# Relativistic difference of angle between two inertial frame

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.

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

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.

• One of these tangent half-angle formulas may be useful. Feb 8 '20 at 22:25
• And indeed it was! See my answer. Feb 9 '20 at 6:04
• indeed.thank you. 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

$$\tan\theta=\frac{\sin\theta'}{\gamma(\beta+\cos\theta')}\tag{1}.$$

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

$$\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\tag{2}$$

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

$$\cos\theta=\frac{1}{\sqrt{1+\tan^2{\theta}}}\tag{3}.$$

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

$$\tan{\frac{\theta}{2}}=e^{-\xi}\tan{\frac{\theta'}{2}}\tag{7},$$

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

• wow!! just amazing.i havent looked up the whole solution,will try to derive after eq 2.thanks a lot sir. Feb 9 '20 at 17:59
• I must be missing something in the problem statement because when I set $\gamma=1$ it implies $\theta=\theta^{'}$ regardless of the relative velocity. Feb 10 '20 at 3:27
• @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. Feb 10 '20 at 3:29
• @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. Mar 9 '20 at 5:26