Does a working pocket watch have more matter than a broken pocket watch?

Question

Say I have two exactly identical pocket watches. Say one pocket watch works and the other does not (the broken one does not work because a gear broke). Does the working pocket watch have more mass than the broken pocket watch?

Answer 1

I am assuming the pocket watches are the old fashioned kind that has a spring that needs to be wound up periodically. I am also assuming that the "broken" watch simply means that the spring has not been wound up so it doesn't function.

In this case, yes, the "working" watch will have a very tiny amount of extras mass when compared to the broken watch. According to this answer on Physics Stackexchange, a watch spring could contain approximately 0.3 joules of energy. By $m=E/c^2$, that means the wound up "working" watch would be only $3 \times 10^{-15} \ grams$ heavier than the "broken" watch. This mass difference is equivalent to the mass of about 3 billion protons. Obviously, this is far to small to be measured by any laboratory scale.

Answer 2

You are worrying about special relativity effects. These effects appear when objects are moving with very high velocities to each other. For everyday objects like watches , Newtonian mechanics is enough to make sense of energy requirements.

In Newtonian mechanics mass is conserved, that is how the Archimedes principle works. So if your two watches are identical, and no pieces have been removed, it is only the kinetic and potential energy in the system that may be different. I will assume that just before breaking the batteries or the springs have the same potential energy. A working watch has kinetic energy in the gears, and potential energy in the spring that was wound, or in the battery if it is electric.

The working clock spends potential energy that goes into motion of gears etc, and then into friction, and then into infrared radiation , while the stopped one keeps the potential energy.Thus it is the working clock that has less energy, even from the first second of breakage of the stopped one.

Special relativity does not enter into this problem.

Looking at the other answer you got, you must realize that it is the starting hypothesis that changes the answer. For a problem to be unique the boundary values have to be given, as changing them changes the answer.