# Is the assumption that gravity acts on everything fundamental to GR?

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## Main Question or Discussion Point

Apologies in advance if this is a naive question. From what I've understood, it's not particularly meaningful to talk about gravity as a force, since it induces the same acceleration (classically speaking) in everything.

Whatever device or accelerometer we use, every component of that too will be accelerated the same way as the observer will (at least classically speaking), since classical gravitation force is proportional to the inertial mass and everything will thus be subject to the same force/mass relation. The observer will not be able to measure (provided they or the measuring device are under no other influence other than gravity) any acceleration whatsoever. This makes it meaningless to talk about gravity-induced "acceleration" since we can't measure it.

One central idea underlying this is the equivalence principle (gravitational mass = inertial mass) that makes sense to me. But I wonder - is the fact that gravity acts on everything also fundamental to this? If there were anything at all that's not acted on by gravity, would the whole "gravity as a force doesn't make sense" premise (or maybe GR itself) break down? Is this question related to Mach's principle in any way?

(I'm not claiming any such thing might or should exist - just trying to clarify my understanding)

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anuttarasammyak
Gold Member
GR says gravity comes from spacetime features. So your question might be answered positively because everything shares spacetime feature.

Ibix
But I wonder - is the fact that gravity acts on everything also fundamental to this?
General relativity describes gravity in terms of geometry. That means that a body's freefall path is predictable solely from knowing the geometry of spacetime. That, in turn, means that it can't depend on intrinsic properties of the body such as its mass. Thus, gravity must affect everything. Furthermore, if we did find something that wasn't affected by gravity, or was affected in a different way, we'd have to abandon GR, yes.

Of course, a major problem at the moment is that we don't believe GR is entirely correct but we can't find anything acting in a way not explainable by GR.

There are two caveats to my first paragraph. First, real bodies are not pointlike, and different parts of them may experience different parts of the gravitational field. That can lead to two bodies not following quite the same path as each other because their mass distribution or internal stresses are different. Second, sufficiently large mass will contribute significantly to spacetime curvature and you don't have the same gravitational field and you get different paths - I'm talking only about test masses above.

Shirish
General relativity describes gravity in terms of geometry. That means that a body's freefall path is predictable solely from knowing the geometry of spacetime. That, in turn, means that it can't depend on intrinsic properties of the body such as its mass. Thus, gravity must affect everything. Furthermore, if we did find something that wasn't affected by gravity, or was affected in a different way, we'd have to abandon GR, yes.

Of course, a major problem at the moment is that we don't believe GR is entirely correct but we can't find anything acting in a way not explainable by GR.

There are two caveats to my first paragraph. First, real bodies are not pointlike, and different parts of them may experience different parts of the gravitational field. That can lead to two bodies not following quite the same path as each other because their mass distribution or internal stresses are different. Second, sufficiently large mass will contribute significantly to spacetime curvature and you don't have the same gravitational field and you get different paths - I'm talking only about test masses above.
Really helpful as usual! One thing though - and I have no technical knowledge whatsoever about this but still - in relation to dark energy, I've heard (mainly via pop-sci youtube videos) that it has an "anti-gravitational" effect, causing expansion of the universe. So isn't that something on which gravity doesn't act the usual way? [and you can call me out if that last line is preposterous or doesn't make any sense]

vanhees71
Gold Member
2019 Award
The source of the gravitational field (which you can reinterpret as the pseudometric of a pseudo-Riemannian spacetime manifold if you are inclined to geometrical interpretations, which are almost all physicists since Einstein) are the energy-momentum-stress distributions of matter (and radiation). This is an integral part of GR. There's no other possibility than this to choose for mathematical reasons. In this sense the gravitational field couples to everything since everything carries energy and momentum.

One other possibility than energy-momentum-stress distributions from matter and radiation is the socalled cosmological constant. With the right sign it's indeed leading to an accelerating expansion of the universe. Any other usual source of energy and momentum tends to slow the expansion of the universe down, because for these sources the gravitational interaction is universally attractive in contradistinction to the other three forces which are both repulsive and attractive. E.g., the electromagnetic interaction is attractive for two charged particles with opposite signs of their charges and repulsive for like signs of their charges. Gravity for any usual "matter and radiation" sources is always attractive. That's why it was such a surprise for cosmologists when they found out that the Hubble expansion is accelerated, and the only way to get such a repulsive gravitational effect is the cosmological constant, which can be reinterpreted as another energy-like source to the gravitational field. That's why it's called "dark energy". It's called "dark", because it's something that does not interact electromagnetically and thus cannot produce or absorb electromagnetic radiation (particularly light).

