<p>The question doesn't specify the distance and relative position between the star and the observer. It might be relevant to resolve the apparent contradiction.</p>
<p>The reason is that you have to consider that not every ray emitted from the star will hit the observer. In order to hit, not only the ray must be emitted at the right direction, but also at the right time, otherwise it will miss the moving observer.</p>
<p>Let's say that at <span class="math-container">$t=t'=0$</span> the star is located at <span class="math-container">$(x=0,z=-d)$</span> and the observer is at <span class="math-container">$(x=0, z=0)$</span>. It emits the first ray. At time <span class="math-container">$t={d \over c}$</span> the ray has reached <span class="math-container">$(x=0,z=0)$</span>, but the observer is situated at <span class="math-container">$(x=vd/c,z=0)$</span>. So the ray misses the observer and no aberration occurs.</p>
<p>Consider now the second ray. It is emitted at time <span class="math-container">$t=t'=0$</span>. At time <span class="math-container">$t=\gamma {d \over c}$</span> the ray is at <span class="math-container">$(x=v\gamma d/c, z=0)$</span>, and the observer is also at <span class="math-container">$(x=v\gamma d/c, z=0)$</span>. So the ray hits the observer, as shown in the pictures.</p>
<p><a href="https://i.sstatic.net/7d786.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7d786.png" alt="enter image description here" /></a></p>
<p><a href="https://i.sstatic.net/8atVl.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8atVl.png" alt="enter image description here" /></a></p>
<p>As you proved, the moving observer sees the star right along the axis <span class="math-container">$z'$</span>. While another observer, which is standing at position <span class="math-container">$(x=v\gamma d/c, z=0)$</span>, not moving with respect to the star, sees the ray at an angle <span class="math-container">$\alpha$</span>. This fact is the aberration.</p>
<p>You can see it also in another way. Imagine to be the moving observer. At time <span class="math-container">$t=\gamma {d \over c}$</span> you look at the star and see it right on the <span class="math-container">$z'$</span> axis. In a very short time, you decide to decelerate and stop with respect to the star. During the brief deceleration you see the star shifting in the sky until, when you are still, the star is at an angle <span class="math-container">$\alpha$</span>.</p>
<p>The apparent paradox of the OP is resolved because the two rays cannot both reach the observer if they are emitted at the same time.</p>
<p>If instead they are emitted at different times, the can both reach the observer, but there is no contradiction. The observer is expected to sees the star in two different directions, because it has moved between the detection of the two rays.</p>

