Originally posted by Tyro
Suppose we have a weightless light source that is moving. Its output light will therefore undergo a Doppler shift. Now, if the lightsource had weight but were static, its light would undergo a red shift. If the light source had weight and were moving, would it undergo both a red shift and a Doppler shift, with the end frequency being the linear superposition of both effects?
Also, if:
- The relative velocity of the transmitter and recepient is the same, will the apparent Doppler shift be zero?
- If the weight of both the transmitter and the recepient is the same, will the apparent red/blue shift be zero?
there are two formulas, one for the gravitational redshift and one for the doppler, and they each give you a ratio of wavelengths----you can just multiply the two factors together.
"superposition" suggests adding two effects together, so it is not quite the right word-----combining the effects in this case means to multiply the two factors
if the gravitational effect is to extend wavelengths by a factor of 1.5
and on top of that the thing is speeding away so fast that the doppler effect extends wavelengths by a factor of 2
then the combined effect is to make the wavelengths 3 times longer (multiply them first by 1.5 and then by 2 gives the same as multiplying them by 3)
the two formulas are each for a factor 1+z
1 + z = received wvlngth/emitted wvlngth = extension factor
so if the factor by which wavelengths get longer is 1.5
then the "z" astronomers talk about is 0.5
it is just a custom that that they have to subtract 1.0 from the
ratio and call it "z"-----it is handy for some purposes
but the fundamental thing is the factor 1+z, and that is what the formulas give
doppler:
wavelngth ratio = 1+z = sqrt((1 + v/c)/(1 - v/c))
Here v is the radial velocity of one relative to the other (by convention, speed towards is negative, speed away is positive).
For small velocities the formula gives approximately 1 + v/c
grav:
wavelngth ratio = 1+z = 1/sqrt(1 - R
S/r)
Here r is the radius of the body and R
S is the Schwarzschild radius 2GM/c
2
WHAT YOU SAY IS TRUE except maybe for the way you said it. If someone one one planet sends a signal to someone on another planet, then the redshift of rising out of the potential well of the sender will just cancel the blueshift of falling down into the potential well of the receiver (if the planets are the same size and mass) other things being equal.
You said the same "weight" and I guess it would be a bit clearer to say the same size and density, but I understood what you meant it I believe its quite true
ABOUT THE DOPPLER if the two are not moving relative to each other----yes, what you say is true, the doppler shift would be zero. They can both be moving relative to some third-party observer but that does not matter. In the doppler formula the only thing that matters is the motion of one relative to the other