# Relativity and particle collision.

1. Jan 28, 2014

### Herbascious J

I have a question; If two particles (let's say protons) collide at high speed, they can produce heavy particles, like the Higgs. My understanding is that the high velocities make the particles more massive, allowing them to produce higher mass particles (relativity).

So... If I fly past the particle collision experiment at high speed, perpendicular to the paths of the particles. the particles would appear heavier to me, than to someone standing still relative to the point of impact. Would it be possible for me to observe a higher energy, or more massive outcome to the collision result? Would I just see the same particles, just more massive according to relativity, or could I actually observe a different particle specie outcome?

P.S. - It has occured to me that the collision would appear to happen at non-head-on angle for the moving observer, perhaps this changes things as well?

2. Jan 28, 2014

### WannabeNewton

The scattering amplitude between an initial Fock (multiparticle) state and a final Fock state is constructed as a Lorentz-invariant quantity. You can't change the scattering amplitude just by boosting to another frame.

Here's a Griffiths level discussion of the above: http://www-pnp.physics.ox.ac.uk/~barra/teaching/feynman.pdf

3. Jan 28, 2014

### Staff: Mentor

That's probably not the best way to look at it, because it is easy to draw the incorrect inference that you draw here:

No, because you're not involved in the collision; you can't change what happens just by changing the speed at which you fly past it. What matters is the relative motion of the particles actually involved in the collision.

A better way of looking at it is to realize that, in relativity, mass and energy are different forms of the same thing: mass can be converted to energy, and energy can be converted to mass. When two particles smash together at high speed relative to each other, their kinetic energy relative to each other can be converted into mass, allowing particles of higher mass (but moving more slowly) to come out of the collision. The best way to analyze such experiments is in the center of mass frame, since that's the frame in which everything adds up easily: mass + kinetic energy in = mass + kinetic energy out. So if kinetic energy out is lower (particles created in the collision are moving more slowly than the particles that collide), mass out can be higher (heavier particles produced).

4. Jan 28, 2014

### Herbascious J

I see... So, the correct way to see the energy of the event is based on the kinetic energy of the particles being converted to matter in the form of new particles. This is what is meant by available energy? Regardless of how I move, identical results should be observed.

5. Jan 29, 2014

### BruceW

Let's say you start off with a bunch of particles, with total four-momentum $P^\mu$. Then the most massive particle you can get is if all of these particles get destroyed, and only one particle is created. agreed? and so this new particle will have four-momentum $P^\mu$. and the mass of this new particle is simply $P^\mu P_\mu$ so the mass of the new particle is equal to the invariant mass of the bunch of particles we had before. The key is that it is invariant, it doesn't matter how we Lorentz transform our system, the invariant mass of the previous bunch of particles is still the same, and therefore the mass of the one created particle will also be the same.

In some ways, I'm just repeating what PeterDonis was saying about the centre of mass frame. Also, I think what WannabeNewton was saying is most likely a much more complete answer. my answer is just a specific example, rather than a general statement about relativistic collisions.

6. Jan 29, 2014

### Staff: Mentor

Yes; as BruceW pointed out, the system as a whole (i.e., all the particles involved in the collision, considered together) has an invariant mass, which is all the masses plus all the kinetic energies in the center of mass frame, added together. That is the total energy available for particles coming out of the collision.

7. Jan 29, 2014

### .Scott

Let's say that in the first experiment, we had a particle coming from the west, another heading straight towards it from the east, and you were approaching from the south. You would see the same spray of particles as everyone else.

In the second experiment, you get to stand still - but the two protons will be approaching each other at an angle, one from the northeast, the other from the north west. The protons are accelerated to a speed that matches (relatively) what you saw in the first experiment.

In this second experiment you'll see the same thing as in the first experiment. Even though everyone standing with you see that the protons are more massive than in the first experiment, the fact that they struck each other at an angle compensates for the difference in energy and results in the same spray of particles.

So were does that extra mass go. The spray will be direct to the south, and because that spray of particles is moving faster, each particle in the spray will appear to be more massive.

8. Jan 29, 2014

### Herbascious J

Ok, that makes sense to me intuitively. Everything compensates with kinetic energy and makes observations equivalent. Thank you for the great thought experiment!