# Why a vector field can not be used to describe gravity?

1. Dec 9, 2015

### Chhhiral

In many books of general relativity I have found the following statment:
a vector field produces repulsive forces between like charges so can not be used to describe gravity.
But they do not show it. How can i show it or where can I find a book or a paper in which it is shown?

Please, to avoid misinterpreting my question read the post published in the session of General Physics: "Vector field: one aspect that everyone knows but nobody shown"

References:
Hobson, Michael Paul, George P. Efstathiou, and Anthony N. Lasenby.General relativity: an introduction for physicists. Cambridge University Press, 2006. pag:191
"A gravitational theory based on a vector field can be eliminated since such a theory
predicts that two massive particles would repel one another, rather than attract."

Gasperini, Maurizio. Relatività Generale e Teoria della Gravitazione. Springer Milan, 2015. pag:27

the English version should be: Gasperini, Maurizio. Theory of Gravitational interactions. Springer Science & Business Media, 2013. at the chapter: "Towards a relativistic theory of gravitation".

I found the same assertion in other books...

2. Dec 9, 2015

### Demystifier

Note, first, that classical electrodynamics is a relativistic vector theory and the force between charges of equal sign is repulsive. So in essence, vector theory of gravity would be like classical electrodynamics with all charges (interpreted as masses) positive, hence with repulsive forces.

Or is there a way to avoid that conclusion? There are 2 ways, but both lead to problems.

First, it s possible to write down "natural" equations of motion for attractive vector gravity, but such equations of motion cannot be derived from a Lagrangian and do not have conservation of energy-momentum.

Second, it s possible to write down a Lagrangian that leads to attractive vector gravity with conserved energy-momentum, but in such a theory kinetic energy is negative.

It all boils down to first writing the equations of classical electrodynamics (for both EM field and the matter), then artificially changing some sign to get a force of the opposite sign, and finally exploring the other consequences of that change of sign. Depending on your imagination, you can make such a change of sign at different places, leading to different theories with different properties.

Misner Thorne Wheeler - Gravitation
Exercise 7.2 (page 179).

Last edited: Dec 9, 2015
3. Dec 9, 2015

### Chhhiral

Thank you very much, I'll read it immediately

4. Dec 9, 2015

### robphy

Possibly useful: https://www.quora.com/Quantum-Field-Theory/Why-do-particles-of-odd-integer-spin-generate-forces-which-can-be-both-attractive-and-repulsive-whereas-particles-of-even-integer-spin-only-attract [Broken]
(I haven't read through it yet... and probably won't have time to do so for a while.)

Last edited by a moderator: May 7, 2017
5. Dec 9, 2015

### Chhhiral

@robphy
Very interesting, thanks!

6. Dec 9, 2015

### bcrowell

Staff Emeritus
Since there are valid classical explanations, I don't see a strong motivation for looking to QFT for an explanation.

Here's are a couple of very simple explanations that may be little hand-wavy, but that I think probably work.

Explanation #1: Suppose we had a localized source of gravitational radiation, and the radiation was monopole in character, like some sound waves. The source would have to be an oscillating mass, but that would violate conservation of mass-energy. So we can't have gravitational monopole radiation. Nor do we expect to have gravitational dipole radiation from a localized source, because that would violate conservation of momentum. So the lowest multipole we expect is quadrupole.

Explanation #2: We define gravitational mass as the thing that's measured by the observation that a cloud of test particles, initially at rest with respect to one another, begin to converge, so that the cloud's volume decreases. If we see the cloud contract like this, we say that there is a mass somewhere inside the cloud. Since gravitational waves can propagate in a vacuum, the effect of a gravitational wave as it passes through a cloud of test particles must be to distort the cloud without changing its volume. This would presumably cause an initially spherical cloud to undergo an ellipsoidal distortion. Such a distortion acts like a quadrupole mode of polarization, not a dipole mode. One way to see this is that if you rotate the polarization by 180 degrees, it's the same.

