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If an observer accelerates he must rotate in space-time?

  1. Feb 14, 2007 #1
    It is my understanding that if an observer accelerates he must rotate in space-time.
    But then why is that factor generally omitted in the calculations related to accelerations?
     
  2. jcsd
  3. Feb 15, 2007 #2
    Look up hyperbolic motion. Nothing is "omitted", everything is calculated correctly.
     
  4. Feb 15, 2007 #3
    MeJennifer wrote:
    Not true. Where did you get such an idea from ?
     
  5. Feb 15, 2007 #4

    Demystifier

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    That is true and it is clear where did you get such an idea from.

    Not true. Where did you get such an idea from?
     
  6. Feb 15, 2007 #5

    robphy

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    This is not off to a good start.
    Can we have a little more clarification [especially unambiguous definition of terms] from the OP?
    ..and hopefully a little more physics to back up the opposing [and somewhat cryptic] "That is true"/"Not true" declarations?
    (Is there a story behind these cryptic remarks?)
     
  7. Feb 15, 2007 #6
    Which term is ambigious?

    An acceleration is a rotation in space-time, with Coriolis forces being the effect.
    Are those effects being properly taken into account when we make calculations using hyperbolic geometry?
     
  8. Feb 15, 2007 #7

    robphy

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    acceleration... are you referring to a 4-acceleration vector? the spatial 3-acceleration in some observer's subspace of spacetime? some coordinate acceleration? or something else?

    rotation... are you referring to Euclidean rotations in an observer's spatial-subspace of spacetime? or Minkowski-boosts (which are sometime thought of as "[pseudo-]rotations" in spacetime)? or both? or something else?

    I'm looking for a clear definition of terms... to help clarify the question for me.
     
  9. Feb 15, 2007 #8
    By acceleration I mean a plain and simple proper acceleration of an observer.
    A rotation in space-time or a Minkowski boost or, more common, a Lorentz boost, it is the same thing.
     
    Last edited: Feb 15, 2007
  10. Feb 15, 2007 #9
    It seems she's referring to the concept of "Rapidity", not spatial rotations. Ever seen the Minkowski diagram showing that a moving observer's natural coordinate frame is rotated such that its time axis direction has a spatial component, and likewise the spatial axis (in the spatial direction of movement) is no longer simultaneous? Instead of parameterising coordinate frames by relative velocity, they can be described by the relative angle that these axes rotate. Also, the Lorentzian signature of the metric renders it a hyperbolic "rotation". I think the main convenience is that these rapidities are simple to add (unlike usual velocitiy addition in SR), since the hyperbolic part hides the convergence.
     
    Last edited: Feb 15, 2007
  11. Feb 15, 2007 #10

    Garth

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    An accelerated gyroscope suffers Thomas Precession.

    This is the gyroscopic reaction to it "leaning over in space-time" when accelerated.

    Garth
     
  12. Feb 15, 2007 #11
    Correct.
    So in calculations don't we eliminate this factor by assuming this is not happening? It seems we are making (hidden) provisions to allow for Fermi-Walker transport. Or am I off the mark here?
     
    Last edited: Feb 15, 2007
  13. Feb 16, 2007 #12
    I recall the original question was about rotation "in space-time" which does not arise purely from acceleration. The issues of rotation in abstract "rapidity space", Thomas precession etc. are IMO rather well covered here:
    http://abacus.bates.edu/~msemon/RhodesSemonFinal.pdf
     
  14. Feb 16, 2007 #13
    You are wrong it does.

    But feel free to explain what else is needed to make it into a rotation.
    So we have, acording to you, acceleration + X, that makes a rotation in space-time. So what is X?
    The floor is yours. :smile:
     
    Last edited: Feb 16, 2007
  15. Feb 16, 2007 #14
    Following robphy's comment in post #5 - could you explain why and how your rotation comes about ? I don't see what I have to explain - it's like asking me to explain why there isn't a leprechaun on the lawn.
     
    Last edited: Feb 16, 2007
  16. Feb 16, 2007 #15
    See for instance:

    o Einstein - "The Meaning of Relativity (Little Lectures of Princeton University)" page 19,
    o Wheeler Thorne Misner - "Gravitation": §6.5,
    o Weinberg - "Gravitation and Cosmology, Principles and Applications of the General Theory of Relativity": Chapter 2, pages 28 and 29
    o Feynman - "Lectures on Physics": Chapter 15-7.
     
  17. Feb 16, 2007 #16
    Your first and third references, at least, do not appear to support your contention. I had thought you might include some brief gist in your own words.
     
    Last edited: Feb 16, 2007
  18. Feb 16, 2007 #17
    Jennifer, if you have all these specific references, can you maybe restate your question less ambiguously? Is it actually Thomas precession that you're talking about?
     
  19. Feb 19, 2007 #18
    MeJennifer - your second reference (MTW) uses the terms 'pseudo-rotation' and "rotation" in quotation marks which by convention indicates that the word is being used differently than it's normal meaning. In this case to mean in Lorentz hyperbolic space where the analogy is only partial since 'normal' rotations will eventually replicate original orientation, unlike those in question. Also might not the equivalence principle lead us to wonder in what way an observer stationary in a gravitational field can be said to be rotating in spacetime ?
     
  20. Feb 19, 2007 #19
    It is clear that prose doesn't convey your ideas in a precise way, neither do the quotes. So how about if you tried to write down the math that goes with your ideas?
     
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