CC BY-SA 4.0
2021-08-16T17:58:10.707Z
Please check again the definitions in my question: the primed observer is located on the z'-axis, so if a light signal is sent from the origin, he could only see it if the signal propagates on the z'-axis. Yet a  light signal travelling along the z-axis in the unprimed frame does not do this according to the LT formula, so it misses the observer on the z'-axis altogether. A light signal emitted along $x=vt\: , z=ct/\gamma$ in the unprimed frame travels along the z'-axis however and thus will be seen without aberration by the primed observer.(the z'-axis being the reference direction)..
Thomas
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2021-08-16T18:17:07.207Z
@Thomas you are right, you had specified that the observer is in the z' axis. I will try to understand better what you are saying and then edit my answer accordingly. Do you agree with me that the two rays cannot both reach the observer? Are you saying that the first ray misses, while the second ray hits, but the observer doesn't see the aberration?
Prallax
CC BY-SA 4.0
2021-08-16T18:28:16.660Z
Yes, exactly, that's what I am saying. By the way, your own equations seem to contain many typos. You have for instance both the star and observer at the origin (x=0, z=0). So it is not really clear what your argument is.
Thomas
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2021-08-16T18:44:17.187Z
Thank you for spotting the typo, I will edit the answer and add also a picture, to make things clearer.
Prallax
CC BY-SA 4.0
2021-08-16T20:00:50.493Z
@Thomas edited, let me know what you think
Prallax
CC BY-SA 4.0
2021-08-19T21:27:59.127Z
OK, it is lot clearer now and I can see where you are coming from. I have the impression though that your argument has more to to with the star's parallax (i.e. positional effects) than with aberration (velocity effects). For instance, aberration should be independent of the distance of the star, but in your example the angle $\alpha$ changes its distance if seen from the same position on the x-axis. Also, note by the way that for a star with a distance of let's say 100 lightyears, the earth would never get to the required point on the x-axis, which would be way outside the solar system.
Thomas
CC BY-SA 4.0
2021-08-15T21:25:28.050Z
2021-08-16T20:00:24.957Z
2021-08-16T20:00:24.957Z
Prallax
<p>The aberration of starlight is claimed to result from the Lorentz transformation of Special Relativity, but I am having problems to understand this.  Assume we have two reference frames <span class="math-container">$x,y,z$</span> (the reference frame of the star) and <span class="math-container">$x',y',z'$</span> (the reference frame of the observer (assumed to be located on the <span class="math-container">$z'$</span> axis)). The reference frames move with constant velocity <span class="math-container">$v$</span> relative to each other along their <span class="math-container">$x$</span> axes. At <span class="math-container">$t=t'=0$</span>,  the origin of both reference frames shall coincide and at this moment a light ray sent out along the <span class="math-container">$z$</span> axis i.e. along
<span class="math-container">$$x=0$$</span>
<span class="math-container">$$y=0$$</span>
<span class="math-container">$$z=ct$$</span></p>
<p>According to the Lorentz transformation
<span class="math-container">$$x'=\gamma (x-vt)$$</span>
<span class="math-container">$$y'=y$$</span>
<span class="math-container">$$z'=z$$</span>
<span class="math-container">$$t'=\gamma (t-vx/c^2)$$</span></p>
<p>this light ray travels in the observer's reference frame along the coordinates
<span class="math-container">$$x'=-\gamma vt=-vt'$$</span>
<span class="math-container">$$y'=y=0$$</span>
<span class="math-container">$$z'=z=ct=ct'/\gamma$$</span></p>
<p>This is generally interpreted to the effect that the apparent position of the star shifts by an angle <span class="math-container">$\alpha$</span> where
<span class="math-container">$$tan(\alpha)=(-x')/z'=\gamma v/c$$</span></p>
<p>However, the star emits light rays in all directions, therefore also in the direction
<span class="math-container">$$x=vt$$</span>
<span class="math-container">$$y=0$$</span>
<span class="math-container">$$z=ct/\gamma$$</span></p>
<p>and this ray travels in the observer's frame along
<span class="math-container">$$x'=0$$</span>
<span class="math-container">$$y'=0$$</span>
<span class="math-container">$$z'=z=ct/\gamma=ct'$$</span></p>
<p>The observer on the <span class="math-container">$z'$</span> axis sees the star therefore exactly at the origin i.e. the same position as for the stationary case. The only difference is that the star appears to be rotated about its axis, but there should not be any positional change of the star due to this.</p>
<p>Is there any way to resolve this apparent contradiction?</p>

CC BY-SA 4.0
2021-08-14T22:10:01.520Z
If you've made some progress on this challenging problem you can post it as an answer, even if incomplete. Just write "partial answer" at the top. Or if you're not so sure yet, as an update to the question.
uhoh
CC BY-SA 4.0
2021-08-15T14:09:34.783Z
The second ray of light that the star emits is in the x-z plane of the star (unprimed frame). So why would you expect the observer to see this light ray traveling in the same direction as the original light ray, which was only in the z direction of the star (unprimed frame)? What is the contradiction here? Can you make it more clear?
Daddy Kropotkin
CC BY-SA 4.0
2021-08-15T15:12:21.773Z
@DaddyKropotkin The point here is that a ray sent along the z-axis in the unprimed frame does not travel along the z'-axis in the primed frame as it has an x'-component. However a different ray sent in the x-z plane in the unprimed frame *does* travel along z'-axis in the primed frame. This means a moving observer does actually not see the star at a different position but he merely sees a rotated star.
Thomas
CC BY-SA 4.0
2021-08-15T15:48:44.247Z
Does the photon that is emitted first have a velocity with x and z components?
Daddy Kropotkin
CC BY-SA 4.0
2021-08-15T15:54:03.947Z
@DaddyKropotkin Look at the transformation equations I wrote in my question: the first photon has only a z-component in the unprimed frame (observer at rest relatively to the star) but an additional x-component in the primed frame (moving observer). It is the other way around for the second photon.
Thomas
CC BY-SA 4.0
2021-08-15T17:45:55.887Z
"Both reference frames move with constant velocity $v$ relative to each other along their x axes." This sentence is confusing to me: is one frame moving at velocity $v$ with respect to the other? OR are both frames (primed and unprimed) moving at speed $v$ and thus are at rest with respect to each other? The math you've written suggests the former.
Daddy Kropotkin
CC BY-SA 4.0
2021-08-15T17:56:39.827Z
@DaddyKropotkin The two reference frames move with velocity $v$ relatively to each other.
Thomas
CC BY-SA 4.0
2021-08-15T19:03:44.180Z
So when you say "but he merely sees a rotated star," you're referring to the formula $\tan\alpha$, correct?
Daddy Kropotkin
<observational-astronomy><positional-astronomy><special-relativity>
Stellar Aberration and Lorentz Transformation
Thomas
# Stellar Aberration and Lorentz Transformation

