# Proton mass at rest?

1. Mar 22, 2015

### grandpa2390

When we say 938 MeV/c^2, is the speed of light already factored in? because I have read in some places where if you multiply this by c^2 you get the energy, and that is 938 MeV rather than 938* (2.998e8)^2

so when considering the rest mass of the proton, is the c^2 a variable, or a unit?

2. Mar 22, 2015

### stevebd1

Unless I'm mistaken, 938 MeV is the energy. If you divide by c2, you get the mass. For the mass in kgs, 1 electron volt (eV) = 1.60217657×10-19 joules

3. Mar 22, 2015

### grandpa2390

eV/c^2 is in fact a unit of mass. eV is a unit of energy. this isn't answering my question. the question is whether that c^2 is a unit, or if the divide by 2.998^2 is never actually performed so that it is a variable. b

4. Mar 22, 2015

### DrGreg

As stevebd1 says, if something has a rest mass of 938 MeV/c2, it has a rest energy of 938 MeV. Surely that is answering your question?

5. Mar 22, 2015

### Staff: Mentor

You can consider c as a unit, specifically a unit of speed, it is a constant not a variable.

If you would like to change to SI units then you would use the conversion factor 1 c = 299792458 m/s. But it is rarely necessary to change to SI units and you you would never just divide by 2.998^2.

Are you comfortable with how to do unit conversions correctly?

Last edited: Mar 22, 2015
6. Mar 22, 2015

### jbriggs444

The c^2 amounts to a conversion factor. It is needed in a system of units where the speed of light is not one unit distance per unit time. You do not just divide by $(2.98 \times 10^8)^2$ without also dividing by (meters/second)2

Edit: Dale is just too fast.

7. Mar 22, 2015

### grandpa2390

so if the c^2 is a unit, why then in this problem I am doing, they plug in 938 MeV / c^2 into m for mc^2 and the result is equal to 938 MeV. shouldn't they have multiplied by 2.998^2 in order to do the conversion to MeV? if it is 938 MeV / c^2, then converting back to energy should require multiplication by c^2. it should be 938 *2.998^2. which is not equal to 938 MeV unless the division by c^2 is never actually performed and it remains variable so that plugging in to mc^2 cancels it out with the c^2 in mc^2.

I consider myself very comfortable with conversions and that is why the solution to this problem is confusing me.because if they are units, the units work out, but not the numbers.

Last edited: Mar 22, 2015
8. Mar 22, 2015

### grandpa2390

no you missed the point of my question. No worries though.

9. Mar 22, 2015

### Staff: Mentor

$c$ is (approximately) equal to $2.998\times{10}^8$ only if you are measuring distance in meters and time and seconds. If you also measure energy in Joules you can divide the energy by $(2.998\times{10}^8)^2$ to get the mass in kilograms. However, here we are using units in which energy is measured in MeV, distances are measured in light-seconds, and time is measured in seconds, and the unit mass is whatever value is determined by these choices. Clearly you don't want to divide by $2.998\times{10}^8$ meters per second, you want to divide by one light-second per second.

10. Mar 22, 2015

### grandpa2390

I am not sure I understand what you are saying, let me repeat it in my own words to make sure. to convert MeV to MeV/c^2, the units of c must be in light seconds per second rather than meters per second. and since light travel 1 light second per second, then this is equal to 1. so multiplying by c^2 in mc^2 would be equal to multiplying by 1. If that is correct, it makes perfect sense. thankyou!

11. Mar 22, 2015

### Staff: Mentor

Yes $E = m c^2$, so if $m = 938\;MeV/c^2$ then by substitution $E = (938\;MeV/c^2) c^2 = 938\;MeV$.

Can you not see that the $c^2$ factors cancel out?

No. That wouldn't be mathematically correct. See the math above. The $c^2$ terms cancel out, and even if they didn't you would never just randomly multiply by 2.998^2. I am not sure what would lead you to believe that this would be correct.

Yes, it does require multiplication by c^2. See above. I multiplied m by c^2 to get energy.

Why do you think this?

I don't understand this comment. It doesn't matter if it is a constant or a unit or a variable. If you divide and then multiply by it then it will cancel out. I think that you are mentally making an artificial distinction between the mathematical treatment of variables and units, and this is likely leading to your confusion. I don't know why you think this, but it is not correct. When you divide by $c^2$ and multiply by $c^2$ it cancels out regardless of whether it is a variable or a unit or whatever.

12. Mar 22, 2015

### grandpa2390

Don't worry about. I understand now.
You could look at this as ls/s
or as saying the number is part of the unit. like if you said he travel 60 miles per 60 minutes, where the division is never actually performed, the number becomes part of the unit. 2.998e8 m/s ^2 is the unit. the actual division is never actually carried out (just like when saying 60miles per 60 minutes, the division 60/60 is never actually performed, and to convert back to miles, I multiply by 60 minutes). That is the answer I needed help getting to because I didn't understand the nature of this new unit. I have never dealt with units replaced by variables before. in my mind, I was mistaking c for the unit, when the real unit is m/s. c is a representation of saying 938 MeV per 2.998e8 m/s squared. the division is not to be performed because then it would no longer be in per c^2, it would just be in per m/s)^2.

@Nugatory thank you

Last edited: Mar 22, 2015