Uniform gravity's space-time curvature

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SUMMARY

The discussion centers on the concept of a uniform gravity field and its implications for space-time curvature within the framework of General Relativity (GR). Participants clarify that a uniform gravity field, defined as having the same value and direction throughout space, leads to non-zero curvature according to the findings in relevant literature. The Bell congruence is introduced as a set of worldlines that describe a uniform gravity field, but it is noted that these worldlines do not maintain constant distances apart, indicating that observers cannot be at rest relative to each other in flat spacetime. The conversation emphasizes the complexity of translating Newtonian concepts of gravity into the relativistic context.

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  • Knowledge of the Bell spaceship paradox and Bell congruence
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max_zhou
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In an ideal world with uniform gravity field, what does its space-time curvature look like? Is it non-zero? If not, how a free particle would be accelerated with the point view of space-time curvature?

By uniform gravity field, I mean a gravity field with same value, same direction everywhere in the space.
 
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max_zhou said:
uniform gravity field

What do you mean by a "uniform gravity field"?
 
PeterDonis said:
What do you mean by a "uniform gravity field"?
Yep, sorry for the ambiguity. I have edited my question.

By uniform gravity field, I mean a gravity field with same value, same direction everywhere in the space.
 
How familiar are you with general relativity?
 
max_zhou said:
In an ideal world with uniform gravity field, what does its space-time curvature look like? Is it non-zero? If not, how a free particle would be accelerated with the point view of space-time curvature?

See case B in the diagram by DrGreg below.

DrGreg said:
This is my own non-animated way of looking at it:

attachment-php-attachmentid-56007-stc-1-d-1361576846-png-png.png


  • A. Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  • B1. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.

    B2. Take the flat piece of paper depicted in B1, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B2 shows exactly the same thing as B1, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  • C. Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B2. This is the equivalence principle in action: if you zoomed in very close to B2 and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

Case B is basically the local situation here on Earth, in the frame of the surface. Locally the tidal effects due to space-time curvature that you see in case C are negligible:

 

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max_zhou said:
By uniform gravity field, I mean a gravity field with same value, same direction everywhere in the space.

This still doesn't answer the question, because now you have to define what you mean by "gravity field". For example: I am in a rocket accelerating at 1 g in flat spacetime. Is there a "gravity field" inside the rocket?
 
Orodruin said:
How familiar are you with general relativity?
Not too much. Still learning. This is like a pre-reading question to me. I googled around about this problem. And some people differentiate 'real' gravity as a gravity cannot be eliminated by transformation of coordinates. That just confuses me more...
 
PeterDonis said:
This still doesn't answer the question, because now you have to define what you mean by "gravity field". For example: I am in a rocket accelerating at 1 g in flat spacetime. Is there a "gravity field" inside the rocket?
Good point. The scenario you described is one of follow-up question I want to follow up. In my question, I take the 'pseudo gravity' due to acceleration as a gravity field.

But my original question scenario would be more like, in an ideal world, there is an infinite slab, the gravity due to this slab is same value, same direction everywhere. I take this as the uniform gravity. Then in the far field from this slab, what does the space-time curvature look like?
 
max_zhou said:
In my question, I take the 'pseudo gravity' due to acceleration as a gravity field.

Then the answer is that you can find a set of worldlines in flat spacetime (called the "Bell congruence" since it occurs in the analysis of the Bell spaceship paradox) that describe a "uniform gravity field" by your definition. However, this congruence has the counterintuitive property that the worldlines do not stay the same distance apart (the expansion scalar of the congruence is positive, not zero). So this set of worldlines does not describe a family of observers that are all at rest with respect to each other. So if your definition of a "uniform gravity field" also requires that the observers who see the uniform gravity field are all at rest relative to each other, then there is no such set of observers in flat spacetime. I'm also not aware of any curved spacetime that admits such a set of observers everywhere.

max_zhou said:
in an ideal world, there is an infinite slab, the gravity due to this slab is same value, same direction everywhere. I take this as the uniform gravity

I don't think there is a solution that has all of these properties, but there are solutions that come close. We had a thread on this a while back that referenced a couple of papers that are relevant; I'll see if I can find it.
 