(Asked by: username_0 on 2021-01-03)

The aberration of starlight is claimed to result from the Lorentz transformation of Special Relativity, but I am having problems to understand this. Assume we have two reference frames $x,y,z$ (the reference frame of the star) and $x',y',z'$ (the reference frame of the observer (assumed to be located on the $z'$ axis)). The reference frames move with constant velocity $v$ relative to each other along their $x$ axes. At $t=t'=0$, the origin of both reference frames shall coincide and at this moment a light ray sent out along the $z$ axis i.e. along
$$x=0$$
$$y=0$$
$$z=ct$$

According to the Lorentz transformation
$$x'=\gamma (x-vt)$$
$$y'=y$$
$$z'=z$$
$$t'=\gamma (t-vx/c^2)$$

this light ray travels in the observer's reference frame along the coordinates
$$x'=-\gamma vt=-vt'$$
$$y'=y=0$$
$$z'=z=ct=ct'/\gamma$$

This is generally interpreted to the effect that the apparent position of the star shifts by an angle $\alpha$ where
$$tan(\alpha)=(-x')/z'=\gamma v/c$$

However, the star emits light rays in all directions, therefore also in the direction
$$x=vt$$
$$y=0$$
$$z=ct/\gamma$$

and this ray travels in the observer's frame along
$$x'=0$$
$$y'=0$$
$$z'=z=ct/\gamma=ct'$$

The observer on the $z'$ axis sees the star therefore exactly at the origin i.e. the same position as for the stationary case. The only difference is that the star appears to be rotated about its axis, but there should not be any positional change of the star due to this.

Is there any way to resolve this apparent contradiction?

**Tags:** observational-astronomy, positional-astronomy, special-relativity
> **Comments:**
> **username_1**: If you've made some progress on this challenging problem you can post it as an answer, even if incomplete. Just write "partial answer" at the top. Or if you're not so sure yet, as an update to the question.  
> **username_2**: The second ray of light that the star emits is in the x-z plane of the star (unprimed frame). So why would you expect the observer to see this light ray traveling in the same direction as the original light ray, which was only in the z direction of the star (unprimed frame)? What is the contradiction here? Can you make it more clear?  
> **username_0**: @DaddyKropotkin The point here is that a ray sent along the z-axis in the unprimed frame does not travel along the z'-axis in the primed frame as it has an x'-component. However a different ray sent in the x-z plane in the unprimed frame *does* travel along z'-axis in the primed frame. This means a moving observer does actually not see the star at a different position but he merely sees a rotated star.  
> **username_2**: Does the photon that is emitted first have a velocity with x and z components?  
> **username_0**: @DaddyKropotkin Look at the transformation equations I wrote in my question: the first photon has only a z-component in the unprimed frame (observer at rest relatively to the star) but an additional x-component in the primed frame (moving observer). It is the other way around for the second photon.  
> **username_2**: "Both reference frames move with constant velocity $v$ relative to each other along their x axes." This sentence is confusing to me: is one frame moving at velocity $v$ with respect to the other? OR are both frames (primed and unprimed) moving at speed $v$ and thus are at rest with respect to each other? The math you've written suggests the former.  
> **username_0**: @DaddyKropotkin The two reference frames move with velocity $v$ relatively to each other.  
> **username_2**: So when you say "but he merely sees a rotated star," you're referring to the formula $\tan\alpha$, correct?  

