# How is proton energy measured (not calculated) for the protons accelerated near speed of light

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.

• 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. – dmckee Jan 29 '17 at 9:33
• @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 – kpv Jan 29 '17 at 19:20

• 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. – kpv Jan 29 '17 at 5:42
• Anna, this would also prevent particles to be accelerated to $c$ in same manner as the momentum formula does. $$p = \gamma m v = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}$$ – kpv Jan 29 '17 at 6:28