# A Natural units and electron mass

1. Sep 16, 2016

### spaghetti3451

I am trying to express some physical constants in natural units of $c=1$ and $\hbar=1$.

Let's start with the electron mass. In SI units, the electron mass is $0.511 \text{MeV}/c^{2}$. I understand that in natural units, the electron mass is simply $0.511\ \text{MeV}$. Now, is the electron mass given by $\frac{\hbar c}{0.511 \text{MeV}} = 1.01 \times 10^{-21}\ \text{cm}$?

Last edited: Sep 16, 2016
2. Sep 16, 2016

### Simon Bridge

Notice that expressing mass in MeV alreafy has c=1 factored in - so why would the conversion factor have an explicit c in it?
How often do you normally convert units by dividing by the quantity you are converting?
When you convert quantity x from unit a into unit b, wouldn't you just multiply it by a conversion factor?

Observations:
To convert MeV into natural units of length, you need to know how many natural length units there are in 1MeV.
In general - setting $\hbar = c = 1$ still leaves some arbitrary units.
A "centimeter" is not usually a natural unit.

See, for example, plank units.
https://en.wikipedia.org/wiki/Natural_units#Planck_units

It looks like the calculation you did just normalizes the electron rest mass and tells you how big the unit of length is, in terms of other systems of units, for the particular set of natural units chosen.

3. Sep 16, 2016

### spaghetti3451

The mass expressed in $\text{MeV}$ needs to be converted to $\text{cm}^{-1}$. Now, $\frac{\text{MeV}}{\hbar c}$ has dimensions of length. Therefore, $\frac{1 \text{MeV}}{\hbar c}$ gives the conversion factor from $\text{MeV}$ to $\text{cm}^{-1}$ in SI units.

There is a typo in my calculation. It is actually $\frac{0.511 \text{MeV}}{\hbar c} = 2.58 \times 10^{12} \text{cm}^{-1}$.

It does, but this system of units is heavily used in particle physics and cosmology.

4. Sep 16, 2016

### spaghetti3451

What do you think?

5. Sep 17, 2016

### Simon Bridge

You seem to be thinking of plank units ... to deal with mass-energy you usually need to also normalize the gravitational constant.
Other schemes normalize the mass of the electron or some other particle commonly dealt with... which amounts to picking a different value for G.

If m is the electron mass, then $(mc^2)/\hbar c$ gives dimensions of [E]/[E.L] = L^-1 ... so that comes out right for you... just makes sure you express $\hbar c$ in units of MeV.cm