- #1

Chuck37

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Here's my simple example. There is a transmitter and a receiver, moving toward each other at constant speed. The receiver measures the frequency of what he receives using his own clock. I believe that the answer (the measured frequency) should be the same regardless of the frame of the observer since it is an objective and co-located thing: number of cycles of received clock per cycle of local clock.

Case 1 is that I observe from the transmitter frame. Without invoking any relativity, the wavelength is [itex]c/f[/itex] and these pass the receiver at a speed of [itex]c+v[/itex]. The observed received frequency is [itex]\frac{c+v}{c/f}[/itex] = [itex](1+v/c)f[/itex], where f is the transmit frequency. From the frame of the transmitter though, the receiver clock looks slow by [itex]\sqrt{1-(v/c)^{2}}[/itex], so in this frame I see him measuring [itex]\sqrt{\frac{1+v/c}{1-v/c}}[/itex]. This is the non-relativistic Doppler divided by his relativistic clock slowing.

Case 2 is that I observe from the receiver frame. Now the wavelength is (c-v)/f and these pass the receiver at c. The observed (non-relativity) frequency is [itex]\frac{c}{(c-v)/f}[/itex] = [itex]\frac{f}{1-v/c}[/itex]. Due to relativity the transmit frequency looks slow by [itex]\sqrt{1-(v/c)^{2}}[/itex], so I multiply the non-relativity version by this factor and get the same answer [itex]\sqrt{\frac{1+v/c}{1-v/c}}[/itex].

The case that messes me up is if I pick a middle frame where they are each observed to be moving toward each other at v/2. By the same rationale as each of the previous, the non relativistic version is [itex]\frac{1+v/2c}{1-v/2c}[/itex]. Special relativity would make both the transmit and receive clock look slow by [itex]\sqrt{1-(v/2c)^{2}}[/itex], but since the transmit and receive are the same (I'd multiply numerator and denominator by this factor) this effect would cancel. I'm left with the non-relativistic version which is very very close numerically to the previous two, but not exact. What am I missing?