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The energy of a proton (7 TeV) is given by special relativity formula in terms of $v$ and $c$. Which approaches infinite as $v$ approaches $c$.

Is this energy actually measured, How?

I know another way it is calculated is via bending moment in the accelerator. But that is also a calculation.

Measuring means in terms of real measurement for example - how much temperature change it causes.

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  • $\begingroup$ The beam dumps at powerful accelerators are engineered to handle the thermalized energy of the beam, and instrumented for monitoring. I haven't seen the data, but it does exactly what you suggest. Not that anyone thinks of it as a test of the beam energy: these are people for whom relativity is a work a day reality. I have personally worked on issues related to the thermal load imposed by a multi GeV, tens of microamp electron beam passing through a 6% radiation-length cryoliquid target. Again, no one bothered to think "Oh, this tests the energy formula", but results ere as expected. $\endgroup$ – dmckee Jan 29 '17 at 9:33
  • $\begingroup$ @dmckee: That is what I was looking for. It does not have to be an accurate measuremnent, only an order of magnitude verification is good enough to indicate whether energy is relativistic, or force. Thx $\endgroup$ – kpv Jan 29 '17 at 19:20
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What is the need to transform the kinetic energy of a single proton to thermal energy in a large ensemble of particles?

Temperature , a thermodynamic measure, is defined only statistically, not for a single particle. An experiment could be set up but it has no meaning since the momentum of a charged particle can be easily measured by its bending in a magnetic field, which is the way charged particle energies are measured in particle experiments.

A second way is using calorimetry which measures both charged and neutral energies, in calorimeters, with much larger errors. Note "calorimetry". It is the depositing of energy in large number of particles in a medium, calibrated to give the energy of the impinging particle.( I guess one could get a temperature from the energy deposited but that would be putting the cart before the horse).

In any case, if bending a track in a magnetic field needs a simple calculation to get the energy of the proton, getting it from the calorimeter needs a lot more calculations. Everything in particle physics depends on calculations.

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  • $\begingroup$ I was thinking of verification of actual energy. Suppose (by a rare chance) the force behaves relativistic rather than momentum and energy. Say a particle is exposed to an electromagnetic force F and suppose the relativistic effect of F1 is given by $$ F = \gamma F1 = \frac{F1}{\sqrt{1 - \frac{v^2}{c^2}}} $$ Then this effect would apply to bending moment as well. And it can give an impression of infinite momentum/energy unless the momentum/energy is measured in some other way then EM. Because in this case EM verification becomes cyclic. $\endgroup$ – kpv Jan 29 '17 at 5:42
  • $\begingroup$ Anna, this would also prevent particles to be accelerated to $c$ in same manner as the momentum formula does. $$ p = \gamma m v = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$ $\endgroup$ – kpv Jan 29 '17 at 6:28
  • $\begingroup$ You could use the relativistic formulas you give in a time of flight experiment for a single proton. This is not a useful method for an experiment studying interactions but has been used in beams google.gr/… $\endgroup$ – anna v Jan 29 '17 at 7:00

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