# Doppler, mirror vs lamp

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1. Jul 25, 2015

### Stephanus

Dear PF Forum,
I have a question again about Doppler effect in relativity.
Before I ask about mirror and lamp, I want to refresh the discussion again
The formula for Doppler factor for light is.
$f = \sqrt{\frac{1+v}{1-v}}f0$
Is this for the frequency of light or for the frequency of receiving?
Can anyone give me the answer, so I'll never misunderstood it again.
B moves toward A at 0.3c
Doppler factor is $\sqrt{\frac{1.3}{0.7}}=1.36$
B shines a red light 400 THz: https://en.wikipedia.org/wiki/Visible_spectrum
Will A see the light as somewhat violet, 742 THz?

2. Jul 25, 2015

### Mentz114

In the formula you wrote $f$ is the received frequency and $f0$ is the sending frequency if the receiver and sender are approaching ( moving towards each other ).

Yes, calculation looks OK. The light has been 'blue-shifted'.

No mirror is involved - the light is being emitted from the source and received without any reflection.

Last edited: Jul 25, 2015
3. Jul 25, 2015

### Stephanus

Thanks Mentz114
Ok, now this.
Let $k = \sqrt{\frac{1+V}{1-V}}$

A see the light blue-shifted, and re emitted the blue-shifted light to B.
Let's say this.
B sends an F frequency light to A. A see the light frequency as F*k.
Now A sends the light with $F*k$ frequency to B again.
B will see the light as $F*k^2$ is that right?
No wait,... if A moves, then every atom in A also dilated.
So actually A sends $\frac{F*k}{\gamma}$
So B will see the light as $\frac{F*k^2}{\gamma}$
No wait, ... the situation is similar. We should use the formula as it is, because A moves toward B and B moves toward A should use the same formula.
1. Is this right that B will see $F*k^2$ if A bounce the signal back to B as it is.
2. What if A bounces the signal not by lamp, but by mirror, what will B see?

Thanks for any help.

4. Jul 25, 2015

### PAllen

An ideal mirror will act like your re-emitter, and the light will be more and more blue shifted ... but the A and B relative speed will be decreasing as energy and momentum are transferred to the light at each reflection.

5. Jul 25, 2015

### Stephanus

Yes, that makes sense.
Light has momentum? Hmhh, reminds me to previous thread about photon with mass.
Okay, confirmation: So it's F*k, F*k2, F*k3, etc...? Please confirm.