---

## 1 answers

### Answer by username_3 on 2021-08-15. Score: 1

The question doesn't specify the distance and relative position between the star and the observer. It might be relevant to resolve the apparent contradiction.

The reason is that you have to consider that not every ray emitted from the star will hit the observer. In order to hit, not only the ray must be emitted at the right direction, but also at the right time, otherwise it will miss the moving observer.

Let's say that at $t=t'=0$ the star is located at $(x=0,z=-d)$ and the observer is at $(x=0, z=0)$. It emits the first ray. At time $t={d \over c}$ the ray has reached $(x=0,z=0)$, but the observer is situated at $(x=vd/c,z=0)$. So the ray misses the observer and no aberration occurs.

Consider now the second ray. It is emitted at time $t=t'=0$. At time $t=\gamma {d \over c}$ the ray is at $(x=v\gamma d/c, z=0)$, and the observer is also at $(x=v\gamma d/c, z=0)$. So the ray hits the observer, as shown in the pictures.

[![enter image description here](https://i.sstatic.net/7d786.png)](https://i.sstatic.net/7d786.png)

[![enter image description here](https://i.sstatic.net/8atVl.png)](https://i.sstatic.net/8atVl.png)

As you proved, the moving observer sees the star right along the axis $z'$. While another observer, which is standing at position $(x=v\gamma d/c, z=0)$, not moving with respect to the star, sees the ray at an angle $\alpha$. This fact is the aberration.

You can see it also in another way. Imagine to be the moving observer. At time $t=\gamma {d \over c}$ you look at the star and see it right on the $z'$ axis. In a very short time, you decide to decelerate and stop with respect to the star. During the brief deceleration you see the star shifting in the sky until, when you are still, the star is at an angle $\alpha$.

The apparent paradox of the OP is resolved because the two rays cannot both reach the observer if they are emitted at the same time.

If instead they are emitted at different times, the can both reach the observer, but there is no contradiction. The observer is expected to sees the star in two different directions, because it has moved between the detection of the two rays.

> **Comments:**
> **username_0**: Please check again the definitions in my question: the primed observer is located on the z'-axis, so if a light signal is sent from the origin, he could only see it if the signal propagates on the z'-axis. Yet a  light signal travelling along the z-axis in the unprimed frame does not do this according to the LT formula, so it misses the observer on the z'-axis altogether. A light signal emitted along $x=vt\: , z=ct/\gamma$ in the unprimed frame travels along the z'-axis however and thus will be seen without aberration by the primed observer.(the z'-axis being the reference direction)..  
> **username_3**: @Thomas you are right, you had specified that the observer is in the z' axis. I will try to understand better what you are saying and then edit my answer accordingly. Do you agree with me that the two rays cannot both reach the observer? Are you saying that the first ray misses, while the second ray hits, but the observer doesn't see the aberration?  
> **username_0**: Yes, exactly, that's what I am saying. By the way, your own equations seem to contain many typos. You have for instance both the star and observer at the origin (x=0, z=0). So it is not really clear what your argument is.  
> **username_3**: Thank you for spotting the typo, I will edit the answer and add also a picture, to make things clearer.  
> **username_3**: @Thomas edited, let me know what you think  
> **username_0**: OK, it is lot clearer now and I can see where you are coming from. I have the impression though that your argument has more to to with the star's parallax (i.e. positional effects) than with aberration (velocity effects). For instance, aberration should be independent of the distance of the star, but in your example the angle $\alpha$ changes its distance if seen from the same position on the x-axis. Also, note by the way that for a star with a distance of let's say 100 lightyears, the earth would never get to the required point on the x-axis, which would be way outside the solar system.  

---

