<p>Yes, there is a transverse relativistic Doppler shift. You can think of it as being caused by time dilation. <a href="https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect</a></p>

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

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

<p>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.</p>

CC BY-SA 4.0
2020-03-03T00:58:20.990Z
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
CC BY-SA 4.0
2020-03-03T01:00:09.287Z
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
CC BY-SA 4.0
2020-03-03T09:48:07.910Z
@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.
ProfRob
CC BY-SA 4.0
2020-03-03T23:45:37.690Z
Thanks, I'll go off in a cornder and m̶e̶d̶i̶t̶a̶t̶e̶ ̶o̶n̶  think about it further...
uhoh
CC BY-SA 4.0
2020-03-03T00:44:47.727Z
2020-03-03T09:46:05.600Z
2020-03-03T09:46:05.600Z
ProfRob
<p>Since an object moving perpendicular to a given "line of sight" has a constantly changing <strong>range</strong> , there is a Doppler shift,  blue when approaching and red when leaving.   </p>

<p>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) . </p>

CC BY-SA 4.0
2020-03-02T21:25:23.437Z
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
CC BY-SA 4.0
2020-03-03T07:49:23.147Z
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
CC BY-SA 4.0
2020-03-02T20:42:12.480Z
2020-03-02T20:42:12.480Z
Carl Witthoft
<p><a href="https://astronomy.stackexchange.com/a/34344/7982">This answer</a> 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.</p>

<p>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 <a href="https://en.wikipedia.org/wiki/Aberration_(astronomy)" rel="nofollow noreferrer">astronomical aberration</a>.  </p>

<p><strong>Question:</strong> 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 <span class="math-container">$|v/c|$</span> is it always a red shift and simply</p>

<p><span class="math-container">$$\Delta f/f = -\frac{v^2}{c^2}?$$</span></p>

CC BY-SA 4.0
2020-03-02T09:39:01.417Z
[How to best format the question mark that immediately follows an equation in a question sentence?](https://math.meta.stackexchange.com/q/31283/284619)
uhoh
CC BY-SA 4.0
2020-03-02T10:09:26.770Z
There should not be a shift that is not ultimately related to the distance. There is certainly a Doppler shift whatever small in aberration. Like for an airplane passing on front of you, if it's this that you mean.
Alchimista
CC BY-SA 4.0
2020-03-02T10:18:33.487Z
@Alchimista I mean a "frequency shift when moving perpendicular" so yes I think I'm asking exactly about "a Doppler shift whatever small in aberration" and looking forward to a posted answer about it. Thanks!
uhoh
CC BY-SA 4.0
2020-03-02T12:52:16.797Z
I might be thinking of the [Transverse Doppler effect](https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect)? For example [Are there any examples where the transverse doppler effect is applied in astronomy?](https://astronomy.stackexchange.com/q/28429/7982) and also [Time dilation on an object circling earth](https://astronomy.stackexchange.com/a/1534/7982)
uhoh
<redshift><doppler-effect>
Is there still a red shift when moving perpendicular to the direction of incidence?
uhoh
# Is there still a red shift when moving perpendicular to the direction of incidence?

(Asked by: username_0 on 2020-03-02)

[This answer](https://astronomy.stackexchange.com/a/34344/7982) 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](https://en.wikipedia.org/wiki/Aberration_(astronomy)).

**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

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

**Tags:** redshift, doppler-effect
> **Comments:**
> **username_0**: [How to best format the question mark that immediately follows an equation in a question sentence?](https://math.meta.stackexchange.com/q/31283/284619)  
> **username_1**: There should not be a shift that is not ultimately related to the distance. There is certainly a Doppler shift whatever small in aberration. Like for an airplane passing on front of you, if it's this that you mean.  
> **username_0**: @Alchimista I mean a "frequency shift when moving perpendicular" so yes I think I'm asking exactly about "a Doppler shift whatever small in aberration" and looking forward to a posted answer about it. Thanks!  
> **username_0**: I might be thinking of the [Transverse Doppler effect](https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect)? For example [Are there any examples where the transverse doppler effect is applied in astronomy?](https://astronomy.stackexchange.com/q/28429/7982) and also [Time dilation on an object circling earth](https://astronomy.stackexchange.com/a/1534/7982)  

---

## 2 answers

### Answer by username_2 on 2020-03-03. Score: 4 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  
> **username_0**: 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.  
> **username_2**: @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.  
> **username_0**: Thanks, I'll go off in a cornder and m̶e̶d̶i̶t̶a̶t̶e̶ ̶o̶n̶  think about it further...  

---

### Answer by username_3 on 2020-03-02. Score: 2

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) .

> **Comments:**
> **username_4**: 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.  
> **username_1**: 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.  

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