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Accelerated observors in flat space

  1. Oct 31, 2011 #1
    Is the following true ?
    In flat space, we have a familiar doppler shift formula for two observers moving relative to each other. Suppose they had a relative ,constant acceleration between them.Then their relative velocity is changing with time. Will the doppler shift formula still hold with a time dependent boost velocity ,say 3+ at where a is the relative constant acceleration between the frames and 3 is the initial relative velocity?
    if not, then how do we tackle this?
  2. jcsd
  3. Oct 31, 2011 #2


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    I'm not quite sure what the question is - but the worldline of an observer with a constant proper acceleration is given by the relativistic rocket equation, see for instance


    A frame or observer that is accelerating is not in an inertial frame, the above article goes into this a bit.
  4. Oct 31, 2011 #3
    I never worked that out myself, just my 2cts:

    The "relativistic Doppler" shift consists of classical Doppler for a medium in rest times the time dilation factor. Probably the acceleration of each matters, so I don't think that it's safe to assume that only "relative acceleration" matters; I'd tackle that by splitting up the cases, starting with the case that only the other one accelerates. And surely also the distance matters, as the received shift at a certain time depends on the propagation time.

  5. Oct 31, 2011 #4


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    The Doppler shift depends only on the difference between the instantaneous velocity of the emitter at the moment of emission and the instantaneous velocity of the receiver at the moment of reception. Acceleration does not affect it.
  6. Oct 31, 2011 #5
    You make it sound it is very simple, while it is actually rather complicated.
    To even measure a Doppler shift one needs one complete period, when the emitter and/or receiver accelerates this brings all kinds of complications if we are talking about relativistic speeds. For instance do all frequencies shift the same way?

    So let's assume O1 and O2 are a distance d away and both start accelerating in opposite directions at the same time from the perspective of an observer in the middle.
    O1's proper acceleration is a1 and O2's proper acceleration is -b1.

    Then what is formula for the Doppler shift?
    Last edited: Oct 31, 2011
  7. Oct 31, 2011 #6


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    I've noticed that topics on this forum are often made to sound much more complicated than they actually are.
    The reception of a photon is a single spacetime event. The frequency is 1/ħ times the photon's energy.
  8. Oct 31, 2011 #7


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    Use null coordinates u and v, where c=1 and u = t-x and v = t+x

    Find u1(tau), v1(tau), u2(tau), v2(tau) for worldline 1 and worldline 2, tau being the proper time along the worldline

    For light travelling in one direction, u will be constant. For light travelling in the other direction, v will be constant. When u is constant, specifying a value of u will specify some particular light wave. The doppler shift for that wave will be

    (du1 / dtau) / (du2 / tau) (or perhaps its reciprocal, depending on which one is the emitter, and how you define the doppler shift. (Was it interval of reception / transmission, or frequency of reception / transmission?).

    Both derivatives most be evaluated at the same value of u, which specifies the worldline of some particiular lightwave, the lightwave whose doppler shift is being calculated.

    The details of working out the numbers do get involved, but it sees to me that the problem specification is not optiimzed for comprehension of the physics involved, and I wonder how helpful this detailed answer will really be to the original poster.

    Perhaps we have another poster who has the false impression that acceleration needs to be referred to another observer (as velocity does), when in fact one can measure one's acceleration without reference to another observer (for example by an onboard accelerometer).

    For the details, using the relativistic rocket equations, we can write a generic expression for u(tau) and v(tau), substituting a positive value of a in for one observer and a negative value for the other. It's conveneint to make the distance between them zero at tau=0, rather than some fixed value. Then

    u(tau) = (1/a) sinh(a*tau) - (1/a) [cosh(a*tau)-1 ]
    v(tau) = (1/a) sinh(a*tau) + (1/a) [cosh(a*tau)+1]

    Then du/tau = cosh(a*tau) - sinh(a*tau), but this is tabulated as du/dtau as a function of tau, and we reallly want du/dtau as a function of u. Fiding du/dtau as a funciton of u requires inverting the function u(tau) to find tau(u), which doesn't have a closed form solution as far as I know.
  9. Oct 31, 2011 #8


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    You'll kick yourself when I remind you that[tex]
    \cosh(a\tau) - \sinh(a\tau) = e^{-a\tau}
    \cosh(a\tau) + \sinh(a\tau) = e^{a\tau}
    [/tex]So, quite easy to invert.:smile:
  10. Oct 31, 2011 #9


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    Heh. Nice to see that it works out easier, not so nice that I missed it. But a good catch.

