5
$\begingroup$

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?

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – uhoh
    Aug 14, 2021 at 22:10
  • $\begingroup$ 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? $\endgroup$ Aug 15, 2021 at 14:09
  • $\begingroup$ @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. $\endgroup$
    – Thomas
    Aug 15, 2021 at 15:12
  • $\begingroup$ Does the photon that is emitted first have a velocity with x and z components? $\endgroup$ Aug 15, 2021 at 15:48
  • $\begingroup$ @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. $\endgroup$
    – Thomas
    Aug 15, 2021 at 15:54

1 Answer 1

1
$\begingroup$

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

enter image description here

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.

$\endgroup$
6
  • $\begingroup$ 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).. $\endgroup$
    – Thomas
    Aug 16, 2021 at 17:58
  • $\begingroup$ @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? $\endgroup$
    – Prallax
    Aug 16, 2021 at 18:17
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Aug 16, 2021 at 18:28
  • $\begingroup$ Thank you for spotting the typo, I will edit the answer and add also a picture, to make things clearer. $\endgroup$
    – Prallax
    Aug 16, 2021 at 18:44
  • $\begingroup$ @Thomas edited, let me know what you think $\endgroup$
    – Prallax
    Aug 16, 2021 at 20:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .