Energymomentum tensor for a scalar field (sign problem!)by knobelc Tags: energymomentum, field, scalar, sign, tensor 

#1
Jun2308, 04:05 AM

P: 14

Hi
I have a small subtle problem with the sign of the energymomentum tensor for a scalar field as derived by varying the metric (s.b.). I would appreciate very much if somebody could help me on my specific issue. Let me describe the problem in more detail: I conform to the sign convention [itex]g_{\mu \nu} = (+,,,)[/itex]. The Lagranagian for a real scalar field is [tex] \mathcal{L} = \frac{1}{2} \dot{\Phi}^2 (\nabla \Phi)^2  V(\Phi ) = \frac{1}{2} g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi V(\Phi ).[/tex] From Noether Theorem we find the energymomentum tensor [tex]T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \Phi)} \: \partial^\nu \Phi  \mathcal{L} g^{\mu \nu} = \partial^\mu \Phi \partial^\nu \Phi  \mathcal{L} g^{\mu \nu}.[/tex] Now I want to derive this via varying the action [tex]S = \int \mathcal{L} \sqrt{g}\; dx^4[/tex] in respect to [itex]g_{\mu \nu}[/itex]. In particular it holds [tex]\delta S = \delta\int \mathcal{L} \sqrt{g}\; dx^4 = \frac{1}{2}\int T_{\mu \nu} \delta g^{\mu\nu} \sqrt{g}\; dx^4.[/tex] [itex]T_{\mu \nu}[/itex] is defined so that varying the action derived from the total Lagrangian [tex] \mathcal{L_{\rm tot}} = \frac{1}{16\pi G} R + \mathcal{L}[/tex] yields the Einstein field equations [tex]G_{\mu \nu} = 8\pi G T_{\mu \nu}.[/tex] (Note that [tex]\delta\int\frac{1}{16\pi G} R \sqrt{g}\; dx^4 = \int G_{\mu \nu} \delta g^{\mu \nu}\sqrt{g}\; dx^4, [/tex] therefore the  sign in the definition of [itex]T_{\mu \nu}[/itex].) Now let's vary the lagrangian of the scalar field: [tex]\delta \int \mathcal{L} \sqrt{g}\; dx^4[/tex] [tex] = \int \delta(\mathcal{L}) \sqrt{g} + \mathcal{L} \delta(\sqrt{g})\; dx^4[/tex] [tex] = \int \delta \left( \frac{1}{2} g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi V(\Phi ) \right) \sqrt{g} + \mathcal{L} \left(\frac{1}{2} g_{\mu \nu} \delta g^{\mu \nu}\right) \sqrt{g}\; dx^4[/tex] [tex] = \frac{1}{2}\int \left( \delta g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi  \mathcal{L} g_{\mu \nu} \delta g^{\mu \nu} \right) \sqrt{g}\; dx^4[/tex] [tex] = \frac{1}{2}\int \left(\partial_\mu\Phi \;\partial_\nu\Phi  \mathcal{L} g_{\mu \nu} \right) \delta g^{\mu \nu} \sqrt{g}\; dx^4.[/tex] Comparing this with the definition of the [itex]T_{\mu \nu}[/itex] yields [tex]T_{\mu \nu} = \partial_\mu \Phi \partial_\nu \Phi + \mathcal{L} g_{\mu \nu}[/tex] leading to the opposite sign as derived by the Noether Theorem. I would appreciate very much if somebody could explain why I get the sign wrong. I know this is a subtle (and possibly unimportant) issue but getting the wrong sign without understanding why gives a bad feeling. Thank you for any help! 



#2
Jun2308, 06:57 AM

Mentor
P: 6,040

According to Wald the KleinGordon energymomentum tensor from Noether's theorem agrees with the KleinGordon energymomentum tensor from varying the metric "up to a numerical factor." I do not know if the numerical factor is 1.
Wald says that in others cases, there is less agreement, and it is the energymomentum arrived at by varying g that appears on the right of Einstein's equation. If you have Wald, look near the bottom of page 457. I first ran into differences between the canonical and symmetric energymomentum tensors in section 12.10 of Jackson. 



#3
Jun2308, 02:52 PM

P: 14

I think, I got the reason for the wrong sign. Since I used the signature [itex]g_{\mu \nu} = (+,,,)[/itex] my definitions of [itex]T^{\mu \nu}[/itex] and [itex]\mathcal{L_{\rm tot}}[/itex] are not correct. With my signature the correct expressions read as
[tex]\delta S = \delta\int \mathcal{L} \sqrt{g}\; dx^4 = +\frac{1}{2}\int T_{\mu \nu} \delta g^{\mu\nu} \sqrt{g}\; dx^4.[/tex] and [tex] \mathcal{L_{\rm tot}} = \frac{1}{16\pi G} R + \mathcal{L}.[/tex] With this I get everything right. :) 


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