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**Stress-Energy-Momentum Tensor from Lagrangian: Technical Question**

I've been reading about how to generate the stress-energy-momentum tensor [itex]T^{\mu \nu}[/itex] from the action

[tex]S = \int d^{4}x \sqrt{|g|} \mathcal{L} [/tex]

[tex]T^{\mu \nu} = \frac{2}{\sqrt{|g|}} \frac{\partial}{\partial g_{\mu \nu}} \left( \sqrt{|g|} \mathcal{L} \right)[/tex]

My impression is that it should not matter whether we're differentiating with respect to upper indices [itex]g^{\mu \nu}[/itex] or lower [itex]g_{\mu \nu}[/itex] but in actual fact it seems to:

Compare

[tex]\frac{\partial}{\partial g_{\mu \nu}} \left( \sqrt{|g|} \mathcal{L} \right) = \frac{1}{2} \sqrt{|g|} g^{\mu \nu} \mathcal{L} + \sqrt{|g|} \frac{\partial \mathcal{L}}{\partial g_{\mu \nu}} \right)[/tex]

with

[tex]\frac{\partial}{\partial g^{\mu \nu}} \left( \sqrt{|g|} \mathcal{L} \right) = -\frac{1}{2} \sqrt{|g|} g_{\mu \nu} \mathcal{L} + \sqrt{|g|} \frac{\partial \mathcal{L}}{\partial g^{\mu \nu}} \right)[/tex]

the difference is sign comes from the fact that

[tex]\delta \sqrt{|g|} = \frac{1}{2} \sqrt{|g|} g^{\mu \nu} \delta g_{\mu \nu} = -\frac{1}{2} \sqrt{|g|} \delta g^{\mu \nu} g_{\mu \nu}[/tex]

But how can the stress-energy-momentum tensor be dependent on whether we're differentiating with respect to lower or upper indices? I am most likely making an error somewhere.

Also, what about the overall sign? I see Weinberg and Carroll's GR book/notes defining the tensor with a -2 instead of my +2 -- but when I use it on the EM free-lagrangian [itex]-\frac{1}{4} F^{\mu \nu} F_{\mu \nu}[/itex] it gives me negative energy.

Is there an un-ambiguous manner to determine both the overall sign and whether to take derivatives with respect to metric tensor elements with upper or lower indices?

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