- #1

Chetlin

- 36

- 0

## Homework Statement

It's not really an assigned problem, but it was part of one, and it's a concept that I'm not quite getting.

## Homework Equations

In special relativity, energy can be expressed using the equation [itex]E^2 = (pc)^2 + (mc^2)^2[/itex], or [itex]E = \sqrt{p^2c^2 + m^2c^4}[/itex]. Apparently, according to my textbook and professor and other things I've read, people often set the value of

*c*to 1, leaving the energy equation as [itex]E = \sqrt{p^2 + m^2}[/itex]. And, often,

*p*and

*m*are expressed in electron-volts, even though the "true" units would be

*e*V/

*c*for

*p*and

*e*V/

*c*

^{2}for

*m*.

## The Attempt at a Solution

In a question I had, I was given the mass of something in electron-volts, and eventually found its momentum, also in electron volts using 4-momentum techniques. It asked for the energy. Let's be general and say that the mass was

*m*

*e*V and the momentum was

*p*

*e*V (the variables have the units of electron-volts). Are you able to just say that the energy is [itex]\sqrt{p^2 + m^2}\:\:e\text{V}[/itex]? I did not feel comfortable doing that and at the end of class today I tried asking my professor if that was actually what we were supposed to do, but it was rushed since it was the end of class and everyone was moving and the only thing I think he understood was that I wasn't comfortable with relativistic units (and it's true, I'm really not, and that's what he told me). As the solution to the problem I gave the very inelegant-looking answer [itex]\sqrt{(p\:\:e\text{V}\!/\!c)^2c^2 + (m\:\:e\text{V}\!/\!c^2)^2c^4}[/itex] which I notice does seem to equal the same thing as the other answer, with all the c instances cancelling. But ... it just doesn't "seem" right to say that mass and momentum are in energy units and then add them to get energy units, since mass and momentum don't have the same units really and can't add. Is this the actual "right" answer or is the first way just as fine?