# E=mc^2 particle physics

1. Dec 4, 2014

### bert2002

Since in particle physics c can be take to equal 1, does this mean that Energy can equal mass if we use electronvolts ?

2. Dec 4, 2014

### Einj

Yes, exactly (of course this is an artifact of setting c=1 and you have to be careful in restoring the right units when needed). However, be careful on the definition of m. One usually says that m is the rest mass, i.e. a constant that only depends on the species of the particle and not on its momentum. In this case the equation becomes $E=\sqrt{m^2+\vec{p}^2}$, where $\vec p$ is the three-momentum of you particle.

3. Dec 4, 2014

### e.bar.goum

Sure. This is why you often see tables of the standard model with mass reported in eV/c^2. And why "mass excess" is normally in units of MeV in nuclear physics. Often, you leave out the /c^2. This frequently confuses students who are studying nuclear/particle physics for the first time - you've got to remember to restore units.

4. Dec 5, 2014

### Staff: Mentor

Similarly, many physicists say things like "the electron's momentum is 300 keV" which technically has the wrong units. What they really mean is either p = 300 keV/c or pc = 300 keV.

My personal convention is to write all relativistic equations in such a way that m always appears together with c2 as mc2, and p always appears together with c as pc. Then I can calculate the energy of the electron above as $$E = \sqrt{(mc^2)^2 + (pc)^2) } \\ E = \sqrt {(511 \text{ keV})^2 + (300 \text{ keV})^2)} \\ E = 583.1 \text{ keV}$$

Last edited: Dec 5, 2014