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lonelyphysicist
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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?
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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