A demonstration of momentum conservation and the relativistic mass–energy equivalence \(E=mc^2\), shown through what happens to the center of mass.
Two identical spaceships A (left) and B (right) float at rest in deep space, separated by a distance \(L = 60\) m. Each has rest mass \(M = 1000\) kg. Ship A then sends something — a baseball, or a pulse of light — toward ship B, which absorbs it.
Two notes about the animation. (1) Real light travels at \(c \approx 3 \times 10^8\) m/s, which is far too fast to see on a screen, so the photon below uses a "movie speed" of \(c = 40\) m/s. The algebra is identical to using the real value — only the time scale is rescaled. (2) The mass loss \(\Delta m\) is treated as instantaneous: the moment A emits the pulse, A's mass drops by \(\Delta m\); the moment B absorbs it, B's mass jumps by \(\Delta m\). In the algebra-based picture you can think of "emit" and "absorb" as discrete events, just like a perfectly inelastic collision.
Since nothing outside the system pushes on it, the total momentum stays zero:
$$ p_A + p_\text{projectile} + p_B \;=\; 0 $$And the position of the center of mass cannot move, because the COM only moves if the total momentum is nonzero:
$$ x_\text{COM} \;=\; \frac{m_A x_A + m_B x_B + m_\text{proj}\, x_\text{proj}}{m_A + m_B + m_\text{proj}} \;=\; \text{constant.} $$For a pulse of light carrying energy \(E\), the momentum is
$$ p_\gamma \;=\; \frac{E}{c}, $$and by Einstein's mass–energy equivalence, that energy is also equivalent to an inertial mass
$$ \Delta m \;=\; \frac{E}{c^{2}}. $$So when A emits light it must lose mass \(\Delta m\), and when B absorbs it B must gain that same mass. If you forget that step, the center of mass appears to wander on its own — which would violate Newton's first law for the system.
In each animation we draw two centers of mass: the red marker at the top tracks the full system (A, B, and the projectile), and the teal marker at the bottom tracks just the sender subsystem of A and the projectile while it's in flight. Both should sit on top of their starting positions if the bookkeeping is right.
The familiar classical case. Ship A throws a ball of mass \(m\) at speed \(v\). Both collisions (the throw and the catch) are perfectly inelastic, so by conservation of momentum:
$$ \underbrace{(M-m)\, v_A}_{\text{A's recoil}} \;=\; \underbrace{m\, v}_{\text{ball}} \;=\; \underbrace{(M+m)\, v_B'}_{\text{B after catch}} $$A emits a pulse of light with energy \(E\) and momentum \(E/c\), and recoils. Now we pretend the light has no mass and that A and B keep their mass \(M\). Momentum conservation still gives a sensible-looking pair of equations:
$$ M\, v_A \;=\; \frac{E}{c} \;=\; M\, v_B' \qquad (\text{wrong assumption}) $$But while the light is in flight, A is moving left and only B is on the right — and B isn't moving yet. With only the two ships counted as having mass, the center of mass becomes
$$ x_\text{COM} \;=\; \frac{M\,x_A + M\,x_B}{2M}, $$and since \(x_A\) is shrinking while \(x_B\) is fixed, \(x_\text{COM}\) drifts to the left. That can't happen for a system with zero total momentum, so something is wrong.
Now we give the light pulse the inertial mass \(\Delta m = E/c^{2}\) that Einstein's equivalence requires. A loses that mass on emission, B picks it up on absorption. The momentum bookkeeping is the same form as a baseball, just with \(c\) in place of \(v\):
$$ (M-\Delta m)\, v_A \;=\; \Delta m\cdot c \;=\; (M+\Delta m)\, v_B' $$Now the center of mass has three pieces — A, the light pulse, and B — and the leftward push from A is exactly cancelled by the light traveling right:
$$ x_\text{COM} \;=\; \frac{(M-\Delta m)\, x_A \;+\; \Delta m\, x_\text{light} \;+\; M\, x_B}{2M} \;=\; \text{constant.} $$About the teal marker. The teal "A + photon" COM stays at A's original position (right on the gray dashed line), not at A's current position. A has moved leftward and the photon has moved rightward, so the COM of just those two ends up between them — it just happens to coincide with where A used to be, because that's the only place the masses balance. Once the photon is absorbed by B and the subsystem is no longer closed, the marker disappears.