Natural unit conversion

Question

I'm a bit confused about different notions of "natural units" that I encounter occasionally. I'm familiar with Planck units, and in particular I can understand the conversion between, say, metres and Planck length, because by noting that

$$[\hbar] = ML^2 T^{-1}\\ [c] = LT^{-1}\\ [G] = M^{-1} L^3 T^{-2}$$

we can get

$$\left[\sqrt{\frac{\hbar G}{c^3}}\right] = L$$

which means that $\sqrt{\frac{\hbar G}{c^3}}$ is our conversion factor between SI (metres) and Planck length.

However, I've recently read papers where the authors only say that $c = \hbar = 1$, leaving out $G$. I'm a bit confused by that, because that means that taking combinations of $\hbar$ and $c$ will leave factors of $M$ in the dimensional analysis which won't cancel with anything. How can I then obtain the conversion factor between metres and the chosen units, analogously to $\sqrt{\frac{\hbar G}{c^3}}$ in the Planck case?

I'm sure that I'm just being stupid here but I'd appreciate a quick explanation.

Answer 1

In natural units you fix only two dimensional constants, which means that you still have one 'unfixed' dimension with which to measure units. Certain physical quantities (velocity, action and their powers and products, which include e.g. $[e^2/4\pi\varepsilon_0]=[\alpha\hbar c]$) are dimensionless, but in general most physical quantities are not dimensionless and require some form of unit to express them.

Energy, for example, can't be expressed in terms of $\hbar$ and $c$. (For a quick argument, $[E/\hbar]=1/T$ is not a power of $[c]$, and other powers of $\hbar$ would leave an unbalanced mass dimension.) This means that you need a dedicated energy unit, which is usually taken to be the electron-volt since the context is normally particle physics.

Natural units are useful because once you've chosen one dimensionful unit, you can express any other quantity in terms of it. Thus, to measure length in terms of an energy unit, for example, you write $$ [L]=[\hbar^\alpha c^\beta\mathrm{eV}^\gamma] $$ and you solve to get $[L]=\hbar c/\mathrm{eV}$. An easier example is mass, which is measured in $\mathrm{eV}/c^2$. Time is measured in $\hbar/\mathrm{eV}$, momentum in $\mathrm{eV}/c$. And so on.

Conceptually, most high-energy theorists choose mass as the fundamental independent dimension, though somewhat confusingly they tend to measure it in $\mathrm{eV}/c^2= \mathrm{eV}$. (In some senses this is perfectly natural: the $\mathrm{eV}$ is the most convenient unit for particle-physics work, and mass is a "fixed" dynamical quantity that depends on the particle but not on its state, and they do have the same dimension in natural units. It's still a tiny bit weird though.) In fact, they will often hijack the square-bracket notation to simply indicate the power of mass in the dimension, as explained here, so that $[L]=[T]=-1$, $[\hbar]=[c]=0$ and $[p]=[m]=1$.

Answer 2

$\hbar=c=k_B=1$ is what we normally refer to as natural units in particle physics. If we additionally set $G=1$ we are ending up with only dimensionless quantities. Sticking with the first case, we have one unit to choose left . Normally, one chooses the unit of energy to be $[E]=eV$ electron volts. From $c=1$ for instance we know that time and length now have the same unit, which is $eV^{-1}$. My way to go on is the following: I am looking for easy equations like $\Delta E \Delta t<\hbar=1$ to see for instance $[t]=eV^{-1}$

Very useful is the fact that $\hbar c=197 $ $MeVfm$, which can be used for various unit conversions. The best way I know is to add appropiate factors of $\hbar=c=k_B=1$ in order to recover SI units.

Answer 3

The easiest way is to add factors of $\hbar$ and $c$ as needed. For instance you can measure energy in $eV$ (electron volts). And then you measure mass in units of $eV/c^2$, for instance you can say the mass of an electron is $511\times 10^3eV/c^2$.

And for other units you also measure them as some power of eV multiplied by some power of $c$ and some power of $\hbar$. For instance $E=\hbar \omega$, so angular frequency can be measured in units of $eV/\hbar$.

Answer 4

Conversions between arbitrary physical quantities will also involve powers of $G$, but these powers are already specified in $\hbar = c = 1$ units. You can convert everything to some power of a length, but you cannot convert between different powers of a length.

If you are working in Planck units, then it's trivial to convert the expression to some arbitrary SI units.Suppose you have an expression $f(x_1,x_2,x_3,\cdots,x_n)$ in Planck units, and you want $f$ to have the unit of, say energy, then you proceed as follows.

Since Planck units have been used, $\hbar = c = G = 1$, so, you are free to divide or multiply each of the variables by arbitrary powers of these constants as that amounts to multiplying by 1. In particular, you are free to divide each of the $x_i$ by it's Planck unit expression and you can multiply $f$ by the Planck unit for energy. But if you do that, then the resulting expression happens to be formally dimensionally correct in SI units, so it is the desired expression valid in SI units.