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

Hermitian Metric - Calculating Christoffel Symbols

  1. Jun 24, 2011 #1

    I am trying to understand what the differences would be in replacing the symmetry equation:

    g_mn = g_nm

    with the Hermitian version:

    g_mn = (g_nm)*

    In essence, what would happen if we allowed the metric to contain complex elements but be hermitian? I am not talking about moving to a fully complex space. We still have the 4 real values of ct, x, y, and z. I have searched the interweb for help, but cannot find anything talking about this particular issue. I only know differential geometry (basically, calculus), so the articles using groups etc. a far over my head, and most articles posted that I found just move to the more complicated "complex space" which I am trying to avoid.

    What I really want to know is how the Christoffel Symbols and the Ricci Tensor would differ for this kind of metric. Any help at all would be greatly appreciated!
    Last edited: Jun 24, 2011
  2. jcsd
  3. Jun 24, 2011 #2


    Staff: Mentor

    Re: Hermitian Metric

    Nothing would be any different. Because all real numbers are their own conjugate transpose.
  4. Jun 24, 2011 #3
    Yes I know, but that assumes the metric has only real values. I am not assuming that, so in essence the metric, Christoffel Sybols, and Ricci tensor would all allow complex values.

    Solving for the Christoffel symbols (from examples I've seen) involves writing out the equation for a zero covariant derivative of the metric, rotating indices, and then using the metric's symmetry to solve it out. Since the symmetry condition is nolonger the same, I can't use that method anymore and am stuck.

    g_mn || k = g_mn | k - g_jn G^j_mk - g_mj G^j_nk = 0

    My inkling is that this should be replaced with:

    g_mn || k = g_mn | k - g*_jn G*^j_mk - g_mj G^j_nk = 0

    ButI'm not entirely sure...
  5. Jun 24, 2011 #4
    Here is the equation using the editor:

    [itex]\frac{\partial g_{\mu \nu}}{\partial x^{\tau}} - g_{\alpha \nu} \Gamma^{\alpha}_{\mu \tau} - g_{\mu \alpha} \Gamma^{\alpha}_{\tau \nu} = 0[/itex]

    which I propose changing to:

    [itex]\frac{\partial g_{\mu \nu}}{\partial x^{\tau}} - g_{\alpha \nu} \Gamma^{\alpha}_{\mu \tau} - g^*_{\mu \alpha} \Gamma^{* \alpha}_{\tau \nu} = 0[/itex]

    in the general Hermitian case. Is this correct?
  6. Jun 24, 2011 #5
    You would end up with physical predictions that contain complex entities. We have no way to interpret this sort of information... What does it mean to have [itex]3i[/itex] apples?
  7. Jun 24, 2011 #6
    While tempting to debate this statement, it's really off topic. I'm more asking about how the other components of differential geometry change when we change the symmetry condition on the metric tensor.
  8. Jun 24, 2011 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let's start with a simpler example. Suppose you have a Newtonian simple harmonic oscillator with equation of motion [itex]d^2x/dt^2+bx=0[/itex], where b is a constant. One way to solve that is to guess a solution of the fotm [itex]x=e^{rt}[/itex], where r is a constant. This give solutions [itex]r=\pm i\sqrt{b}[/itex]. The most general solution can be found by taking a linear superposition of the two solutions, [itex]x=c_1e^{rt}+c_2e^{-rt}[/itex]. When you pick the c's to match real-world initial conditions, you get a real-valued x.

    The Einstein field equations are different, because they're nonlinear. Therefore you can't find a family of complex-valued solutions for the metric and take linear superpositions of them to get a real-valued result. There are certain tricks that can sometimes be used to "realify" a complex-valued metric, but they aren't tricks that work in general. If you want to see an example of such a technique, I believe that's how Kerr originally found the Kerr spacetime.
  9. Jun 24, 2011 #8

    Thank you for the reply. I am aware of what you are talking about, but honestly I'm not concerned with creating a "real only" metric. I guess my question is more of a pure mathematical question than physics, and i was hoping that someone out here could point me to some references that would answer my particular questions and even help me learn more about the math involved. That is why I went to the length to describe the resources I had already found and why there were bad (not of help). I treat asking questions on these forums as a last resort, as I don't like taking you guy's time just to help me in my own personal research.

    I'm sure this avenue, as it relates to physics, of using complex metrics has been tried before (and likely failed). I am not trying to create some crackpot theory, merely trying to understand it better (and learn what has worked, what hasn't, and why). I'm here though because I'm at a loss to find anything out there that has been helpful, and was hoping someone here may have themselves gone down this road and exploration and found good references.
  10. Jun 24, 2011 #9


    User Avatar
    Science Advisor

    Maybe http://arxiv.org/abs/gr-qc/9312032" [Broken]'s Chapter 3?
    Last edited by a moderator: May 5, 2017
  11. Jun 25, 2011 #10


    User Avatar
    Science Advisor

  12. Jun 25, 2011 #11

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    If you make the metric complex, then you ARE doing complex geometry. You need to take care in computing the curvature, because you can't start naively sticking complex conjugates everywhere.

    Note that complex geometry really only works for Euclidean-signature manifolds; not Lorentzian-signature. The reason is simple:

    a^2 + b^2 = (a + ib) (a - ib)

    but for Lorentzian signature, there is no convenient way to write (a^2 - b^2) in terms of complex numbers.

    On the other hand, if you want to see an example of the use of complex functions to give you real metrics, look up the Israel-Wilson metrics. These are metrics for multiple extremal charged black holes. Extremal black holes, where the electric charges are properly balanced against the masses, experience no net attraction or repulsion, so it is easy to write down metrics for any number of such black holes, placed anywhere in space.
  13. Jun 25, 2011 #12

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    I forgot to add, another thing you can do is to use complex geometry to solve Einstein's equations in Euclidean signature for 2 complex (i.e. 4 real) dimensions. Then you can Wick rotate the result to get a Lorentzian signature manifold. This is also a common technique, used to find "gravitational instantons" such as the Gibbons-Hawking metrics.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook