curvature of space and lorentz transformation

Discussion in 'Physics & Math' started by devire, Jul 6, 2007.

  1. devire Registered Member

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    if space is curved, would that mean that over very long distances a lorentz transformation is inaccurate, since it was derived using the pythagorean theorem?
     
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  3. GhostofMaxwell. Banned Banned

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    I wouldn't have thought so because Lorentz just says your linear travel from point A to point B is contracted the closer to C. It says nothing about the shape of that line, and im sure it would even apply to any squiggly line you care to take between two points.
     
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  5. GhostofMaxwell. Banned Banned

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    Are you mixing up small angle approximation being derived from pythag?
     
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  7. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Tha familiar Lorentz Transformation is only good in flat space. That is, you must specify a metric in order to derive the Lorentz Transformations rigorously. If you used the Pythagorean Theorem to derive them (I'm not sure I've ever seen it done this way), then you have implicitly assumed a flat space-time, and thus the Minkowski metric: http://en.wikipedia.org/wiki/Minkowski_metric.

    If you're working in a curved space-time, the Lorentz Transfomation is a bit more subtle.
     
  8. GhostofMaxwell. Banned Banned

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    Impressive but above me.
    So say you take a trip to apha-century measuring your speed and distance travel, are you saying that you cant derive a precise figure from the Lorentz formula for the actual distance because space is curved?
     
  9. devire Registered Member

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    79
    that is what i suspected, but since i've only seen it derived from the pythagorean theorem and i am not too familiar with how metrics work, i wasn't too sure.
     
  10. devire Registered Member

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    well even if space were curved, the distance u speak of would not be enough to have any significant effect.
     
  11. GhostofMaxwell. Banned Banned

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    Indeed. A line on a piece of paper remains the same length even if you roll the paper up.
     
  12. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    How familiar with calculus are you? Essentially, the metric is the Jacobian. People always get pissed when I say that, but it contains the same information. The Jacobian is the factor you have to put in the volume or surface integral to account for the fact that the space isn't flat. So, for example

    \( \int d^3x \longrightarrow \int \sqrt{- \det M} d^3r\).

    M is the metric, and r is some new set of coordinates. If you want spherical coordinates, you (maybe) know that

    \( \int d^3x \longrightarrow \int r^2 \sin\theta dr d\theta d\phi\).
     
  13. devire Registered Member

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    79
    isn't the pythagorean theorem invalid in curved space however?
     
  14. GhostofMaxwell. Banned Banned

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    OK I stand corrected.
     
  15. GhostofMaxwell. Banned Banned

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    I believe so, but I dont see the relevance.
     
  16. przyk squishy Valued Senior Member

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    Er, I thought the Jacobian was a local approximation for the transformation, and the (transformed) metric was:
    \(g'_{\mu \nu} = g_{ \alpha \beta } \Lambda^{ \alpha }_{ \; \mu } \Lambda^{ \beta }_{\; \nu }\)​
     
    Last edited: Jul 6, 2007
  17. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Hmmm. I'm pretty sure that the square root of the (negative) determinant is the Jacobian. I could be wrong, though.
     
  18. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    I don't think so, but there is probably some analog. For example, one can draw a triangle on a globe whose angles add up to 270 degrees. Perhaps there is some generalization, but I am not aware of it...
     
  19. temur man of no words Registered Senior Member

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    Pythagorean theorem for curved space:

    \( (ds)^2=g_{ik}dx^idx^k. \)

    For \(g_{ik}=\delta_{ik}\) you get the usual Pythagorean theorem.
     
  20. temur man of no words Registered Senior Member

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    You can say that to a beginner but metric is not only Jacobian (or measure to be accurate), it defines smoothly varying inner product in the tangent bundle.
     
  21. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Like I said---I always piss off the mathematicians when I say that.
     
  22. przyk squishy Valued Senior Member

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    It is, but not by definition. If \(J\) is the Jacobian matrix of the transformation, \(\eta\) the metric in the "old" basis and \(\eta'\) the metric in the "new" basis, then:
    \(\eta' = J^{t} \eta J\)​

    So:
    \(\det(J) = \sqrt{ \frac{ \det( \eta' ) }{ \det(\eta) } }\)​

    If \(\eta\) is the Minkowski metric, then \(\det( \eta ) = -1\) and:
    \(\det(J) = \sqrt{ - \det( \eta' ) }\)​

    I'm not sure how much of this you already know, so...
     
  23. temur man of no words Registered Senior Member

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    1,330
    You can also say that metric defines a homeomorphism between the tangent and cotangent spaces.

    This Jacobian thing is connected to the fact that metric defines a unique n-form on n-dimensional manifold. You can only integrate forms on manifold so you multiply any scalar by this unique form to get something you can integrate.
     

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