# I Are there any alternative interpretations to GR?

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1. Mar 2, 2016

### olgerm

Are there any interpretation to general relativity that described gravity as field (which do not have to be vectorfield, but may have 10 components) and physical-space as classical euclidean space?
Can it be mathematically proven that such interpretation can not exist?

I am not talking about approximation ,but an interpretation ,which where exactly in line with the general relativity.

2. Mar 2, 2016

### PAllen

First, if you were going to use a flat spacetime background, it would have to be Minkowski spacetime not Euclidean. Secondly, many interesting GR solutions have non-trivial topology, so a flat background could not be ordinary Minkowski spactime. However, given Minkowski spacetime with appropriate topology, you could do this. None of the observables would be related to the background (flat) metric, but it could still be used as the basis for interpretation. There are authors who take this approach.

3. Mar 2, 2016

### olgerm

Would it be possible to describe spatial coordinates in Euclidean space and time separately (like in classical physics)? Time dilation could then be explained by forces ,that gravity-field causes. For example: 2 pointmasses are orbiting circulary around their center of mass so that their z-coordinate is constant.
$T=\frac{\omega}{2 \cdot \pi}$
$\omega=\sqrt{\frac{a}{r}}$
$a=\frac{m \cdot G}{4\cdot r^2}$
so $T=\sqrt{\frac{m \cdot G}{r^3}}\cdot \frac{1}{4\cdot \pi}$
Now lets watch these pointmasses from frame of reference which is moving with velocity v in direction of z-axis. Period of orbiting must be $\sqrt{1-\frac{v^2}{c^2}}$ times bigger than in last frame of reference, according to SR. So gravitational force must be $1-\frac{v^2}{c^2}$ times smaller, than in last frame of reference.

Since force may depend of velocity in this interpretation, this stipulation may be met.

4. Mar 2, 2016

### PAllen

No, because there is invariant differential aging in SR with no gravity, and GR must approximate SR everywhere, locally.

Also, you ignored the points I made about topology.

5. Mar 2, 2016

### olgerm

Locally means in area ,which volume approaches to 0?
which conditions, more specify, must be met locally to satisfy SR?

6. Mar 2, 2016

### PAllen

Locally means in a small volume, for a short time.

The metric must locally approach (in appropriate coordinates) (+1,-1,-1,-1) or (-1,+1,+1,+1). This ensures that locally, the Lorentz transform applies between local inertial frames.

7. Mar 3, 2016

### olgerm

In force-based interpretation locally must gravityfield be constant and homogeneous.

Could not it be ensured by forces in force-baced interpretation?

Last edited: Mar 3, 2016
8. Mar 3, 2016

### Staff: Mentor

I can't think of a force that would lead to time dilation.

9. Mar 5, 2016

### olgerm

I do not have any general overview of such force in force-based interpretation, but it does not mean, that such force can not exist.
In special case of 2 pointmasses having a circular orbit I know a condition that the force must satisfy.
In first frame of reference (notated without ´) masscentre of pointmasses is in rest.
Second frame of reference (notated with ´) is moving with speed v.
In both interpretations:
$a=\frac{F}{m}$
$\omega=\sqrt \frac{a}{r}$
$T=\frac{2 \cdot \pi}{\omega}$
so $T=2 \cdot \pi \cdot \sqrt\frac{F}{m \cdot r}$​

in standard interpretation:
$\frac{_Δt}{_Δt´}=\sqrt{1-\frac{v^2}{c^2}}$
so $\omega´ =\omega \cdot \sqrt{1-\frac{v^2}{c^2}}$
so $T´=T \cdot \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$​

in force-based interpretation:
$T´=T \cdot \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$T´=2 \cdot \pi \cdot \sqrt\frac{F´}{m \cdot r}$
$T=2 \cdot \pi \cdot \sqrt\frac{F}{m \cdot r}$
so $F´=\frac{F}{1-\frac{v^2}{c^2}}$
so $F´=\frac{F \cdot c^2}{(c+v)\cdot(c-v)}$​

such condition is easy to find for any 2-body system.

10. Mar 5, 2016

### Staff: Mentor

It isn't the condition that is in doubt. It is the behavior of forces. In your derivation of this condition you used relativity, not Newtons laws. Starting from Newtons laws you don't get time dilation.

Besides, your math really neglects the fact that forces are vectors.

11. Mar 5, 2016

### olgerm

I know, I have not claimed ,that Newton physics are exactly in line with the general relativity. Time dilation is not explained by Newtonian physics. I have asked about interpretation to general relativity. Of course in that interpretation gravity would not be described be Newton´s law of universal gravitation.

