# Is there still a red shift when moving perpendicular to the direction of incidence?

This answer suggests that when there is zero radial velocity... there will be no Doppler shift but it's not likely to be intended as precise and absolute.

I remember reading about something like this in a post here but I can't find it. While to first order there is no frequency shift when moving perpendicular (we wouldn't expect there to be be due to symmetry i.e. which way? Up or down?) I think there is a second order effect which is a Doppler shift associated with the small change in angle due to astronomical aberration.

Question: Is this a thing? If so, what would be an expression for the frequency shift and what is it called? If the magnitude of each effect is $$|v/c|$$ is it always a red shift and simply

$$\Delta f/f = -\frac{v^2}{c^2}?$$

Yes, there is a transverse relativistic Doppler shift. You can think of it as being caused by time dilation. https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

There can be a redshift or a blueshift depending on when, where and who does the measurement.

e.g. a receiver with a source going around it in a circular orbit. The receiver sees a lower frequency (redshift), by a factor $$\gamma = (1 -v^2/c^2)^{-0.5}$$. On the other hand, a receiver orbiting the source would receive a blueshifted signal by the same factor. When $$v \ll c$$ then $$\gamma \simeq 1 + \frac{v^2}{2c^2}\ ,$$ so whilst the standard (longitudinal) Doppler shift is of order $$v/c$$, the transverse Dopper shift is of order $$v^2/c^2$$.

NB: This scenario is chosen so that the relative motion of the source and receiver are perpendicular to the line between them. All other scenarios are complicated by the usual Doppler shift you would see because there is a component of the velocity along the line joining the source and receiver.

• Thanks, but is it possible to explain this without invoking an orbital configuration, or is that a necessary aspect of the effect you're describing? – uhoh Mar 3 at 0:58
• unrelated comment: I'm still wondering about motion that's simply perpendicular to the direction of a source and astronomical aberration causing it to appear to have a slight radial component and then a shift, based on the finite speed of light but not necessarily on relativistic effects. And I guess that would always be a blue shift and not red. – uhoh Mar 3 at 1:00
• @uhoh The exampe I chose has motion that is "simply perpendicular to the direction of a source" and is thus a purely relativistic effect. All other examples do not share this property. – Rob Jeffries Mar 3 at 9:48
• Thanks, I'll go off in a cornder and m̶e̶d̶i̶t̶a̶t̶e̶ ̶o̶n̶ think about it further... – uhoh Mar 3 at 23:45

Since an object moving perpendicular to a given "line of sight" has a constantly changing range , there is a Doppler shift, blue when approaching and red when leaving.

The shift drops to zero at the point of crossing the line of sight because at that instant the radial speed is zero, as you suggested. So, the general magnitude is calculated using trigonometry: the range is effectively the hypotenuse and the cosine leg is the distance to the crossing point. The sine leg is the distance along the object's line of travel between the object and the crossing point. Differentiate to get the delta(range)/delta(time) .

• Very true, although as the Wikipedia article on the transverse Doppler effect explains, this gets slightly little complicated due to the finite time it takes for the light to travel from the source to the observer. – PM 2Ring Mar 2 at 21:25
• This is what I referred to in my comment. Relativistic or not, I see it just as a Doppler shift due to the fact that the distance between source and receiver changes. – Alchimista Mar 3 at 7:49