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  • #10
max_zhou said:
Not too much. Still learning. This is like a pre-reading question to me. I googled around about this problem. And some people differentiate 'real' gravity as a gravity cannot be eliminated by transformation of coordinates. That just confuses me more...
I do not think this is a pre-reading question. You need to understand GR to at least sone extent in order to even formulate the question properly (as you have seen by the replies you have gotten).

I also suggest that you do not use the A-level tag for this type of thread. Its intended meaning is that you have an understanding of the subject at graduate level or higher.
 
  • #11
PeterDonis said:
Then the answer is that you can find a set of worldlines in flat spacetime (called the "Bell congruence" since it occurs in the analysis of the Bell spaceship paradox) that describe a "uniform gravity field" by your definition. However, this congruence has the counterintuitive property that the worldlines do not stay the same distance apart (the expansion scalar of the congruence is positive, not zero). So this set of worldlines does not describe a family of observers that are all at rest with respect to each other. So if your definition of a "uniform gravity field" also requires that the observers who see the uniform gravity field are all at rest relative to each other, then there is no such set of observers in flat spacetime. I'm also not aware of any curved spacetime that admits such a set of observers everywhere.
I don't think there is a solution that has all of these properties, but there are solutions that come close. We had a thread on this a while back that referenced a couple of papers that are relevant; I'll see if I can find it.
Thank you!
 
  • #12
Orodruin said:
I do not think this is a pre-reading question. You need to understand GR to at least sone extent in order to even formulate the question properly (as you have seen by the replies you have gotten).

I also suggest that you do not use the A-level tag for this type of thread. Its intended meaning is that you have an understanding of the subject at graduate level or higher.
Fair enough. Looks like it is not straightforward to transfer a question from Newton's world to Einstein's world. I would like to change the level tag. But do not know how...
 
  • #13
A.T. said:
See case B in the diagram by DrGreg below.
Case B is basically the local situation here on Earth, in the frame of the surface. Locally the tidal effects due to space-time curvature that you see in case C are negligible:


Will take a careful look later. Thanks!
 
  • #14
max_zhou said:
I would like to change the level tag. But do not know how...

You wait for a Mentor to change it for you, as I just did. :wink:

You can ask a Mentor to have the level changed if you think it should be; we have the magical admin powers to do that. But I saw the posts by @Orodruin and you and just went ahead and changed it.
 
  • #15
Thank everyone for the discussion so far. I found a paper: https://core.ac.uk/download/pdf/25331744.pdf . From this paper, the conclusion is a uniform gravity corresponds to a non-zero curvature. I will understand it better after learning more about general relativity.
 
  • #16
max_zhou said:
From this paper, the conclusion is a uniform gravity corresponds to a non-zero curvature.

You might be missing a crucial point: what this paper calls the "Relativistic Theory of Gravitation" is not General Relativity. See here for some background (and bear in mind that Wikipedia cannot be relied on to give an accurate picture of the actual status of a theory--the article conspicuously fails to mention any issues with Logunov's theory, of which there are plenty, but most importantly it makes predictions that do not agree with experiment):

https://en.wikipedia.org/wiki/Anatoly_Logunov#Relativistic_theory_of_gravitation
 
  • #17
PeterDonis said:
You might be missing a crucial point: what this paper calls the "Relativistic Theory of Gravitation" is not General Relativity. See here for some background (and bear in mind that Wikipedia cannot be relied on to give an accurate picture of the actual status of a theory--the article conspicuously fails to mention any issues with Logunov's theory, of which there are plenty, but most importantly it makes predictions that do not agree with experiment):

https://en.wikipedia.org/wiki/Anatoly_Logunov#Relativistic_theory_of_gravitation
Thank you, Peter! I guess I can only learn how to reframe my question after learning more about GR. But I would intuitively think my question could be somehow decomposed and consumed in GR and has a simple answer...
 
