Answer to Question #88554 in Mechanics | Relativity for Mike Gale

Question

What is the elapsed time for a 3rd observer in the Twin Paradox who remains stationary with respect to the centre of mass? My conjecture is t3 = gamma1 * t2 = gamma2 * t2, with velocities reckoned with respect to the 3rd observer.

Answer 1

Assume that the moving observer A has mass $m_A$ and moves with speed $v$. Then let the stationary observer S which is on the Earth have mass $m_S$ at 0 speed and let the beginning of x-axis be at the Earth as well. The position of center of mass is thus

$$x_C=\frac{x_Am_A+0\cdot m_S}{m_A+m_S},$$

where $x_A$ - position of the moving observer A. This point is a function of time:

$$x_A=v_A\cdot t,$$

where $t$ - time for the stationary observer S. Remember now that our observer A moves with very high speed, thus his mass will increase and finally the position of center of mass will be the following distance from earth:

$$x_C=\frac{v_At\frac{m_A}{\sqrt{1-v^2_A/c^2}}}{\frac{m_A}{\sqrt{1-v^2_A/c^2}}+m_S},$$

to simplify this expression, use Lorentz factor:

$$\gamma_A=\frac{1}{\sqrt{1-v^2_A/c^2}},$$

which gives

$$x_C=\frac{v_At\cdot m_A\gamma_A}{m_A\gamma_A+m_S}.$$

Now put our middle observer C at the center of mass. As we see, the center of mass is a function of time. The more time the experiment lasts - the further from earth not only A, but C also becomes. What is the speed of C?

$$v_C=\frac{x_C}{t}=v_A\frac{m_A\gamma_A}{m_A\gamma_A+m_S}.$$

So for this observer the elapsed time will be

$$t_C=t_3=t\sqrt{1-\frac{v_C^2}{c^2}}=t\sqrt{1-\frac{v_A^2}{c^2}\Big(\frac{m_A\gamma_A}{m_A\gamma_A+m_S}\Big)^2}.$$

In case $m_S=m_A$:

$$t_C=t_3=t\sqrt{1-\frac{v_A^2}{c^2}\frac{\gamma_A^2}{(\gamma_A+1)^2}}.$$

Now express the ratio $v_A^2/c^2$ in terms of $\gamma_A$ and substitute this to time C equation above:

$$t_C=t\sqrt{1-\frac{\gamma_A^2-1}{\gamma_A^2}\frac{\gamma_A^2}{(\gamma_A+1)^2}}=t\sqrt{1-\frac{\gamma_A^2-1}{(\gamma_A+1)^2}}=\sqrt{\frac{2}{\gamma_A+1}}.$$