# Carroll: Energy-momentum tensor for a scalar field theory

In Carroll's Introduction to General Relativity: Spacetime and Geometry, there is a section titled Classical Field Theory in chapter 1. There, he mentions that:

"The action leads via a direct procedure (involving varying with respect to the metric itself) to a unique energy-momentum tensor. Applying this procedure to $$L = -\frac{1}{2}\eta^{\mu \nu}(\partial_\mu \phi)(\partial_\nu \phi) - V(\phi)$$ leads straight to the energy momentum tensor for a scalar field theory, $$T^{\mu \nu}_{scalar} = \eta^{\mu \lambda}\eta^{\nu \sigma}\partial_{\lambda}\phi \partial_{\sigma}\phi - \eta^{\mu \nu}[\frac{1}{2}\eta^{\lambda \sigma}\partial_{\lambda}\phi \partial_{\sigma}\phi + V(\phi)]."$$

How is this last expression obtained?

• – SG8
Apr 19, 2021 at 12:54
• Apr 6, 2022 at 12:23

You need to start by replacing the flat space $$\eta_{\mu\nu}$$ by the general metric $$g_{\mu\nu}$$ so that
$$S[\phi]\to S[g,\phi]= \int d^dx \sqrt g \left\{-\frac 12 g_{\mu\nu} \partial^\mu \phi \partial^\nu \phi -V(\phi)\right\}.$$ Now, using $$\delta \sqrt g/\sqrt g =\frac 1 2 g^{\mu\nu}\delta g_{\mu\nu}$$, we have the variation $$\delta S[g,\phi]=\int d^dx \sqrt g \left\{-\frac 12 \partial^\mu \phi \partial^\nu \phi + \frac 12 g^{\mu \nu} \left(-\frac 12 g_{\alpha\beta } \partial^\alpha \phi \partial^\beta \phi -V(\phi)\right)\right\}\delta g_{\mu\nu}$$ The Hilbert energy momentum tensor is defined by either of $$\delta S= \frac 12 \int d^dx \sqrt{g}\left\{T_{\mu\nu}\delta g^{\mu\nu}\right\}= -\frac 12 \int d^dx\sqrt{g}\left\{ T^{\mu\nu} \delta g_{\mu\nu}\right\}.$$ (the relative minus sign comes from $$\delta g^{\mu\nu}= - g^{\mu\sigma}\delta_{\sigma\tau} g^{\tau\nu}$$) so we read off Caroll's result.

The answer lies in your text "The action leads via a direct procedure (involving varying with respect to the metric itself) to a unique energy-momentum tensor.". You need to vary with respect to the metric tensor $$\eta^{\mu\nu}$$. You have to do ($$\sqrt{\eta}= \sqrt{-\text{det}(\eta_{\mu\nu}}))$$:

Construct the action: $$S = \int d^4x \sqrt{\eta}L$$

Perform the variation with respect to the metric tensor:

$$\cfrac{\delta S}{\delta \eta^{\mu\nu}}=0$$

• You need to remember the $\sqrt g$ factor in the measure to get the second term. Apr 19, 2021 at 11:44
• Does a variation imply what we in effect do with $\phi$ to obtain the Euler-Lagrange equations for CFT? If so, does it just amount to the Euler-Lagrange equations but with the 'coordinate' being the metric? If so, how do you differentiate with respect to a tensor with two indices? Apr 19, 2021 at 11:47
• @PhutureFysicist i edited Apr 19, 2021 at 11:54
• @PhutureFysicist Carroll covers variation with respect to the metric in chapter 4.3: Lagrangian formulation. Apr 19, 2021 at 12:00