  • #18
max_zhou said:
But I would intuitively think my question could be somehow decomposed and consumed in GR and has a simple answer...
You really cannot think ”intuitively” about a theory that you have not worked with extensively. Intuition comes from prolonged and repeated exposure to a subject. A more appropriate description would be ”naively think” as your expectations are based on the ideas available to you in Newtonian gravity that you might expect to find equivalences to in GR. However, it is not always true that such an equivalence makes sense andcan be made.
 
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  • #19
max_zhou said:
Thank you, Peter! I guess I can only learn how to reframe my question after learning more about GR. But I would intuitively think my question could be somehow decomposed and consumed in GR and has a simple answer...

Unfortunately there is no easy way to do this and there is no "simple" answer. GR is fundamentally different than Newtonian mechanics - GR has no notion of a "gravitational field" in the Newtonian sense anymore. Instead it has spacetime curvature. The two are analogous but they are not the same and so statements about one don't necessarily translate to the other.

Constructing your infinite slab in GR and finding the solutions also seem problematic to me and I have not seen a solution for such a matter distribution. Exact solutions in GR are hard to come by.
 
  • #20
Orodruin said:
You really cannot think ”intuitively” about a theory that you have not worked with extensively. Intuition comes from prolonged and repeated exposure to a subject. A more appropriate description would be ”naively think” as your expectations are based on the ideas available to you in Newtonian gravity that you might expect to find equivalences to in GR. However, it is not always true that such an equivalence makes sense andcan be made.
Fair corrections.
 
  • #21
Matterwave said:
Unfortunately there is no easy way to do this and there is no "simple" answer. GR is fundamentally different than Newtonian mechanics - GR has no notion of a "gravitational field" in the Newtonian sense anymore. Instead it has spacetime curvature. The two are analogous but they are not the same and so statements about one don't necessarily translate to the other.

Constructing your infinite slab in GR and finding the solutions also seem problematic to me and I have not seen a solution for such a matter distribution. Exact solutions in GR are hard to come by.
I see. I guess I cannot be satisfied with the answer that my scenario cannot work in a GR sense until I have a real understanding of GR.
 
  • #22
max_zhou said:
I see. I guess I cannot be satisfied with the answer that my scenario cannot work in a GR sense until I have a real understanding of GR.

I did not say your scenario cannot work in a GR sense - at least not the one with the infinite slab of matter. I only said I do not know of any such solutions in GR. It may not be nearly as simple as you think it is. The original question regarding a uniform gravitational field wouldn't work in GR since there's no more gravitational field in GR.
 
  • #23
Matterwave said:
I did not say your scenario cannot work in a GR sense - at least not the one with the infinite slab of matter. I only said I do not know of any such solutions in GR. It may not be nearly as simple as you think it is. The original question regarding a uniform gravitational field wouldn't work in GR since there's no more gravitational field in GR.
Sorry. I mis-interpreted your response. I think I will come back to this question in the future. I would hope we could have a solution for this scenario.
 
  • #25
max_zhou said:
I would intuitively think my question could be somehow decomposed and consumed in GR and has a simple answer...

See the papers linked to in the post I just referenced. I won't say the answers given there are "simple", but they at least give some good info on how a scenario like the one you are thinking of can be formulated in GR.
 
  • #26
PeterDonis said:
See the papers linked to in the post I just referenced. I won't say the answers given there are "simple", but they at least give some good info on how a scenario like the one you are thinking of can be formulated in GR.
Thank you, Peter! I also found the mathpages.com page you mentioned in that thread. I would mark you answer as the correct answer if I could.
I will return after I have enough training in understanding the two papers.
 
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  • #27
max_zhou said:
Thank you, Peter! I also found the mathpages.com page you mentioned in that thread. I would mark you answer as the correct answer if I could.
I will return after I have enough training in understanding the two papers.
Be careful not to get totally absorbed in mathpages, there is a lot of unique (really priceless) stuff there! ;)
 
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