# Unifying proper acceleration and gravity?

1. Apr 14, 2013

### SpiderET

Was there any serious attempt (peer reviewed of course :) to unify proper acceleration and gravity?
From my view as hobby physicist I have the feeling, that equivalency priciple in General relativity is not including proper acceleration, but better and extended theory should explain/unify all types acceleration and gravity together.

2. Apr 14, 2013

### WannabeNewton

Unify in what sense? The theory tells us that a particle in free fall has no proper acceleration. If the particle is indeed subject to 4-forces, say due to interaction with some field, then we can directly relate the resulting proper acceleration to quantities derived from the space-time metric. For example if we have a charged particle in curved space-time (with the space-time metric being a solution to Einstein's equations) interacting with an electromagnetic field then we can write $a^{b} = \frac{q}{m}F^{b}_{}{}_{c}u^{c} = u^{c}\nabla_{c}u^{b}$ where $F_{ab}$ is the electromagnetic field strength tensor. Could you explain to me what you mean exactly by unify? Thanks.

3. Apr 14, 2013

### Staff: Mentor

The best current theory for reconciling gravitational forces and the accelerations they produce, non-gravitational forces and the accelerations they produce, and proper acceleration is General Relativity. It is so mathematically elegant and works so well in its domain of applicability that there is not a lot of interest in finding a "better" alternative - but if someone did want to go looking for one, the first step would be to seriously learn General Relativity, so as to be able to recognize something better when it came along.

Most people who have put in this level of effort will be working on unsolved problems in cosmology or in reconciling GR and quantum mechanics.

(In this context, "seriously learn" means enough study to be able to, for example, solve the problems in a textbook like Misner, Thorne, and Wheeler's "Gravitation").

4. Apr 14, 2013

### WannabeNewton

Or use it long enough to be able to lift off a table with ease

5. Apr 14, 2013

### Staff: Mentor

Hah - I remember when I bought my copy back in my senior year of college. My roommate picked it up, set it back down, and said "How convenient - A textbook AND a demonstration, all in one".

6. Apr 14, 2013

### SpiderET

OK, I see that my question has to be clarified.

I mean, that "acceleration" caused by centrifugal forces was within GR explained as fictious forces and gravity itself has become a fictitious force,[26] as enunciated in the principle of equivalence. This could be seen as sort of unification of centrifugal force and gravity.

But has there been an attempt to explain also proper acceleration in similar way and unify it with gravity?

More detailed (from http://en.wikipedia.org/wiki/History_of_centrifugal_and_centripetal_forces )

Ultimately this notion of the transformation properties of physical laws between frames played a more and more central role.[22] It was noted that accelerating frames exhibited "fictitious forces" like the centrifugal force. These forces did not behave under transformation like other forces, providing a means of distinguishing them. This peculiarity of these forces led to the names inertial forces, pseudo-forces or fictitious forces. In particular, fictitious forces did not appear at all in some frames: those frames differing from that of the fixed stars by only a constant velocity. In short, a frame tied to the "fixed stars" is merely a member of the class of "inertial frames", and absolute space is an unnecessary and logically untenable concept. The preferred, or "inertial frames", were identifiable by the absence of fictitious forces.[23][24][25]

Later the general theory of relativity further generalized the idea of frame independence of the laws of physics, and abolished the special position of inertial frames, at the cost of introducing curved space-time. Following an analogy with centrifugal force (sometimes called "artificial gravity" or "false gravity"), gravity itself became a fictitious force,[26] as enunciated in the principle of equivalence.[27]

Last edited: Apr 14, 2013
7. Apr 14, 2013

### Staff: Mentor

There is a fundamental difference between the"fictitious" forces like gravity (in general relativity but not classical physics and special relativity), centrifugal force, or Coriolis force, and the "real" forces like electromagnetism or what you feel when you push on an object: You can always find a coordinate system in which the fictitious forces disappear and the acceleration they produce disappears; and you can't do that with the "real" forces.

Proper acceleration is (loosely) defined as that acceleration that can't be made to go away with a change of coordinates, so is by definition not produced by the so-called fictitious forces.

Hence, the unification has already proceeded about as far as it can.

8. Apr 14, 2013

### PAllen

The joke I heard was the MTW falsifies the EP - it obviously falls faster than everything else.

