# I GR as a Graded Time Dilation Field in Euclidean Space?

1. Mar 11, 2014

### rjbeery

The title says it all, really. Are we able to describe GR in terms of a Graded Time Dilation Field in Euclidean space?

From here we can see that light curvature can be analogously described via a material with a graded index refraction, so my question is really whether or not the following is is capable of encompassing GR:

$$t_0 = t_f \sqrt {1 - \frac{r_0}{r}} \frac{t_f}{t_0} = \frac{1}{\sqrt{1 - \frac{r_0}{r}}} =$$analogy to "n" in optical medium

2. Mar 11, 2014

### phyzguy

The metric tensor is a symmetric tensor with 10 degrees of freedom. Some of these are removed by the physical constraints of the theory, but it is certainly not possible to describe GR with a single scalar field as you propose.

3. Mar 11, 2014

### Staff: Mentor

As phyzguy said, the answer is no, there are not enough degrees of freedom in a scalar field.

Note that the result from the paper you cited applies specifically to a "static spherically symmetrical gravitational field". This particular spacetime has reduced degrees of freedom due to the assumed symmetry, so it can be represented by a scalar field whereas general spacetimes cannot.

4. Mar 11, 2014

### Bill_K

A few things this "theory" does not describe correctly:

• cosmological solutions
• gravitational waves
• black holes
• perihelion precession

5. Mar 11, 2014

### rjbeery

Yes! I was formulating a response to phyzguy along these lines before I read your response, DaleSpam. I'd like to understand "where" the analogy breaks down. It's difficult for me to extend the analogy into anything other than a gravitational field (e.g. time dilation caused by relative motion, etc), but I'm left wondering if the pure simplicity of a 3-D Euclidean space could possibly be reduced this far; dimensional symmetry, zero curvature...I thought perhaps the excess degrees of freedom could be eliminated...but I don't know which is of course why I'm asking.

Hi Bill_K!
I would say that the analogy could be extended to some or all of these in addition to DaleSpam's rotating masses, etc, by replacing the calculated GR curvature with the time dilation field. Note that the "formula" I give for calculating this field is derived from the GR time dilation itself.

6. Mar 11, 2014

### WannabeNewton

How would you predict frame dragging using this theory?

7. Mar 11, 2014

### PAllen

NO, it is derived from one solution of GR that is static and has perfect spherical symmetry. There is no part of the universe that exactly meets these conditions. In any realistic solution, curvature cannot be represented by a single function of position and time. This is a mathematical theorem, not subject to debate or interpretation. Specifically, curvature of 2-surface can be represented by a scalar. For more than 2 dimensions, it cannot.

8. Mar 12, 2014

### Staff: Mentor

In the vacuum region I think it breaks down as soon as you break spherical symmetry. In the matter region it breaks down as soon as you break spherical symmetry, or have a non-static distribution of matter, or have a matter field which is not a perfect dust.

A rotating mass is not spherically symmetric, it is at best axisymmetric.

9. Mar 12, 2014

### phyzguy

For example, consider the case of two orbiting neutron stars, losing energy due to gravitational radiation and spiraling in to form a black hole. We know they lose energy due to gravitational radiation (see Hulse-Taylor binary or this paper). There is no way to describe a situation like this with a single scalar field. As Bill_K said, it will not even describe gravitational waves.

10. Mar 12, 2014

### rjbeery

Hi PAllen! I appreciate what you're saying but the OP calls for Euclidean space with zero curvature. We model EM activity as a field in flat 3-D space, yes? Restricting the conversation solely to time dilation caused by gravity could we do the same thing here?

I not claiming that you or others are wrong on this point, I just want to verify that everyone is stripping time out from the geometry before denouncing the idea.

11. Mar 12, 2014

### Staff: Mentor

This leaves out a lot of "GR", since there are many scenarios covered by "GR" that do not even have a meaningful concept of "time dilation caused by gravity". If you want to restrict consideration only to those scenarios that do, then you need to ask about something more restricted than "GR". (Which then leads to the question, why would you be restricting yourself to just those scenarios?)

12. Mar 12, 2014

### rjbeery

The reason is: baby steps. I'm just exploring the idea. Could you give me some examples of scenarios covered by GR in which "time dilation caused by gravity" does not apply? Are you referring to time dilation caused by relative motion? Or something else?

13. Mar 12, 2014

### PAllen

Any situation that isn't stationary. That is, anything more complex than an isolated body in an unchanging state. It really is only the case of isolated, unchanging, body for which you could define a time dilation field. In all other cases, GR does not specify gravitational time dilation. That is, the whole notion of gravitational time dilation is not a general feature of GR; it is a specific derived feature applicable only in the most restricted cases.

14. Mar 12, 2014

### Staff: Mentor

Previous posters have already listed a number of them, but to summarize:

(1) The concept of "time dilation due to gravity" only applies if the spacetime is stationary, which basically means that one can find a family of worldlines in the spacetime along which the geometry does not change with time. (The technical way to say this is that the spacetime has a timelike Killing vector field.)

Examples of spacetimes that are not stationary: any spacetime in which objects have orbits that are not perfectly circular (such as the planets in the solar system, including effects like the perihelion precession of Mercury); any spacetime where gravitational waves are emitted (such as binary pulsars); black holes (because the region inside the event horizon is not stationary even if the region outside the horizon is); cosmology (because the universe is expanding).

(2) The concept of "time dilation due to gravity" is only *sufficient* to describe all the effects of gravity if the spacetime is static, which means that there is a family of spacelike hypersurfaces that are orthogonal to the family of worldlines along which the geometry does not change with time. The orthogonality property is necessary for us to be able to view the spacelike hypersurfaces as "space at an instant of time".

