# A Question on Relativistic Mass (SR)

1. Jul 18, 2012

### ASmc2

Hello. I am independently studying modern physics. I have a basic special relativity question.
Consider a completely inelastic collision Of course, objects are moving at a constant velocity before and after the collision with respect to a stationary observer. Let's call him Rick. Suppose that Rick calculates the difference between the sum of the masses of the objects when they are apart and the mass of the composite object they make after they stick together. In his frame, he finds M. Here is the question. I know other inertial observers in different frames will measure different values of momentum and kinetic energy for the objects. I think, though, they will measure the same mass defect (M)? If not, why?

P.S. No force fields around.

2. Jul 18, 2012

### Staff: Mentor

If by "mass" you mean "rest mass", then since that is an invariant, all observers will calculate the same rest mass for each object involved, before and after the collision, so they will all calculate the same "mass defect".

3. Jul 18, 2012

### pervect

Staff Emeritus
Realistically, if you have a relativistic collision between two solid objects, you'll wind up with a hot ball of expanding plasma and perhaps the occasional exotic particle, depending on exactly how the collision was done.

However, you can visualize the two solid objects sticking together and colliding nicely, but you should remember that the collision product is going to be relativistically hot. It's not particularly plausible that any known form of matter would stay together in a nice package for you when it was that hot, but it doesn't break any physical laws so you can ask what happens if it does.

If you don't have any - or at least much - energy radiating from your relativistically hot final system, you can use the law of conservation of energy to say that the total energy of the final system is the same as the total energy of the initial system before the collision. I'll add in some notes as to what happens when you have radiated energy as well.

I will adopt the modern convention here of using "mass" to mean "invariant mass", so anywhere you see the word "mass", you can assume that I mean "invariant mass" and not something else.

If we call the invariant masses of the two particles before collision m1 and m2, the energies before the collision will be gamma1*m1 for particle 1 and gamma2*m2 for particle 2, where gamma = 1/sqrt(1-(v/c)^2). The two particles won't necessarily have equal speeds, so we need a velocity, and hence a gamma, for each of them.

So the energy and the invariant mass of the system at rest after the collision in its own frame will be m_tot = gamma1*m1+gamm2*m2.

If some energy Erad is radiated from the composite system after the collision, the total mass will be equal to m_tot = Erad = gamma1*m1+gamma2*m2 - Erad. If Erad is a function of time, m_tot will be a function of time. Erad will also in general depend on the observer, as well as time, due to issues involving the relativity of simultaneity

Note that in the rest frame of the composite particle, it's invariant mass will be equal to its energy, since the velocity of the composite particle is zero in its own frame of reference, and E = gamma * m.

In some other frame, the total energy of the final system will be the gamma factor * invariant mass.

In any frame, the invariant mass well be the same as long as Erad = 0. The invariant mass of a closed system is a constant which is independent of the observer doing the measuring. That's one of the reasons it's such a useful concept.

Sometimes people use "relativistic mass". It's mostly depreciated, though you'll find the occasional die-hard who insists that it isn't, and it's hardly worth arguing about. If you do run into the term, just remember that it's another name for energy.

If your confused about "which mass causes gravitation - invariant mass or energy", the most correct answer is neither one. Plugging either sort of mass into Newton's gravity law is not going to give relativistically correct answers. Answering the question of what causes gravity would drift from the equestion at hand and require a rather long post, so I won't do it here (besides, I'm out of time).

Last edited: Jul 18, 2012