Problem 1: Displacement of Tube due to Photon Emission
Analysis: Since the system acts in free space without external forces, the Center of Mass (COM) must remain stationary. The emission of a photon from one end and its absorption at the other corresponds to the transfer of an “effective mass” from left to right.
The effective mass of the photon ($m_{ph}$) is derived from Einstein’s mass-energy equivalence:
$$ m_{ph} = \frac{E}{c^2} = \frac{h\nu}{c^2} $$Let the entire tube system (with total mass $M_{sys} = m + m_1 + m_2$) displace by a distance $\Delta x$ to the left to compensate for the photon’s effective mass moving a distance $l$ to the right. With respect to system’s mass Mass of photon can be ignored.
Using the principle of moments for the Center of Mass ($\sum \Delta m_i x_i = 0$):
$$ M_{sys} (\Delta x) + m_{ph} (l) = 0 $$ $$ (m + m_1 + m_2) \Delta x = – \left( \frac{h\nu}{c^2} \right) l $$The magnitude of the displacement is:
$$ |\Delta x| = \frac{h\nu}{c^2} \left( \frac{l}{m + m_1 + m_2} \right) $$