# Relativistic doppler shift

1. Feb 15, 2017

### AishaGirl

1. The problem statement, all variables and given/known data
A probe is launched with velocity $v=0.8c$. A beacon emits a light with wavelength $\lambda=500nm$ in its rest frame. Years later the probe is located by NASA using a telescope, When they measure the light they find the wavelength $\lambda=500nm$ in their rest frame. Is this possible? What is the explanation for this observation?

2. Relevant equations
Dopple shift equation.

3. The attempt at a solution
Essentially the problem states that $v/v\prime = \lambda\prime/\lambda=1$ I can use the doppler shift equation to find the relationship between $\beta$ and $\theta$

I'm inclined to think that this is not possible and the light should be redshifted but after working through the question there seems to be some angle whereby the shift balances or cancels out and I don't understand why.

Using the doppler shift equation I can say that $\frac{\lambda\prime}{\lambda} = \frac{\sqrt{1-\beta^2}}{1+\beta \cos\theta} \implies \cos\theta=\frac{1}{\beta}\Big( \sqrt{1-b^2}-1\Big)$

It's clear that the square root term will always be less than or equal to 1 and so the right side will always be negative given that $\cos\theta < 0$ for $\pi/2 < \theta \leq \pi$

Then using the taylor expansion it becomes $\beta \rightarrow 0 \implies \cos\theta \approx \frac{1}{\beta}\Big( 1-\frac{1}{2}\beta^2-1\Big)=-\frac{\beta}{2}$

So there is a time when the probe travels nearly perpendicular to pi/2. If I just plug in $\beta=4/5$ I get $\theta=120^\circ$.

All well and good but I still fail to see where the speed of the probe is taken into account... I cannot see a time where the angle of a probe moving at 0.8c will ever balance to create no shift, the Earth is moving at a fraction of the speed of probe. I don't understand...

Last edited: Feb 15, 2017
2. Feb 15, 2017

### Andrew Mason

What effect would a change in the direction of the probes' relative velocity have on the wavelength of the light signal observed by NASA? What if it is moving on a path that is tangential to its displacement relative to the NASA observer?

AM

Last edited: Feb 15, 2017
3. Feb 15, 2017

### Staff: Mentor

What is your $\gamma$? You don't use it in the way it is typically used in special relativity (Lorentz factor).

If something is moving directly towards you, you'll see it blueshifted. If something moves directly away from you, you see it redshifted. As the shift is continuous, there has to be an angle where you see zero shift. It would require the probe to come closer to the receiver (the telescope), which doesn't really match the description (unless the probe is launched from Proxima Centauri or whatever).

That equation is good.

$\beta$ is large, you shouldn't use a taylor expansion. You also do not need it because you know its value.

4. Feb 15, 2017

### PeroK

If the spaceship was moving directly away, then the light would be red-shifted. If it turned round in a long loop and headed towards Earth, then it would be blue-shifted. As the change from red to blue should be continuous, then somewhere on this loop you would get the original wavelength.

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5. Feb 15, 2017

### AishaGirl

Yeah sure if it changes speed or direction then I would expect to see some change in its wavelength but there is no mention of this in the question... If the probe fell into orbit around a planet or something and its velocity was significantly reduced then I'd agree it might balance out.

*EDIT* Sorry the gammas should be lambdas (wavelength), been typing too many gammas recently, sorry.

6. Feb 15, 2017

### Staff: Mentor

Technically the probe could have been launched from somewhere else. Looks odd, but I would mention it as possibility.

7. Feb 15, 2017

### PeroK

The answer is simply that the probe has changed direction. It's no longer moving directly away from Earth.

8. Feb 15, 2017

### AishaGirl

OK thanks. So just the clarify if the probe continued its velocity of 0.8c away from Earth its wavelength will remain redshifted to an observer?

9. Feb 15, 2017

### PeroK

Yes, you can see that from your equations.