Question on Relativistic Doppler Shift

arunma
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I have a quick relativity question. Say an observer is moving at a relativistic velocity with respect to some light source that emits light at one particular frequency (the observer's path does not intersect the emitter). At the point of closest approach, will light from the emitter be Doppler shifted in the observer's reference frame? To illustrate what I'm talking about, I've attached an image, where the observer's velocity is parallel to the x-axis, and at closest approach, its separation from the emitter is parallel to the y-axis.

Anyway, I'm tempted to say that there's no Doppler shift, since the emitter's velocity with respect to the observer is parallel to the observer's velocity. But I'm told that this isn't correct. Does anyone know what's going on here?
 

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The relativistic Doppler shift is omega'=gamma(omega+v.k),
where v is the source velocity vector in the oberver's rest frame,
w is the angular frequency and k the wave vector of the light
in the source rest frame.
Even when v and k are perpendicular so v.k=0,
there is a Doppler shift, called the "transverse Doppler shift".
This is derived in many textbooks.
 
arunma said:
Anyway, I'm tempted to say that there's no Doppler shift, since the emitter's velocity with respect to the observer is parallel to the observer's velocity. But I'm told that this isn't correct. Does anyone know what's going on here?
Even when the relative velocity of source to observer is perpendicular (transverse) to the line between them, one still has an effect due to time dilation. As Meir Achuz says, this is called the transverse Doppler effect.
 
Thanks, I'll look up the transverse Doppler shift.
 
Look it up in a textbook. Wikipedia is full of confusion on this
(and other things too).
 
I don't have a textbook to look it up in, but i was thinking that time in the moving object is going slower (assuming that the light source is the reference frame). Therefore, you would have more periods (?) passing each second in the moving object than at the source. Therefore, a doppler shift towards higher frequency. However, when you are moving away from an object, there is a doppler shift down. So, is there a point at which total doppler shift is zero?

and also what about aberration? is the doppler shift based on the apparent source of light or the actual (relative to the moving object)?aberration, time dialation, doppler shift, so many things to keep track of...
 
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mr200backstrok said:
I don't have a textbook to look it up in, but i was thinking that time in the moving object is going slower (assuming that the light source is the reference frame). Therefore, you would have more periods (?) passing each second in the moving object than at the source. Therefore, a doppler shift towards higher frequency.

No, it is the observer's reference frame that we are dealing with here. According to time dilation, when a clock and observer are moving relative to each other, the clock runs slower in the reference frame in which the observer is at rest (which is often called the "observer's reference frame"), than in the reference frame in which the clock is at rest (which is often called the "clock's reference frame"). Therefore in the "observer's reference frame" in a given amount of time (as measured by his own clock), fewer periods elapse on a moving clock or moving light source or whatever, and the observed frequency is lower than the frequency that would be observed by someone who is moving along with the source (and for whom the source is at rest).
 

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