7. Dec 9, 2015

### bcrowell

Staff Emeritus
Don't you mean that the energy of gravitational waves is negative? In the MTW exercise you refer to, the kinetic energy of the material particles is the same, including sign, as in ordinary E&M.

Although MTW does this in a fancypants formulation using Lagrangians, I think the basic idea can be expressed much more simply. The theory of a vector field is Maxwell's equations, and all you can really do is flip signs in certain places. (And by the equivalence principle, all particles have the same "charge to mass ratio.") To make the force attractive, you have to make the field's energy density negative, just as it is in Newtonian gravity. This is so that when you bring one mass on top of another mass, the total energy stored in the field goes down. But this means that the energy of gravitational waves has to be negative.

I also don't think the MTW-style argument is really that clearcut. Their point in that section of the book is not that everything comes out OK if you use a symmetric tensor. They develop the theory with a symmetric tensor and show that it lacks self-consistency. The whole section is about attempts to treat gravity as a field on a background of flat spacetime, and the real point is that none of the attempts work.

Last edited: Dec 9, 2015
8. Dec 10, 2015

### Demystifier

I have done my analysis (a few mounts ago) independently of MTW, and then found out that something similar has also been done in MTW. By "negative kinetic energy" I had in mind negative kinetic energy of the gravitational field, which (in the vector theory) corresponds to negative energy of gravitational waves.

Related to this, I would like to add a few notes.

First, one can define a new Lagrangian by multiplying the Lagrangian above (the one with negative kinetic energy of gravitational field) by -1. The resulting equations of motion are the same, so it is essentially the same physical theory. Yet, in this form the kinetic energy of the gravitational field becomes positive, while kinetic energy of matter becomes negative. This is one of the reasons that I said that "kinetic energy is negative", without specifying do I mean kinetic energy of gravitational field or that of matter; the two possibilities are really equivalent.

Second, in general relativity, energy of gravitational field and energy of gravitational wave do not have the same sign. If it looks paradoxical (which certainly looked so to me until I resolved it), see
especially post #8.

9. Dec 10, 2015

### Demystifier

Since you write textbooks on general physics (which, by the way, are really great), it is not surprising that you find an explanation without Lagrangians simpler. But for me, it is the Lagrangian formulation that looks simpler. Just as Lagrangians look "too fancy" to some, non-Lagrangian approaches look "too baroque" to me. I guess it has to do with someone's general style of thinking; even in architecture I prefer modern elegant style over baroque style - but that's just me.

10. Dec 10, 2015

### bcrowell

Staff Emeritus
I think we should distinguish two approaches. Let's call the one you've been referring to #1. This involves doing gravity as a field on a background of flat spacetime. #2 would be something more like the ADM approach, which doesn't use a flat background.

In approach #1, the theory of a vector field is isomorphic to Maxwell's equations, although we can do a sign flip to make the force attractive. In that theory, there is only one natural way to define energy, because the field equations lead uniquely to one specific conservation law. Of course with any conservation law, we can rescale the conserved quantity by an arbitrary factor, and the factor can be negative if we wish, but this has no physical significance. But in this approach, we have no other wiggle room in terms of how to define the energy of the field. It's a specific local expression involving the square of the field, which just has some arbitrary scaling factor, call it $\alpha$. Therefore all we need is the elementary argument to establish that gravitational waves carry negative energy. That is, if we choose the sign of $\alpha$ so that the force is attractive, then the field's energy density is negative, and gravitational waves carry negative energy. However, approach #1 is IMO of limited interest for our present purposes. We can it to prove a no-go theorem for a vector theory of gravity, but we can also use it to prove a no-go theorem for a tensor theory of gravity. These really amount to no-go theorems for gravity as a theory of a field on a flat background. They tell us nothing conclusive about whether gravity is a scalar, vector, or tensor field.

In approach #2, the equivalence principle guarantees that we will never have a local expression for the energy density of the gravitational field. This leaves us a lot of wiggle room in defining energy, because we can come up with an expression that is nonlocal, as in ADM. This wiggle room is what allows us to have negative energy for the field of the earth, but positive energy for a gravitational wave.