Natural Units used in Nuclear and Particle Physics -

In our last post, we discussed the unit of energy, mass, cross-section, radioactivity, spin, and electric quadrupole moment. We discussed why $\rm MeV$ is the preferred unit of energy by nuclear physicists? We calculated the mass of the electron, proton, and neutron in $\rm MeV/c^2$ . We calculated the value of the Planck constant in $\rm MeV.s$ and the value of $\rm \hbar c$ in $\rm MeV.fm$ .

Natural Units

In this post, we will talk about a special kind of unit used by researchers in nuclear and particle physics, called “Natural Units”.

Natural units are physical units of measurement based only on universal physical constants.

Here units are redefined so that the speed of light and reduced Plank’s constant become equal to one. That is,

$c =1$ and $\hbar =1$

Once we fixed $c =1$ and $\hbar =1$ , all other kinematical units can now be expressed in terms of units of energy. If we choose $\rm MeV$ as the unit of energy, which is obvious in Nuclear physics then all the basic physical quantities like length, area, time, mass, momentum, decay rate, cross-section, etc. can be expressed in terms of the powers of $\rm MeV$ . So let’s find the Natural units of these quantities one by one.

Natural Unit of Energy, Momentum, and Mass

For any kind of system (either relativistic or non-relativistic), we have a famous formula that can nicely relate the total energy, momentum, and mass of the system. That formula is;

E^2 =(pc)^2+(mc^2)^2

Here $E$ is total energy, $p$ is momentum, $m$ is mass and $c$ is the speed of light. From the dimensional analysis, one can easily say that the dimension of $E$ = dimension of $pc$ = dimension of $mc^2$ . But if you choose a unit in which $c=1$ then the above equation becomes;

E^2 =p^2+m^2

And you can measure the momentum and mass in $\rm MeV$ . So, the Natural Unit of Energy, momentum, and mass is $\rm \ MeV$ .

Example 1:

There is a system where energy, momentum, and mass are given in Natural units. If energy, $E=10\rm \ MeV$ , momentum, $p=8\rm \ MeV$ , and mass, $m=6\rm \ MeV$ then how do you convert these quantities into SI units?

Solution

We know that

$\begin{aligned}\rm 1\ MeV &=10^6 \times (1.6\times 10^{-19}\rm \ J)\\&=1.6\times 10^{-13}\rm \ J\\ \end{aligned}$

So, the value of energy in an SI unit can be calculated as;

$\begin{aligned}E&=\rm 10\ MeV\\ &=10 \times (1.6\times 10^{-13}\rm \ J)\\&=1.6\times 10^{-12}\rm \ J\\ \end{aligned}$

The momentum can be calculated as;

$\begin{aligned}p&=\rm 8\ MeV/c\\ &=8 \times \dfrac{1.6\times 10^{-13}\rm \ J}{3\times10^8\rm \ m/s} \\&=4.27\times 10^{-21}\rm \ kg.m/s\\ \end{aligned}$

The mass can be calculated as;

$\begin{aligned}m&=\rm 6\ MeV/c^2\\ &=6 \times \dfrac{1.6\times 10^{-13}\rm \ J}{(3\times10^8\rm \ m/s)^2} \\&=1.07\times 10^{-29}\rm \ kg\\ \end{aligned}$

Unit of Length and Time in Natural Unit

We know that $\hbar$ has unit of energy $\times$ time. So the unit of time will be the unit of $\hbar /$ Unit of Energy. But in natural units, $\hbar=1$ so the unit of time will be 1/Unit of Energy. That is, $\rm 1/MeV$ .

Also, we know that c has units of Length/time. So the unit of time will be the unit of Length/Unit of c. But in natural units, c = 1 so the unit of time will be the unit of Length. That is, $\rm 1/MeV$ .

Unit of the Cross-section in Natural Unit

As we know, the cross-section has units of Area so, in Natural Units, the cross-section is measured in $\rm 1/MeV^2$ .

The reaction rate has units of 1/Time and hence the Natural Units of Reaction rate is $\rm MeV$ .

Unit of Velocity in Natural Unit

Velocity can be measured in terms of $c$ and in natural units, $c=1$ which is a constant, so velocity is dimensionless in Natural Units.

Unit of charge in Natural Unit

For this let’s start with the fine structure constant, which is dimensionless also. The fine structure constant is given by;

\alpha =\dfrac{e^2}{4\pi\epsilon_0\hbar c}

And it is nearly equal to $\frac{1}{137}$ .

From electrodynamics, we have $c^2 =\dfrac{1}{\epsilon_0\mu_0}$ c. So, if you choose the constant, $\epsilon_0=1$ then, it automatically sets $\mu_0=1$ . And if you choose the constant, $\epsilon_0 =\dfrac{1}{4\pi}$ then, $\mu_0$ will be $4\pi$ .

Actually, we need to choose the value of $\epsilon_0$ and $\mu_0$ in such a way that the value of $c$ becomes 1.

Let’s choose $\epsilon_0 =\dfrac{1}{4\pi}$ and $\mu_0 =4\pi$ . So, the fine structure constant will be

\begin{aligned}\alpha &=\dfrac{e^2}{4\pi\epsilon_0\hbar c}\\ \Rightarrow \dfrac{1}{137}&=\dfrac{e^2}{4\pi\times\frac{1}{4\pi}\times1\times1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\hbar = 1\ \text{and}\ c = 1)\\ \Rightarrow e&=0.085436\\ \end{aligned}

So, the fine structure constant is dimensionless in natural units.

If you want to take $\epsilon_0 =1$ and $\mu_0 = 1$ then you will have $e = 0.302862 \approx 0.303$ . And again it is dimensionless. Therefore, the charge is dimensionless in natural units.