Is the Doppler Effect Formula Different for Sources in Different Regions?

[etotheipi]

If the source is at $(t,0)$ in $S$ and the receiver is at $(t',0)$ in $S'$ which moves at $\beta_x$ w.r.t. $S$, then by considering two crests at $(0,0)$ and $(T_s, 0)$ in the source frame $S$ and transforming these events into $S'$ we can derive that $\lambda_{r} = \lambda_{s} \sqrt{\frac{1+\beta_x}{1-\beta_x}}$.

That's the formula that I see everywhere, and it seems okay if the origin of $S'$ is in the region $x>0$, but if the origin of $S'$ is in the region $x<0$ then wouldn't we have $\lambda_{r} = \lambda_{s} \sqrt{\frac{1-\beta_x}{1+\beta_x}}$, because of the left-right symmetry?

I've never seen this mentioned anywhere, so I wondered if someone could clarify? Thanks

 

[Nugatory]

Changing the origin desn't change the sign in front of $\beta_x$, but when you choose to place the location of of the source and receiver at the respective origins of the two frames, you are also choosing whether the source and receiver are moving towards or away from one another at time $t=0$.h

 

[PeterDonis]

If the source is at $(0,0)$ in $S$ and the receiver is at $(0,0)$ in $S'$

Neither of these make sense: $(0, 0)$ is a point in spacetime, not a worldline. I think what you mean here is that the source is at $x = 0$ in $S$ and the receiver is at $x' = 0$ in $S'$.

 

[etotheipi]

Changing the origin desn't change the sign in front of $\beta_x$, but when you choose to place the location of of the source and receiver at the respective origins of the two frames, you are also choosing whether the source and receiver are moving towards or away from one another at time $t=0$.h

But if $\beta_x < 0$ and the origin of $S'$ has an $x$-coordinate less than zero, then the source and receiver are separating and the equation should be $\lambda_{r} = \lambda_{s} \sqrt{\frac{1-\beta_x}{1+\beta_x}}$, right?

Neither of these make sense: $(0, 0)$ is a point in spacetime, not a worldline. I think what you mean here is that the source is at $x = 0$ in $S$ and the receiver is at $x' = 0$ in $S'$.

You're right, yes, sorry

 

[Ibix]

In this case, $\beta$ is the radial speed of the source with respect to the receiver. The sign convention affects whether you consider speed to be positive when the two are approaching or receding from each other.

I have to say it's one of the places I don't bother remembering a convention from one problem to the next. I either use what I'm given or just write one formula or other and then deduce which convention I'm using post hoc.

 

[PeterDonis]

I have to say it's one of the places I don't bother remembering a convention from one problem to the next. I either use what I'm given or just write one formula or other and then deduce which convention I'm using post hoc.

My usual strategy in these cases is to, in the words of John Baez, "attempt to make an even number of sign errors".

 

[etotheipi]

In this case, $\beta$ is the radial speed of the source with respect to the receiver. The sign convention affects whether you consider speed to be positive when the two are approaching or receding from each other.

I have to say it's one of the places I don't bother remembering a convention from one problem to the next. I either use what I'm given or just write one formula or other and then deduce which convention I'm using post hoc.

Fair enough, it's just I thought that $\beta_x$ is the signed $x$ component of $\vec{\beta}$. But in that case maybe it has a slightly different meaning here. Thanks

 

[Ibix]

My usual strategy in these cases is to, in the words of John Baez, "attempt to make an even number of sign errors".

As an undergraduate we nicknamed these "Steve Gull errors". He'd taken on the EM course he taught at short notice and his notes were rather hurriedly written and incompletely proof read - but he knew perfectly well where the derivations should lead and the errors tended to mysteriously correct themselves a line or two before the final result...

 

[Ibix]

Fair enough, it's just I thought that $\beta_x$ is the signed $x$ component of $\vec{\beta}$. But in that case maybe it has a slightly different meaning here. Thanks

Doppler has radial and tangential effects, not $x$, $y$, and $z$ effects, so it mostly makes sense to think about polar coordinates centered on the receiver for Doppler. My guess is that whatever you are reading is considering a specific scenario where the emitter happens to be crossing the $x$ axis with velocity towards or away from the receiver - probably so they can easily apply the Lorentz transforms in their standard form.

 

[etotheipi]

Doppler has radial and tangential effects, not $x$, $y$, and $z$ effects, so it mostly makes sense to think about polar coordinates centered on the receiver for Doppler.

Yeah, a little bit of extra reading tells me that apparently the constant of proportionality in the general case is $\gamma (1+\beta_r) = \frac{1+\vec{\beta}\cdot \hat{r}}{\sqrt{1-\vec{\beta}^2}}$ if the polar coordinate system is established with an origin at the receiver. But this equation won't be the same if we switched to the source coordinate system, because that pesky aberration means $\cos{\theta_r} \neq \cos{\theta_s}$.

I think that's probably as far as I'll take this today, I'm too tired to look at the circular motion case right now... thanks for the help ☺

 

[vanhees71]

It's more clear if you formulate the Doppler effect in terms of the frequency and wave number, making use of the fact that
$$(k^{\mu})=\begin{pmatrix} k \\ \vec{k} \end{pmatrix}$$
is a four-vector with $k=|\vec{k}|=\omega/k$.

Say, these are the components in the rest frame of the light source. For an observer moving with four-velocity
$$u^{\mu}=\gamma (1,\vec{\beta}), \quad \gamma=\sqrt{1-\vec{\beta}^2}$$
the frequency is given by
[EDIT: Corrected in view of #12]
$$k'=\omega'/c =k_{\mu} u^{\mu} =\gamma (k-\vec{v} \cdot \vec{k}/c).$$
Now take the extreme cases

(a) $\vec{v}=c \vec{\beta}$ is in the direction of $\vec{k}$. Then you have
$$k'=\gamma k (1-\beta) = k \sqrt{\frac{1-\beta}{1+\beta}},$$
i.e., you get a red-shift, because you are moving away from the light source.

(b) If $\vec{\beta}$ points in the opposite direction you have
$$k'=\gamma k(1+\beta) = k \sqrt{\frac{1+\beta}{1-\beta}},$$
i.e., you have a blue shift, because you move towards the light source.

(c) A really relativistic effect is that there's a "transverse Doppler shift", i.e., also for $\vec{\beta} \perp \vec{k}$,
$$k'=\gamma k =\frac{k}{\sqrt{1-\beta^2}},$$
i.e., you always get a red shift, which is due to time dilation between the rest frame of the light source and the observer's frame.

 

[etotheipi]

Thank you @vanhees71, I haven't seen that approach before. I just had one question; I thought the phase velocity was given by $\frac{\omega}{k}$, and wouldn't that mean that the four vector is$$(k^{\mu})=\begin{pmatrix} k \\ \vec{k} \end{pmatrix} = \begin{pmatrix} \frac{\omega}{c} \\ \vec{k} \end{pmatrix}$$with $k=|\vec{k}|=\frac{\omega}{c}$? Sorry if I am missing something

 

[vanhees71]

Of course, you are right. I always get the $c$'s wrong, when typing to quickly :-((. I'll correct the original post.