just starting from the expression for ##T^{\mu\nu}## in terms of the fields and trying to show ##\partial_\mu T^{\mu\nu} = 0##?

    Regarding the summation indices, just rename one of your summation indices before insertion of the Lagrangian into the expression for T.

    Note: I wrote this some time ago butseemingly never sent it. My apologies if I repeat something nrqed already said.
     
  8. Nov 12, 2014 #7
    I havent got any working for T. The previous question requires me to prove the expression for M using Noether's theorem and the infinitesimal Lorentz transformation [itex]\delta x^{\mu} = \alpha^{\mu}_{\nu}x^{\nu}[/itex]. Here is my working so far, I get stuck towards the end I think:

    http://i.imgur.com/aHW3UXA.png
     
  9. Nov 12, 2014 #8

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Remember that the current is normally defined as the derivative with respect to the infinitesimal symmetry parameter (in your case ##\alpha##, infinitesimal Lotentz transformations). I suggest you use this and keep in mind that ##\alpha_{\mu\nu}## has some special property ...
     
  10. Nov 12, 2014 #9
    Yes I know [itex]\alpha_{\mu\nu}=-\alpha_{\nu\mu}[/itex] but I dont know what you mean by the derivative being defined with respect to alpha. Also is my expression for [itex]\delta\phi[/itex] even correct...?
     
  11. Nov 12, 2014 #10

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Take the current you have found and differentiate wrt ##\alpha_{\mu\nu}## (but pick indices carefully!!). The resulting current is that generated by that particular Lorentz group generator (the expression you have is for a general Lorentz transformation).
     
  12. Nov 12, 2014 #11
    I still dont see what you mean by differentiate with respect to alpha. Yes I have a general current, so are you saying I plug in my delta x and then divide through by the infinitesimal parameter? basically the following:
    [itex]\delta J^{\mu}=-T^{\mu\rho}\delta x_{\rho}[/itex] is my general current.

    Plug in [itex]\delta x_{\rho}=\alpha^{\nu}_{\rho}x_{\nu}[/itex] to get

    [itex]\delta J^{\mu}=-T^{\mu\rho}\alpha^{\nu}_{\rho}x_{\nu}[/itex].

    Then divide through by alpha? I have no idea what to do with the indices if this is what I'm meant to do :S

    EDIT : i think I have to contract T first by splitting it into symmetric + antisymmetric parts after I plug in alpha...then divide through by alpha?
     
  13. Nov 12, 2014 #12

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, you have a current which is proportional to the infinitesimal boost. You can now select which boost to make, essentially letting ##\alpha_{\mu\nu} = \delta^\rho_\mu \delta^\sigma_\nu - \delta^\sigma_\mu \delta^\rho_\nu## where ##\sigma## and ##\rho## are some fixed indices. The second term is required by the anti-symmetry of ##\alpha##.

    This essentially is equivalent to differentiating your current wrt ##\alpha##.

    Again, be careful not to introduce the same name for two different indices!

    Edit: By the way, the very last step in the image you linked is not valid - so start from ##J^\mu = -\alpha_{\rho\nu} x^\nu T^{\mu\rho}##.
     
    Last edited: Nov 12, 2014
  14. Nov 12, 2014 #13
    Okay so I have two questions - firstly, how did you swap the nu-indices positions in your expression for J? I mean I had

    [itex]J^{\mu}=-\alpha_{\rho}^{\nu} x_{\nu}T^{\mu\rho}[/itex] whereas you have [itex]J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu \rho}[/itex]? How did you do that?

    Secondly, is this statement now correct:

    [itex] J^{\mu}=-\frac{1}{2}\alpha_{\rho\nu}x^{\nu}(T^{\mu\rho} - T^{\mu\nu})[/itex] ? I Anti-symmetrised T like my lecturer suggested.

    However from this point I dont know how to move forward...so im still stuck. I really am confused about your expression for alpha in terms of those deltas. I understand the antisymmetry of alpha and I'd be tempted to use [itex]\alpha_{\nu\rho}=\frac{1}{2}(\alpha_{\nu\rho}-\alpha_{\rho\nu})[/itex] but dont really know what im doing as far as the differentiation you suggested is concerned.
     
