What happens to the velocity of a radioactively decaying object?

Question

Suppose that I have a lump of a radioactive material, like Uranium-235. I put it in an ideal box, which perfectly isolates the inside from the outside - no radiation escapes the box and the outer surface of the box does not change in any way. The box is thrown by an astronaut with velocity $v_0$. The mass of the box and the lump is $m_0$.

As time goes by, the radioactive material decays. After time $t$ has passed, the total mass of the box and the partially decayed lump has fallen to $m_1$. Some mass has "evaporated" and turned into other forms of energy, for example by alpha decay of Uranium-235.

Given that the conservation of momentum mandates that $m_0v_0=m_1v_1$ and that the mass has decreased, it seems like the velocity of the box had to have increased.

This seems strange because for an outside observer it looks like the black box has spontaneously accelerated without any external force acting on it. This would even allow for creating a simple, yet effective, rocket engine - just take some uranium onboard and see the rocket start to fly faster all by itself.

Answer 1

These two statements are incompatible with each other

an ideal box, which perfectly isolates the inside from the outside - no radiation escapes the box and the outer surface of the box does not change in any way

And

the total mass of the box and the partially decayed lump has fallen to $m_1$

If nothing escapes the box, specifically is no energy or momentum leaves the box, then the mass of the box is constant regardless of what happens inside the box.

The mass of the box is given by $$m^2 c^2=E^2/c^2-p^2$$ so if the box is isolated then $E$ and $p$ are constant by the conservation of energy and momentum respectively. Therefore $m$ is also constant.

I suspect that the reason you believe that the mass decreases is because of the very common misunderstanding that $E=mc^2$ represents a “conversion” between mass and energy. This is an incorrect understanding because if $m$ is “converted” into $E$ then $m$ decreases and $E$ increases which would violate the conservation of energy.

Also, if scientists had wanted to write a formula stating that mass could be converted into energy it would be $\Delta E=-\Delta m c^2$. The form of $E=mc^2$ is intended to convey a relationship that holds at each instant of time, just like $F=ma$ is intended to hold at each instant of time. So the $E$ at rest at some instant in time is equal to $mc^2$ at that same instant of time.

Note that the famous $E =mc^2$ is a special case of the more general $m^2 c^2=E^2/c^2-p^2$ for $p=0$. In radioactive decay (in the center of momentum frame) a particle “splits” into two particles with equal and opposite momentum. Since the momentum is equal and opposite, the total $p$ is unchanged (it remains 0 in the center of momentum frame). Some of the rest energy is converted into kinetic energy, but by conservation of energy the total $E$ is unchanged. Therefore the total mass of the box is unchanged.

Answer 2

You have to decide whether you want to model the box or not; on the left of your picture you don't assign it any mass, but on the right it scatters the alpha particles. Momentum conservation holds also for radioactive decay of course. You can model it as a mass in a box with a loaded spring that ejects a marble, same thing. Momentum of the combined system mass plus ejecta plus box is conserved.

Answer 3

You may understand this easily using $E=mc^2$ and the old concept of relativistic mass. The mass doesn't decline for the simple reason that you've not allowed any energy to leak from the box. The mass is always simply $E/c^2$.

Dale, above, uses a different, revisionist definition of mass, but that just complicates problems of this sort.