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A Inertial frames: from GR to SR

  1. Feb 28, 2017 #1
    Hello everyone, here I come with a question about inertial frames as defined in General Relativity, and how to prove that the general definition is consistent with the particular case of Special Relativity.

    So to contextualize, I have found that one can define inertial frames in General Relativity, as follows: given a 4-dimensional lorentzian manifold with metric tensor g and signature (-1,1,1,1), a frame field is defined to be a set of four vector fields e0,e1,e2,e3 such that:

    g(ei,ej) = ηij (η being the Minkowski matrix)

    Further, one says that the frame field in question is inertial and nonrotating if the following condition is satisfied:

    ∇_{e0}ei = 0

    where ∇ is the Levi-Civita connection on the manifold (let me call this "Definition A"). All this straight out of Wikipedia, and quite elegant and nice.

    Now, my question is how does this correspond to the definition of inertial frames in Special Relativity as usually found in textbooks. So now assume that g is the Minkowski metric and that we are in Minkowski spacetime, where there is a given global coordinate chart (t,x,y,z) (I am setting c=1 for simplicity), so that:

    g = -dt⊗dt + dx⊗dx + dy⊗dy + dz⊗dz

    Then it is straightforward to see that the vector fields ∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z are indeed an inertial nonrotating frame field in the abstract sense of Definition A. But in Minkowski spacetime one has Definition B (the one usually found in textbooks), where we are given an initial "precursor" inertial frame (the one used by the scientist writting the textbook), which in this context is a coordinate chart with certain physical properties, namely that there are no inertial forces (i.e. the Christoffel symbols vanish identically), and can of course be identified with the coordinate chart (t,x,y,z) I mentioned earlier. And then we are told that "inertial frames" are all those coordinate charts related to this precursor coordinate chart by a Lorentz transformation, which is a linear coordinate transformation Λ such that η = transpose(Λ) * η * Λ (this guarantees that the spacetime interval is preserved).

    So, here comes my question: take an arbitrary nonrotating inertial frame in the sense of Definition A, that is a set of four vector fields e0,e1,e2,e3 satistying the requisites of that abstract definition. Is it the case, then, that there exists a coordinate chart x0,x1,x2,x3 and a suitable Lorentz transformation Λ from (t,x,y,z) (the "precursor coordinate chart" inherent to the construction of Minkowski spacetime) to (x0,x1,x2,x3) such that:

    ei = ∂/∂xi for i=0,1,2,3 ?
     
    Last edited: Feb 28, 2017
  2. jcsd
  3. Feb 28, 2017 #2

    PeterDonis

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    Not in curved spacetime in general, no. See the section "Comparison with coordinate basis" on this page:

    https://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

    For an explicit example of an inertial nonrotating frame field that has no corresponding coordinate basis in the sense you describe, see the Lemaitre observers example later in the same article.
     
  4. Mar 1, 2017 #3
    Yes, I imagined this would not be possible in curved spacetime... but what about flat Minkowski spacetime?
     
  5. Mar 1, 2017 #4
    Note: I have edited the question accordingly.
     
  6. Mar 1, 2017 #5

    PeterDonis

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    The only inertial nonrotating frames in flat Minkowski spacetime are the standard inertial frames, which obviously correspond to the standard inertial coordinate charts.
     
  7. Mar 1, 2017 #6
    Yes, I imagined so. But why? (mathematically):

    -) given the arbitrary frame field e0,e1,e2,e3, what is the mathematical construction of the corresponding coordinate chart?
    -) what is the mathematical construction of the Lorentz transformation that relates that chart to the chart that I used to construct Minkowski spacetime in the first place?

    I understand that this is the physically sensible answer, but I want to know the exact mathematical construction that guarantees that this is the case.
     
  8. Mar 1, 2017 #7

    PeterDonis

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    Look at the mathematical criteria for an inertial non-rotating frame. The easiest coordinates to use when expressing those criteria are Minkowski coordinates--note that you can express any set of frame vectors in those coordinates--because in those coordinates all of the connection coefficients are zero and covariant derivatives are just partial derivatives. The criteria then are:

    $$
    (e_0)^\mu \partial_\mu (e_i)^\nu = 0
    $$

    for all ##e_i##. You should be able to convince yourself that the only way to satisfy these equations is to have ##\partial_\mu (e_i)^\nu = 0## for all ##e_i##, i.e., for each of the frame vectors to have constant coefficients in Minkowski coordinates. And of course these vectors are just the coordinate basis vectors of all possible inertial coordinate charts (one of which will be the chart in which we expressed the above equations).

    There isn't one in general because not all frame fields correspond to coordinate charts. To check whether a particular set of vectors can be a coordinate basis, you need to check that their Lie brackets pairwise vanish, per the Wikipedia article I linked to in post #2.

    There isn't one in general because of the answer to #1 above.

    Even in the case where there is a transformation, "Lorentz transformation" is not really a good term for it, because that term implies a transformation between two global Minkowski coordinate charts. But you are talking about a transformation between a Minkowski chart and one that isn't in general Minkowski.
     
  9. Mar 1, 2017 #8
    Yes! Exactly as you say, using the fact that Minkowski coordinates have all their Christoffels equal to zero, and forcing the fact that my arbitrary frame field must be inertial and non-rotating, I arrive at precisely the system of equations that you mention. So I see now that I was on the right track, thank you ;)

    I struggle, though, to convince myself that said system can only be satisfied if all the partials of the tetrad are zero... I suppose that some algebraic manipulations of the equations must make this evident, but so far I haven't been able to figure how to do this... could you point out how such a fact could be made evident?

    Many thanks again

    EDIT: I had no hope of building coordinates if the ei's are an arbitrary frame field (I was in fact referring to the arbitrary inertial frame field, but forgot to mention this). But then, suppose you have a frame field ei such that [ei,ej] = 0 for all i,j. Is this a sufficient condition for there to be coordinates such that the ei's are the coordinate vector fields? Do the ei's have to be an inertial frame field in order for this to be the case?
     
  10. Mar 1, 2017 #9

    PeterDonis

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    Yes, it's equivalent to the Lie bracket being zero.

    No. For example, the coordinate basis vectors in Rindler coordinates in Minkowski spacetime form a frame field, but it is not inertial (though it is non-rotating).
     
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