# Thermodynamic - conservation of energy/mass problem, I can't figure it out

1. May 17, 2006

### BrianConlee

Ok, I was trying to think of a first problem/question to pose to everyone for my first pose, so here it goes....

Situation:
One Antimatter Proton and One normal matter proton are on a collision coarse. Each one is traveling at about 3/4 the speed of light.

Upon collision, where does the energy that was propelling the atoms go?
Was their increased mass due to their speed forced into the 2nd matter to energy conversion? (the first being the initial force to mass) Was it merely expelled as heat in some form of friction? Or did it violate thermodynamics?

Brian

2. May 17, 2006

### JesseM

When particles and antiparticles annihilate, high-energy photons are released (I think it's always two photons, but I'm not sure). From the point of view of any inertial frame, the energy of the photons after the collision should equal the energy of the particle and antiparticle beforehand.

Also, it doesn't really make sense say that any energy was needed to "propel" the particles, since objects in motion naturally tend to continue moving at constant velocity, it takes an application of energy to change their velocity. The energy of a given particle does increase in your frame as its velocity in your frame increases (according to the formula $$E^2 = m^2 * c^4 + p^2 * c^2$$, where p is the relativistic momentum $$mv/\sqrt{1 - v^2/c^2}$$), but it's not as if energy is being expended to keep the particle moving.

3. May 19, 2006

### pervect

Staff Emeritus

It's an easy answer. Energy is conserved. You don't even have to know (or care) what the "mass" of the particles involved is. All that is important as far as working the problem goes is the energy of the particles.

The important thing is that the particles initally had some energy, and that when they annhilate, energy (and momentum) must both be conserved. Therfore the energy of the two particles after annhilation is the same as the energy of the particles before annhilation.

ps - while the "mass" of the particles increases if one uses "relativistic mass", the mass does not increase if one uses the more modern "invariant mass". However, the key point is really that mass is totally irrelevant to this problem.

You do however need to know the formula for the energy of a particle moving at a specified velocity v. That formula is:

$$E = m_0 c^2/ \sqrt{1 - v^2 / c^2}$$

People who use invariant mass (as I do) will frequently write m instead of $m_0$ in the above formula.

Regardless of how it is written, the $m_0$ in this equation is the invariant mass of the particle which is sometimes called the 'rest mass'.

4. May 20, 2006

### BrianConlee

Hey,
thanks for the reply. I think this forum is a godsend. Can you tell me more about invariant mass. I must admit it is a new concept.

Brian

5. May 20, 2006

### pervect

Staff Emeritus
I'd suggest reading the sci.physics.faq :

Does mass change with velocity?

This is just the bare-bones of the FAQ, there is more information there.

The reason I prefer invariant mass for use with special relativity is that the mass of an isolated system is the same for all observers, the invariant mass does not depend on the coordinate system used (i.e. the velocity of the mass relative to the observer).

This is not the case for "relativistic mass" - here the mass depends on the observer (what coordinate system is used, the velocity of the mass with respect to the observer).

Thus invariant mass IMO best meets the original intent of mass, which is a "measure of the quantity of matter".