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

A Invariant metric

  1. Dec 16, 2016 #1
    Hello, I am looking for some nontrivial metric on ℝ^2 invariant under the coordinate transformations defined by the 2x2 matrix
    [1 a12(θ)]
    [a21(θ) 1],
    where aik is some real function of θ. In the same way that the Minkowski metric on ℝ^2 is invariant under Lorentz transformations.
    Does this metric exist? If not does it exist for some related type of transformations? And why? Are there some other nice features about this kind of transformations/matrices?
  2. jcsd
  3. Dec 17, 2016 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Okay, this is something that can be completely solved using matrix algebra.

    First of all, every nonsingular matrix can be diagonalized. So if we assume it's nonsingular, then there is a matrix [itex]U[/itex] such that [itex]U g U^T = \tilde{g}[/itex], where [itex]\tilde{g}[/itex] has the form:

    [itex]\tilde{g} = \left( \begin{array} \\ g_1 & 0 \\ 0 & g_2 \end{array} \right)[/itex]

    and where [itex]U^T U = 1[/itex] ([itex]U^T[/itex] means the transpose of [itex]U[/itex]).

    So let's look for transformations [itex]\tilde{T}[/itex] that preserve [itex]\tilde{g}[/itex]. That means that for any column matrices [itex]u[/itex] and [itex]v[/itex],

    [itex](\tilde{T} v)^T \tilde{g} (\tilde{T} u) = v^T \tilde{g} u[/itex]

    which means that [itex]\tilde{T}^T \tilde{g} \tilde{T} = \tilde{g}[/itex]

    You can find the form of [itex]\tilde{T}[/itex] by using matrix algebra, but I'll skip to the answer:

    [itex]\tilde{T} = \left( \begin{array} \\ cos(\theta) & \sqrt{\frac{g_2}{g_1}} sin(\theta) \\ \sqrt{\frac{g_1}{g_2}} sin(\theta) & cos(\theta) \end{array} \right)[/itex]

    (This matrix has to be real, which means that if [itex]\frac{g_2}{g_1} < 0[/itex], then you have to choose [itex]\theta[/itex] to be imaginary, which means using [itex]sinh[/itex] and [itex]cosh[/itex] instead of [itex]sin[/itex] and [itex]cos[/itex]).

    Now, to get back to the original problem, if [itex]\tilde{T}[/itex] preserves [itex]\tilde{g}[/itex], then [itex]T \equiv U^T \tilde{T} U[/itex] is a transform preserving the original [itex]g[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted