# Factors of $c$ when giving masses in natural units?

I am starting a course on particle physics, and have been introduced to natural units. I am slightly confused, because we are using 'natural units', and yet masses are stated as, for instance, $$139.6\ {\rm MeV}/c^2$$ for a pion. For some reason I thought it would be sensible for the $$c$$ here to represent a dimensionless, numerical factor (although I realise now that is contrary to the whole point of natural units- not to have the factors!). I attempted a question on natural unit conversion- specifically, calculating the Compton wavelength in natural units and then convert to SI units, for this pion. The answer gave the Compton wavelength ($$\lambda = 1/m$$ in natural units) as $$7.16\times10^{-3}\ {\rm MeV}^{-1}$$. i.e. the factors of $$c$$ in the statement of the mass seem to have been completely ignored!

So my question is: why do factors of $$c$$ appear in the statement of masses in natural units, and are they purely symbolic or need they be taken into account in any calculation in natural or SI units?

• They did it to confuse you, or out of sloppiness. It is understood the cs in $MeV/c^2$ are equal to 1 in natural units, so you ignore them. But you instantly know how to convert those in any other system of units, by plugging in c's value in your favorite system. Feb 1, 2019 at 20:22

The most famous formula ever is probably $$E=mc^2$$. We use this fact to express mass in terms of energy.

But what are those energies and masses? You know, $$m$$ in that formula is not the classical mass, it depends on the velocity. We should write $$E(v)=m(v)\cdot c^2$$.

In particular, when we set $$v=0$$, we have the rest mass, $$m_0$$, to which we associate a rest energy $$E_0$$. $$E_0$$ is defined as

$$E_0=m_0\cdot c^2$$

Where $$m_0$$ is the "usual" mass, the rest mass.

Now, since measuring energies is pretty much easier than measuring masses, we tend to measure rest energies. So we usually know $$E_0$$. That's why we would write the mass as

$$m_0=\frac{E_0}{c^2}$$

This has an advantage: when $$E_0$$ is given in multiples of electronvolts, the result is quite beautiful. For example, for an electron, $$E_0\simeq 511 keV$$, so we can write that the (rest) mass of the electron is $$m_0\simeq 511 keV/c^2$$.

And of course the $$c^2$$ is there. If you perform that division, you will get $$m_0\simeq 56,78 \ \frac{keV}{(1 m/s)^2}$$; and if you transform $$keV$$ to Joules, you'll get $$9,1\cdot 10^{-31} \frac{J}{(m^2/s^2)}$$, which are the same number in kilograms.

So, answering to your first question, yes: the $$c$$ is actually there.

Now, what really happens is an inconvenient truth: the pysicist are lazy. We don't like writing the same things so many times, so we just relax and don't care haha.

The mass appears in so many equations, many times indeed. For each time the mass appears, there is a factor $$E_0/c^2$$.

So, what we do is saying: let's stop writing $$/c^2$$. We do know mass has units of mass, so, whenever we see a mass in $$MeV$$, we must deduce that it is because there is a hidden $$c^2$$. It's hidden, but it is there.

So when you read $$m_0=0,511 MeV$$, you must do a dimensional analysys and say: "hey, units don't match; that means I have to fill the equation with $$c$$'s until they match".

There's always the same question: where do I add them? I've got different options. Well, both are equivalent. You can fill them multiplying the LHS, or dividing the RHS:

$$m_0=0.511 MeV/c^2$$, or $$m_0c^2=0.511MeV$$. They are equivalent.

So the answer is that the $$c$$ is hidden, omitted for lazyness, but it is actually there. Students hate this because it is really confusing at the beginning.

Personally, I keep hating it, but I can't swim against the stream all the time. I keep writing $$c$$'s in my notes, and people actually likes it. Everybody prefers seeing all the $$c$$'s, because reading them is free and helpful, they just don't want to bother to write them, haha.

Natural units

But this is quite little rigurous, so we've invented a good argument: I am using another unit of lenght, let's call it $$L$$, and I define it as $$L=c\cdot(1s)$$.

Like this, light travels $$3\cdot 10^8$$ meters in one second, or $$1L$$, which is the same.

So I can write that the speed of light is $$c=1L/s$$. Like this, $$c$$ has the numerical value of $$1$$.

This is good, because ¡, liek this, the energy $$E_0$$ (which is energy), and the mass, expressed in "energy / c² in this new unit" will have the same number: $$511 keV$$.

The only difference now is that $$E_0=511keV$$ is a complete datus, and $$m=511 keV$$ is lacking a division of $$c^2$$, but that would be dividing by $$(1 L/s)^2$$, so the only problem now is units, and it is no longer the numerical value.

When we used the international system, $$m_0=511 keV/c^2$$, and that $$c^2$$ was completely there.

When we use these "natural units", $$m_0=511keV$$, and the only difference is that those keV are actually divided by a new unit, but the numebr doesn't change.

Both interpretations are equivalent.

In natural units you set $$\hbar = c = 1$$ (dimensionless), so you don't have to carry the $$c$$'s even when writing down any unit. Say, by the relativistic energy relation, $$E^2 = (pc)^2 + (mc^2)^2$$ setting $$c=1$$ makes energy, momentum and mass have the same unit, which the standard is to pick it as $$\text{MeV}$$ or other energy unit. So, although in SI we measure the electron mass as $$0.511 \text{ MeV}/c^2$$, in natural units we simply say the electron mass is $$0.511 \text{ MeV}$$ (no need for $$c$$). And by the commutation relation, $$[x,p] = i \hbar$$ making $$\hbar = 1$$ makes distance have the inverse unit of momentum, which is $$\text{MeV}^{-1}$$, so your result is correct. If you want to convert it to SI you plug in the necessary $$\hbar$$'s and $$c$$'s to get the correct unit. Example: multiplying the Compton wavelength you calculated by the numerical values of $$\hbar c$$ - which in SI has units of $$\text{eV} \cdot \text{nm}$$ - will give you the correct value: $$\lambda = \hbar c (7.16 \times 10^{-3} \text{ MeV}^{-1}) = (197 \text{ eV}\cdot\text{nm})(7.16 \times 10^{-9} \text{ eV}^{-1}) = 1.41 \times 10^{-6} \text{ nm}$$