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Conservation of Stress-Energy tensor

  1. Oct 1, 2014 #1
    I have a few questions regarding the solution to this problem. First of all I have the Stress-Energy tensor for [itex]a[/itex] scalar fields [itex]\phi^a[/itex]

    [tex]T_{Noether}^{\mu\nu} = \displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a - g^{\mu \nu}\mathcal{L}
    [/tex]

    To ensure symmetry I need to add a total divergence
    [itex]\partial_{\lambda}K^{[\lambda\mu]\nu}[/itex] which is anti-symmetric in [itex]\mu[/itex] and [itex]\nu[/itex]

    So I have a new tensor [tex]T^{\mu\nu}=T_{Noether}^{\mu\nu} + \partial_{\lambda}K^{[\lambda\mu]\nu}[/tex] I need to show that this thing is conserved.

    [itex]\partial_{\mu}T^{\mu\nu}=\partial_{\mu}T_{Noether}^{\mu\nu}+\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}[/itex]

    The solution then says that because [itex]\partial_{\mu}\partial_{\lambda}[/itex] is symmetric in the indicies and because [itex]K^{[\lambda\mu]\nu}[/itex] is anti-symmetric in those indicies then [itex]\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}=0[/itex]

    I don't see why this is. I tried writing [itex]2K^{[\lambda\mu]\nu}=K^{[\lambda\mu]\nu}+K^{[\mu\lambda]\nu}[/itex] but that didn't get me anywhere.


    Now, after that, I have [itex]\partial_{\mu}T_{Noether}^{\mu\nu}=\partial_{\mu}\left[\displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a - g^{\mu \nu}\mathcal{L} \right]

    = \displaystyle \sum_a \partial_{\mu}\left[\frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a\right] - \partial^{\nu}\mathcal{L}
    [/itex]

    [itex] =\displaystyle \sum_a \left[\partial_{\mu}\frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial^{\nu}\phi^a + \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial_{\mu}\partial^{\nu}\phi^a\right]- \partial^{\nu}\mathcal{L}[/itex]

    [itex]=\displaystyle \sum_a \left[\frac{\partial \mathcal{L}}{(\partial\phi_a)}\partial^{\nu}\phi^a + \frac{\partial \mathcal{L}}{(\partial_{\mu}\phi_a)}\partial_{\mu}\partial^{\nu}\phi^a\right]- \partial^{\nu}\mathcal{L}[/itex] after imposing the equations of motion. All fine up to here. Our solution then says that this thing is equal to

    [itex]\displaystyle \sum_a \frac{\partial \mathcal{L}}{(\partial\phi_a)}\partial^{\nu}\phi^a- \partial^{\nu}\mathcal{L}[/itex]

    I don't understand how this is arrived at. It then goes on to say that this is equal to [itex]\partial^{\nu}\mathcal{L}-\partial^{\nu}\mathcal{L}[/itex]. Again, I don't understand how this was achieved. Any pointers would be very appreciated. Thanks.
     
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  3. Oct 1, 2014 #2

    Orodruin

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    Is this really true for an anti-symmetric tensor?

    Is it really saying this exactly? In the first of these expressions, I would switch the order of the ##\mu## and ##\nu## derivatives in the second term and then apply Leibniz rule backwards.
     
  4. Oct 2, 2014 #3
    Sorry, typo, that should be a minus sign between the two terms.

    That's what I have in the notes, yeah. No explanation, it just jumps from one to the other. I'll try that, thanks.
     
  5. Oct 2, 2014 #4

    Orodruin

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    So then you are essentially done. Switch the differentiation order, then switch the naming of the summation indices.
     
  6. Oct 3, 2014 #5
    Do I only switch the order on the second term or on both? So I have [itex]2\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}=\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu} - \partial_{\mu}\partial_{\lambda}K^{[\mu\lambda]\nu}[/itex]
    [itex]=\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu} - \partial_{\lambda}\partial_{\mu}K^{[\mu\lambda]\nu}[/itex]
    Then renaming indicies in the second term (I'm really not sure if this is justified, it seems very shaky)
    [itex]=\partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu} - \partial_{\mu}\partial_{\lambda}K^{[\lambda\mu]\nu}=0[/itex]
    If I rename the indicies wouldn't I have to do it in both terms and I'd arrive back exactly where I started?


