Uniform gravitational field possible in GR?

In summary: It...In summary, the metric of an infinite massive plane is not zero, and the tidal force is temporal.
  • #1
JandeWandelaar
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Mentors’ note: this thread is forked from https://www.physicsforums.com/threads/free-fall-in-curved-spacetime.1016510/

PeroK said:
But, if they place small objects at rest at different locations within the spacecraft , then over time those objects will drift apart. Which demonstrates that the spacetime is curved
But what if the gravity field is homogeneous? Like that of an infinite massive plane? The objects in the ship will stay where they are. An infinite massive plane is quite unrealistic, but an observer in a spaceship can think they are in free space or falling freely in such a homogenous field.
 
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  • #2
JandeWandelaar said:
what if the gravity field is homogeneous? Like that of an infinite massive plane?
There is no such thing as a flat spacetime with a gravity field. The so-called "homogeneous gravity field" still has a nonzero Riemann curvature tensor; the spacetime is still not flat. (Actually, there is more than one solution to the Einstein Field Equations that can be described this way, but all of them have nonzero curvature.)

We have had previous threads on this, though it's been a while.
 
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  • #3
PeterDonis said:
There is no such thing as a flat spacetime with a gravity field
There isn't indeed. A distribution of testmasses will stay as it is when falling freely in it. You wouldn't be able to tell if you are falling freely in it, or that you are in empty space. The Riemann tensor is not zero, but isn't the Weyl tensor?
 
  • #4
JandeWandelaar said:
The Riemann tensor is not zero, but isn't the Weyl tensor?
OK, I’ll bite. What metric are you considering?
 
  • #5
Nugatory said:
OK, I’ll bite. What metric are you considering?
I think I found out already. I mean the metric of the infinite massive plane. On both sides. If a spaceship falls freely inside of it and we place clocks in it at different locations, the clocks will stay stationary but show diverging times while falling. In free space this is not the case. So Weyl is not zero. The tidal force is temporal.
 
  • #6
JandeWandelaar said:
It has nothing to do with the black hole scenario but you wouldn't be able to tell if you were in flar spacetime or not. Or would the objects closer to the infinite plate age slower?
If no possible measurement can tell that you are not in flat spacetime then in what sense are you not in flat spacetime?

This is setting to one side the question of whether a universe-spanning, non-zero, homogenous gravity field is possible under the model of General Relativity.
 
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  • #7
JandeWandelaar said:
I mean the metric of the infinite massive plane.
So what is that metric? That is, what are the metric coefficients ##g_{\mu\nu}## as functions of the ##x_\mu##?
 
  • #8
Can't we say it's something like Rindler?

$$ ds^2 = -(1 + \frac{a}{c^2}z)^2 c^2 dt^2 + dz^2 $$

So it's time that will be tidal when falling freely? Ignoring the other two spatial coordinates.

Or should it be:

$$ ds^2 = -(1 + \frac{a}{c^2}z^2) c^2 dt^2 + dz^2 $$

Or neither?
 
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  • #9
Well, plug it into the Ricci tensor and check whether it's zero, Johnny Walker.
 
  • #11
Uniform gravitational field possible in GR?

It's not even possible in Newtonian gravity (without infinite sheets and their ilk).
 
  • #12
JandeWandelaar said:
Can't we say it's something like Rindler?
...
Or should it be:
...
Or neither?
And whatever it is, is it a vacuum solution of the EFE?
 
  • #13
Nugatory said:
And whatever it is, is it a vacuum solution of the EFE?
Is an infinite space with an infinite massive plane a vacuum? Outside it yes. The clocks tick differently wrt each other at different heights. But space is flat.
 
  • #14
JandeWandelaar said:
I mean the metric of the infinite massive plane.
This is not a satisfactory answer. You need to actually write down the metric, and it would be very nice if you would give a reference for where you got it. You do not seem to realize that there are a lot of complications lurking beneath what appears to you to be a simple phrase. It's nowhere near as simple as you think.

