Can we stop protons from moving in LHC?

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Hi my question is simple, can we make the protons stay in a fixed position (not accelerating them) in LHC by using the powerful magnets there?

Thanks
 
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Probably, but why?
 
You can't change a proton's momentum with magnets. In principle, I suppose you could run the accelerating RF in reverse, which would take a minute or two to get the protons moving slowly.
 
No, the magnets are not designed for that. Penning traps can do that, and various experiments at CERN do store protons or even antiprotons.

Vanadium 50 said:
In principle, I suppose you could run the accelerating RF in reverse, which would take a minute or two to get the protons moving slowly.
You would lose them in the process - slowing down means blowing up their bunch size and then they don't fit in the narrow parts of the beamline any more.
 
That's why I said "in principle". In practice there isn't a way to do a gamma_T jump in the LHC, so you can't decelerate below transition energy.
 
I realize this post is almost 1 year old, but I'm still curious (with regards to proton speed), as to what would happen to a proton if its speed = 0. In a perfect vacuum, would its mass be different? And if not in a vacuum (for example if one shone a torch on it), would it be quickly be bombarded by other protons back to its original speed?
 
The mass of a proton is always the same.
wifirex said:
And if not in a vacuum (for example if one shone a torch on it), would it be quickly be bombarded by other protons back to its original speed?
?

If a proton at rest collides with a proton of the LHC beam, they either scatter a bit (the proton at rest gets a small amount of kinetic energy, the beam proton gets deflected a tiny bit) or both get destroyed and produce new particles.
 

Does this include protons with kinetic mass? For example a proton that has been ejected from a neutron that decayed? My understanding was that not all protons were created equal due to Lorentz transformations. But my real question behind all of this is - I wanted to know if protons behave differently when they are at rest compared to when they are moving.

Second point - disregarding the LHC for a minute; if a proton at rest gets hit by a second proton on a perfect tangent and is traveling at the speed of light, then wouldn't they split the kinetic energy and then both be traveling at 1/2 the speed of light?

Lastly back to the LHC - can a proton at rest truly be shattered by a beam of protons? Would it not have to be traveling in the opposite direction at a similar speed to achieve this?
 
There is no "kinetic mass".
wifirex said:
I wanted to know if protons behave differently when they are at rest compared to when they are moving.
They are the same particles, but of course there are differences when they are moving. If a moving car hits you it is the same object as a car at rest, but the result is different.

What does "on a perfect tangent" mean? Protons are not billard balls.
Protons cannot move at the speed of light.
wifirex said:
then wouldn't they split the kinetic energy and then both be traveling at 1/2 the speed of light?
No.
Also note that splitting the kinetic energy (which does not happen) would work differently. If you half the kinetic energy of a nonrelativistic object (a car, for example), you reduce its speed to ~70% its initial value.
If you half the kinetic energy of a proton moving at 99.9999% the speed of light, it moves at about 99.9% the speed of light (didn't calculate it, could be slightly more or less, but it should give an idea).

wifirex said:
Lastly back to the LHC - can a proton at rest truly be shattered by a beam of protons?
Not by a beam, by a single collision. Yes. Again, this has a classic analog: a car at high speeds crashing into a car at rest leads to an accident as well.
 
  • #10
Shame. I just spent an hour elaborating on a group of ideas/theorems that I hoped you would all be privy to - only to have them wiped by the stroke of a simple button. Forget the physics and get into the sharing and storing of I information - or we all lose.

All the best to your venture!
 

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