Yes, I did.

Can anybody prove that forming such force-based interpretation to theory of relativity is impossible?
Can anybody form such interpretation to theory of relativity?

12. Mar 5, 2016

### olgerm

Just wanted to show basic idea of how forces may describe time dilation in another interpretation.

13. Mar 5, 2016

### Staff: Mentor

Newtonian physics is the theory of physics that defines forces, so...

This becomes almost meaningless. You cannot use Newtonian forces to replicate time dilation, so then what do you mean by "forces". You would need to make up a new definition of "force". In that definition maybe you could put all of relativity, but then you would just wind up with normal relativity except that you are labeling something as "force". This "force" in relativity would not be force as we know it from Newtonian physics, so in what way is this a force-based interpretation?

I don't think that it is impossible, just useless. It would be nothing more than arbitrarily sticking the label "force" somewhere in relativity.

By treating a vector as a scalar you didn't show anything.

14. Mar 5, 2016

### olgerm

The pointmasses are orbiting circularly so they´r net force is always directed to they´r center of mass. F length of force-vector.
The pointmasses are orbiting circularly so they´r acceleration is always directed to they´r center of mass. a is length of acceleration-vector.
In the example situation the second frame of reference is moving in direction ,in which the pointmasses speeds are always 0, in respect of first frame of reference.

15. Mar 5, 2016

### olgerm

I try to make more clear what I am asking.
Is it possible to use such interpretation to theory of relativity in which:
• physical-space is ordinary 3-dimensional Euclidean space.
• time is same for all observers.
• Newton´s second law remains in force.
• Since it is interpretation to theory of relativity ,it must be in accordance with the theory of relativity.
In that interpretation obviously:
• Newton´s law of universal gravity does not hold true.
• gravityfield is not 3-dimensional vector field.
nota bene: by saying that the interpretation is force-based I do not mean that gravity is only a force. I mean that gravityfield affects bodies by impressing force on them.

16. Mar 5, 2016

### PAllen

What you seek is already impossible without any gravity because what you suggest violates SR for empty space with no charges and nothing except tiny test bodies. Your only possible way out is the LET interpretation of SR where there is an undetetectable preferred frame whose time you declare to be absolute, but it does not correspond to any possible measurements. However, this cannot, in principle, be generalized to match GR for reasons I have mentioned twice and you have comletely ignored. GR predicts non-trivial topology. There is no possible interpretation that can be layed on flat space + time that can modify its topology.

So that answer is that it is trivially provable that what you seek is impossible. You could accomplish it for weak field predictions of GR, but not for the whole of GR. (Weak field assumption allows you to ignore the cases where GR predicts non-trivial topology).

17. Mar 5, 2016

### Staff: Mentor

Exactly. Is there any linear transformation which has these properties? Specifically that the length of the vector changes as described and the direction of the vector remains pointing towards the center of mass in both frames.

By treating F as a scalar and trying to derive a necessary condition on its transformation you did not show anything because F is a vector and so you need to derive conditions on the transformation of a vector. This is why it is important to do your math carefully. As written your equations 2, 4, 9, and 10 are simply wrong.

18. Mar 5, 2016

### Staff: Mentor

In addition to PAllen's point, these two desiderata are incompatible. Newton's second law is incompatible with relativity.

Whatever idea of "force" you might create in order to make a force-based interpretation would have to be a new ad-hoc concept of "force" created expressly and solely for the purpose of making the interpretation work. It cannot be the standard concept of force from Newton's laws.

19. Mar 5, 2016

### Staff: Mentor

Bullet #1: No, because "space" has no invariant definition in relativity, it depends on your choice of coordinates. The only invariant geometry is that of spacetime.

Bullet #2: No, because "time" has no invariant definition in relativity, it depends on your choice of coordinates. The only invariant is that of proper time along a particular worldline, and that depends on the worldline.

Bullet #3: No, because, as Dale already pointed out, Newton's second law is incompatible with relativity.

Bullet #4: No, because, as shown above, the other three points can't be made consistent with relativity.

20. Mar 6, 2016

### olgerm

Okay, as I understand it is impossible even for special relativity. Probably I should have asked about SR first. I have been learning gravitoelectromagnetism and got an idea that maybe GR is similar - just more parameters for one point and interpreted differently, but I was wrong. Did I get it right: not-classical topology and geometry of relativity theory are physical reality, not part of interpretation?
Can´t any interpretation with Euclidean space characteristic give same predictions as non-trivial topology interpretation gives?

Last edited: Mar 6, 2016
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