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Homework Help: Stress energy tensor transformation

  1. Sep 15, 2018 #1
    1. The problem statement, all variables and given/known data
    Show that if you add a total derivative to the Lagrangian density ##L \to L + \partial_\mu X^\mu##, the energy momentum tensor changes as ##T^{\mu\nu} \to T^{\mu\nu}+\partial_\alpha B^{\alpha\mu\nu}## with ##B^{\alpha\mu\nu}=-B^{\mu\alpha\nu}##.

    2. Relevant equations


    3. The attempt at a solution
    So we have ##T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}L##, where ##\phi## is the field that the Lagrangian depends on. If we do the given change on the Lagrangian, the change in ##T_{\mu\nu}## would be ##\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha =\partial_\alpha \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha##. From here I thought of using this: ##g_{\mu\nu}\partial_\alpha X^\alpha=g_{\mu\nu}\partial_\alpha \phi \frac{\partial X^\alpha}{\partial \phi}## But I don't really know what to do from here. Mainly I don't know how to get rid of that ##g_{\mu\nu}##. Can someone help me?
     
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  3. Sep 16, 2018 at 5:01 AM #2

    Orodruin

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    This is not correct. First of all, you are making the assumption that you can just insert ##\partial_\mu X^\mu## in the expression for ##T_{\mu\nu}## and obtain the correct ##T_{\mu\nu}##. This is not so because the expression assumes a dependence only on derivatives up to first order existing in ##L##. If your ##X^\mu## contains any derivatives of ##\phi## (as it would have to to make it Lorentz invariant), then ##\partial_\mu X^\mu## will contain derivatives of second order and higher. Second, you are assuming that
    $$
    \frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu \phi)} = \partial_\alpha \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)},
    $$
    which is simply not true.

    Instead, I suggest that you start from the expression for ##\delta \mathscr S## in terms of the Lagrangian for the transformation ##\delta y^\mu = k^\mu## and ##\delta \phi = 0##, which is what gives the energy-momentum tensor as the current ##J^\mu = T^{\mu}_\nu k^\nu## and takes the form
    $$
    \delta \mathscr S = \int [\{L(\phi(y-\delta y))-L(\phi(y))\} + \partial_\mu(L \delta y^\mu)] d^4x.
    $$

    Edit: Also note that you have some errors in your index placements.
    Edit 2: You should not have to make any assumptions on the actual field content.
     
