I was studying Prof. Padmanabhan's book on general relativity; "Gravitation: Foundations and Frontiers, which I found very well written. I just started the book and the exercises are also challenging.

I came across this equation in section 1.3.2, where Padmanabhan derives the direction of motion of a moving particle in different inertial frames.

The formula (eq 1.22) reads

$$ \tan \theta=\gamma^{-1}\frac{\sin \theta'}{\beta+\cos \theta'}, $$

and then he asked to show that this can be written as

$$ \tan \frac{\theta'}{2}=e^{-\xi}\tan \frac{\theta}{2}, $$

where $\xi$ is the rapidity defined as

$$ \tanh \xi=\beta,\quad \gamma=\cosh \xi, \quad \beta\gamma=\sinh\xi, \quad \beta=\frac{V}{c}, $$

where $V$ is the magnitude of relative velocity between the two inertial frames.

It looks like a very clever manipulation of trigonometric identities which I can't figure. Basically, I can't come up with $\tan \frac{\theta}{2}$ or $\tan \frac{\theta'}{2}$. From the definition of $\xi$ some simple algebra gives you

$$ e^{-\xi}=\frac{(1-\beta)\sin\theta'}{\tan \theta(\beta+\cos \theta')}, $$

but I cant go any further. Help please.

  • $\begingroup$ One of these tangent half-angle formulas may be useful. $\endgroup$
    – G. Smith
    Feb 8 '20 at 22:25
  • $\begingroup$ And indeed it was! See my answer. $\endgroup$
    – G. Smith
    Feb 9 '20 at 6:04
  • $\begingroup$ indeed.thank you. $\endgroup$ Feb 9 '20 at 18:01

In deriving this, I found that you, or the book, have $\theta$ and $\theta'$ reversed in your second formula.

We start with the given formula for $\tan\theta$ in terms of $\theta'$, namely


We can compute $\tan{\frac{\theta}{2}}$ using the trigonometric identity


if we know $\cos\theta$, and we can compute $\cos\theta$ from $\tan\theta$ using another trigonometric identity,


Substituting (1) into (3) gives

$$\begin{align}\cos\theta&=\frac{1}{\sqrt{1+\left(\frac{\sin\theta'}{\gamma(\beta+\cos\theta')}\right)^2}}\\ &=\frac{\beta+\cos\theta'}{\sqrt{(\beta+\cos\theta')^2+(1-\beta^2)\sin^2\theta'}}\\ &=\frac{\beta+\cos\theta'}{1+\beta\cos\theta'} \end{align}\tag{4}$$

and substituting (4) into (2) gives

$$\begin{align}\tan{\frac{\theta}{2}}&=\sqrt{\frac{1-\frac{\beta+\cos\theta'}{1+\beta\cos\theta'}}{1+\frac{\beta+\cos\theta'}{1+\beta\cos\theta'}}}\\ &=\sqrt{\frac{1+\beta\cos\theta'-\beta-\cos\theta'}{1+\beta\cos\theta'+\beta+\cos\theta'}}\\ &=\sqrt{\frac{(1-\beta)(1-\cos\theta')}{(1+\beta)(1+\cos\theta')}}\\ &=\sqrt{\frac{1-\beta}{1+\beta}}\tan{\frac{\theta'}{2}} \end{align}\tag{5}.$$

Expressing $\beta$ in terms of $\xi$, we have

$$\begin{align}\sqrt{\frac{1-\beta}{1+\beta}}&=\sqrt{\frac{1-\tanh\xi}{1+\tanh\xi}}\\ &=\sqrt{\frac{\cosh\xi-\sinh\xi}{\cosh\xi+\sinh\xi}}\\ &=\sqrt{\frac{e^{-\xi}}{e^\xi}}\\ &=e^{-\xi} \end{align}\tag{6}.$$

Substituting (6) into (5), we get


which is the reverse of what you stated. For positive $\beta$, according to (1), $\theta$ should be less than $\theta'$.

  • $\begingroup$ wow!! just amazing.i havent looked up the whole solution,will try to derive after eq 2.thanks a lot sir. $\endgroup$ Feb 9 '20 at 17:59
  • $\begingroup$ I must be missing something in the problem statement because when I set $\gamma=1$ it implies $\theta=\theta^{'}$ regardless of the relative velocity. $\endgroup$ Feb 10 '20 at 3:27
  • $\begingroup$ @CinaedSimson $\gamma=1$ implies $\beta=0$. They aren’t independent. The relationship is $$\gamma=\frac{1}{\sqrt{1-\beta^2}}.$$ Of course the angles are the same when the frames have no relative motion. $\endgroup$
    – G. Smith
    Feb 10 '20 at 3:29
  • $\begingroup$ @G.Smith: there's only one velocity vector in equation $1.2$ - the relative velocity. The author "derives the direction of motion of a moving particle in different inertial frames." Where's the second inertial velocity vector? When $\gamma=1$, the relativistic velocity addition formula should reduce to the Galilean addition of velocities. $\endgroup$ Mar 9 '20 at 5:26

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