A little problem in general relativity;momentum

  • Thread starter haushofer
  • Start date
  • Tags
    General
In summary: So if you have two observers (say on Earth and on a spaceship) that measure different impulses, then the space-time of the spaceship would be curved differently?Yes, that is correct.
  • #1
haushofer
Science Advisor
Insights Author
2,944
1,475
Hi there, I'm a physics student who has recently finished a course on general relativity, and quite new here. I'm also Dutch, so excuse me if my English is messed up :) But I have a dillema, which keeps me out of my sleep.

The field-equations link space-time curvature to the energy-momentum tensor. What bothers me: an external observer measures the impuls of an moving object of course differently than the object itself does. The question is: Does this mean that curvature of space-time is frame dependant? I mean, several observers would measure different impulses, and this would curve the space-time differently, measured by each observer. I'm not sure if I interpret the energy-momentum tensor and curvature correctly, and my intuition says the curvature is not frame- dependant. My guess is that I'm overlooking something, but what?

Many thanks in forward :)
 
Physics news on Phys.org
  • #2
haushofer said:
Hi there, I'm a physics student who has recently finished a course on general relativity, and quite new here. I'm also Dutch, so excuse me if my English is messed up :) But I have a dillema, which keeps me out of my sleep.

The field-equations link space-time curvature to the energy-momentum tensor. What bothers me: an external observer measures the impuls of an moving object of course differently than the object itself does. The question is: Does this mean that curvature of space-time is frame dependant? I mean, several observers would measure different impulses, and this would curve the space-time differently, measured by each observer. I'm not sure if I interpret the energy-momentum tensor and curvature correctly, and my intuition says the curvature is not frame- dependant. My guess is that I'm overlooking something, but what?

Many thanks in forward :)


"Space-time curvature" is potentially ambiguous - we have the Riemann tensor, R_abcd, and the Einstein tensor, G_ab. Of the two, the Riemann is the most general description of space-time curvature.

But as far as answering your question goes, the important thing is that they are both tensors.

The fact that they are both tensors describes how they transform between inertial frames. You can boost either one from an unprimed frame to a primed frame with the transformation matrix [tex]L^a{}_{a'}[/tex], which should be very familiar to you as a "Lorentz boost" - the same Lorentz boost you learned in special relativity.

Specifically

[tex]
G_{a' b'} = G_{ab} L^a{}_{a'} L^b{}_{b'}
[/tex]
[tex]
R_{a' b' c' d'} = R_{abcd} L^a{}_{a'} L^b{}_{b'} L^c{}_{c'} L^d{}_{d'}
[/tex]

The specific numerical values of the components of the tensor depend on the frame and the coordinate system used. In that sense, the components of the tensor depend on the frame used, and different observers describe the same curvature with different numerical values.

But because the transformation properties are defined, the tensor as a whole is considered to be a "geometric object", one that exists independent of any particular frame or coordinate system.
 
  • #3
pervect said:
"Space-time curvature" is potentially ambiguous - we have the Riemann tensor, R_abcd, and the Einstein tensor, G_ab. Of the two, the Riemann is the most general description of space-time curvature.

But as far as answering your question goes, the important thing is that they are both tensors.

The fact that they are both tensors describes how they transform between inertial frames. You can boost either one from an unprimed frame to a primed frame with the transformation matrix [tex]L^a{}_{a'}[/tex], which should be very familiar to you as a "Lorentz boost" - the same Lorentz boost you learned in special relativity.

Specifically

[tex]
G_{a' b'} = G_{ab} L^a{}_{a'} L^b{}_{b'}
[/tex]
[tex]
R_{a' b' c' d'} = R_{abcd} L^a{}_{a'} L^b{}_{b'} L^c{}_{c'} L^d{}_{d'}
[/tex]

The specific numerical values of the components of the tensor depend on the frame and the coordinate system used. In that sense, the components of the tensor depend on the frame used, and different observers describe the same curvature with different numerical values.

But because the transformation properties are defined, the tensor as a whole is considered to be a "geometric object", one that exists independent of any particular frame or coordinate system.

