# “Observed” angular velocity

I have a homework question based on the following diagram:

I need to find the angular velocity of the object as seen by an observer at the origin of the frame. The question says that the observed angular velocity is given by $\omega_0 = \Delta \varphi_0 / \Delta t$, where $\Delta t = t_2 - t_1$ and $\Delta \varphi_0 = \varphi_0(t_1^*) - \varphi_0(t_2^*)$ is the change of the angle $\varphi$, the polar angle which the object had at times $t_1^*$ and $t_2^*$. These are the times when the light detected by the observer (at $t_1$ and $t_2$) was emitted. The object is located at $(x_1, y_1)$ at time $t_1^*$ and it is located at $(x_2, y_2)$ at time $t_2^*$.

I started to answer this question by just finding an expression for $\Delta \varphi_0$ in terms of the coordinates ($\Delta \varphi_0 = \arctan{(y_1/x_1)} - \arctan{(y_2/x_2)}$) and dividing this by an expression for $\Delta t$. However, I was told that this was not correct. Apparently, I have to get an expression for the angular momentum in terms of the angle $\theta$ in the diagram. My lecturer told me that this was could be done with some basic geometry. I can't quite see what to do though. Also, I was told that the angle $\Delta \varphi_0$ was assumed to be very small. I have a feeling that what the lecturer wants us to derive is the expression $\omega = \frac{|\vec{v}|\sin{(\theta)}}{|\vec{r}|}$, but I can't see how to get this. Can anyone help?

Edit: This question is from a special relativity course, but I believe this can be answered without using any knowledge of special relativity. As I said, the lecturer told me that it was basically just geometry.

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## 2 Answers

Here's something that might help you out: this problem is motivated by astronomical observations of distant stars or galaxies. If you think about it, when you look through a telescope at an astronomical object, the only thing you can measure (without a spectroscope) is its angular position $\varphi_0$. So the way you obtain information about the object's motion is to record its angular position at one time $t_1$, and then later at another time $t_2$. You can then calculate the ratio $\Delta \varphi_0/\Delta t$, and since for an astronomical object $\Delta \varphi_0$ is very small (just think about how little the stars appear to move through the sky over any human-length timescale), this ratio is a pretty good approximation to the derivative $\frac{\mathrm{d}\varphi_0}{\mathrm{d}t} = \omega$, the observed angular velocity of the object.

However, what you see through a telescope is only part of the story. $\Delta \varphi_0/\Delta t$ is not necessarily the actual angular velocity of the object, because even though you observed two light rays a time $\Delta t$ apart, it doesn't mean that they were emitted $\Delta t$ apart. The only way the time between their emission would be the same as the time between their observation is if both light rays had traveled the same distance before reaching you. But for many astronomical objects, that's not the case. Since they have motion parallel to your line of sight as well as perpendicular to it, it's very likely that one of the light rays was emitted closer to you than the other, which means it spent less time in transit. You need to take that difference of transit time into account, which is why the answer you should be getting is a little more complicated than $\omega = \frac{v\sin\theta}{r}$.

As far as actually solving the problem, I'd suggest that you start by finding expressions for the the travel time of each light ray - that is, find an expression for $t_1 - t_1^*$, and similarly for $t_2 - t_2^*$. Also, because the angle $\Delta\varphi_0$ is so small, you can do calculations parallel to and perpendicular to the line of sight separately. I would definitely suggest that you ignore the $x$ and $y$ coordinates entirely - think of it the way someone would be observing this from Earth, where they wouldn't know anything about those coordinates.

By the way, here's the reason you're doing this in a special relativity course: it's a common calculation in astrophysics to work this problem backwards, i.e. to determine the actual velocity of the distant object given observations of $\Delta\varphi_0$ and $\Delta t$, along with spectroscopic data that gives you the parallel component of the velocity. In certain cases, the velocity works out to be faster than light if you use the naive formula $\frac{v\sin\theta}{r}$. You need to account for the difference in light travel time to show that these objects actually do obey special relativity.

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Thank you very much for this informative answer. I assume the travel time of each light ray would just be the distance from the object to the origin divided by $c$ (i.e. $d_1/c$ and $d_2/c$) but I'm not sure. Also, I was wondering if you could clarify the following comment: "because the angle $\Delta \varphi_0$ is so small, you can do calculations parallel to and perpendicular to the line of sight separately". What kind of calculations would these be? – saurs Jul 30 '11 at 10:20
You get to figure out what kind of calculations they are ;-) Although in retrospect, perhaps that particular hint caused more confusion than it's worth. All I'm saying is, there isn't anything particularly tricky about this problem. – David Zaslavsky♦ Jul 30 '11 at 19:51

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David Zaslavsky has laid out the main ideas very well. One additional hint about relating $\Delta\varphi_0$ to $\theta$: There are two right triangles in the diagram that share a common side. Express the length of that common side in terms of $\theta$ and given quantities, and also in terms of $\Delta\varphi_0$ and given quantities. (This is a slightly more explicit variation on David's suggestion about treating parallel and perpendicular components separately.) – Ted Bunn Jul 30 '11 at 22:26
@Ted Bunn: Thank you for that hint. Now that I think about it, however, I'm quite confused by the diagram. What does $\Delta r$ represent? If $\vec{r}$ is the position vector of the particle, then shouldn't $\Delta \vec{r}$ be pointing in the direction of $\vec{v}$? So I suppose the common side is given by $\Delta r / \cos{(\theta)}$. I'm not sure how to get an expression for that side in terms of $\Delta \varphi_0$ - would I use the sin rule or something like that? Sorry for all the questions, I know this should be an easy problem. – saurs Jul 31 '11 at 2:25

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$\Delta r$ is the change in the magnitude of the vector $\vec r$. It's not the magnitude of the vector difference -- that is, $\Delta r\ne |\Delta\vec r|$. If you decompose $\vec r$ into radial and tangential components, then $\Delta r$ is the difference in the radial components. For small $\Delta\varphi_0$, that's the length of the line segment shown in the diagram. – Ted Bunn Jul 31 '11 at 2:46
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If you split the velocity into two components (as show above), one being radial (I call it $v_r$) and the other in the hoop direction (I call it $v_{\varphi}$), then by inspection (trig properties) you can see the hoop velocity is $$v_{\varphi} = v\,\sin\theta$$ now by nature of angular velocity the hoop velocity is also $$v_{\varphi} = \omega\;r$$

Combine the two to get your answer.

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