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Energy-momentum tensor: metric tensor or kronecker tensor appearing?

  1. Jul 2, 2011 #1
    Hi

    This might be a stupid question, so I hope you are patient with me. When I look for the definition of the energy-momentum tensor in terms of the Lagrangian density, I find two different (?) definitions:
    [tex]{T^\mu}_\nu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi - {\delta^\mu}_\nu \mathcal{L}[/tex]
    [tex]{T^\mu}_\nu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi - {g^\mu}_\nu \mathcal{L}[/tex]
    Which of the two is correct? Is this somehow a matter of convention or something like that? I have seen both more than once.
     
  2. jcsd
  3. Jul 2, 2011 #2
  4. Jul 3, 2011 #3
    Thanks, this makes sense. Am I right to say that
    [tex]T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial^\nu \phi - \delta^{\mu\nu} \mathcal{L}[/tex]
    and
    [tex]T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial^\nu \phi - g^{\mu\nu} \mathcal{L}[/tex]
    are the same only if one makes the convention that [tex]{\delta^\mu}_\nu[/tex] is no longer just a symbol for the kronecker delta but a tensor, namely [tex]{\delta^a}_b = g^{a c} g_{c b}[/tex] and that then, [tex]\delta^{ab} = {\delta^a}_b g^{bc} = g^{ab}[/tex] but that the two are not the same if [tex]\delta^{\mu \nu}[/tex] is understood as the kronecker delta?

    If I see things correctly, one has to look carefully if an appearing delta is just a symbol for the kronecker delta in components or really a (raised or lowered) version of the metric tensor. This is slightly confusing.
     
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