Problem — Doppler-factor velocity composition

Problem

Spaceship $B$ is heading towards Earth ($A$) with velocity $v_{B(A)}=0{,}6\,c$; spaceship $C$ is heading towards $B$ with $u_{C(B)}=0{,}6\,c$ as measured by $B$. Calculate $u_{C(A)}$ using the multiplicative form (product of Doppler factors) and verify with the algebraic form of velocity composition.

Solution

Method 1 — product of Doppler factors. The composition law in multiplicative form is $\sqrt{\frac{1+u_{C(A)}/c}{1-u_{C(A)}/c}} = \sqrt{\frac{1+v_{B(A)}/c}{1-v_{B(A)}/c}}\cdot\sqrt{\frac{1+u_{C(B)}/c}{1-u_{C(B)}/c}}$ With both velocities $=0{,}6\,c$ (approaching), each factor equals $\sqrt{\frac{1+0{,}6}{1-0{,}6}} = \sqrt{\frac{1{,}6}{0{,}4}} = \sqrt{4} = 2$ So the direct factor is the product: $\sqrt{\frac{1+u_{C(A)}/c}{1-u_{C(A)}/c}} = 2\cdot 2 = 4$ Squaring and solving, setting $\beta = u_{C(A)}/c$: $\frac{1+\beta}{1-\beta} = 16 \quad\Rightarrow\quad 1+\beta = 16 - 16\beta \quad\Rightarrow\quad 17\beta = 15 \quad\Rightarrow\quad \beta = \frac{15}{17}$

$\ev{u_{C(A)} = \frac{15}{17}\,c \approx 0{,}882\,c}$

Method 2 — algebraic form. Verification with the composition law: $u_{C(A)} = \frac{u_{C(B)} + v_{B(A)}}{1 + u_{C(B)}\,v_{B(A)}/c^2} = \frac{0{,}6\,c + 0{,}6\,c}{1 + 0{,}6\cdot 0{,}6} = \frac{1{,}2\,c}{1{,}36} = \frac{15}{17}\,c \approx 0{,}882\,c$

Interpretation. The two methods agree. The naive sum $0{,}6\,c + 0{,}6\,c = 1{,}2\,c$ would exceed $c$; the denominator $1{,}36$ “brakes” the result down to $0{,}882\,c$, below the speed of light as required by the second postulate.

Links

Topics: Special relativity Concepts: Relativistic velocity composition · Doppler effect Skills: Symbolic setup