# Scalar gravity -Feynman lectures on gravitation

"scalar gravity" -Feynman lectures on gravitation

Hi all,

I'm trying to understand the following claim from Feynman's lectures on gravitation, section 3.1 (p.30 in my edition). He's considering how heating or cooling two clouds of gas would change their mutual gravitational attraction.

Feynman said:
Electric forces are unchanged by random motions of the particles. Now the interaction energy is proportional to the expectation value of $\gamma=1/\sqrt{1-v^2/c^2}$. Since the resulting potential is not velocity dependent the proportionality factor must go as $\sqrt{1-v^2/c^2}$. This means that the interaction energy resulting from the operator 1, corresponding to a scalar field, must go as $\sqrt{1-v^2/c^2}$. This means that the spin-zero theory predicts that the attraction between masses of hot gas would be less than that for cool gas.

I don't understand this statement. The electric scalar potential is the time component of a four vector which gets dotted into another four vector, so that the resulting physics is invariant. I don't see anything in the expression for a Yukawa potential that I can identify with the velocity dependence he's talking about. Can someone please explain?

Bill_K
In tensor gravity the source of the field is T00, the 00 component of a rank 2 tensor. Under a Lorentz transformation, T00 → γ2 T00. But in the meanwhile, the volume element contracts, d3x → (1/γ) d3x. So the integrated source M = ∫ T00 d3x → γ M grows only as one factor of γ. Thus for tensor gravity the source of the gravitational field is the total energy, or "relativistic" mass, which grows with velocity like γ.

(Be clear that we are talking about random internal velocities, whose directions average out, not an overall velocity of the object.)

A vector theory of gravity would be like electromagnetism, in which the source density is J0, the 0th component of a vector. Under a Lorentz transformation, J0 grows like J0 → γ J0. But this is exactly compensated by the contraction of the volume element, and so the source of the gravitational field in this case is independent of internal velocities, just like the total electric charge is.

For a scalar theory of gravity, the source density is a scalar, presumably the trace of the stress-energy tensor, and is a Lorentz invariant. But the volume contraction is still present to contribute a (1/γ), and so the effective total strength of the source will decrease with increasing internal velocities.

• 1 person
Thanks for your reply. How exactly is the volume contraction affecting the scalar mediated interaction? If the source of the gravitational field were some sort of scalar density to which the putative scalar graviton coupled, wouldn't the lorentz contraction of the volume increase the density?

Bill_K
The density is not changing because it is being squeezed! It is changing due to its own behavior under a Lorentz transformation.
Hmm. I think I may have been putting too much emphasis on the word "density" (thinking of it as the amount of some stuff per volume, which by definition would appear to increase to a relatively moving observer) and not enough on the word "scalar" 