# The Large Hadron Collider - Where is the particle?

1. Dec 17, 2009

### mintparasol

Ok, apologies if my understanding is limited or even wrong but I know I'll learn something from your replies even if my question turns out to be idiotic!

Relativity gives us length contraction as we approach the speed of light but quantum mechanics gives us uncertainty as to a particles velocity vs. it's position.

So on the one hand you have an accelerated particle contracting under increased velocity in relativity and on the other, you have the same particle's position taking the form of an ever lengthening probability wave in quantum mechanics with increased velocity.
Which is the more relevant theory with regard to the LHO? More importantly, does it even matter? Am I wrong or is this another one of those cases where the 2 theories fail spectacularly to agree?

2. Dec 17, 2009

### diazona

Why would the probability wave be lengthening?

And anyway, for situations like high-energy collisions you need to use quantum field theory - neither basic quantum mechanics nor special relativity on their own are quite appropriate. I don't know enough about QFT yet to give you a proper answer to this question, but I'm sure it's not as simple as it looks.

3. Dec 17, 2009

### rhenretta

The point of the LHC is energy, not speed (or the relativistic effects). What they are doing is taking a proton, and by accelerating it to near the speed of light, they increase its energy. When the opposing beams are allowed to collide, the totality of their energy both from mass and from the kinetic energy of velocity turns into brand new particles, even some that are much more massive than the original particles.

Last edited: Dec 17, 2009
4. Dec 17, 2009

### rhenretta

I should clarify the mass + kinetic energy...

$$E=mc^2$$ is actually wildly inaccurate in this context, and wouldn't explain the collisions taking place at the LHC. According to this, a particle going any speed (even close to the speed of light) would have the same energy. The formula is actually:

$$E=\gamma mc^{2}$$

Where
$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

The thing to notice about $$\gamma$$ is that it is close to 1 at low speeds relative to light. Thus, $$E=mc^{2}$$ is only accurate for something at rest. Now, when the velocity increases towards the speed of light, then $$v^2/c^2$$ gets closer and closer to 1, the denominator will get smaller and smaller, $$\gamma$$ gets bigger and bigger, and energy becomes more and more.

It is really a very crude and expensive way of increasing the energy of an object with mass, but it is the only way we have right now.

If this was more math than you were looking for, sorry, but this was my first opportunity to really play around with the math tags on here ;-)

5. Dec 17, 2009

### mintparasol

Well in my (limited) understanding, doesn't the particle's position become less certain with increased velocity? Isn't the position of a particle a probability wave in quantum physics?

6. Dec 17, 2009

### rhenretta

The accuracy of the particle's position becomes less known, the more accurately you know the velocity, it is completely indifferent towards what the actual velocity is

7. Dec 17, 2009

### mintparasol

Yup, too much math. I'm just a lay nut with too much time on my hands, haha!

So it doesn't really matter where the particle is or how we define it's position providing we can cause a high energy collision to generate 'new' detectable particles?

I'm feeling a bad case of 'it's a wave, no, it's a particle' coming on!

8. Dec 17, 2009

### mintparasol

Well exactly, they're gonna have the velocity of the particle nailed to quite a high degree of accuracy are they not? Otherwise they can't really say much about the energy of the collision, no?

9. Dec 17, 2009

### rhenretta

thankfully, in this case, not many waves to have to worry about. It is really a very simple take object 1, bam it into object 2 and see what comes out.

10. Dec 17, 2009

### rhenretta

keep in mind, the position of the particles doesn't really matter. It's actually a narrow stream of particles, sometimes they collide, sometimes they don't. You'd probably be shocked to learn that there are over 1 billion collisions every second at the LHC, they aren't taking a single proton, and accurately positioning it to collide with a specific 2nd proton

11. Dec 17, 2009

### mintparasol

Hmmm, are we bamming waves or particles tho?!!!!!

I'm starting to see that my question doesn't really make any difference but your answers are giving me a better understanding. In news articles I'd read (and maybe on here, it's too late at night to go searching..) there have been claims made that they actually isolate single particles and accelerate them. I suppose I should look at the mechanics as being more along the lines of what's happening with the electrons in the tube of my old tv...

12. Dec 17, 2009

### rhenretta

in this case it is the particle-like behavior of the particles that matter. In wave particle duality, it isn't that the physical particle is either a wave or a particle, but exhibits both particle like and wave like properties. In order to fully explain phenomena, you have to use both sets of properties.

You are referring to Heisenberg's uncertainty principle when you say the more we know about the position, the less we know about velocity. Keep in mind, this problem only exists for the very intricate measurements you might want to take in quantum mechanics. As far as the electron accelerator in a CRT television, even the size of each pixel on a television is not enough to make Heisenberg's uncertainty principle become important. It becomes important when you are trying to determine an electron's position around a nucleus.

13. Dec 18, 2009

### mintparasol

OK, I'm gonna stop drawing erroneous conclusions based on my reading of pop. science material!

14. Dec 18, 2009

### Staff: Mentor

Be careful here, the uncertainty relationship is between MOMENTUM and position, not velocity and position. For highly relativistic particles a large uncertainty in momentum may be only a small uncertainty in velocity.

15. Dec 18, 2009

### DARKSYDE

isnt the partical in question an atom? isnt that what the lhc is excelerating? if so the the uncertainty principle doesnt really apply like it would a proton or electron.

16. Dec 18, 2009

### rhenretta

the LHC is accelerating large hadrons (IE the LH part of LHC). These are particles consisting of 3 quarks (uud = proton, ddu=neutron). the uncertainty principle doesn't apply only because they are not interested in the actual positions of the particles

17. Dec 18, 2009

### humanino

Yes. Protons at those energies turn out to look like pancakes in the lab.
No. In Heisenberg's inequality it's not $p\Delta x \sim \hbar$ but $\Delta p\Delta x \sim \hbar$

I'd say naively one could expect the contrary from what you describe, in agreement with the previous observation. You expect the monochromaticity of your beam to be relatively constant, that is to say $\Delta p / p$ remains constant. As momentum increases, if this monochromaticity remains the same, $\Delta p$ keeps increasing so that $\Delta x$ follows the same as we can expect from length contraction. Remember that QM and special relativity are consistent into QFT.