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How to diagonalize a warped metric

  1. May 26, 2010 #1
    In GR it is often desireable to find coordinates in which the metric is flat, at least locally.

    My question is, given a general, non-diagonal metric tensor, how to diagonalize it, and how this process relates to a new coordinate system in which the metric is locally flat.

    The metric g can be written as a real, symmetric 4x4 matrix. Theorems of linear algebra state the existence of matrix P allowing the similarity transformation PT * g * P = L, where P is an orthogonal matrix, PT is its transpose (and inverse), and L is the diagonal matrix of eigenvalues of g. Further theorems claim that P will have the above properties if it is constructed by defining its columns as the normalized eigenvectors of the matrix g, in whatever column order. If you want the diagonalized matrix to have specific elements, like Diag(-1,1,1,1), I know of no way to do this via transformation P. Also note that the above similarity transformation defines what the theorems of linear algebra mean by "diagonalization". There could be other definitions.

    Now we step into physics. Given a general metric g, there is a theorem used in physics which claims that there exists a local coordinate transformation from original coordinates {x} to new coordinates {x'} in which g expressed in the new coordinates as g' is diagonal and is flat. That is, g' = Minkowsky metric = Diag(-1,1,1,1). The theorem, quoted in Hartle's Gravity, an Introduction.... (equation 7.13) goes beyond this to claim derivatives of g' are also locally zero. But we are not interested in that part for our purposes here. It is emphasized that the transformation to flat spacetime is local. It is good only for the immediate and mathematically vanishing neighborhood of some event E. For another event there will be another transformation to flat spacetime. The physicist's coordinate transformation which diagonalizes g seems to present a different definition of "diagonalization" from that of linear algebra.

    So, the question arises, Is there a connection between these two, and how does the matrix of eigenvalues of linear algebra become that of flat space = Diag(-1,1,1,1).
    A simple example of the problem is found in the 2D Warp Drive metric due to Alcubierre:
    dS^2 = -dt^2 +[dx - K dt]^2, where K is an event-dependent physical number. Since we are interested only in a locally valid transformation we treat K as a constant. The new coordinates which transform g --> g' = Diag(-1,1) are given by:
    dt' = dt
    dx' = dx-Kdt.

    The transformation matrix M that one associates with this transformation (v' = Mv, where v is the 2D spacetime vector (t,x)) is clearly not orthogonal (the transpose does not equal the inverse). So, it does not comport with the linear algebra theorems. Also, the eigenvalues of the matrix for g corresponding to dS^2 above, are not {-1,+1}. Clearly M is not the same as P above.

    Is there a good and mathematically rigorous analysis of this issue in the literature? I can't find it in the several GR texts I have looked at.
  2. jcsd
  3. May 26, 2010 #2

    Ben Niehoff

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    Simple. For clarity, let's do this on a 2-dimensional Riemannian manifold. Extending to the n-dimensional pseudo-Riemannian case is straightforward.

    Suppose you diagonalize G to get diag(a, b), where both a, b > 0. To convert this to diag(1, 1), just factor it:

    [tex]\mathrm{diag}(a, b) = \mathrm{diag}(\sqrt{a}, \sqrt{b}) \; \cdot \; \mathrm{diag}(1, 1) \; \cdot \; \mathrm{diag}(\sqrt{a}, \sqrt{b})[/tex]


    [tex]\Lambda = R^T S R[/tex]

    where [itex]\Lambda[/itex] is the diagonal matrix of eigenvalues, S is the diagonal matrix of [itex]\pm 1[/itex], and R is the diagonal matrix of the square roots of the absolute values of the eigenvalues. Then the full decomposition is written

    [tex]G = P^T R^T S R P = (RP)^T S (RP)[/tex]

    and so G can be brought to the desired form. Note that RP is NOT orthogonal in the matrix-algebra sense.

    However, the columns of [itex](RP)^{-1}[/itex] are orthonormal when measured in the metric G, as can be verified by checking [itex]((RP)^{-1})^T G (RP)^{-1}[/itex].
  4. Jun 8, 2010 #3
    Ben, This analysis seems to work fine if the diagonalized form of the metric is the matrix identity = diag(1,1). But for S=diag(-1,1) it fails. To make it work one must introduce the additional matrix factor diag(i,1), where i=sqrt(-1).
    Then, another question remains: assuming that you have found a complex similarity transformation C such that diag(-1,1) = C(transpose) G C, then how do you find the coordinate transformation (t,x) <--> (t',x') that produces a local inertial frame?
  5. Jun 9, 2010 #4


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    Ben did say he was talking about a Riemannian manifold, not a pseudo-Riemannian. In the case of spacetime, you diagonalise using Sylvester's law of inertia and you'll get both positive & negative eigenvalues (3 of one & one of the other), so no imaginary numbers needed.
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