# Using Relativisitic Units (where c = 1)

1. Mar 12, 2014

### Chetlin

1. The problem statement, all variables and given/known data
It's not really an assigned problem, but it was part of one, and it's a concept that I'm not quite getting.

2. Relevant equations
In special relativity, energy can be expressed using the equation $E^2 = (pc)^2 + (mc^2)^2$, or $E = \sqrt{p^2c^2 + m^2c^4}$. 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 $E = \sqrt{p^2 + m^2}$. And, often, p and m are expressed in electron-volts, even though the "true" units would be eV/c for p and eV/c2 for m.

3. 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 eV and the momentum was p eV (the variables have the units of electron-volts). Are you able to just say that the energy is $\sqrt{p^2 + m^2}\:\:e\text{V}$? 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 $\sqrt{(p\:\:e\text{V}\!/\!c)^2c^2 + (m\:\:e\text{V}\!/\!c^2)^2c^4}$ 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?

2. Mar 12, 2014

### BruceW

both ways are fine. When you use something like relativistic units, you just need to remember what you used to turn the units from the usual SI units into the natural units. In this case, you just used one 'conversion' which is c=1. So if you have any equation in the natural units, then you can get back to the SI units by getting the dimensions correct by inserting 'c' into the places where it needs to be.

In other words, there is a unique way to convert between the two systems of units. So therefore, you can think of any equation, written in the two different set of units as being 'essentially the same equation'. However, if you forgot that you used c=1. (for example, you might have used c=2 but can't remember either way), then you cannot get back to your original equation. So it is important to remember what rule you used to do the conversion in the first place.

3. Mar 13, 2014

### Chetlin

Thanks! I understand this a lot more now.

That's something I considered too, setting c = 2 or something. It seems that if you do that, you can't express p and m with the same units now. The units for p become eV/2 and those for m become eV/4. The only way I can see to work with these kinds of units is to basically multiply things again to get it to the same form we had when c = 1. I know it's not "normal" to set c equal to anything else, but I know it's still something that can be done. However, it's tough to see how doing that wouldn't make the calculation messier than just leaving every c in from the beginning.

4. Mar 13, 2014

### BruceW

yeah, In this case, c=1 is definitely the nicest option, but as you say, we are still free to make whatever choice we want. As an example of when the conversion involves a numerical factor, there are Gauss-cgs units, which have a factor of $\sqrt{4\pi \epsilon_0}$ to convert the electric field from Gaussian-cgs units to SI units. So, in this case, they have a numerical factor of $\sqrt{4\pi}$ which is just used to make the equations a bit nicer.