OK, so different observers describe the same curvature, independant of their velocities, because we're talking tensorequations here? Is it right to compare it with a rotation of a vector, in which the length is conserved, and the components are changed? Well, I've had some problems with tensors, like describing the properties of the Riemann tensor, with a switch to the frame in which the metric takes its canonical form, and then take your conclusions in general because the properties of tensors are frame-independant...

But what you are practically saying is, that the energy-momentum tensor of a moving object is frame independant, just like the metric is in special relativity? I'm still a little confused here, because even without relativity the impuls of an object depends on the observer...
 
  • #4
haushofer said:
The field-equations link space-time curvature to the energy-momentum tensor. What bothers me: an external observer measures the impuls of an moving object of course differently than the object itself does. The question is: Does this mean that curvature of space-time is frame dependant?
Yes.
I mean, several observers would measure different impulses, and this would curve the space-time differently, measured by each observer.
The strength of the gravitational field will be observer dependant regardless of whether the spacetime is curved or not.
I'm not sure if I interpret the energy-momentum tensor and curvature correctly, and my intuition says the curvature is not frame- dependant. My guess is that I'm overlooking something, but what?
pervect said:
"Space-time curvature" is potentially ambiguous..
Huh? :confused: Why? Since when?

Pete
 
  • #5
When people say "space-time curvature", they can mean a lot of things - that's why it's potentially ambiguous. It's a rather vague phrase, it's really not very specific.
 
  • #6
haushofer said:
OK, so different observers describe the same curvature, independant of their velocities, because we're talking tensorequations here? Is it right to compare it with a rotation of a vector, in which the length is conserved, and the components are changed? Well, I've had some problems with tensors, like describing the properties of the Riemann tensor, with a switch to the frame in which the metric takes its canonical form, and then take your conclusions in general because the properties of tensors are frame-independant...

But what you are practically saying is, that the energy-momentum tensor of a moving object is frame independant, just like the metric is in special relativity? I'm still a little confused here, because even without relativity the impuls of an object depends on the observer...

Rotation is a reasonable example. If a vector is a unit vector in the x-direction, and we rotate the coordinate system, it might now be a unit vector in the y direction. We don't think of the vector as being any different, though, we think of the coordinate system as having changed. The components of the vector are definitely NOT constant when we change coordinate systems. We have to distinguish between the components of the vector, and the vector as an abstract entity. The former change as we change coordinates - the later idea is an abstraction, and we say that the vector itself as an abstract entity does not change when we change the coordinates, only our description of it changes.

You'll have to separate the abstract idea of a vector or tensor from its representation as components for this point to make sense. I think that's where your problem arises - I don't know of any clearer way to explain the difference.
 
  • #7
I like Pete's explanation, it was simpler :smile: .
 
  • #8
pervect said:
When people say "space-time curvature", they can mean a lot of things - that's why it's potentially ambiguous. It's a rather vague phrase, it's really not very specific.
If a person means anything else but a non-zero Riemann tensor (i.e. no tidal forces) then they've made an error. What is so ambiguous about that?

Pete
 
  • #9
pmb_phy said:
If a person means anything else but a non-zero Riemann tensor (i.e. no tidal forces) then they've made an error. What is so ambiguous about that?

Pete

We're arguing semantics. I have seen several instances where authors use curved space-time to mean something other than a non-zero Riemann tensor. You loudly decry such usage as an error, for instance in statements that various authors make about how gravity implies space-time curvature. Yet such usage is very common. Rather than conclude that all these authors are wrong, I conclude that the phrase "space-time curvature" does not necessarily mean that the Riemann tensor is zero. By common usage, iit also can mean that the metric coefficients aren't those of a flat Minkowski space-time.
 
  • #10
pervect said:
We're arguing semantics.
I don't think so myself.
I have seen several instances where authors use curved space-time to mean something other than a non-zero Riemann tensor.
Can you give me an example?
You loudly decry such usage as an error, ..
Not true. I've never said or implied that in my entire life.
for instance in statements that various authors make about how gravity implies space-time curvature.
I never said such a thing. I've always held to (1) gravitational force implies non-vanishing Christoffel symbols and (2) tidal force implies spacetime curvature.

