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How does you weigh a quark?

  1. Jul 10, 2005 #1
    I refer to the high energy collisions of particles. The more energy you put in, the more mass you get out... Or at least heavier particle/antiparticle pairs. But, I only know of measuring the angles of trajectory around a magnetic fields and using the mass/charge ratio assuming the charge is the same. Yeah, Chemistry student here. So is there another method of weighing broken particles that I'm just oblivious to?
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  3. Jul 10, 2005 #2

    James R

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    You can use conservation of momentum in particle collisions, which tells you something about the masses without needing to know anything about the charges.

    This is how the neutrino was discovered. Momentum didn't seem to be conserved in beta decays. People realised that an extra particle must be involved.
  4. Jul 10, 2005 #3
    So charges are completely ignored? What if a change in charge happens instead of energy turning into mass? The particle, through the conservation of momentum, would appear to be heavier/lighter than it actually is. I'm not dissing Einstein, I just like to explore open avenues to try and close them.

    Arthur Conan Doyle would be proud...
  5. Jul 10, 2005 #4
    wouldn't you have to factor in charge to get the right answer for conservation of momentum?
  6. Jul 10, 2005 #5
    I would think that a change in charge would result in...well...a change in charge. Since a single quark would either decay or enter into a bound state, you should be able to reconstruct the charges of all the resultant particles. If you're colliding p-pbar, then you better get a total charge of zero. Plus, since there is no mechanism for charge of a particle to change during collision, it would probably produce several more unexpected results.

    The bottom line I think is that particle physicists count the charge of all the particles that come out of their collision. Any shift in charge would show up when they added everything together. Since this has never been observed, it seems that charge is being conserved during the interactions in the collision.
  7. Jul 10, 2005 #6
    "since there is no mechanism for charge of a particle to change during collision"

    Interesting. Going back to the point of topic, if you use the same method of measuring charge to measure mass, how do you know that you're not changing the charge? And adding the charges up wouldn't show up as Particle/Antiparticle pairs have the same mass and charge when created so the net result would still be the same.

    "What I'm trying to ask is does the actual apparatus to measure the mass of heavy particles created during high energy collisions use the same process to measure charge?" and assuming it is, "What evidence is there that this extra mass isn't a change in charge?"

    And don't anyone quote E=mc^2, please... I'm delving into the value of "m" here.
  8. Jul 10, 2005 #7

    James R

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    Momentum is totally independent of charge, and so is energy. For example, momentum is mv, mass times velocity. No mention of charge.

    You can't change energy into charge, or charge into mass, or momentum into energy. You CAN change mass into energy, and vice-versa.

    In working out what is going on in particle collisions, physicists make use of a whole bunch of known conservation laws which basically say what goes in is what comes out. For example, total energy must be conserved in a collision. If the incoming particles have a certain total amount of energy (including binding energy, potential energy etc.), then all of that energy must still be present in one form or other after the collision. Same goes for the total momentum of a system of particles. If there are no external forces on the system, the initial and final momentum total must be the same.

    Total charge is also conserved in all but a rare few particle interactions.

    These kinds of rules mean that certain processes are physically impossible. For example, a proton can never decay into two photons, because charge would not be conserved.
  9. Jul 10, 2005 #8
    i thought [tex]\rho=\frac{h}{\lambda}[/tex], not mv
  10. Jul 10, 2005 #9
    Just giving a long-winded version of what James R said:

    Asking whether or not the charge changes is a valid question, and I'm not sure how best to explain an answer. Hopefully on of the forum regulars will save me once I get into enough trouble, but I'll give it my best shot.

    If you measure charge in a cloud chamber type environment, then yes, you are measuring charge and mass at the same time, because you're measuring the curvature of the track. However, particles like quarks sooner or later decay into stable particles that we're familiar with; electrons, protons, neutrinos and the like. Unless you're saying that the "extra" charge has disappeared by that time (in which case it reverts back to mass the moment it's measured, which makes the theory very strange), those particles should also exhibit unusual charge. So what we're really looking for is an electron or proton with very strange, possibly non-integer charge. And we've never found one: so far high-energy stable particles correspond to the appropriate recoil behaviors due to their mass. High energy nuclei do not experience measurable alterations to their electromagnetic behavior. An electron is an electron is an electron. Their electric field may distort if you view them at high velocities, but so far their charge is always their rest frame charge.

    Which leads to one of the conclusions of quantum mechanics. The charge of a particle is an inherent physical property of the particle involved. Like spin, it does not change. It is a single number. It has no relation with the particle's energy or with its momentum. If this were not true, then measuring the charge/mass ratio of an electron by accelerating it through a simple accelerator would give you different values as you cranked the voltage up (adjusting for SR as usual). If this is true than the charge of a quark is fixed. If this is not, then we'll need to find out where these off-kilter daughter particles have gone.

    If someone has a better argument, I think I need one.
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