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I How do we get invariant curvature from momentum and energy?

  1. Dec 9, 2017 #1
    Hi all,

    I understand the mathematics behind special relativity pretty well, but I only have a bare conceptual understanding of general relativity. My understanding is that energy, momentum and stress (as described in the energy-stress tensor) are what contribute to space-time curvature and hence gravity. (Shortest summary of GR ever? :P)

    However, my understanding from special relativity is that the energy and momentum possessed by a system are not invariant quantities but are measured differently by different observers. Special relativity has also taught me that space-time intervals are invariant/absolute. That suggests to me that space-time curvature would be absolute too, as well as the logic telling me that gravitational fields should be absolute too - an object can't be a black hole for one observer but not for another... can it?

    And yet if space-time curvature/gravity is something we can all agree on... how can it come from relative quantities like energy and momentum? Is this a situation like space and time combining to form space-time intervals in special relativity, where relative quantities change for different observers together in such a way that they can be combined into one absolute quantity for all observers? Or is space-time curvature not invariant and something that can be measured differently depending on who's measuring?

    Thanks!
     
  2. jcsd
  3. Dec 9, 2017 #2

    PeterDonis

    Staff: Mentor

    Yes.

    Individually, they are not. But when properly "packaged" into the stress-energy tensor, they are--more precisely, observables calculated by combining the stress-energy tensor with appropriate other vectors and tensors are.

    That's correct; spacetime geometry is not observer-dependent.

    Because it's not energy or momentum or stress individually, but an appropriate combination of all of them that is frame-invariant. It's similar to the way energy and momentum form a 4-vector in SR, whose length is invariant (it's the rest mass), even though energy and momentum individually are not.
     
  4. Dec 9, 2017 #3
    The curvature can be represented by tensors of different rank: it can be scalar (rank 0 tensor), matricial (rank 2 tensor), etc.
    In general relativity all curvatures are related to a rank 4 tensorial curvature (a four-index-thing), the riemann curvature tensor, that measures the change of a vector that makes an infinitesimal loop at the plane i j as a linear transformation [ tex ] R_{ikjl}=R(x_i,x_j)_{kl} [ /tex ]. With it, and the metric (a matrix field/rank-2 tensor field that tells you how to measure distances and similar things), you can form many invariants (scalars that don't change with coordinate transformations).

    The stress, energy and momentum are encoded in the stress-energy tensor (it's definition comes from Noether theorem, the action functional, and invariance of the physics under infinitesimal translations), the stress-energy tensor is the source of the Einstein Curvature Tensor, a particular mix of riemann curvature with the metric that has the property that it's conserved in the same sense in which the stress-energy tensor is conserved in a curved spacetime. The Riemann curvature get's almost uniquely defined by this equation, the invariants that are constructed from it doesn't deppend on who meassures them. Being a black hole is related to the invariants, i don't know well how, that's outside my knowledge.
     
  5. Dec 10, 2017 #4

    Dale

    Staff: Mentor

    This is correct energy and momentum are not invariant. However, were you aware that energy and momentum can be combined into a four-dimensional vector that is covariant? Individually they are not invariant, but together they are covariant meaning that they transform in the same way as time and space do. These are called four-vectors.
     
  6. Dec 10, 2017 #5
    Thanks, that makes sense! So just to make sure I've got this - the stress-energy tensor (and hence space-time curvature) is the same for all observers, despite the fact that energy and momentum are not invariant - a little like how distances in space and lengths in time are relative in special relativity but space-time intervals are invariant?
     
  7. Dec 10, 2017 #6
    Ehm... The invariants that I told you that existe remaint invariants un the sence that you are telling as invariant. The energy momentum tensor in a given point of spacetime contracted with two given vectors <v,Tw> will be an invariant, so the matter contento isn't subjetive... But T isn't invariant, as a vector isn't.
     
  8. Dec 10, 2017 #7

    Dale

    Staff: Mentor

    The stress energy tensor is covariant. That means that it refers to the same geometric object in all frames. But different frames will assign different values to the components. Energy and momentum are components of the stress energy tensor.
     
  9. Dec 10, 2017 #8

    PeterDonis

    Staff: Mentor

    Yes, but, to add to @Dale's response, it's more complicated for the stress-energy tensor because there are more components.

    As I noted before, energy and momentum in SR form a 4-vector, which has one invariant, its length (the rest mass). More generally, you can form the inner product of this 4-vector with any other 4-vector to obtain an invariant. (In fact, one way of looking at the components of the energy-momentum 4-vector is as inner products of that 4-vector with unit timelike and spacelike 4-vectors that define an inertial frame.)

    The stress-energy tensor is a 2-index tensor, so, if you want to get an invariant from it, you have to combine it with two vectors to form an invariant which is the analogue of an inner product (this process is called "contraction"). That means there are more ways to form invariants from the stress-energy tensor, and you can't just think of it as having one invariant "length" like a 4-vector does.
     
  10. Dec 10, 2017 #9

    Orodruin

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    It would be more precise to say that the components of T are not invariant. The tensor T as an object really is not connected to any coordinate system or basis and remains the same whatever basis you choose. It is the components of T relative to a basis that are not invariant. The same thing goes for vectors. The vectors themselves do not care for particular bases, it is the same vector whatever basis you choose, but choosing different bases will lead to different vector components.
     
  11. Dec 11, 2017 #10
    If i remember well, invariant is a precise term, it means an object that is in the scalar representation of the diff group (the 'group of coordinate transformations'). The lenght of a vector, the angle between two vectors, etc are invariants.
     
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