  15. Nov 12, 2014 #14

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Raising and lowering of indices is done as usual, with the metric tensor. If you are contracting a covariant index with a contravariant one, you can switch them without thinking twice (although maybe you should).

    I cannot tell exactly what your lecturer said, but T in general is not anti-symmetric and you cannot simply remove the symmetric part. Your approach to alpha should also work, but the easiest by far is just differentiating or picking a particular alpha.
     
  16. Nov 12, 2014 #15
    So by differentiating wrt alpha, you mean something like this, right?
    [itex]\frac{J^{\mu}}{\alpha_{\mu\nu}}[/itex]? But you're saying that due to the antisymmetry of alpha, I have two possibilities - the one I just stated and [itex]-\frac{J^{\mu}}{\alpha_{\nu\mu}}[/itex]?

    Doing this gives me 2 separate equations and I dont even know how to handle the alpha's indices in the denominator. I guess I dont know enough to apply your approach :( or there's something obvious that I'm missing. Even with my approach I am yet to reach a solution...

    EDIT: My lecturer said that the symmetric part vanishes under contraction.
     
  17. Nov 12, 2014 #16
    okay I've done what I understand from your approach. Please tell me if this is what you meant:

    [itex]J^{\mu}=-\alpha_{\rho\nu}x^{\nu}T^{\mu\rho} = + \alpha_{\nu\rho}x^{\rho}T^{\mu\nu}[/itex]

    Differentiating the first equality gives:

    [itex]\delta J^{\mu}=\frac{\partial J^{\mu}}{\partial \alpha_{\rho\nu}}=-x^{\nu}T^{\mu\rho}[/itex]

    Differentiating the second equality gives:

    [itex]\delta J^{\mu}=\frac{\partial J^{\mu}}{\partial \alpha_{\rho\nu}}=\frac{\partial J^{\mu}}{\partial \alpha_{\nu\rho}}=+x^{\rho}T^{\mu\nu}[/itex]

    So in total I have

    [itex]\delta J^{\mu}=\frac{1}{2}(x^{\rho}T^{\mu\nu}-x^{\nu}T^{\mu\rho})[/itex]?

    This is different from what I'm required to derive just by a minus sign - but I dont know if this is right mathematically at all.
     
  18. Nov 12, 2014 #17

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    So the point is that, due to the anti-symetry, you get the anti-symmetry directly when you differentiate since the components of ##\alpha## are related and you get must get something anti-symmetric out (in the indices of ##\alpha##). This you can anti-symmetrize (wrt one of the T indices and the x index), which hopefully is what your lecturer was referring to. The tensor ##T^{\mu\nu}## is not anti-symmetric. In fact, for the KG Lagrangian:
    $$
    T^{\mu\nu} = (\partial^\mu \phi) \frac{\partial \mathcal L}{\partial(\partial_\nu \phi)} - g^{\mu\nu}\mathcal L
    = (\partial^\mu\phi)(\partial^\nu \phi) - g^{\mu\nu}\mathcal L,
    $$
    which is manifestly symmetric and would disappear on anti-symmetrization.
     
  19. Nov 12, 2014 #18
    He might've been but that goes way beyond me...are you saying that I did it wrong...?

    I think I've got a solution but it depends on this being true: [itex]\alpha_{\mu\nu}x^{\nu}=-\alpha_{\nu\mu}x^{\mu}[/itex]. Basically if I take the transpose of alpha and put in the - sign due to anti-symmetry, do i have to swap the index that's being summed with x? I dont think I should, in which case my solution is invalid and I am officially stumped to no end...could you clarify please?
     
  20. Nov 12, 2014 #19

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This equation cannot be true. On one side you have ##\mu## as a free index and on the other ##\nu## ... I think the most straight-forward way is to simply note that you can pick a transformation such as the one I suggested in post #12. The other is noting that:
    $$
    \frac{\partial \alpha_{\mu\nu}}{\partial \alpha_{\rho\sigma}} = \delta^\rho_\mu \delta^\sigma_\nu - \delta^\sigma_\mu \delta^\rho_\nu
    $$
    for fixed indices, which, naturally, gives exactly the same result.
     
  21. Nov 12, 2014 #20
    I just dont see how the index of x changes...can you please explain?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Conservation of Noether current
  1. Noether Currents (Replies: 13)

  2. Noether Current (Replies: 1)

Loading...