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The following snippet is from Wald:

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The factor $\alpha_M$ seems to be unmotivated here, moreover, as Wald clearly shows, the KG-field and the EM-field have different $\alpha_M$s.

However, from Carrol:

enter image description here

Here $S_H$ is the standard Einstein-Hilbert action $S_H=\int R\sqrt{-g}\ d^4x$, and apparantly Carrol applies a constant factor in the coupled action to the gravitational action, instead of the matter action. Moreover, it is clear that if we define the stress-energy tensor without the rather arbitrary factor of $2$ but compensate for it on the left hand side, we get $$ S=\frac{1}{8\pi G}S_H+S_M. $$

I am confused, because according to Carrol, that's it. There is no ambiguity in the SEM-tensor. I am inclined to "side with" Carrol here for the following line of thought:


Ignoring the well-known boundary action problem of GR, it is clear the if we want gravity to satisfy that

  • gravity is a purely metric theory;
  • gravity is diffeomorfism-invariant, eg. the action is independent of the coordinates;
  • the field equations for gravity are second order differential equations

then the most general form of the action is $$ S_G=\int (\alpha R+\beta)\sqrt{-g}\ d^4x $$ where $\alpha$ and $\beta$ are constants (I guess this might not be the most general after all, I am not sure if it is possible to find such a theory without any additional fields that have higher order field equations, but all higher order degrees of freedoms can be gauged away). Factoring out $\alpha$, setting $\alpha=1/\kappa$ and $\beta/\alpha=-2\Lambda$ nets $$ S_G=\int \frac{1}{\kappa}(R-2\Lambda)\sqrt{-g}\ d^4x. $$

Now if $S_M$ is the action for the matter fields, then the combined action is $$ S=S_G+S_M $$ and setting the variation of $S$ with respect to $g$ to zero gives $$ 0=\int\left(\frac{1}{\sqrt{-g}}\frac{\delta S_G}{\delta g^{\mu\nu}}+\frac{1}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu\nu}}\right)\delta g^{\mu\nu}\sqrt{-g}\ d^4x, $$ but $$ \frac{1}{\sqrt{-g}}\frac{\delta S_G}{\delta g^{\mu\nu}}=\frac{1}{\kappa}\left(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}\right) $$ so setting the variation to 0 gives $$ \frac{1}{\kappa}\left(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}\right)=-\frac{1}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu\nu}}, $$ naming the RHS as $T_{\mu\nu}$ gives $$ R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}.$$ If we set the cosmological constant to zero, and take the Newtonian limit, we get $\kappa=8\pi G$ (in $c=1$), which is the usual form of the EFE.

I see no constant factor ambiguity here. However, if $S_M$ is the matter action and $\lambda$ is a constant then if $\phi_{(i)}$ (the matter fields) extremize $S_M$, then they will also extermize $\tilde{S}_M=\lambda S_M$, so I could technically sneak in a constant factor into the matter action too.

The end result is... I am a confused!

Question: Is the stress-energy tensor uniquely determinedby the action principle or not? Wald seems to suggest not. Carrol seems to suggest yes. My own line of thought suggests... maybe (?). If not, then is there any purely theoretical (and preferably rather mechanical) method to determine the ambigous constant?

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  • $\begingroup$ Are you only asking about possible conventional multiplicative constant factors of the standard actions (such as, Einstein-Hilbert, Klein-Gordon, etc), or are you wondering about the possibility of completely different actions, say, along the lines of this Phys.SE question? $\endgroup$ – Qmechanic Dec 9 '16 at 18:10
  • $\begingroup$ @Qmechanic I am only asking about multiplicative constants. It is clear to me that the individual actions are insensitive to multiplications by constants, but adding different actions with different constants seems to be at the heart of this issue. The SEM-tensor (at least in inertial frames) has well-defined interpretations for each of its components, which are in principle measurable, so I don't think there should be a freedom for multiplicative factors, yet apparantly there is. I seek to understand how this is resolved and how the actual SEM-tensors are found. $\endgroup$ – Bence Racskó Dec 9 '16 at 18:13
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    $\begingroup$ My 2 cents here : the action (partial parts and also the total one) is defined up to a multiplicative constant. The constant could be absorbed into the definition of the field that enters the action. What should remain are only coupling constants. Whatever how you place your constants, the EMT is unique in General Relativity. I don't see any ambiguity here. $\endgroup$ – Cham Dec 9 '16 at 19:44

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