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## Main Question or Discussion Point

Suppose you are given the Lagrangian of a scalar field [tex]\Phi(t)[/tex]

[tex] \mathcal{L} = \frac{1}{2} \dot{\Phi}- \nabla \Phi - V(\Phi ).[/tex]

By introducing covariant notation with [tex]\eta_{\mu \nu} = (1,-1,-1,-1)[/tex] this reads as

[tex]\mathcal{L} = \frac{1}{2} \eta^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- V(\Phi ).[/tex]

Let us now switch to GR, i.e. $\eta^{\mu \nu} \rightarrow [tex]g^{\mu \nu}[/tex]. The Lagrangian remains the same since covariant derivative of a scalar field is the same as normal derivative, i.e. [tex]\nabla_{\mu} \Phi = \partial_{\mu} \Phi[/tex]. I derive the energy-momentum-Tensor [tex]T^{\mu \nu}[/tex] by varying [tex]g_{\mu\nu}[/tex] in the action

[tex]S = \int \mathcal{L} \sqrt{-g}\; dx^4,[/tex]

i.e.

[tex]\delta S = \frac{1}{2}\int T^{\mu \nu} \delta g_{\mu\nu} \sqrt{-g}\; dx^4.[/tex]

So we obtain with the variation [tex]\delta g_{\mu\nu}[/tex]

[tex] \mathcal{L} = \frac{1}{2} \dot{\Phi}- \nabla \Phi - V(\Phi ).[/tex]

By introducing covariant notation with [tex]\eta_{\mu \nu} = (1,-1,-1,-1)[/tex] this reads as

[tex]\mathcal{L} = \frac{1}{2} \eta^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- V(\Phi ).[/tex]

Let us now switch to GR, i.e. $\eta^{\mu \nu} \rightarrow [tex]g^{\mu \nu}[/tex]. The Lagrangian remains the same since covariant derivative of a scalar field is the same as normal derivative, i.e. [tex]\nabla_{\mu} \Phi = \partial_{\mu} \Phi[/tex]. I derive the energy-momentum-Tensor [tex]T^{\mu \nu}[/tex] by varying [tex]g_{\mu\nu}[/tex] in the action

[tex]S = \int \mathcal{L} \sqrt{-g}\; dx^4,[/tex]

i.e.

[tex]\delta S = \frac{1}{2}\int T^{\mu \nu} \delta g_{\mu\nu} \sqrt{-g}\; dx^4.[/tex]

So we obtain with the variation [tex]\delta g_{\mu\nu}[/tex]

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