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I'm researching applications of relativistic beaming and I want to derive a formula for the aberration effect but I am stuck (I am off by a factor of 1/c). Here's what I have:

Consider a star in its rest frame, $S^\prime$, moving at speed $v$ that emits a photon at an angle $\theta^\prime$. In the $S$ frame, the observer measures the angle to be $\theta$. I made a picture below to illustrate this

enter image description here

We have from the picture the identities $\cos(\theta)=\frac{x}{ct}$ and $\cos(\theta^\prime)=\frac{x^\prime}{ct^\prime}$. The Lorentz transformations are from the prime coordinate system are the following:

$$x^\prime=\frac{x-vt}{\sqrt(1-\beta^2)}=t\frac{\cos(\theta) -\beta}{\sqrt(1-\beta^2)}$$

$$y^\prime =y$$

$$t^\prime=\frac{t-\frac{vx}{c^2}}{\sqrt(1-\beta^2)}=t\frac{1-\beta\cos(\theta)}{\sqrt(1-\beta^2)}$$

Now, $$\cos(\theta^\prime)=\frac{x^\prime}{ct^\prime}=\frac{t\frac{\cos(\theta) -\beta}{\sqrt(1-\beta^2)}}{ct\frac{1-\beta\cos(\theta)}{\sqrt(1-\beta^2)}}$$

After simplifying we get $$\cos(\theta^\prime)=1/c \frac{\cos(\theta)-\beta}{1-\beta\cos(\theta)}$$

Which is almost the relativistic aberration formula but again I am off by $c^{-1}$. Does anyone know what is wrong with this derivation?

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  • $\begingroup$ I guess they will close this question as off topic and my answer will be deleted too, so just in case, the answer is you have missed a $c$ in your very first question $x=ctcos\theta$. $\endgroup$
    – Paradoxy
    Jul 26, 2019 at 1:59
  • $\begingroup$ Why is this considered off topic? $\endgroup$
    – hwhorf
    Jul 26, 2019 at 14:31
  • $\begingroup$ It's not considered as off-topic yet, I'm just guessing it will be. Because although you have shown some effort in solving your question, "you are not asking about a physical concept" and "giving the full solution to question like this is also forbidden", several answers of mine were deleted like this. So always check your topic before they close it. $\endgroup$
    – Paradoxy
    Jul 26, 2019 at 15:21

1 Answer 1

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$$\cos(\theta)=\frac{x}{ct}\rightarrow x=ct\cos(\theta)$$ $$c\beta=v$$ $$x'=\frac{x-vt}{\sqrt{1-\beta^2}}=\frac{ct \cos(\theta)-c\beta t}{\sqrt{1-\beta^2}}=ct(\frac{ \cos(\theta)-\beta }{\sqrt{1-\beta^2}})$$

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  • $\begingroup$ This answer is great thank you very much! $\endgroup$
    – hwhorf
    Jul 26, 2019 at 14:31
  • $\begingroup$ @Paradoxy, can you please continue the derivation. You have a radical in your denominator; yet, hwhorf does not. Or else, please explain the discrepancy. $\endgroup$ Aug 12, 2022 at 3:49
  • $\begingroup$ @MichaelLevy note that in the right side of my last equation, there is a "t". If you use Lorentz transformation of time to change it to " t' " no longer you will see such discrepancy. $\endgroup$
    – Paradoxy
    Aug 13, 2022 at 4:41
  • $\begingroup$ @Paradoxy, the transform $t' = \frac{t-\frac{v\,x}{c^2}}{\sqrt{1-\beta^2}}$, when substituted, does not yield what is to be shown. $\endgroup$ Aug 13, 2022 at 18:37

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