Inertial & Non-Inertial Frames: Light Wavelengths

[HALON]

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 $1/γ$. Conversely, if the central body sends a signal to the rotating body, the rotating body will receive the wavelength as $γ$ times shorter. Is this correct?

 

[ghwellsjr]

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 $1/γ$.

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

Conversely, if the central body sends a signal to the rotating body, the rotating body will receive the wavelength as $γ$ times shorter. Is this correct?

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

But this may just be semantics.

 

[WannabeNewton]

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.

 

[HALON]

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


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

But this may just be semantics.

I see your point about the semantics, but essentially we are in agreement.

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

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.

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.