The primary example of a spacetime that is stationary but not static is any spacetime in which the central object is rotating. The rotation causes effects such as "frame dragging" that cannot be modeled by treating gravity as a scalar field. So even though there is a meaningful concept of "time dilation due to gravity", that by itself is not sufficient to describe all the effects of gravity for a rotating source.

15. Mar 12, 2014

### rjbeery

If we could account for motion-induced time dilation in addition to gravity-induced time dilation would you consider the theory more complete? Or are you suggesting that there are additional elements of GR which would not be accounted for?

16. Mar 12, 2014

### Staff: Mentor

This wouldn't change anything. "Motion-induced time dilation" can't be modeled as a scalar field anyway; it's not an invariant, it's a frame-dependent quantity, so it's not even the same kind of thing as "gravity-induced time dilation" to begin with.

17. Mar 12, 2014

### WannabeNewton

Could you account for all of electromagnetic theory using only a scalar electric potential? Could a scalar potential describe the circulating magnetic fields generated by currents and the circulating electric fields generated by time-varying magnetic fields?

18. Mar 12, 2014

### PAllen

How about 6, which is the actual number required. Then just call it the metric tensor after specification of arbitrary coordinate conditions.

19. Mar 12, 2014

### rjbeery

PAllen, you've already made the journey through GR (I presume), but don't you agree that teaching gravity as a time dilation field analogous to an EM field in a flat Euclidean space would be...more aesthetically pleasing if nothing else? Why single out gravity as the only force determining the geometry of our surroundings?

20. Mar 12, 2014

### PAllen

You don't have to call it geometry. You can treat it as fields on an abstract background. Weinberg, among others, has worked on expressing it this way. The point is, that if you want it to match the predictions of GR or spin-2 field theory, you need 6 functions not one or two.

21. Mar 12, 2014

### WannabeNewton

The EM field is an antisymmetric 2-tensor field so that doesn't help your case. Secondly, we can describe EM geometrically in a manner very much like GR. This is (part of) gauge field theory. And in my opinion there is nothing more aesthetically pleasing than a geometrical theory of physics, be it gravity or EM or any other gauge field.

22. Mar 12, 2014

### pervect

Staff Emeritus
I'm not sure I understand the proposed theory, but it seems to me like it isn't even consistent with special relativity, having a Euclidean spatial structure rather than a Lorentzian space-time structure.

If it had a Lorentzian structure, we could use the PPN formalism to show the shortcomings of such a theory, but I won't add more remarks on that until we determine more about the theory.

My impression of the theory is that rather than being based on special relativity and the Lorentz transform, it's based on pre-relativity ideas of absolute time, with the addition of a scalar field that controls the "rate" at which the absolute time flows.

I hope my concerns are clear, I fear there may be a language barrier related to the underlying concepts related to absolute/relative time and the relativity of simultaneity, based on previous discussions with the OP.

23. Mar 13, 2014

### rjbeery

Hi Wannabe!

I'm not trying to make a case, really, I'm just exploring this idea. Can you or PAllen help me appreciate why the EM field can be described as a two scalar potential but gravity requires 6? Does it have something to do with the generality of GR vs a presumption of dimensions in studying EM waves?

24. Mar 13, 2014

### WannabeNewton

Yes poor choice of words on my part, I apologize.

I'm not sure what you mean by "why the EM field can be described as a two scalar potential but gravity requires 6"...what is a "two scalar potential" to start with? And in what sense does gravity instead require 6 of them?

The electromagnetic field is entirely specified by a given 4-potential $A_{\mu}$ (equivalent up to a gauge $A_{\mu} \rightarrow A_{\mu} + \partial_{\mu}\varphi$) whereas the space-time geometry associated with the gravitational field is entirely specified by a metric tensor $g_{\mu\nu}$ (equivalent up to a gauge = diffeomorphism). In the case of EM, we define a gauge covariant derivative $D_{\mu} = \partial_{\mu} + ie A_{\mu}$ and the curvature of $D_{\mu}$ is the electromagnetic field $F_{\mu\nu}$. In the case of gravity, we have the covariant derivative $\nabla_{\mu}$ defined in terms of $g_{\mu\nu}$ in the usual way through the Christoffel symbols and the curvature of $\nabla_{\mu}$ gives rise to the various fields describing tidal gravitational effects (expansion, shear, vorticity etc.). So this is essentially how we geometrically describe EM except the geometrical object that $D_{\mu}$ is associated with is more abstract than the physical space-time that $\nabla_{\mu}$ is associated with.

Now, why do we need a 4-tensor theory of gravity whereas we only need a 4-vector theory of EM? Well the 4-vector theory of EM follows straight from Maxwell's equations so that's a done deal. We could certainly try to posit a congruent 4-vector theory of gravity. In fact in a good GR textbook you will find a detailed walk-through of scalar and vector theories of gravity in an attempt to show that they are not adequate for describing gravity (see e.g. Straumann section 2.2.4). The key underlying mechanism that allows for a 4-vector theory of EM but rejects a 4-vector theory of gravity is the fact that like electric charges repel whereas all masses attract-it turns out that a Lorentz covariant i.e. relativistic vector field theory necessarily gives rise to repulsion between like "charges". Such a classical field theory cannot describe gravity. We are thus necessarily led to a 4-tensor theory of gravity (as a theory of a dynamical space-time geometry).

To answer your other question, no it doesn't have to do with the degrees of freedom. Both the gravitational and electromagnetic fields have two degrees of freedom at each point. This is the case for massless fields.

25. Mar 13, 2014