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

[Ameno]

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:
$${T^\mu}_\nu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi - {\delta^\mu}_\nu \mathcal{L}$$
$${T^\mu}_\nu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu \phi - {g^\mu}_\nu \mathcal{L}$$
Which of the two is correct? Is this somehow a matter of convention or something like that? I have seen both more than once.

 

[LostConjugate]

See this article which states that any mixed version of the metric tensor will be equal to the mixed Kronecker delta.

http://en.wikipedia.org/wiki/Mixed_tensor

 

[Ameno]

Thanks, this makes sense. Am I right to say that
$$T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial^\nu \phi - \delta^{\mu\nu} \mathcal{L}$$
and
$$T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial^\nu \phi - g^{\mu\nu} \mathcal{L}$$
are the same only if one makes the convention that $${\delta^\mu}_\nu$$ is no longer just a symbol for the kronecker delta but a tensor, namely $${\delta^a}_b = g^{a c} g_{c b}$$ and that then, $$\delta^{ab} = {\delta^a}_b g^{bc} = g^{ab}$$ but that the two are not the same if $$\delta^{\mu \nu}$$ 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.