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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?

DentPanic42
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2 Answers2

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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.

mike stone
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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:

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

Noone
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  • You need to remember the $\sqrt g$ factor in the measure to get the second term. – mike stone Apr 19 '21 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? – DentPanic42 Apr 19 '21 at 11:47
  • @PhutureFysicist i edited – Noone Apr 19 '21 at 11:54
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    @PhutureFysicist Carroll covers variation with respect to the metric in chapter 4.3: Lagrangian formulation. – AccidentalTaylorExpansion Apr 19 '21 at 12:00