JandeWandelaar said:
Can't we say it's something like Rindler?
No, because the Rindler metric is the metric of a region of flat Minkowski spacetime with no stress-energy anywhere. It is not the metric of a region of spacetime containing "an infinite massive plane".

JandeWandelaar said:
The clocks tick differently wrt each other at different heights. But space is flat.
You need to either back up claims like this with math and a reference, or stop making them.
 
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  • #15
Will the infinite massive plane contract? If it's an infinite slab of mass, what will happen? There is no center where too to contract.
 
  • #16
vanhees71 said:
This paper isn't very helpful for the present topic because it does not investigate what kind of stress-energy is required to produce the solution given in section IV (the "uniform gravitational field" solution in GR).

Here is a better reference for the present thread topic:

https://arxiv.org/abs/0708.2906

It gives a good discussion of the complications that I have been telling the OP are lurking beneath what he appears to believe is a simple phrase. It isn't.
 
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  • #18
JandeWandelaar said:
The Riemann tensor is not zero, but isn't the Weyl tensor?
JandeWandelaar said:
Will the infinite massive plane contract? If it's an infinite slab of mass, what will happen? There is no center where too to contract.
To answer all of these questions, you need to first write down the metric. Then you need to compute the necessary tensors from it. The paper I linked to in post #16 should be helpful.
 
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  • #19
PeterDonis said:
To answer all of these questions, you need to first write down the metric. Then you need to compute the necessary tensors from it. The paper I linked to in post #16 should be helpful.
Thanks. I just downloaded a paper on Researchgate. It said that particles separated in x- and y-direction will move towards each other when in free fall. Which is absent in the Newtonian case. It's more complicated than I thought...
 
  • #20
JandeWandelaar said:
I just downloaded a paper on Researchgate.
Please give a reference.
 
  • #21
PeterDonis said:
Please give a reference.

The general relativistic infinite plane​

  • September 2007
  • https://www.researchgate.net/journal/American-Journal-of-Physics-0002-9505 76(1)
DOI:10.1119/1.2800354
  • Source
  • https://www.researchgate.net/deref/http%3A%2F%2Farxiv.org%2Fabs%2F0708.2906
Ah...it's the same...
 
  • #23
Vanadium 50 said:
Uniform gravitational field possible in GR?
It's not even possible in Newtonian gravity (without infinite sheets and their ilk).
Try an off-center spherical cavity within a sphere of uniform density.
 
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1. What is a uniform gravitational field in the context of General Relativity (GR)?

A uniform gravitational field in GR refers to a gravitational field that has a constant strength and direction throughout a given region of space. This is in contrast to a non-uniform gravitational field, where the strength and direction of the gravitational force varies at different points in space.

2. Is a uniform gravitational field possible in GR?

Yes, a uniform gravitational field is possible in GR. In fact, the theory of GR allows for a wide range of possible gravitational fields, including uniform fields. However, in reality, truly uniform gravitational fields are rare and are only found in hypothetical scenarios or in regions of space very far away from any massive objects.

3. How does GR explain the existence of a uniform gravitational field?

In GR, the presence of a uniform gravitational field is explained by the curvature of spacetime caused by the presence of massive objects. This curvature affects the motion of objects in the field, causing them to experience a uniform acceleration towards the source of the gravitational field.

4. Can a uniform gravitational field exist in the absence of massive objects?

No, a uniform gravitational field cannot exist in the absence of massive objects. In GR, the presence of a gravitational field is always tied to the presence of mass or energy, as described by Einstein's famous equation E=mc^2. Without any mass or energy, there would be no curvature of spacetime and therefore no gravitational field.

5. How does the concept of a uniform gravitational field in GR differ from that in Newtonian gravity?

In Newtonian gravity, a uniform gravitational field is described as a field that has a constant strength and direction at all points in space, regardless of the presence of mass. However, in GR, a uniform gravitational field is only possible in the presence of mass or energy, and its strength and direction may vary depending on the distribution of mass in the region of space. Additionally, the concept of a gravitational field itself is fundamentally different in GR, as it is described as the curvature of spacetime rather than a force acting between masses.

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