    There's also another line in the solution I don't understand. It says
    [itex] \displaystyle \int d^3x \partial_{i}K^{[i 0]\mu}=0 [/itex] when integrated over a surface. Here [itex]i=1,2,3[/itex] is the space index. I tried writing it out explicitly but to be honest I don't really know how to integrate tensors.
     
  7. Oct 3, 2014 #6

    Orodruin

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    One term only. Otherwise you end up exactly where you started.

    So ##\mu## here is just an index, meaning that this represents four different relations. If you let ##\vec K = (K^{10\mu},K^{20\mu},K^{30\mu})##, then what you have is just something of the form
    $$
    \int d^3x \partial_{i}K^{[i 0]\mu} = \int (\nabla \cdot \vec K) d^3x
    $$
    to which you can just apply Gauss' theorem.
     
  8. Oct 4, 2014 #7
    To be honest, just renaming the indicies seems a little "unsafe" to me? What is the justification for it? Now if I had something like ##\partial_{\mu}\phi^{\mu}## I would have no problem in writing this as ##\partial_{\nu}\phi^{\nu}## since we're just summing over repeated indicies and the end result will be the same, but in this case I don't see why we can do it, especially due to the antisymmetry of the tensor.


    Ahh okay I see, and then this equates to zero because we're integrating over the surface at spatial infinity?
     
  9. Oct 4, 2014 #8

    Orodruin

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    You can always rename your summation indices like this, they are just summation indices and it does not matter what you call them (just don't call them the same thing as another index). In this case you could also use the anti-symmetry property more directly: ##K^{[\mu\nu]\sigma} \equiv - K^{[\nu\mu]\sigma}##, before renaming the summation indices.

    This would be the line of argumentation, yes.
     
  10. Oct 4, 2014 #9
    I understand now. Thank you very much for your help! :)
     
  11. Nov 6, 2014 #10
    I've come back to this after a while off and I'm pretty sure that the solution taking the 4-divergence of the stress-energy tensor and getting zero is incorrect, at least in the way it goes about it. Surely if we take the 4-divergence of the expression given for ##T_{Noether}^{\mu\nu}## and get zero that would imply that every Lagrangian has a conserved stress-energy tensor, but we know it's only true for systems with invariance under space-time translations.

    The start of the questions states "According to Noether's theorem for a system with invariance under space-time translations ##T_{Noether}^{\mu\nu}## (followed by the expression for it) is conserved".

    So realistically there's nothing to show as you're already told it's conserved, unless you have to actually derive the conservation directly from Noether#s theorem. But I don't see how directly taking the 4-divergence of the tensor will give you the correct result, based on my reasoning above.
     
  12. Nov 6, 2014 #11

    Orodruin

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    True, the stress-energy tensor resulting as the Noether current is only conserved if the Lagrangian is invariant under space-time translations. However, the way I read your problem, you are supposed to show that (given that this is true) you can symmetrise it by adding K and you still end up with a stress-energy tensor with zero divergence. This follows from the properties of K.
     
  13. Nov 6, 2014 #12
    I know that - that isn't the problem.

    What the question explicitly says to do is show that
    ##\partial_\mu T^{\mu\nu} = \partial_\mu T_{Noether}^{\mu\nu} = 0##

    As you said, by the anti-symmetry property of K, we will have ##\partial_\mu T^{\mu\nu} = \partial_mu T_{Noether}^{\mu\nu}##, and we know that this is zero for a translationally invariant system so we have that the 4-divergence of the new stress energy tensor is also zero. My problem here is the fact that in the solution I was given it took the 4-divergence of ##T_{Noether}^{\mu\nu}## and, using no properties of the Lagrangian, arrived at zero, which I do not believe to be possible.
     
