Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confused by proof of Lorentz properties from invariance of interval

  1. Feb 29, 2012 #1
    I've seen a few short proofs that if that some transformation [itex]\Lambda[/itex] preserves the spacetime interval, then

    [itex]\Lambda^\top g \Lambda = g[/itex]

    where g is the spacetime metric.

    They have all relied on an argument using some simple algebra to show that

    [itex](\Lambda^\top g \Lambda) x \cdot x = g x \cdot x[/itex]

    and since this is true for *any* x, it must be true that

    [itex]\Lambda^\top g \Lambda = g[/itex]

    This confuses me. I don't see how this "since it's true for any x" step is justified.

    For example,

    [itex]\left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right) x \cdot x = \left( \begin{array}{ccc} 0 & 0 \\ 0 & 0 \end{array} \right) x \cdot x[/itex]

    for any x, and by the same argument, since it is true for *any* x, it must be true that the arrays are equal and -1 = 0 = 1. Did I just break math?

    Intuitively, I don't think I could do the same trick to produce a counterexample in 3+ dimensions, but this seems like kind of a subtle point to sweep under the rug in a proof.

    Does anyone have a more detailed version of this argument that would (hopefully) make more sense to me?
  2. jcsd
  3. Feb 29, 2012 #2


    User Avatar
    Gold Member

    You cannot cancel a factor of zero from an equation because it can give contradictions

    0 x = 0 y


    x = y

    for all x, y.
  4. Feb 29, 2012 #3
    If it's the zero matrix that is making you think of division by zero, it's also true for a 90 degree rotation in the opposite direction.

    [itex]\left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right) x \cdot x = \left( \begin{array}{ccc} 0 & 1 \\ -1 & 0 \end{array} \right) x \cdot x[/itex]

    Those matrices aren't equal, either.

    To be clear, my confusion is that in several proofs the claim is made that two matrices are equal because of their relation via the dot product of any vector. It's clearly not true for all matrices as the above counterexample shows, so what properties of these particular matrices are exploited to make it true for them?
  5. Feb 29, 2012 #4


    User Avatar
    Gold Member

    The equation you've written is an example of the length of vectors being invariant under spatial rotations, which form a group. The rotation matrices are unimodular with orthogonal rows and columns.

    That is confusing.
  6. Feb 29, 2012 #5
    Here's an example of this from "Problem Book in Quantum Field Theory" by Voda Radovanovic...

    He does specifically mention x is in M4, which rules out my counterexamples, since they only work in 2D. I presume this is a well-known and obvious argument in order to be the solution to the very first problem of the book, but it seems pretty subtle to me if I want to prove the result in more detail...
  7. Feb 29, 2012 #6


    User Avatar
    Gold Member

    That seems logical. If [itex]\Lambda[/itex] is length preserving then g, which defines the interval should be invariant under it.
  8. Feb 29, 2012 #7
    Actually, thinking about this a bit, I think it's different than a statement that vector length or dot product is preserved.

    For example, for any orthogonal matrices A and B, [itex](Ax) \cdot (Ax) = x \cdot x = (Bx) \cdot (Bx)[/itex], which is directly related to the length being preserved.

    But [itex] (Ax) \cdot x = (Bx) \cdot x[/itex] fails for most orthogonal matrices. (As an example, check the identity versus a 90 degree rotation where you end up with x * x = 0 which clearly can't be true for all x).

    In the 2D rotational case, for example, A and B would have to rotate x by the same absolute value of theta for their projections onto the original x to be equal. I can't think of any examples of nonequal rotations A and B from higher dimensions where this would hold, since any 3D+ rotation A has an eigenvector on its axis (with A x * x = x * x) which is not an eigenvector of any other rotation B (so B x * x < x * x). I think this must ultimately be related to why this argument works, but there's a fair amount of linear algebra behind it.

    I agree it's an intuitive result, and I have no doubt it's true. I just think the argument used to get there (If A x * x = B x * x for all x in M4, then A = B) is not so obvious, and in fact it's not even true in M2.

    Maybe it's just part of me getting used to physics to accept the more hand-wavy "it seems reasonable" kind of argument even if I don't fully understand the mathematics behind it. It tends to leave me terribly confused, though.
  9. Feb 29, 2012 #8


    User Avatar
    Gold Member

    If Ax=Bx then (Ax).x = (Bx).x. I don't know if it's a sufficient condition though.
  10. Mar 1, 2012 #9
    Well, I managed to figure this out after a bit more pondering. It's necessary to use the properties of g on both sides of the equation.

    (My thought that maybe it worked because of some nutty property of rotations in dimensions greater than 2 was crap, since e.g. [itex]\left( \begin{array}{ccc} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array} \right) x \cdot x = 0[/itex] for all x. [Stupid even dimensions.])

    I did it in more detail this way...

    For arbitrary matrices A and B, where [itex](Ax) \cdot x = (Bx) \cdot x[/itex] for all x, you can show that their entries on the main diagonal must be equal by picking each unit vector in the basis and substituting it for x.

    For example, picking [itex]x = \left( \begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix} \right)[/itex] in [itex]A_{ij} x_i x_j = B_{ij} x_i x_j[/itex] yields [itex]A_{00} = B_{00}[/itex]

    Now that we know the diagonals are equal, you can show by picking each vector in the basis with two 1's in it (e.g. (0, 1, 1, 0)) that [itex]A_{ij} + A_{ji} = B_{ij} + B_{ji}[/itex].

    For example, picking [itex]x = \left( \begin{matrix} 1 \\ 0 \\ 1 \\ 0 \end{matrix} \right)[/itex] in [itex]A_{ij} x_i x_j = B_{ij} x_i x_j[/itex] yields [itex]A_{00} + A_{02} + A_{20} + A_{22} = B_{00} + B_{02} + B_{20} + B_{22}[/itex] which implies [itex]A_{02} + A_{20} = B_{02} + B_{20}[/itex] since [itex]A_{nn} = B_{nn}[/itex] for all indices n.

    Now going from the general case to the special case where B = g, and i ≠ j, then [itex]A_{ij} + A_{ji} = g_{ij} + g_{ji} = 0[/itex], so the entries off diagonal are antisymmetric.

    Finally, we know that [itex]\Lambda^{\top} g \Lambda[/itex] is symmetric, since g is symmetric and [itex] (\Lambda^{\top} g \Lambda)^{\top} = \Lambda^{\top} g^{\top} \Lambda = \Lambda^{\top} g \Lambda[/itex].

    The only way that the off-diagonal entries can be both symmetric and antisymmetric is if they are zero, and the on-diagonal entries must be equal, so [itex]\Lambda^{\top} g \Lambda = g[/itex] as expected.

    The same proof seems like it would work for any diagonal matrix substituted for g, so there's probably some slightly more sophisticated linear algebra that would make short work of this through some decomposition or something.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook