Do physicists use particle "energy" to refer to kinetic energy?

Question

In 1963, this paper was written about the effects of radiation on solar panels. The paper states that:

When electrons at energies greater than 145 KeV and protons at energies greater than 98eV bombard a silicon crystal, they can displace an atom from the crystal lattice, producing a lattice vacancy and a recoil atom which comes to rest as an interstitial atom.

However, the resting energies of electrons and protons are far greater than this, at roughly 511 KeV and 938 MeV respectively. I concluded that the paper was referring to kinetic energy rather than total energy, and adjusted my calculations based on this conjecture.

So: Was I correct to assume that the paper referred to kinetic energy, or was it instead some other measure of the particles' energy?

More generally, is there a standard meaning for a particle's "energy" when referring to such particles moving at relativistic speeds?

Answer 1

Yes, in this case what's meant by "energy" is the kinetic energy $K = (\gamma-1)mc^2$ that the incident particle can transfer to the target system. As you point out, it wouldn't make sense to talk about an electron, which has rest energy $E_0=mc^2=511\rm\,keV$, to have a total energy $E=\gamma mc^2 = mc^2 + K$ of only 100 eV.

For ultra-relativistic particles with $E\gg E_0$ it's a reasonable approximation to think of the kinetic energy and the total energy as being identical, and for nonrelativistic systems the meaning is unambiguous from context, so it's only in a fairly narrow energy region around $\gamma\approx2$, or $K\approx E_0$, that you have to be careful to distinguish between kinetic and total energy. That makes us sloppy. Sorry.

Answer 2

Special relativity theory gives you $(mc^2)^2 = E^2 - (pc)^2$, where $m$ is the rest mass of particle (511 keV for electrons), $p$ is the momentum and $E$ the total energy. When you have the natural unit system where $c = 1$, the equation becomes $m^2 = E^2 - p^2$, which might be strange looking for you.

When an electron has an energy of 145 keV, it must be the kinetic energy only, there is just no way that you can choose $p$ such that $E$ could be smaller than $mc^2$. For higher energies, it could become ambiguous. If you have $E \gg mc^2$, then the actual difference will be small because the particles are ultra-relativistic anyway.

In a particle physics context, where particles can annihilate each other, energy is usually meant as the total energy including rest mass, just like the above equation says.

Answer 3

Not only is it common to quote kinetic energy as the energy in contexts other than high-energy physics, it's basically also where the whole convention of writing energies in electron-volts comes from: $1\:\mathrm{keV}$ is the kinetic energy that a singly-charged particle, regardless of mass, picks up if you allow it to drop through the potential of a linear accelerator, i.e. capacitor charged to $1\:\mathrm{kV}$. And this is the energy such an e.g. electron can then put to use when colliding with the cathode of an X-ray tube. X-rays normally don't get close to electron rest mass, so as said before it's unambiguous here which energy is meant.

(For the X-rays themselves it's of course unambiguous anyway, thanks to zero rest mass.)

Answer 4

Yes. Bombardment implies kinetics. The (rest) mass of electrons and protons is fixed, it would make no sense to discuss it as a variable. More generally, a particle's energy can be considered to be composed of its rest mass, its kinetic energy, and it's potential energy. There's no single meaning to the term, in fact, energy is an abstraction meaning there is no substance or thing called "energy".