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Energy-Momentum Tensor for the Klein-Gordon Lagrangian

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data

    The energy-momentum tensor ##T^{\mu\nu}## of the Klein-Gordon Lagrangian ##\mathcal{L}_{KG} = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}## is given by

    $$T^{\mu\nu}~=~\partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG}.$$
    Show that ##\partial_{\mu}T^{\mu\nu}=0##.

    2. Relevant equations

    3. The attempt at a solution

    $$\partial_{\mu}T^{\mu\nu} \\
    =\partial_{\mu}[\partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG}]\\
    =\partial_{\mu}(\partial^{\mu}\phi\partial^{\nu}\phi-\partial^{\nu}\mathcal{L}_{KG})\\
    =(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)-\partial^{\nu}(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2})\\
    =(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)-\partial^{\nu}(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi)+m^{2}\phi(\partial^{\nu}\phi)$$

    Where do I go from here? I know I need to use the Klein-Gordon equation, but using the KG equation cancels the first and third terms and leaves the second term.
     
  2. jcsd
  3. Apr 22, 2016 #2

    TSny

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    The right parenthesis is misplaced in last line above.

    Did you miss a term when writing out ##\partial_{\mu}(\partial^{\mu}\phi\partial^{\nu}\phi)##?
     
  4. Apr 22, 2016 #3
    μνμνφ + m2φ = 0
     
  5. Apr 28, 2016 #4
    Ok. Let me start again.

    ##\partial_{\rho}T^{\mu\nu} = \partial_{\rho}(\partial^{\mu}\phi\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}_{KG})##

    ##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\eta^{\mu\nu}\partial_{\rho}\mathcal{L}_{KG}##

    ##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\eta^{\mu\nu}\partial_{\rho}(\frac{1}{2}\partial_{\sigma}\phi\partial^{\sigma}\phi-\frac{1}{2}m^{2}\phi^{2})##

    ##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\rho}\partial_{\sigma}\phi)(\partial^{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

    Where do I go from here?
     
  6. Apr 28, 2016 #5

    TSny

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    Are the 3rd and 4th terms related?

    You want to investigate the divergence ##\partial_{\mu}T^{\mu \nu}## rather than ##\partial_{\rho}T^{\mu \nu}##
     
  7. Apr 28, 2016 #6
    Yes. The third and fourth terms are identical as follows:

    ##(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\rho}\partial_{\sigma}\phi)(\partial^{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

    ##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\rho}\partial^{\sigma}\phi)(\partial_{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

    ##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)-\frac{1}{2}\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

    ##=(\partial_{\rho}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\rho}\partial^{\nu}\phi)-\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\rho}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\rho}\phi)##

    Okay. Let me relabel ##\rho## to ##\mu##. Then, I have

    ##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial^{\mu}\phi)(\partial_{\mu}\partial^{\nu}\phi)-\eta^{\mu\nu}(\partial_{\sigma}\phi)(\partial_{\mu}\partial^{\sigma}\phi)+\eta^{\mu\nu}m^{2}\phi(\partial_{\mu}\phi)##

    ##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial_{\mu}\phi)(\partial^{\mu}\partial^{\nu}\phi)-(\partial_{\sigma}\phi)(\partial^{\nu}\partial^{\sigma}\phi)+m^{2}\phi(\partial^{\nu}\phi)##

    ##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+(\partial_{\sigma}\phi)(\partial^{\sigma}\partial^{\nu}\phi)-(\partial_{\sigma}\phi)(\partial^{\nu}\partial^{\sigma}\phi)+m^{2}\phi(\partial^{\nu}\phi)##, where the second and third terms will now cancel

    ##=(\partial_{\mu}\partial^{\mu}\phi)(\partial^{\nu}\phi)+m^{2}\phi(\partial^{\nu}\phi)##

    ##=(\partial_{\mu}\partial^{\mu}\phi+m^{2}\phi)(\partial^{\nu}\phi)##, where we will now use the Klein-Gordon equation

    ##=0##.

    I think it's all correct now, isn't it?
     
  8. Apr 28, 2016 #7

    TSny

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    Looks nice.
     
  9. Apr 28, 2016 #8
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