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If I'm observing radiation emitted from a spherical outflow, is it correct to assume that that on the axis of the spherical outflow perpendicular to the observer the doppler boost is similar to the doppler transverse effect? I arrive at this conclusion by just substituting the relativistic aberration formula into the relativistic Doppler effect equation.

From a non-relativistic perspective, the velocity component is zero as its emitting 90 degrees from the line of sight of the observer.

But when considering a relativistic outflow, at 90 degrees to the observer, the doppler factor, delta, is still delta=1/(Gamma), where Gamma is the Lorentz factor. I have a hard to visualising these things, and I'm pretty sure my calculations check out, but I was just curious if I have done this correctly. Does this sound correct?

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  • $\begingroup$ Assuming your observer is at rest relative to the center of the outflow, the Doppler effect already takes into account the direction of motion of the outflow segment (moving source) emitting the radiation. Why would you bring in relativistic aberration on top of that? $\endgroup$ – udrv Mar 14 '17 at 21:04
  • $\begingroup$ @udrv "the Doppler effect" does not yield any information about the relativistic boost to the intensity. $\endgroup$ – ProfRob Mar 14 '17 at 22:39
  • $\begingroup$ @RobJeffries OP asks about Doppler factor, not intensity. $\endgroup$ – udrv Mar 14 '17 at 23:09
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I think it does.

Rotating observer launches a photon at oblique angle backward, and this photon approaches observer in the center at right angle to direction of motion of rotating observer and this photon will be redshifted at gamma.

Vice versa - observer in the center launches (or reflects the same) photon at right angle and rotating one receives it at oblique angle.

Reflected photon has to return to observer who launched it at the same frequency, since energy cannot disappear to nowhere. Sure, we consider "perfect" observers or mirrors that do not consume any energy.

Received by observer frequency is:

$$ f_0 = \frac {f_s} {\gamma (1+ \frac v c \cos \theta_0) }(1) $$

here

$$v=\omega r$$

$$\theta_0$$ is angle of reception.

If this angle is $$\pi/2$$, then

$$f_0=f_s/\gamma$$

But, for rotating observer photon always comes at oblique angle.

Angle of emission and angle of reception are tied by relativistic aberration formula.

$$ \cos {\theta_0} = \frac {\cos {\theta_s} - \frac v c} {1- \frac v c \cos \theta_s} $$

We can rewrite (1)

$$ f_0 = \gamma (1 – \frac {v \cos \theta_s} {c}) f_s (2)$$

Observer in the centre launches photon at right angle to direction of motion of rotating observer, so

$$\cos \theta_s = 0$$

we get

$$f_0= \gamma f_s$$

i.e. photon is blueshifted gamma times at reception.

If rotating observer holds a mirror he reflects photon at oblique angle backward and photon comes back to centre redshifted i.e. at the same frequency it was released.

Good to note, that photon launched from the opposite side of the rim of rotating disc will come at the same frequency as it was launched, since angles of emission and reception will be the same. So, these observers will not measure mutual time dilation.

That was confirmed by Champeney and Moon time dilation test. Nature, 1963.

http://mathpages.com/home/kmath587/kmath587.htm

“This is a stationary configuration (up to spatial isotropy), so we can definitely say the spatial distance travelled by the signal pulses is not changing. Classically it would follow that there was no Doppler effect, but the time dilation of special relativity implies that the proper time of the circling entity is reduced relative to the rest frame time coordinate of the central entity. As a result, the received signal will be either red-shifted or blue-shifted, depending on whether the transmitter or the receiver is moving in a circle.”

R.C. Jennison (Nature, 1964) considered ray path on rotating disc and concluded that light beams take different patches back and forth from the centre to the rim. Thus, by these patches rotating observers can determine their own rotation.

Some references:

[1] Hay H J, Schiffer J P, Cranshaw T E and Egelstaff P A 1960 Phys. Rev. Lett. 4 165-6

[2] Kundig W 1963 Phys. Rev. 129 2371 - 5

[3] Essen J 1964 Nature 202 787

[4] Jennison R C 1964 Nature 203 395-6

[5] E. Zanchini "Correct Interpretation of two experiments on the Transverse Doppler Shift". Phys. Scr. 86 (2012)

[6] Essen L Bearing on the recent experiments on the special and general theories of relativity No 4934 Nature, May 23 1964

[6] Essen L A time dilatation experiment based on the Mossbauer effect. Proc. Phys. Soc.vol.86 1965

Some animation:

https://www.youtube.com/watch?v=hnphFr2Iai4

https://www.youtube.com/watch?v=AGMINcBYojc

Some links

https://www.researchgate.net/publication/304781760_Specific_Features_of_Time_Dilation_During_Circular_Movement

https://www.researchgate.net/publication/304792446_The_Problem_of_Slow-witted_Aliens_or_The_Transverse_Doppler_Effect_Simply_Explained

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  • $\begingroup$ 1) OP never mentioned a rotating spherical outflow anywhere, and don't see why you bring this up. It's a valid, but different problem. 2) "Reflected photon has to return to observer who launched it at the same frequency, since energy cannot disappear to nowhere": No. A perfect reflection from a (heavy) moving mirror conserves energy-momentum, not energy alone. 3) Documenting your claims is good, but since this is a well-settled topic, there is no need to overdo it. A couple of good, reliable texts (and refs therein) and animations suffice. $\endgroup$ – udrv Mar 14 '17 at 23:06
  • $\begingroup$ This was helpful, thank you. Thank you also udrv for clarifying. I mainly use rybicki and lightman as my source; I just had a hard time visualising this, Albert found a good way. $\endgroup$ – ghator Mar 15 '17 at 21:38

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