# Doppler shift clarification

Hello,

I am well and truly stumped on Doppler shift. The complete expression is

$$f'=f \gamma (1-\beta cos(\theta))$$

where:

$$\beta=v/c$$
$$v=source\ velocity$$
$$\gamma=\frac{1}{\sqrt{1-\beta^{2}}}$$
$$\theta$$ is the angle between received photons (lab frame) and the direction of motion of the star.

Now for a directly receding source, where $$\theta=180^{0}$$ this gives the shifted frequency to be, after moving around,

$$f'=f*\sqrt{\frac{1+\beta}{1-\beta}}$$.

Now in a lot of text books this is the other way round ie,

$$f'=f*\sqrt{\frac{1-\beta}{1+\beta}}$$.

Why is this is so?? It's been giving me a headache for ages trying to figure it out.

It is important to define correctly the angle theta. You could have an illujmiknating look at
Robert Resnick "Introduction to Special Relativity" John Wiley 1968 pp.82-87

For this entire post, assume that $\nu = 1$.

The reason for the discrepancy is that one case takes velocity to be negative when motion is in the direction away from the source:

$$v = -\frac{c}{2},$$
$$\nu^\prime = \nu \sqrt{\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}}},$$
$$\nu^\prime = \nu \sqrt{\frac{1.5}{0.5}},$$
$$\nu^\prime \approx 1.73.$$

However, the other case takes velocity to be positive when motion is in the direction away from the source:

$$v = \frac{c}{2},$$
$$\nu^\prime = \nu \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}},$$
$$\nu^\prime = \nu \sqrt{\frac{1.5}{0.5}},$$
$$\nu^\prime \approx 1.73.$$

I prefer to keep the velocity positive (a vector always has positive magnitude or length), which is how it works in the full-fledged version:

$$\nu^\prime = \nu \frac{1 - \cos \theta \cdot \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}}.$$

Using the parameters previously defined:

$$\theta = 180^\circ = \pi{\;}{\rm radians},$$

$$\cos \theta = -1,$$

$$v = \frac{c}{2},$$

it is found, algebraically, that:

$$\nu^\prime = \nu \frac{1 + \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} = \nu \frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}} = \nu \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}},$$

$$\nu^\prime \approx 1.73.$$

I find that it is sometimes best to just forget the fancy algebraic equalities and work only with the full-fledged equation. Having only one equation to remember seems easiest to me.

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You'll have to forgive me, I made a silly error. The angle $\theta = \pi$ means that the observer is moving directly toward the source, not away from it. I'm sure you probably caught that error.

Either way, the remainder of the post stands as is. Just replace any instances of "away" with "toward" and it should be all good.

This exemplifies the confusing nature of the limiting cases for me, and how only the full-fledged version really highlights the physical nature of the relativistic Doppler effect. ex: The numerator is simply a matter of a "cat-n-mouse" chase between the observer and the photon ($1 - \cos \theta \cdot \frac{v}{c}$), and the denominator is simply a matter of the observer's kinematic time dilation ($\sqrt{1 - \frac{v^2}{c^2}}$).

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Dale
Mentor
2021 Award
As shalayka mentioned, the difference is just a sign convention on if a positive v is towards or away. Whenever you make a calculation just check if you have a redshift for velocities away or a blueshift for velocities towards. If you got the wrong one then switch your sign and repeat, that's easier than trying to figure out which convention a given equation uses.

Doc Al
Mentor
I am well and truly stumped on Doppler shift. The complete expression is

$$f'=f \gamma (1-\beta cos(\theta))$$

where:

$$\beta=v/c$$
$$v=source\ velocity$$
$$\gamma=\frac{1}{\sqrt{1-\beta^{2}}}$$
$$\theta$$ is the angle between received photons (lab frame) and the direction of motion of the star.
In that form of the Doppler shift equation, the angle of the photons is measured in the source frame not the lab frame. The angle is with respect to the velocity of the observer (again, in the source frame). Thus a directly receding source will have $\theta=0$, not $\theta=180$.

Hello everyone,

Thank you all for your many and varied answers to my question (especially Shalayka, all that Latex must have taken you ages, always takes me an eon to do!), it's cleared that problem up for me.

Vuldoraq

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Hello everyone,

Thank you all for your many and varied answers to my question (especially Shalayka, all that Latex must have taken you ages, always takes me an eon to do!), it's cleared that problem up for me.

Vuldoraq

It's really nice to be able to help. Plus, this effect is probably my favourite part about special relativity, so it was fun.