    Last edited: Sep 16, 2018 at 5:25 AM
  4. Sep 16, 2018 at 11:21 AM #3
    Thank you for your reply. So I tried this: $$\frac{\delta(L+\partial_\mu X^\mu)}{\delta y^\nu}=\Big(\frac{\partial L}{\partial \phi}+\frac{\partial(\partial_\mu X^\mu)}{\partial \phi}-\partial_\mu\frac{\partial L}{\partial (\partial_\mu \phi)}-\partial_\mu\frac{\partial (\partial_\mu X^\mu)}{\partial (\partial_\mu \phi)}\Big)\frac{\delta \phi}{\delta y^\nu}+\partial_\mu\Big(\frac{\partial L}{\partial(\partial_\mu\phi)}\frac{\delta \phi}{\delta y^\nu}+\frac{\partial (\partial_\mu X^\mu)}{\partial(\partial_\mu\phi)}\frac{\delta \phi}{\delta y^\nu}\Big)$$ So this is the normal derivation from my book, but with the replacement made to the Lagrangian. We also have $$\frac{\partial L}{\partial \phi}-\partial_\mu\frac{\partial L}{\partial (\partial_\mu \phi)}=0$$ as the field satisfies the equations of motion. Then we have $$\frac{\delta \phi}{\delta y^\nu}=\partial_\nu \phi$$ which is the infinitesimal transformation for a scalar field. As the Lagrangian, even wit the new term, is a scalar field we also have: $$\frac{\delta (L+\partial_\mu X^\mu)}{\delta y^\nu}=\partial_\nu L+\partial_\nu(\partial_\mu X^\mu)$$ Equating the 2 expressions we get:
    $$\Big(\frac{\partial(\partial_\mu X^\mu)}{\partial \phi}-\partial_\mu\frac{\partial (\partial_\mu X^\mu)}{\partial (\partial_\mu \phi)}\Big)\partial_\nu\phi+\partial_\mu\Big(\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu\phi+\frac{\partial (\partial_\mu X^\mu)}{\partial(\partial_\mu\phi)}\partial_\nu\phi\Big)=\partial_\nu L+\partial_\nu(\partial_\mu X^\mu)$$ Now from the old stress energy tensor we have $$\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu\phi-g_{\mu\nu}L$$ so the part I need to write in the form $$\partial_\mu (M_{\mu\nu})$$ is what remains from there: $$\Big(\frac{\partial(\partial_\alpha X^\alpha)}{\partial \phi}-\partial_\mu\frac{\partial (\partial_\alpha X^\alpha)}{\partial (\partial_\mu \phi)}\Big)\partial_\nu\phi+\partial_\mu\Big(\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\partial_\nu\phi\Big)=\partial_\nu(\partial_\alpha X^\alpha)$$ I replaced the index for the summation for X with ##\alpha##. Based on what I was told my answer should look like this $$T_{\mu\nu} \to T_{\mu\nu} +\partial_\alpha\Big( \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)}\partial\nu \phi - \frac{\partial X^\mu}{\partial(\partial_\alpha \phi)}\partial_\nu \phi \Big) $$. If we apply the product rule in the second term in our equation, we get rid of one more term and we end up with: $$\Big(\frac{\partial(\partial_\alpha X^\alpha)}{\partial \phi}-\partial_\mu\frac{\partial (\partial_\alpha X^\alpha)}{\partial (\partial_\mu \phi)}\Big)\partial_\nu\phi+\partial_\mu\Big(\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\Big)\partial_\nu\phi+\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\partial_\mu\partial_\nu\phi=\partial_\nu(\partial_\alpha X^\alpha)$$ $$\frac{\partial(\partial_\alpha X^\alpha)}{\partial \phi}\partial_\nu\phi+\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\partial_\mu\partial_\nu\phi=\partial_\nu(\partial_\alpha X^\alpha)$$ Now I am not sure what to do. Is it ok so far? Any suggestions now? Thank you again for help!
     
  5. Sep 16, 2018 at 2:38 PM #4

    Orodruin

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    You really need to stop thinking of the field ##\phi## altogether and start thinking more in terms of how the action changes when you do the ##\delta y^\mu## transformation and how that affects ##\delta L##.

    Also note that some of your terms have four ##\mu## indices. That is never a good sign.
     
  6. Sep 16, 2018 at 2:52 PM #5
    I am a bit confused, don't I need the ##\phi##, as what I am doing is like a chain rule?
     
  7. Sep 16, 2018 at 2:58 PM #6

    Orodruin

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    No, it is irrelevant what is inside the ##X^\mu## as your transformation of the fields is ##\delta \phi = 0##, all changes in the action will be expressible in terms of ##X^\mu## and its derivatives.
     
  8. Sep 16, 2018 at 3:04 PM #7
    I am not sure why the change in ##\phi## is zero. When you derive E-L equations, you actually need ##\delta \phi## to be non-zero.
     
  9. Sep 16, 2018 at 3:06 PM #8

    Orodruin

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    I am not talking about the equations of motion, I am talking about the 1-parameter symmetry related to the conserved current that is the energy-momentum tensor contracted with the spacetime translation vector ##k^\nu##, i.e., ##J^\mu = T^\mu_\nu k^\nu##.
     
  10. Sep 16, 2018 at 3:47 PM #9
    Ok, so for the change in action we have: $$\frac{\delta S}{\delta y^\mu}=\int(\frac{\delta L}{\delta y^\mu}+\frac{\delta \partial_\alpha X^\alpha }{\delta y^\mu})$$ And the first term gives the old result. Is this ok so far?
     
  11. Sep 17, 2018 at 12:55 AM #10
    Any further hint would be greatly appreciated
     
  12. Sep 19, 2018 at 5:03 AM #11

    Orodruin

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    Did you get any constraints on the form of ##X^\mu##? I suggest that you do the case where ##X^\mu## is a function of the fields only (and not derivatives of the fields). If ##X^\mu## depends on the field derivatives, then ##\partial_\mu X^\mu## will depend on the second derivatives of the fields.
     
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