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The angle transformation under Lorentz transformation:

$$ \cos\theta = \frac{\cos\theta' + \beta}{1 + \beta\cos\theta},\quad \sin\theta = \frac{\sin\theta'}{\gamma^2(1 + \beta\cos\theta')} $$

is normally derived from the velocity transformation:

$$ u_\parallel = \frac{u_\parallel' + v}{1+(v/c^2)u_\parallel'},\quad u_\perp = \frac{u_\perp/\gamma}{1+(v/c^2)u_\parallel'} $$

and then

$$ \tan\theta = \frac{u_\perp}{u_\parallel} = \frac{u_\perp'/\gamma}{u_\parallel' + v} = \frac{u'\sin\theta'}{\gamma(u'\cos\theta + v)} $$

letting $u' = c$ and get:

$$ \tan\theta = \frac{\sin\theta'}{\gamma(\cos\theta'+\beta)} $$


I am wondering why we don't start with the length transformation:

$$ x = \gamma(x'+\beta ct'),\quad y = y' $$

and:

$$ \tan\theta = \frac{y}{x} = \frac{y'}{\gamma(x'+\beta ct')} $$

which leads to an annoying $ct'$ here


Can someone tell me why we must use velocity transformation and not length transformation to get the angle transformation?

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  • $\begingroup$ Why not start with "time transformation"? Velocity is not any kind of "associated quantity" with length. $\endgroup$
    – m4r35n357
    Oct 29, 2023 at 11:05
  • $\begingroup$ time transformation is associated with length $\endgroup$ Oct 30, 2023 at 5:47

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I think, since you are only considering 2D (as oposed to 3D), i.e. a component of the velocity that is perpendicular to the Lorentz transformation and another that is parallel with it, you are allowed to say that $$x'=r'\cos\theta'\\y'=r'\sin\theta'$$ for some distance $r'$. The latter distance, then, is going to be the distance that satisfies $ct'=r'$, and therefore you will have the Eq. in your original post $$\tan\theta=\frac{\sin\theta'}{\gamma(\cos\theta'+\beta)}$$

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