Frequency shift and relative velocity

[Saw]

Suppose two observers A and B are moving with regard to each other at relative velocity = v = 0.5c, which they have determined for each other by measuring the time interval that a point of the other frame takes to pass by two points of their own frame, whose clocks are synchronized using the Einstein convention.

Now they try to measure their relative velocity but using the Doppler shift method: in turn, each of them sends a light wave to the other one with a frequency of emission fE = 1; the wave bounces back at the target and returns to the source with frequency of reception fR.

In a classical scenario, the formulas would be (please correct if I am wrong):

a) If one of the observers is stationary with regard to the medium of propagation of light that classical relativity assumed to exist and the other is moving:

For approaching frames:
fR = fE * (1+v/c)/(1-v/c) = 1(1+0.5)/(1-0.5) =1.5/0.5= 3
For receding frames:
fR = fE * (1-v/c)/(1+v/c) = 1(1-0.5)/(1+0.5) =0.5/1.5= 0.33

The formulas would be the same, no matter if the operation is carried out by A or B.

b) If both observers move with regard to the medium:

Here I am not so sure, but at least for an example where A and B recede from each other and A moves at vA with regard to the medium = (+ 1/6) c and B moves also with regard to the medium at vB = (2/3) c (the difference vB – vA being equal to the relative velocity = v = 0.5 c) the formula would be:

fR = fE * [(1+vA/c)(1-vB/c)]/[(1-vA/c)(1+vB/c] = 1[(1+1/6)(1-2/3)]/[(1-1/6)(1+2/3)] = 0.28

Again the result of the operation would be the same, no matter who carries it out. But, according to my calculations, different values of vA and vB, even if the relative difference between the two of them is still 0.5c, lead to different frequency shifts.

Still in the classical context, let us suppose that an observer has measured fR and wants to obtain v:

a) If one observer is stationary with regard to the medium:

For approaching frames:
v = [(fR/fE) - 1]/[(fR/fE) + 1] c = [(3-1)/(3+1)] c = (2/4)c = 0.5 c

For receding frames:
v = (1 - fR/fE)/(1 + fR/fE) c = (1-0.33)/(1+0.33) c = 0.667/0.333 c = 0.5 c

b) If both observers move with regard to the medium:

… Well, I have not seen any formula in the books and in fact it seems it cannot be derived from the above b) formula for fR, since we have two unknown values, vA and vB.

In SR, instead, there is no medium for the propagation of light and the expression “velocity with regard to the medium” makes no sense. Only relative velocity matters. The formula for obtaining fR is:

fR = fE * sqrt [(1+v/c)/(1-v/c)],

but I have some doubts:

First, the sign of v. I have seen contradictory statements. I think that, just like in the classical scenario, v should be:

Positive for approaching frames:
fR = fE * sqrt [(1+v/c)/(1-v/c)] = sqrt[(1+0.5)/(1-0.5)]=sqrt(1.5/0.5) = sqrt(3) = 1.73205 = frequency augments (blue-shifted)

Negative for receding frames:
fR = fE * sqrt [(1-v/c)/(1+v/c)] = 1 * sqrt(0.5/1.5) = sqrt(0.33) = 0.57735027 = frequency decreases (red-shift)

Second, is this the formula for a two-way trip? In other words, if A has emitted the wave, which has reflected at B, would fR be the frequency at which A receives back the waves?

Third, solving for v above, we obtain these formulas:

For approaching frames:
v = [(fRB/fEB)^2 -1]/[(fRB/fEB)^2 + 1] c = [(1.73205)^2-1]/[(1.73205)^2+1]=(3-1)/(3+1)=2/4=0.5 c

For receding frames:
v = [1 - (fR/fE)^2]/[1 + (fR/fE)^2] c = [(1- 0.333)/(1+0.333)] c =0.667/1.333 = 0.5 c

Is this right? Are these formulas (or some variation, depending on the angle of motion of the target) what you would use with a radar signal, if you wish accuracy to a relativistic level?

 

[Dale]

My favorite page on this topic is the mathpages site:
https://www.mathpages.com/rr/s2-04/2-04.htm
This page includes information about the Doppler shift both for signals with a medium like sound as well as for signals like light with no medium. It also does not make any assumption about one being at rest or moving relative to the medium, and even includes some information later on about Doppler shifts in arbitrary directions.

 

[RPinPA]

A comment about the two-way trip: The wave is Doppler-shifted twice. The target receives a Doppler-shifted wave in its frame, reflects it at the shifted frequency in its frame, and that is received double-shifted at the original transmitter. If you look at the theory of Doppler radar (which I've worked with fairly extensively) you'll see that factor of two in the equations.

By "the equations", I mean the usual first-order approximation for $v \ll c$, which you'll see as $\Delta f/f = 2v/c$ for Doppler radar with that factor of 2.

You're using the exact equations, so apply the shift twice, which will result in removing the square root, for instance for approaching target the frequency received back at the transmitter will be:
$$\frac {f_R} {f_E} = \frac {1 + \frac v c} {1 - \frac v c} \approx
\left (1 + \frac v c \right ) \left (1 + \frac v c + \frac {v^2} {c^2} + \cdots \right )
\approx 1 + \frac {2v} {c}$$

For the last step all I did was square $1 + (v/c)$ and drop the second order term.