  14. Nov 6, 2014 #13

    Orodruin

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    I strongly suspect they have implicitly used that the Lagrangian is not explicitly dependent on the coordinates and is only a function of the field itself and its derivatives. In fact, this thing:
    is only equal to ##\partial^\nu \mathcal L## if ##\mathcal L## is not explicitly dependent on the coordinates. If not, it is missing a term
    $$
    \frac{\partial \mathcal L}{\partial x_\nu}.
    $$
    Note! I am here making a difference between ##\partial^\nu## and ##\partial/\partial x_\nu##. The former is the derivative acting on the Lagrangian as a whole, the latter is only acting on the explicit dependence of the Lagrangian on the coordinates.
     
  15. Nov 6, 2014 #14
    I'm afraid you've lost me now. How are you arriving at this?

     
  16. Nov 6, 2014 #15

    Orodruin

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    Let us write out the arguments for the Lagrangian explicitly: ##\mathcal L = \mathcal L(x^\mu,\phi,\partial_\mu \phi)## (I will only use one ##\phi## for brevity). The derivative ##\partial^\nu \mathcal L## is given by the chain rule:
    $$
    \partial^\nu\mathcal L(x^\mu,\phi,\partial_\mu \phi) =
    \frac{\partial \mathcal L}{\partial x^\mu} \partial^\nu x^\mu +
    \frac{\partial \mathcal L}{\partial \phi} \partial^\nu \phi +
    \frac{\partial \mathcal L}{\partial(\partial_\mu\phi)} \partial^\nu \partial_\mu \phi,
    $$
    where the terms originate from differentiating each argument of the Lagrangian in order. This is equal to
    $$
    \partial^\nu\mathcal L(x^\mu,\phi,\partial_\mu \phi) =
    \frac{\partial \mathcal L}{\partial x_\nu} +
    \frac{\partial \mathcal L}{\partial \phi} \partial^\nu \phi +
    \frac{\partial \mathcal L}{\partial(\partial_\mu\phi)} \partial^\nu \partial_\mu \phi.
    $$
    The expression you have involves only the last two terms and the first term is the derivative of the Lagrangian with respect to the explicit coordinate dependence only. This term disappears if the Lagrangian is invariant under translations.
     
  17. Nov 6, 2014 #16

    Orodruin

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    This may be easier to see when your underlying manifold is only 1 dimensional and you can make a difference between ##d/dt## and ##\partial/\partial t##. What you have in that case is the total time derivative of the Lagrangian
    $$
    \frac{d}{dt}\mathcal L = \frac{\partial \mathcal L}{\partial t} + \frac{\partial \mathcal L}{\partial \phi} \dot\phi + \frac{\partial \mathcal L}{\partial \dot\phi} \ddot\phi.
    $$
    The same principle applies here, but we cannot write ##\partial^\nu## as a total derivative because it contains partial derivatives with respect to the coordinates (however, with the meaning that also the ##\phi## should be differentiated, but the Lagrangian could also have an explicit coordinate dependence and we need to account for that when using the chain rule).
     
  18. Nov 6, 2014 #17
    Ahh yes! That's excellent! Thank you so much! A similar argument was running through my head but using the variation of Lagrangian instead of the derivative and considering it's late I didn't put two and two together.

    The Lagrangian is a Lorentz Scalar so of course has no dependence on the coordinates explicitly (please correct me if I'm wrong), only through the fields and their derivatives and of course the question states that the fields are translationally invariant, so the translation wont manifest itself in the Lagrangian through them, and the hence the required result.

    I often feel silly when it is little things like this that I should know which hold me back, but then when everything is explained I feel like I have such a fuller knowledge. You've been a tremendous help. Thank you very much.
     
  19. Nov 7, 2014 #18

    Orodruin

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    Don't feel silly. You are not the first nor last to be confused by this.

    Just one thing, a Lagrangian can still be Lorentz invariant but depend on the coordinates (consider an external background field for example). In these cases, the energy momentum tensor is not conserved, just as energy and momentum are not conserved in classical mechanics if there is an external force.
     
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