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A little problem in general relativity;momentum

  1. Feb 26, 2005 #1

    haushofer

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    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 :)
     
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  3. Feb 26, 2005 #2

    pervect

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    "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.
     
  4. Feb 27, 2005 #3

    haushofer

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    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....
     
  5. Feb 27, 2005 #4
    Yes.
    The strength of the gravitational field will be observer dependant regardless of whether the spacetime is curved or not.
     
  6. Feb 27, 2005 #5

    pervect

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    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.
     
  7. Feb 27, 2005 #6

    pervect

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    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.
     
  8. Feb 27, 2005 #7

    Chronos

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    I like Pete's explanation, it was simpler :smile: .
     
  9. Feb 28, 2005 #8
    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
     
  10. Feb 28, 2005 #9

    pervect

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    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.
     
  11. Feb 28, 2005 #10
    I don't think so myself.
    Can you give me an example?
    Not true. I've never said or implied that in my entire life.
    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.

    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
     
  12. Feb 28, 2005 #11

    haushofer

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    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 lets 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:
     
  13. Feb 28, 2005 #12

    pervect

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    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. Feb 28, 2005 #13
  15. Mar 1, 2005 #14

    haushofer

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    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 :)
     
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