9. Apr 14, 2013

### HomogenousCow

That christofell symbol-proper velocity term is not tensorial, you cannot treat it as a four force

10. Apr 14, 2013

### WannabeNewton

$a = \nabla_{u}u$ is a coordinate independent expression; this is a basic definition from Riemannian geometry that holds for any linear connection $\nabla$. In GR we interpret $a^{b}$ as the 4-acceleration when $u^{b}$ is the 4-velocity; if $a^{b}\neq 0$ then we say there exists a 4-force acting on the particle (Wald page 69). Who said it isn't covariant? $a^{b} = u^{a}\nabla_{a}u^{b}$ is a perfectly covariant expression; it makes no reference to coordinates and only involves $u^{a}$ and quantities derived from the metric i.e. derivative operator $\nabla_{a}$ coming from the levi-civita connection. The equation would not be covariant if the christoffel symbols appeared only by themselves but that is clearly not the case here.

11. Apr 14, 2013

### HomogenousCow

What I meant was, in the geodesic equation, many people regard it as a f=ma deal and interperet the christofell symbol term as a pseudo-force, that would not work because that term is not tensorial

12. Apr 14, 2013

### WannabeNewton

See the calculations here: http://www.mathpages.com/home/kmath641/kmath641.htm
Also take note of what the OP's referenced quote and what Nugatory said: the inertial forces, e.g. inertial forces in rotating reference frames, do not transform properly under coordinate transformations as "real" forces do. The failure of the christoffel symbols to be tensorial is no different and in fact under a coordinate transformation, the inertial forces pick up the same extra terms that christoffel symbols do (see link).

EDIT: Are you perhaps saying that we cannot think of gravity as an inertial force in the same manner as the inertial forces in say rotating reference frames because $u^{a}\nabla_{a}u^{b} = 0$ is covariant whereas if we think of the christoffel symbol terms for free fall motion as an inertial force related to gravity then we run into a problem because it does not transform covariantly? If so then I agree completely and this was also what Nugatory was saying above.

Last edited: Apr 14, 2013
13. Apr 14, 2013

### HomogenousCow

Yes, I think the semantics is making things foggy

14. Apr 14, 2013

### HomogenousCow

I'm not sure what we are discussing here, the geodesic equation is essentially the statement that the proper acceleration of the particle is 0, gravity is not a four force because the RHS in the geodesic equation is not a tensor, I'm not sure how we extend this to classical mechanics because nothing there are four tensors anyways.

15. Apr 15, 2013

### WannabeNewton

That there is no meaningful way to describe gravity in terms of some 4-force in GR is a result of the equivalence principle; one of the basic tenets of GR is that the equations of motion for a freely falling test particle become $u^{a}\nabla_{a}u^{b} = 0$ i.e. the trajectories of freely falling test particles are geodesics of the space-time. This is obviously a tensor equation.

Regarding the last comment, did you look at the link from the previous post?

EDIT: Just to add, if you start from the energy momentum tensor for a single free test particle $T^{ab} = \frac{m}{\sqrt(-g)}\int u^{a}u^{b}\delta ^{(4)}(y - x(\tau))d\tau$ and use local conservation of energy $\nabla_{a}T^{ab} = 0$, you can directly arrive at $u^{a}\nabla_{a}u^{b} = 0$ (see Carroll chapter 4 problem 3).

Last edited: Apr 15, 2013
16. Apr 15, 2013

### SpiderET

Thanks a lot for explanation, so conclusion is that current solution in GR is sufficient and nobody has really need for further implementation of proper acceleration into GR.

17. Apr 15, 2013

### Seminole Boy

In my opinion, we need to understand light and space better before we can improve GR. I think our understanding of light--and how it reaches us--is a huge problem in contemporary physics. We can run mathematics all day long, but if our understanding of light and space is incomplete, accepted mathematical formalism is not as important as we try to make it out to be.

18. Apr 15, 2013

### Staff: Mentor

This is total nonsense. There is no observed behavior of light that QED does not accurately predict. Our understanding of light is complete. There are lots of problems, but that isn't one.

19. Apr 15, 2013

### Seminole Boy

DaleSpam: if you actually believe what you just said, that tells me a lot about your understanding of science. To say that we completely understand light is absolute nonsense. Please tell me that you didn't mean your post in a literal way.

20. Apr 15, 2013

### Staff: Mentor

I mean it completely literally. If you disagree then please post a scientific reference to any EM phenomenon that is not explained by QED. If there is no such phenomenon then our understanding is complete.

I get irritated by silly claims that our understanding of EM is incomplete, like yours. Never once has any of the people making such a claim actually been able to produce any phenomenon that is not explained by modern EM theory.

Last edited: Apr 15, 2013