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Rotating minkowski metric

  1. Nov 26, 2009 #1
    Hi, if you perform this coord. transformation on minkowski metric in cartesian coord. (T,X,Y,Z):
    T=t , X=x cos(w t)-y sin(w t) , Y=x sin(wt)+y cos(wt) , Z=z
    (it is a frame rotating with angul. speed w around the Z axis) then you get this metric (I'm using the +,-,-,- convention):
    ds^2=(c^2-w^2(x^2+y^2))dt^2+2w y dt dx-2w x dt dy - dx^2 - dy^2 - dz^2

    Try calculating some components of the curvature tensor (or the ricci scalar),what I get is different from zero! Even if it is just flat metric in different coordinates, why? what's the mistake? Shouldn't flatness (Riemann=0) be coord. invariant?
    Thanx
    Riccardo
     
  2. jcsd
  3. Nov 27, 2009 #2
    Components of a tensor are, kind of by definition, coordinate-dependent. The tensor itself should be independent of the coordinates used.

    Take for example a Minkowski space with a coordinate transformation x' = 1/x. Now you clearly have a singularity at x' = 0 because you map the spacetime in a funny way, that does not however mean that the spacetime is curved; the singularity is only generated by the funny coordinates.
     
  4. Nov 27, 2009 #3

    haushofer

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    Yes, you should clearly distinguish between physical and coordinate singularities, which is the idea of maximal analytic extensions. But if I have the equation

    [tex]
    R^{\rho}_{\ \mu\nu\rho} = 0
    [/tex]

    for the Minkowski metric, and I perform a coordinate transformation in which I go to a rotating observer, shouldn't this again give the condition above? Spacetime is still not curved, right? And if I perform a linear transformation on the Riemanntensor=0 it should still give 0.
     
  5. Nov 27, 2009 #4
    Yeah I guess that was a bad example, but consider then a change of variables from (dx, dy, dz) to (a(t)dx, a(t)dy, a(t)dz). How is that any different than the coordinate transformation in the original post?
     
  6. Nov 27, 2009 #5
    I don't think I see what you mean. As he said before if every component of Riemann is 0 and you make a linear transformation on it (because in a change of coord. tensors transform that way) it remains 0. Same thing if you consider the curvature scalar, it's a scalar right? it means it is independent of the coordinates you use to express it. How can it be different from 0 in this case?
     
  7. Nov 27, 2009 #6

    bcrowell

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    Hi, Rick89,

    You are correct that it should still be a flat-space metric regardless of the coordinate transformation. It doesn't matter if the transformation is linear.

    Clamtrox wrote
    This is true, but if the tensor is zero, then the linear structure of the tensor transformation equations guarantees that it will still be zero after the transformation. That's a technical thing, but the big picture here is that GR is a totally coordinate-independent theory, so, e.g., any solution to the field equations had still better be a solution if you do an arbitrary change of coordinates.

    Rick89, I'm not sure if your metric is correct. This is the kind of thing that I always use a computer algebra system for if possible, because otherwise you can easily blow an afternoon tracking down some silly sign mistake. Here is some Maxima code that checks the curvature of the polar-coordinates version of the metric you're talking about:

    Code (Text):

    load(ctensor);
    dim:3;
    ct_coords:[t,r,theta];
    lg:matrix([1-r^2,0,-r^2],
              [0,-1,0],
              [-r^2,0,-r^2]);
    cmetric();  
    ricci(true);
     
    I've set omega equal to 1 for convenience. Maxima is free and open-source, so you can just install it on your own machine and try it out. When you run the code above, it verifies that the Ricci tensor is zero. It looks to me like your off-diagonal components in your metric may be incorrect, but I haven't checked carefully. One way to pin down the mistake would be to use Maxima to find the curvature of your Cartesian-coordinates metric. If it's zero, then your metric is probably right, but your calculation of the curvature has to be wrong. If it's not zero, then your metric is wrong.

    There's a discussion of the rotating Minkowski metric here: http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.4 [Broken] The interpretation is kind of subtle. Even though the spacetime is flat, you get a Lorentz contraction that sort of looks like a spatial curvature.
     
    Last edited by a moderator: May 4, 2017
  8. Nov 27, 2009 #7
    Hi bcrowell,
    thanx for your reply, but unfortunately I already checked it with a package for Mathematica I found somewhere on internet.
    Now I downloaded Maxima and tried my form of the metric on it:

    lg:matrix([1-(x^2+y^2),2y,-2x,0],
    [2y,-1,0,0],
    [-2x,0,-1,0],
    [0,0,0,-1]);

    It gives a non-vanishing Ricci as you can check for yourself. I'm not sure why!!
    The point is on those coordinates, don't change them, because otherwise you miss the point, in fact by changing coordinates you can obviously go back to cartesians...
    So the problem is still not solved.
    Thanx anyway, Maxima is really nice, I didn't know it, so it was a great suggestion!
     
  9. Nov 27, 2009 #8
    I forgot to say. I am quite sure that the transformation leads to such a metric, I checked it many times. Even if the coord. transformation didn't have a physical sense , still it should preserve the flatness of minkowski metric. I can't see the mistake, can anyone help?
    Thank you
     
  10. Nov 27, 2009 #9

    bcrowell

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    Hi, Rick89,

    Okay, now we've narrowed it down. If maxima verifies that the metric isn't flat, then it must be the metric that's wrong, not your calculation of the curvature. If you like, you could post your derivation of the metric, and see if someone else can find a mistake that you missed.

    Are you sure the factors of 2 are correct? When you express the metric as a line element ds^2=..., its cross-terms don't correspond directly to the off-diagonal terms of the metric; they differ by a factor of 2.

    -Ben
     
  11. Nov 27, 2009 #10
    Yes! you're right! I was so silly! I didn't notice that of course... I was not used to work with not diagonal metrics and so I didn't notice that the package in Mathematica is also using a different convention for that! In fact it allows you to write the metric as a line element but it does not symmetrize it automatically, you have to introduce both dx dt and dt dx. So that was my original problem! Then I also did it wrong in Maxima again because I didn't notic that there. So that's solved! Thank you very much... a lot of inexperience I must confess...
     
  12. Nov 27, 2009 #11
    I knew it had to be something stupid... it had to... but I couldn't really see it, I needed an external eye on it...Bye
    Riccardo
     
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