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Are inertial and non-inertial frames assymetrical?

  1. Aug 11, 2014 #1
    When neglecting gravity and body size, if a body rotating at uniform angular velocity about a central body sends a light signal to the central body, the central body will receive the wavelength as longer by [itex]1/γ[/itex]. Conversely, if the central body sends a signal to the rotating body, the rotating body will receive the wavelength as [itex]γ[/itex] times shorter. Is this correct?
  2. jcsd
  3. Aug 12, 2014 #2


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    The wavelength is longer but I would say by [itex]γ[/itex] times. Or you could say the frequency of the light is slower by a factor of [itex]1/γ[/itex].

    Again, it is shorter but I would say by [itex]1/γ[/itex] times and the frequency is [itex]γ[/itex] times faster.

    But this may just be semantics.
  4. Aug 12, 2014 #3


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    Didn't we already talk about this exact same thing in a previous thread of yours? You asked literally the same question in https://www.physicsforums.com/showthread.php?t=762176 and got the same answer ghwellsjr gave you above.
  5. Aug 12, 2014 #4
    I see your point about the semantics, but essentially we are in agreement.

    [EDIT] I began with [itex]γ=(1-v/c)^{1/2}[/itex] for an instant of angular velocity, then [itex]f_{orbit}=f_{central}/γ[/itex], which is simply [itex]1/γ[/itex]
    Using [itex]c=fλ[/itex]
    we get Orbit's view of light as [itex]c=(1/γ*f)(γλ)[/itex]
    And the reciprocal is Central's view of light [itex]c=(γf)(1/γ*λ)[/itex]

    Yes but... it was the last question I posted on that thread and I didn’t receive your reply. You implicitly answered it earlier in a very detailed (and for me complicated) way, which I took as agreement. Indeed my last question there was also rather longwinded. So I condensed the question (without all the equations) to seek clarification and confirmation.

    If ghwellsjr is correct then it satisfies me and this thread may be closed.
    Last edited: Aug 12, 2014
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