Shirish and Ibix
Ibix
I don't know much about dark energy, but do note that it isn't a test body. It's more or less uniform throughout the universe and all we know of it is that it's a source of gravity. Note that I said sources of gravity don't really count because they affect spacetime themselves, and you can't really fairly compare them to things that don't have a significant effect.

The source of the gravitational field (which you can reinterpret as the pseudometric of a pseudo-Riemannian spacetime manifold if you are inclined to geometrical interpretations, which are almost all physicists since Einstein) are the energy-momentum-stress distributions of matter (and radiation). This is an integral part of GR. There's no other possibility than this to choose for mathematical reasons. In this sense the gravitational field couples to everything since everything carries energy and momentum.

One other possibility than energy-momentum-stress distributions from matter and radiation is the socalled cosmological constant. With the right sign it's indeed leading to an accelerating expansion of the universe. Any other usual source of energy and momentum tends to slow the expansion of the universe down, because for these sources the gravitational interaction is universally attractive in contradistinction to the other three forces which are both repulsive and attractive. E.g., the electromagnetic interaction is attractive for two charged particles with opposite signs of their charges and repulsive for like signs of their charges. Gravity for any usual "matter and radiation" sources is always attractive. That's why it was such a surprise for cosmologists when they found out that the Hubble expansion is accelerated, and the only way to get such a repulsive gravitational effect is the cosmological constant, which can be reinterpreted as another energy-like source to the gravitational field. That's why it's called "dark energy". It's called "dark", because it's something that does not interact electromagnetically and thus cannot produce or absorb electromagnetic radiation (particularly light).
If possible, could you elaborate on the phrase "can be reinterpreted as another energy-like source"? I seek clarification on this phrase because you mentioned that gravitational field couples to everything since everything carries energy and momentum. If dark energy can be reinterpreted as another energy-like source, shouldn't gravitational field couple to it in the same way as it does to usual matter/energy? Yet apparently that's not true - the coupling in case of dark energy is different from the usual stuff.

vanhees71
Gold Member
2019 Award
I refer to the "derivation" of the Einstein field equations as given in Landau Lifshitz. Having established the idea that the action for the equations of motion should be generally covariant and that you want a 2nd-order equation. The only scalar you can build from the metric components with up to 2nd derivatives, where the coefficients in the expression of the 2nd derivatives are only functions of the metric components but not their derivatives is the Ricci scalar ##R##, i.e., you get ##\mathcal{L}_1 \propto R##. The only other possibility is a constant ##\mathcal{L}_2=\Lambda##. This finally leads to
$$\mathcal{L}=-\frac{1}{\kappa} (R+2 \Lambda)+\mathcal{L}_{\text{matter}}$$
with ##\Lambda=\text{const}##. The action is
$$A=\int \mathrm{d}^4 q \sqrt{-g} \mathcal{L}.$$
Then by variation of the action the Einstein field equation follows
$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=\kappa T_{\mu \nu} + \Lambda g_{\mu \nu}.$$
Bringing the term with the cosmological constant to the right-hand side indeed it looks like an additional contribution to the energy density; ##T_{\mu \nu}## is the energy-momentum-stress tensor of "matter and radiation".

I refer to the "derivation" of the Einstein field equations as given in Landau Lifshitz. Having established the idea that the action for the equations of motion should be generally covariant and that you want a 2nd-order equation. The only scalar you can build from the metric components with up to 2nd derivatives, where the coefficients in the expression of the 2nd derivatives are only functions of the metric components but not their derivatives is the Ricci scalar ##R##, i.e., you get ##\mathcal{L}_1 \propto R##. The only other possibility is a constant ##\mathcal{L}_2=\Lambda##. This finally leads to
$$\mathcal{L}=-\frac{1}{\kappa} (R+2 \Lambda)+\mathcal{L}_{\text{matter}}$$
with ##\Lambda=\text{const}##. The action is
$$A=\int \mathrm{d}^4 q \sqrt{-g} \mathcal{L}.$$
Then by variation of the action the Einstein field equation follows
$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=\kappa T_{\mu \nu} + \Lambda g_{\mu \nu}.$$
Bringing the term with the cosmological constant to the right-hand side indeed it looks like an additional contribution to the energy density; ##T_{\mu \nu}## is the energy-momentum-stress tensor of "matter and radiation".
I have insufficient understanding as of now. I should read up GR before pursuing the line of thought in my original post further - then I can reference your post again. Maybe I'll get back to you in some months. Thanks so much again!