5
$\begingroup$

$\newcommand{\l}{\mathcal L} \newcommand{\g}{\sqrt{-g}}$$\newcommand{\fdv}[2]{\frac{\delta #1}{\delta #2}}$I want to calculate the energy-momentum tensor in curved free space by functional differentiation with respect to the metric. The Lagrangian density that I have with units $c=1$ is the following

$$\l = \g \left(- \frac 1 4 F^{\mu \nu} F_{\mu \nu} \right).$$

I have calculated the variation of the action $\delta S$ to be

$$\delta S = \frac 1 2 \int\g \left(\frac 1 4 F^{\mu \nu} F_{\mu \nu} g_{\rho \sigma} - F^\tau_{\;\;\rho} F_{\tau \sigma} \right) \delta g^{\rho \sigma} - \g F^{\mu \nu} (\delta F_{\mu \nu})\; \mathrm d^4 x. \tag 1 $$

However I cannot get rid of the $\delta F_{\mu \nu}$ term. I've looked Wikipedia and seen that the energy momentum tensor with lower indices is the term in the brackets, which makes me think that the following association is correct

$$F^{\mu \nu} (\delta F_{\mu \nu}) \leftrightarrow T_{\rho \sigma} \delta g^{\rho\sigma} .\tag 2 $$

Using the association $(2)$, I can write equation $(1)$ as

$$\delta S = \frac 1 2 \int\g \left[ \left(\frac 1 4 F^{\mu \nu} F_{\mu \nu} g_{\rho \sigma} - F^\tau_{\;\;\rho} F_{\tau \sigma} \right) - T_{\rho \sigma} \right] \delta g^{\rho\sigma} \; \mathrm d^4 x \quad ,$$

so that

$$\frac 2 \g \fdv \l {g^{\rho \sigma}} = \left(\frac 1 4 F^{\mu \nu} F_{\mu \nu} g_{\rho \sigma} - F^\tau_{\;\;\rho} F_{\tau \sigma} \right) - T_{\rho \sigma} = 0 $$

$$ \implies T_{\rho \sigma} = \frac 1 4 F^{\mu \nu} F_{\mu \nu} g_{\rho \sigma} - F^\tau_{\;\;\rho} F_{\tau \sigma} \quad , $$

which is the correct result but it is unclear to me as to why I the association $(2)$ should be true.

$\endgroup$
6
$\begingroup$

The energy momentum tensor is found by varying the metric and holding all other fields constant. Since clearly $$\frac{\partial F}{\partial g}=0\longleftrightarrow \delta_gF=0$$ we end up with $$\delta_g S=\frac{1}{2}\int\mathrm{d}v\,\left(F^2g_{\mu\nu}/4-F^\tau{}_\mu F_{\tau\nu}\right)\delta g^{\mu\nu}$$ and comparison with $$\delta_gS:=\frac{1}{2}\int\mathrm{d}v\,T_{\mu\nu}\delta g^{\mu\nu}$$ leads to the correct result.

Note that what we are really varying here is just the matter action. The full general relativistic action contains the Einstein-Hilbert gravitational action, the cosmological constant term and a matter term. Put together, we have $$ S_\mathrm{GR}=S_\mathrm{EH}+S_\Lambda+S_\mathrm{M}$$ It is true that the variation of this vanishes wrt. the metric, i.e. $\delta_g S_\mathrm{GR}=0$. However, the energy-momentum tensor is calculated by examining only $\delta_g S_\text{M}$, which is not generally zero.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why do I define $$\delta_g S := \frac 1 2 \int \mathrm d V \; T_{\mu \nu} \delta g ^{\mu \nu}$$? $\endgroup$ – Gonenc Jul 19 '15 at 11:25
  • 1
    $\begingroup$ @gonenc That's the standard definition. Does your book not explain it? $\endgroup$ – Ryan Unger Jul 19 '15 at 11:34
2
$\begingroup$

Comment to the question (v2): The association (2) is not correct. To find the Hilbert SEM tensor, one varies the action wrt. the metric $g_{\mu\nu}$; not wrt. the gauge potential $A_{\mu}$ (or the field strength $F_{\mu\nu}$).

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The Einstein-Hilbert lagrangian coupled to a matter action $$ S_m[\varphi,g] = \int d^Dx\, \sqrt{-g}\mathcal L_m(\varphi,\partial_\mu\varphi), $$ i.e. $$ S[g,\varphi] = \frac{1}{16\pi G}\int d^Dx\,\sqrt{-g} R + \int d^Dx\,\sqrt{-g} \mathcal L_m(\varphi,\partial_\mu\varphi), $$ satisfies $$ \delta S=-\frac{1}{16\pi G} \int d^Dx\, \sqrt{-g} \mathcal E^{\mu\nu}\delta g_{\mu\nu} + \int d^Dx\, \sqrt{-g} \frac{1}{\sqrt{-g}}\frac{\delta\left( \sqrt{-g}\mathcal L_m\right)}{\delta g_{\mu\nu}}\delta g_{\mu\nu} $$ where $\mathcal E_{\mu\nu}$ is the Einstein tensor $R_{\mu\nu}-g_{\mu\nu}R/2$ and hence we define the Rosenfeld energy-momentum tensor $$\boxed{ T^{\mu\nu} = \frac{2}{\sqrt{-g}}\frac{\delta\left( \sqrt{-g}\mathcal L_m\right)}{\delta g_{\mu\nu}}= \frac{2}{\sqrt{-g}}\frac{\delta S_m[\varphi, g]}{\delta g_{\mu\nu}}} $$ getting $$ dS=-\frac{1}{16\pi G}\int d^Dx\,\sqrt{-g}\left(\mathcal E^{\mu\nu}- 8\pi G T^{\mu\nu} \right)\delta g_{\mu\nu}. $$ The action of a massless vector $A_\mu$, where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu=\nabla_\mu A_\nu-\nabla\nu A_\mu$, can be written as \begin{align*} S_m =&\ -\frac{1}{4}\int d^Dx\, \sqrt{-g}F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} \int d^Dx\, \sqrt{-g} A_\mu \left(-g^{\mu\nu}\Box+\nabla^{\nu}\nabla^\mu\right)A_\nu. \end{align*} So the equations of motion for $A_\mu$ are $$ \left(-g^{\mu\nu}\Box+\nabla^{\nu}\nabla^\mu\right)A_\nu=0. $$ The energy-momentum tensor is easier to express from the first form of the action: $$ \delta S_m = - \frac{1}{4}\int d^Dx\, \frac{1}{2} \sqrt{-g} g^{\mu\nu}\delta g_{\mu\nu} F_{\mu\nu}F_{\alpha\beta}g^{\mu\alpha}g^{\nu\beta} +\frac{1}{2}\int d^Dx\, \sqrt{-g} F_{\mu\nu}F_{\alpha\beta}g^{\mu\alpha}g^{\nu\sigma} g^{\beta\tau}\delta g_{\sigma\tau} $$ from which $$\boxed{ T^{\mu\nu} = F^{\alpha\mu} F^{\beta\nu}g_{\alpha\beta}-\frac{1}{4}g^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}}=F^{\alpha\mu} F^{\beta\nu}g_{\alpha\beta}+g^{\mu\nu} \mathcal L_m, $$ and the Einstein equations $$ R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R+\Lambda g^{\mu\nu} =8\pi G T^{\mu\nu}. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.