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Conservation of charge

  1. Feb 14, 2015 #1
    I don't know much about the topic but if I'm correct charge should always be conserved. What happens when two protons collide in the LHC? Don't they split into quarks and then decay, lowering the amount of positive charge in the system?
  2. jcsd
  3. Feb 14, 2015 #2
    Energy is always conserved. Matter and energy are inter-convertible, so it is not necessary for charge (a characteristic of matter/anti-matter) to be conserved.
  4. Feb 14, 2015 #3


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    This is simply wrong. Charge is most definitely conserved. If protons collide, this will result in new particles, but the sum of all of the charges of the outgoing particles must add up to the same total charge as the incoming charge.
  5. Feb 14, 2015 #4
    • Repeated wrong explanation
    The OP was talking about two individual protons colliding in the large hadron collider. When they collide, it is possible that they are converted into energy. The energy is eventually converted back into particles where the total charge will be conserved, and that happens when conditions are feasible. Energy however, is conserved at any stage.
  6. Feb 14, 2015 #5
    I'm sorry, but you're just talking nonsense here.
  7. Feb 14, 2015 #6


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    So is charge. If you think otherwise, please give a specific example of a process in which you think charge is not conserved at some stage.
  8. Feb 14, 2015 #7


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    I am sorry, but this sounds like you have been given an extremely popularised version of what goes on at the LHC and opened it up for your personal interpretation. This is not an accurate description of events and you would do well to realise that. Physics Forums is a great place to learn about and discuss physics, in part because we do have experts here as well as laymen. However, this requires that we all ask questions when we are wondering about things and do not throw out overly popularised theories when the understanding of the underlying physics is not there.
  9. Feb 14, 2015 #8


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    Why would they? If for example you have :
    [itex] p + p \rightarrow p + p + p + \bar{p} [/itex]
    Do you see a violation of charges? In this case the valance quarks [itex]uud + uud[/itex] will give as a product more quarks [itex]uud+ uud + uud+ \bar{u} \bar{u} \bar{d}[/itex]. And the same goes on for all the processes.
    In fact the products can be a lot many particles, especially by hadronization, which result in jets of hadrons (pions and protons/neutrons are measured) coming out of the collision.

    I don't really like the phrase "converted into energy". This doesn't make a real physical sense, except for if someone abuses the meaning of the formula [itex]E=mc^2[/itex] in the same way Einstein used to do [he enjoyed that formula more than [itex]E^2 = p^2 c^2 + m^2 c^4 [/itex] ]. For me, especially for the many body systems the first formula doesn't make a real physical sense, what is the "mass" in that case? Especially when you go on saying that energy is converted back... how do you define energy in your case? In any case the particles colliding don't have to pass from one stage to the other via some energy conversion, they just create new particles [through some of their energy] that fly around, undergo decays or are detected.
    Every kind of symmetry has to hold at any stage if it is to be a symmetry. If for some reason we find that the symmetry is not there, then we must have something that violates the symmetry (for example the violation of parity and time reversal by the weak interactions).
  10. Feb 14, 2015 #9
    I apologize for mixing up concepts here.

    My confusion stemmed from the fact that I'd once watched a documentary on Discovery channel in which it was stated that quantum fluctuations result in the spontaneous formation and disappearance of particles(i.e. virtual particles) in high energy environments, and hence I took this to be a case which was a violation of the conservation of charge. On checking back on it now on the Internet, I read the crucial part which I'd missed - that a particle is always spontaneously formed along with its anti-particle(pair production) , so no actual violation of the law occurs. Thank you for correcting my half knowledge along the way.
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