Yet such usage is very common. Rather than conclude that all these authors are wrong, I conclude that the phrase "space-time curvature" does not necessarily mean that the Riemann tensor is zero. By common usage, iit also can mean that the metric coefficients aren't those of a flat Minkowski space-time.
Can you elaborate and or give an example? The spacetime can be flat and the spatial coordinates be Cartesian and yet there can be zero spacetime curvature.

Pete
 
  • #11
pervect said:
You'll have to separate the abstract idea of a vector or tensor from its representation as components for this point to make sense. I think that's where your problem arises - I don't know of any clearer way to explain the difference.

Yes, I'm familiar with the idea that the objects themself do not change, but the components do so. Like I said, I've had some problems with tensors :)

But let's take an example. A particle is moving with, let's say, 0.9*c relative to Observer 1. Observer 2 travels with the particle, so the relative speed betwoon O2 and the particle is 0. How do O1 and O2 think about the gravitational interaction of the particle with other objects? ( so, how do O1 and O2 think about the way the particle curves space time around itself? Do they measure the same interactions of the particle with neighbouring particles, or not?)

Excuse me if I'm being stupid here. :blushing:
 
  • #12
haushofer said:
Yes, I'm familiar with the idea that the objects themself do not change, but the components do so. Like I said, I've had some problems with tensors :)

But let's take an example. A particle is moving with, let's say, 0.9*c relative to Observer 1. Observer 2 travels with the particle, so the relative speed betwoon O2 and the particle is 0. How do O1 and O2 think about the gravitational interaction of the particle with other objects? ( so, how do O1 and O2 think about the way the particle curves space time around itself? Do they measure the same interactions of the particle with neighbouring particles, or not?)

Excuse me if I'm being stupid here. :blushing:

Let's make the particle a black hole.

One observer is stationary relative to the black hole. He measures some Riemann tensor in a local frame. Make it a local orthonormal frame.

Another observer is moving relative tot he black hole. He measures diffrent *components* of the Riemann tensor in his local frame.

The components of the Riemann tensor are in general different for the two observers, though interestingly enough for someone moving directly towards the black hole, the components of the "boosted" Riemann are the same as the non-boosted Riemann. This is not true, however, if the direction of the boost does not point directly towards the black hole.

You should be able to find a detailed calculation of the components of the Riemann tensor in a section that deals with the tidal forces on an observer falling into a black hole. MTW has such a caclulation in "Gravitation" for instance.

So the components are different in general. It's basically a philosophical viewpoint that says that these different components represent the same physical reality. The numerically differeng components represent the same physical reality because the components transform in a known manner, i.e. because the the transformation of the tensor is covariant.
 
  • #14
Ah, the geodesic equotion was already familiar with me. It becomes more clear now, thanks everybody for your time ! If someone has more information, by all means :)
 

1. What is "A little problem in general relativity; momentum"?

"A little problem in general relativity; momentum" refers to a specific scenario in which the equations of general relativity, which describe the effects of gravity on the movement of massive objects, encounter a difficulty when trying to incorporate the concept of momentum.

2. Why is this problem significant?

This problem is significant because it highlights a limitation in our current understanding of general relativity and its ability to fully explain the behavior of massive objects in the universe. It also provides a platform for further investigation and potential advancements in our understanding of gravity.

3. Can you explain the concept of momentum in general relativity?

Momentum in general relativity refers to the quantity of motion that a massive object possesses. It is described by the product of the object's mass and its velocity, and is a crucial factor in understanding how objects move and interact in the presence of gravitational forces.

4. How does this problem relate to other theories in physics?

This problem is closely related to the theories of special relativity and Newtonian mechanics. It also has implications for other areas of physics such as quantum mechanics and cosmology, as the concept of momentum plays a role in these fields as well.

5. Is there a solution to this problem?

Currently, there is no definitive solution to this problem. Scientists are actively researching and proposing various theories and explanations, but it remains an open question in the field of physics. It is possible that a future breakthrough may provide a resolution to this problem.

Similar threads

  • Special and General Relativity
Replies
19
Views
1K
  • Special and General Relativity
2
Replies
35
Views
388
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
9
Views
881
  • Special and General Relativity
Replies
3
Views
1K
  • Special and General Relativity
Replies
17
Views
440
Replies
62
Views
4K
  • Special and General Relativity
Replies
11
Views
1K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
6
Views
244
Back
Top