- #1
knobelc
- 14
- 0
Hi
I have a small subtle problem with the sign of the energy-momentum 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 energy-momentum 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!
I have a small subtle problem with the sign of the energy-momentum 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 energy-momentum 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!