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Measuring relativistic energy

  1. May 9, 2006 #1
    how do we measure the relativistic energy of a tardyon?
  2. jcsd
  3. May 9, 2006 #2


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    The laboratory answer is via calorimeters,

    http://en.wikipedia.org/wiki/Calorimeter_(particle_physics [Broken])
    Last edited by a moderator: May 2, 2017
  4. May 9, 2006 #3
    A calorimeter will only measure the kinetic energy of a particle, not the total energy.

    Last edited by a moderator: May 2, 2017
  5. May 9, 2006 #4


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    One can also determine rest energy via calorimetry if one looks at particle /antiparticle annhiliation.

    This may not be the most precise way of determining rest energy, though.

    Also, I took the question to be more in terms of measuring the energy of a single particle, i.e. the particle physics version of the calorimeter. The original principle was based on measuring the actual heat energy, but they have since then become more sophisticated and complicated. Some of the more sophisticated versions routinely take into account known rest masses, i.e. they identify that a muon was created, and add in the known energy to do that into the particle energy.

    Besides the Wiki article,


    talks about this a little

    going into a lot of the details (Visit the above links and click on the sub-links - also see electromagnetic vs hadronic calorimeters).

    I hope that this was the answer that was being looked for - i.e. how do particle physicists actually measure the energy of particles.
  6. May 10, 2006 #5
    If the particle is charged then use a cyclotron. Assume B (strength of magnetic field) is a given. Measure r = radius of circle particle is moving in. Measure the speed the particle is moving at. Then use the cyclotron relation p = qBr derived here


    to get m = p/v = relativistic mass. Multiply by c^2 to get E = total inertial energy.

  7. May 10, 2006 #6
    Actually that only tells you the kinetic energy, not the total energy. Are you saying that a calorimeter measures E, not K? In any case the rest energy is so small compared to the kinetic energy there is little difference between the two.

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