# B Envisioning particle collision.

1. Feb 29, 2016

### dbertels

I constantly hear the term 'bits flying off' in a particle collision'. Isn't this somewhat misleading? Wouldn't it be more accurate to say that, in a collision, a 'fountain' of new particles are created and that our main aim is to use the energy resulting from the collision to create new, heavier particles?

2. Feb 29, 2016

### BvU

What actually happens at such small scales eludes our imagination; we should be content that we can describe the events with ever improving accuracy. Personally I like the 'bits flying off' analogon in hadron collisions. And for me the 'fountain' idea evokes a totally different association, one that to me doesn't seem to 'fit'. But it's a personal thing.

And yes, energy is partially used to create new particles. Heavier than before, but not heavier than the good old Heisenberg limit allows.

3. Feb 29, 2016

### Orodruin

Staff Emeritus
Energy does not result from the collisions. It is just redistributed.

4. Feb 29, 2016

### dbertels

@BvU Granted I shouldn't have used the 'fountain' image, it just introduced another analogy that can be misconstrued.
@Orodruin Does this mean that the energy/mass required to create the Higgs boson was present in the colliding hadrons?

5. Feb 29, 2016

### Orodruin

Staff Emeritus
Energy is a frame dependent quantity. The energy needed was mainly kinetic energy in the colliding protons. Still, if you would go to the rest frame of one of the protons its kinetic energy would be zero (although the other would have an enormous kinetic energy).

6. Feb 29, 2016

### dbertels

Is the term 'frame' here used in the sense of 'state'?
Can you point me to any articles that can help me understand this?

7. Feb 29, 2016

### Orodruin

Staff Emeritus
Definitely not. It refers to inertial frames. Energy is observer dependent.

8. Feb 29, 2016

### Dirk Pons

The 'bits flying off' perspective is a disassembly one. It tends to imply that the particle already had all those 'bits' partially pre-formed and assembled inside itself, ready to break off.The problem with that approach is that particles in different situations show different output products, hence implying different input structures.

I find it easier to envisage this process as remanufacture of particle identities. This handily also covers the decay processes where there is only one input particle. By 'remanufacture' I mean that the input particle(s) are induced to change by some initiation event (impact, decay, photon absorption, etc) followed by a process where the energy is partitioned up differently (and excess energy is liberated as a photon). Other quantum numbers (charge, matter-antimatter attributes) are also re-distributed in the process, and usually conserved. These attributes and energies determine the identities of the output particles. Consequently small changes in the process settings can result in different output particles.

9. Feb 29, 2016

### dbertels

Very interesting statements - unfortunately, I'm not in a position to comprehend this. Maybe you can point me to material that may throw light on these statements?

Thanks for taking the time answering this - it gives me material to research.
Interesting how all the 'realities' of the classical world lose their meaning in particle physics, hence any description will fall short of this - and maybe trying to understand the collision using classical world analogies is futile, that envisioning 'bits breaking off' is as meaningless as 'creating new particles' ..

10. Mar 1, 2016

### Orodruin

Staff Emeritus
This is not anything strange or new in relativity. It is true also in classical mechanics. Obviously, if you go to an object's rest frame its kinetic energy is going to br smaller than in a frame where it is moving.

11. Mar 1, 2016

### Staff: Mentor

Step away from particle physics for a moment, and into (hopefully) more familiar terrritory.

Suppose you're standing by a road, and a car goes past you at speed v = 60 m/s. It has kinetic energy $\frac{1}{2}mv^2 = \frac{1}{2}m(60^2)$ joules, provided m is in kilograms.

Now suppose you're in a car that is going 50 m/s, and the car described above passes you in the same direction. In your new reference frame ("point of view"), that car is passing you at a rate of 10 m/s, and its kinetic energy in this reference frame is $\frac{1}{2}m(10^2)$ joules.

12. Mar 1, 2016

### dbertels

Great, I understand what you're saying here and I think I can see where my confusion is coming from - I was thinking particle collision involved quantum physics because of the scale of the particles involved, hence my 'creation of new particles' analogy (that possibly clears up the fact I'm no physicist).
But all your explanations seem to point to relativity (I assume because of the relativistic velocities involved?).

13. Mar 1, 2016

### Orodruin

Staff Emeritus
It involves both relativity and quantum physics.

14. Mar 1, 2016

### ChrisVer

That's what they call "particle physics"... or "high energy physics"... the combination of both special relativity and quantum physics.

That's relativity alone. Jtbell gave a nice way to see how this can work. For special relativity search for 4momentum.
The relation: $E^2 = p^2 c^2 + m^2 c^4$ where E is the energy, p is the momentum and m is the mass of the body (c=speed of light in vacuum), shows the same thing... If you change your momentum, the energy must change so that the mass of the body will be constant [ a proton's mass is always equal to a proton's mass]... the momentum obviously depends on the observer.

Last edited: Mar 1, 2016
15. Mar 1, 2016

### Orodruin

Staff Emeritus
No it is not. It is true in classical mechanics as well. What is important here is the relativity principle, which holds just as true in classical mechanics as it does in relativity.

This is a crucial point which many seem to miss and think there is something strange going on in relativity.

16. Mar 1, 2016

### ChrisVer

thanks for the point, I corrected it.

17. Mar 1, 2016

### dbertels

Still, I'm interested to know what the best way would be to describe what happens in a particle collision, if you had to explain it to non-physicists in the most meaningful way possible. Any takers on this?

18. Mar 1, 2016

### BvU

Yes, the total energy in the center of mass of the colliding hadrons was more than the energy/mass required to create the Higgs boson.
The concentration of energy in a very small space allows the formation of particles.

The famous Carreras would make a tralaTeV accelerator in the Cern lecture room by shooting off a small scrap of paper with a rubber band. That kinetic energy was also more than needed for a Higgs boson, but (un)fortunately not concentrrated in a small enough space.

19. Mar 1, 2016

### Dirk Pons

Not all particle processes involve high velocity impacts. Decay processes would be an example: they happen for static particles.

The reason for high velocities is to artificially add extra energy to the particles. The reason this is useful is that this energy is then converted (think E=mc^2 at this point) into new mass, which is then available to make new particles that otherwise would not have appeared at a slow collision.

20. Mar 1, 2016

### dbertels

I'm liking that - So the kinetic energy created by the collision allows this to happen..