How Does the Relativistic Doppler Effect Apply on a Rotating Disk?

[kreil]

Homework Statement

A source and detector are placed a certain angle \phi apart on the edge of a rotating disk. The source emits radiation in all directions at a frequency \omega_* in its instantaneous rest frame. What frequency is the radiation detected at? (Hint: Little information is provided because little information is needed.)

Homework Equations

If the source moves away from the detector with speed V and emits photons of frequency w at an angle a', the detected frequency is,

\omega' = \omega \frac{\sqrt{1-V^2}}{1-Vcos \alpha'}

The Attempt at a Solution

It seems to me that in the frame of the detector, the source is at rest (V=0) since they are both moving with the same angular speed. Thus, the equation above yields w'=w. I am doubting the problem is this simple, and I'm not sure how else to interpret the situation.

 

[diazona]

Yes, the source and detector are moving with the same angular speed, but are they moving with the same (linear) velocity? Remember that it's velocity that really defines a reference frame.

Besides that, you could account for the time it takes light to travel from the source to the detector; specifically, the detector will have moved slightly around the circle by the time the radiation reaches it. But perhaps you're not required to take the calculation to that level of detail.

 

[kreil]

The magnitude of their linear velocities would be the same, but the direction of the vectors would not be.

I'm not sure how to calculate the directions of these vectors in a useful way (incorporating the angle, phi)

I feel as though since the source itself is moving, in its rest frame relativistic beaming still occurs and it emits higher energy radiation in the direction of motion. However, the problem doesn't state whether the detector is in front of or behind the source in the direction of motion, which seems like a crucial aspect of the problem..

 

[diazona]

I feel as though since the source itself is moving, in its rest frame

Do you notice the contradiction in what you just said there?

Anyway, I would actually suggest considering this problem from the lab frame at first. Take one particular instant of time, at which the source has a particular velocity and the detector has a particular velocity. You can figure out how those velocities are related, although it'll take some careful thinking about the sizes of various angles. I highly recommend drawing a diagram.

 

[94JZA80]

The magnitude of their linear velocities would be the same, but the direction of the vectors would not be.

hence, their angular velocities are not the same. the magnitude of velocity is not the same as velocity itself. the magnitude, or absolute value, of velocity is in fact speed, b/c it disregards direction. their angular speeds may be the same, but their directions at any given instant most surely are not the same. hence, their angular velocities are also not the same.

The Attempt at a Solution

It seems to me that in the frame of the detector, the source is at rest (V=0) since they are both moving with the same angular speed.

again, think of it in terms of instantaneous velocity. if your source and detector were mounted to the outside edge of a tire on a forward moving car, and the source led the detector by \phi = 90°, then at the instant that the source is at the top of the tire (and moving directly forward), the detector at that same instant is moving directly upward - this would be a lot easier to see with a diagram, but your imagination will have to make do. nevertheless, as you can see, although the source and detector are again moving at the same angular speed, they are never moving in the same direction at anyone instance, and so their angular velocities cannot be the same. i know this example is quite different from your problem, as a tire on a car will never approach relativistic velocities. but these objects don't necessarily need to be moving/rotating at relativistic velocities to show that their angular velocities are not the same.

i still do not know how to solve the problem, but i would highly suggest taking diazona's advice and tackle the problem using diragrams.

 

[kreil]

We never said the angular velocities were the same, we said the angular speeds are the same. Clearly the vectors point in different directions. At any rate, I'm still having trouble with this but here is where I am currently:

I noticed that since the length of the velocity vectors are the same, they sweep the same central angle \theta over any period of time, and thus

R tan \theta = |v_D|=|v_S|

If the source emits radiation at time t with frequency \omega^*[/tex] and velocity v_{rad}=\frac{\omega^* \lambda}{2 \pi}, then a short time later at time (t + dt), the source and detector have each moved a small distance ds along the arc of the disk, and <br /> <br /> ds = R d\theta \implies \omega = \frac{ds}{dt}=R \frac{d\theta}{dt}<br /> <br /> if the disk is rotating with angular speed omega. Since the source has moved a distance ds away from the released wave, the wavelength has been increased by this distance, or <br /> <br /> \lambda \rightarrow \lambda + ds<br /> <br /> which implies that the detected frequency will be lower than the original frequency, \omega^*.<br /> <br /> However, it seems all of this is useless since I can&#039;t plug any of it into my Doppler formula in OP

 

[kreil]

"I feel as though since the source itself is moving, in its rest frame relativistic beaming still occurs."

Do you notice the contradiction in what you just said there?

what I meant to say was that in the rest frame of the source an observer sees the source at rest but the radiation itself moving.

 

[diazona]

We never said the angular velocities were the same, we said the angular speeds are the same. Clearly the vectors point in different directions. At any rate, I'm still having trouble with this but here is where I am currently:

I noticed that since the length of the velocity vectors are the same, they sweep the same central angle \theta over any period of time, and thus

R tan \theta = |v_D|=|v_S|

I'm not sure how you got to that formula, but the units aren't correct, so there must be something missing in there.

If the source emits radiation at time t with frequency \omega^*[/tex] and velocity v_{rad}=\frac{\omega^* \lambda}{2 \pi}, then a short time later at time (t + dt), the source and detector have each moved a small distance ds along the arc of the disk, and <br /> <br /> ds = R d\theta \implies \omega = \frac{ds}{dt}=R \frac{d\theta}{dt}<br /> <br /> if the disk is rotating with angular speed omega. Since the source has moved a distance ds away from the released wave, the wavelength has been increased by this distance, or <br /> <br /> \lambda \rightarrow \lambda + ds

<br /> The wavelength wouldn&#039;t change. The phase of the wave at the detector might change, but not the wavelength.<br /> <br /> Anyway, focus on the relative velocity between the source and detector at a single instant. You can assume that the disk is small enough that the travel time of the radiation between the source and detector is negligible.