    I still don't know if working the example out in detail clarified anything to the OP, I suppose we'll see.
  11. Oct 31, 2011 #10
    No, the formula for inertial motion does not hold, you can see the derivation for the accelerated case here.
    An easier way to work this out is to use the equivalence principle and to replace the acceleration with a gravitational field.
    Last edited: Oct 31, 2011
  12. Oct 31, 2011 #11
    I must have missed something as I do not think the example is worked out in detail.
  13. Oct 31, 2011 #12


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    Actually that link shows, like Bill_K said earlier, that only the velocity at emission and velocity at reception matter. Acceleration comes into play only in so far determining the correct velocity of emission and absorption.

    The principle of equivalence will be easy only if both source and target are accelerating in just the right way to emulate stationary source and target in a uniform field (otherwise, you may as well analyze in any inertial frame using the formulas in your link, in which acceleration does not figure). If I understand, that was not the OP's example. The OP was talking about source and target in relative acceleration.
  14. Nov 1, 2011 #13
    This is not exactly what he said. Here is his exact quote:

    Either way, since no one gave the formula and its derivation, I thought that I would try to answer the OP.

    Yes, obviously.
    Last edited: Nov 1, 2011
  15. Nov 1, 2011 #14
    Really, did you make that up yourself or could you point to any literature to back that up?

    If it is a single event then how would you explain the Doppler effect when the target approaches or retreats from a source that fired a single photon?
  16. Nov 1, 2011 #15


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    If thinking in terms of photons, it's all about energy. Take the 4-momentum of the photon in the emitter's (instantaneous) frame, transform this (null) vector to the target's (instantaneous) frame. Then energy / planck = detected frequency.
  17. Nov 1, 2011 #16


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    I think with my comments, and Dr Greg's remarks, the basic equations governing the motions of the body and of calculating the doppler shift have been presented. It's left as an exercise for the interested reader to go through the details of finishing off the solution.

    I think that my derivation, though hastily written and lacking diagrams, should be at least complete enough for a person with a sufficient interest to ask specific intelligent questions about the solution (and possibly complete enough that one could find the solution without questions).

    Of course, I suppose one would need at least calculus to follow the math, if one was trying to solve the problem without calculus, one might get stuck, and I have no idea what the background of the OP is.

    But in any event, I'm in a waiting mode - I'll try and monitor the thread to see if any specific questions arise.
  18. Nov 1, 2011 #17
    Since when is science a secret?
    I think there are too many here who claim to know it all and like to patronize whenever they get the chance but when asked about hard formulas cop out.

    By the way, I have the formulas.

    For instance one thing that is far from worked out are the emission and absorption proper times which are essential in what each observer will measure.
    Last edited: Nov 1, 2011
  19. Nov 12, 2011 #18
    Thanks to all of you for considering my problem.
    Actually while going through an astrophysics book, i read that the total doppler shift is equal to the addition of the shifts due to relative motion and gravitational red shift, UP TO FIRST ORDER. So i thought that there might be a general formula that gives the the exact doppler shift due to arbitrary motions of emitter and observer in a space with gravity. (metric specified). However, i ran into trouble through the calculation because in gravitational red shift calculation we actually cancel out the co-ordinate times of the emitting and reception of signals and thus the ratio of the frequencies is just inversely proportional to the proper time(wristwatch time) of the observer and emitter(both at rest) (which depend on the metric at those places.)however if they are in motion , the integrals become time dependent and we can no longer equate the co - ordinate times.
    That is why i tried another approach.. but first i wanted to see if it works in flat space with accelerations. using the beautiful quantity - the minkowskian dot product between the four velocity of the observor and the photon momentum 4 vector gives me the energy of the photon measured by the observor and hence its frequency. This gives exactly the relativistic doppler shift in flat space.
    But now when i carry over the same method in gravitational field , i again run into trouble because i have absolutely no idea how to compute the change in photon wave vector , given a metric. This may be because of my lack of a proper understanding of general relativity. Actually i dont even know what null co -ordinates mean , so i could not follow pervect's solution to the problem.
    However, i agree to bill k that the reception of photon is a single spacetime event and accelerations dont affect the results. it is only the 4 velocity of the observer at the instant of reception of the photon that matters.
    actually you are right. i really had a misconception about this. acceleration can be measured without reference , say by measuring the angle which water surface in a container makes in an accelerated frame.
    Thanks to all of you for your kind help. I actually live in a place where i do not have internet connection.I sincerely apologize for my late reply. Thanks.
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