Why is gravity sensitive to absolute energies?

Question

In QFT absolute energies play no role in the physical set-up, only relative energies (i.e. energy differences) are important. However, in general relativity this doesn't appear to be the case, I've read that gravity is sensitive to absolute energies. I thought this could be argued by considering the $T_{00}$ component in the stress-energy tensor $T_{\mu\nu}$, i.e. the energy density $\rho$. The problem is, absolute quantities such as mass and energy density appear in QFT as well so how does one explain that QFT is not sensitive to absolute energies whereas general relativity is?

Answer 1

Let's start with qft. Imagine taking the action $S$ and adding a constant. The new action is

\begin{equation} S_{new} = S + M^4 \int d^4 x \end{equation} where M is a mass scale needed for dimensions.

This last term in the action does not depend on any fields so it does not contribute to any dynamics. One way to express this is that the only effect of the last term is to change the path integral by an overall constant \begin{equation} Z_{new} = \int D\phi e^{iS + i M^4 \int d^4 x} =\left( e^{i M^4 \int d^4 x} \right) Z \end{equation} where $\phi$ is a generic label for all the fields and $Z=\int D\phi e^{iS}$. The overall constant always drops out of correlation functions since they will be things like $Z^{-1} \delta Z/ \delta J$ (or if you like the constant can just be absorbed into the measure of the path integral).

General relativity is diffeomorphism invariant which prevents us from just writing down a constant times $d^4x$. The diff invariant generalization of the above relevant for Gravity is \begin{equation} S_{new,GR} = S + M^4 \int d^4 x \sqrt{-g} \end{equation} The factor of $\sqrt{-g}$ changes everything. The "constant" term now depends on the metric, which is a dynamical field. So this term now does contribute to the dynamics.

In path integral terms, the point is that you can't factor $e^{iM^4\int d^4 x \sqrt{-g}}$ outside the path integral because you are integrating over $g_{\mu\nu}$.

Indeed, we usually identify $M^4= M_{pl}^2 \Lambda$ where $M_{pl}$ is the planck scale and $\Lambda$ is the cosmological constant.

To summarize, Gravity sees everything, including constants. If you like, the cosmological constant is an intrinsic energy associated to spacetime ("vacuum energy"). The more spacetime volume you have the more "cosmological constant" energy you have. Since the metric is used to measure volumes, Gravity sees this contribution to the energy.

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Edit in response to comments below.

A mass term in qft is a term quadratic in the fields with no derivatives. For a scalar field a mass term looks like \begin{equation} S_{mass} = -\frac{1}{2} \int d^4 x m^2 \phi^2 \end{equation} This is field dependent so it contributes to the dynamics. You can't factor the mass term out of the path integral. The same goes for a more general potential $V(\phi)$.

Indeed on some level you can think of the cosmological constant as being like a potential for Gravity. Writing $g_{\mu\nu}=\eta_{\mu\nu} + h_{\mu\nu}/M_{pl}$ the cc term can be expanded in powers of $h$ \begin{equation} \sqrt{-g} \sim 1 + h/M_{pl} + h^2/M_{pl}^2 + \cdots \end{equation} where all the coefficients in this "potential for $h$" are fixed by diff invariance. I wouldn't take this analogy too seriously, but it's a way you can look at it.

Finally, it was asked if one could say Gravity sees the cc because of diff invariance. I would say that is indeed a valid way of looking at things based on the argument above.

Answer 2

Andrew's answer is correct on the essential logic of the argument. Here's a more formal version of it that illustrates that the crucial thing is really that the metric is a dynamical field in general relativity, not the invariance of the action under any specific transformation:

When talking about physical theories on arbitrary manifolds ("backgrounds"), we need to be mindful that we cannot just integrate functions over volumes, but what is integrated are n-forms (for n the dimension of the manifold). This is more explicit when we choose to write the action integral in terms of the volume element - for our Lagrangian density $\mathcal{L}$, the Lagrangian is $$ L = \int \mathcal{L}\mathrm{d}V.$$ This volume element - or volume form - always involves the metric $g$ of the underlying space as $\mathrm{d}V[g] = \sqrt{g}\mathrm{d}x^1\wedge\dots\wedge\mathrm{d}x^n$. This is well-known to everyone who has ever done such an integral in spherical coordinates - the "spherical volume element" $r^2\sin(\phi)$ is exactly the determinant of the Euclidean metric in spherical coordinates.

When the metric is not a dynamical field, this is just a constant from the viewpoint of the action. So we can add some constant $C$ to $\mathcal{L}$ and this results in a constant term $C\int \mathrm{d}V$ being added to the Lagrangian, not changing the equations of motion at all and is hence undetectable.

In contrast, when $g$ itself is a dynamical field, then of course the term $C\int \mathrm{d}V[g]$ contributes to the equations of motion of $g$ since the variation of $\mathrm{d}V[g]$ with respect to $g$ is non-zero, so adding constants to the Lagrangian density is detectable in such theories via changes in the equations of motion.

This argument is completely independent of whether or not the theory in which $g$ is a dynamical field possesses "diffeomorphism invariance" in any sense.