# Natural Units

I’m back! I decided to blog this because it didn’t feel substantial enough to have notes on its own and I couldn’t find a proper place for it elsewhere so here I am.

Today, I wanted to discuss natural units i.e. the system under which $c=1$, $\hbar=1$, etc. It became apparent to me that many believe such a system is the result of convenience or laziness but this could not be further from the truth! There are some deep physical insights associated with such a system which I think never weaves its way into standard curriculum and one which I would like to lay down here. First, we must understand how we work with physical units.

## Unit Groups and Physical Regimes

In physics, chemistry, and many other fields, all physical quantities are measured in combinations of the following quantities.

• Mass [M]
• Length [L]
• Time [T]
• Charge [Q]
• Temperature [K]

For example, force is measured in units of $[M]\cdot [L] \cdot [T]^{-2}$. Similarly, all physical quantities can be reduced to products of the above units which we will call our fundamental unit types. This means having a system to measure all physical quantities simply requires us to choose a unit standard for these five unit types.

Note: the fundamental unit types are non-unique. For example, we can have velocities be our fundamental unit type and remove length. We can then express lengths as products of velocities and times. We see this in practice in the SI unit system where the fundamental unit type of charge is replaced with current $[J]$ and charge is simply $[J]\cdot [T]$.

The choice of standard however depends highly on the system/regime we are working in. For example, if we are working with atomic physics, all relevant lengths are in Angstroms which is $10^{-10}$ meters. For universal time scales, we may want to measure in millions of years. For early universe measurements, we might want gigakelvin whereas for everyday life just Kelvin. So for each system, we choose units based on the physical regime we are working in. It happens that humans only really have intuition for numbers in the range of $10^{\pm 5}$ so we usually choose units that keep values in that range. We will call each choice of standard units a unit group (the reason I say group and not system will soon be clear). Each unit group is associated with some physical regime.

The SI unit group is the group that corresponds to the anthropogenic regime i.e. that which corresponds to daily human life. Lengths are measured in meters, time in seconds, current in amperes, temperature in Kelvin, and mass in kilograms. Naturally, this is what we would want as the standard for scientific communication because it expresses physical quantities in an intuitive regime.

## Stoichiometric and Physical Coefficients

Consider the ideal gas law.

$\displaystyle PV=NkT$

where, in order, the variables represent pressure, volume, number of particles, Boltzmann’s constant, and temperature.

We can express $N$, for example, as $10$ particles or $5$ pairs. These are two different unit standards. It often happens to be the case that in the systems we usually work with, $N\approx 10^{24}$ and so it would be absurd to choose to express it with the number of singular particles. Instead, we often like to use moles which represents exactly $6.02214076\cdot10^{23}$ particles. This number is called Avogadro’s number $N_A$ or

$\displaystyle N_A = 6.02214076*10^{23} \cdot \frac{\text{units}}{\text{mol}}$

Usually, Avogadro’s number is the above without the units at the end but I include the units to make a key observation. Namely, that $N_A=1$. Note that the numerator i.e $6*10^{23}$ units is equal to the denominator $1$ mol so the overall values must be $1$.

How is this coefficient useful then? Well, this is exactly how one makes stoichiometric calculations. You have something in one units, say meters, and you want to convert to another, say millimeters. You multiply by some form of 1, namely $1000 \text{ mm}/\text{m}$ which gives you your desired result in the desired units. So now, we have a quick way of converting between singular units and moles using Avogadro’s constant: it is the exact type of “1” we desire.

Because it just means 1, we can simply insert this into our equation anywhere!

$\displaystyle PV=NN_AkT$

Now, if we measure $N$ in moles, they cancel out the with the denominator in $N_A$ and we are left with the final result in our desired units.

As a result, $N_A$ is not really a coefficient with physical meaning, merely a stochiometric coefficient that can be used as a computational aid. These types of coefficients, despite their redundancy, are often kept in equations anyway because they are extremely useful in streamlining computation.

Such coefficients are often realized when the two different regimes overlap for the same system. For example, we often like to look at single particle energies which is what $kT$ refers to but like to only look at moles of particles when discussing $N$. As a result, we add an extra coefficient $N_A$ that connects the two regimes.

There do however exists other coefficients which do have physical meaning like, for example,

$\displaystyle k = 1.380649\cdot 10^{-23} \frac{\text{J}}{\text{K}}$

There is no a priori conversion of temperatures to energies and the coefficient $k$, in that sense, holds real physical value converting between the two. Another example would be $c$, the speed of light, in an equation like $E=mc^2$. Let’s call these physical coefficients.

In that sense, we see that coefficients in equations come in two types: stochiometric which are simply computational aids that always equal $1$ and physical which contain real physical information about the system its used in.

## Physically Relevant Quantities

We now take a slight detour to note the following.

If you look at any equation that involves temperature, you will note that it is always accompanied by $k$. In other words, $k$ and $T$ never appear alone but only as $kT$. Even in equations that seem to have it alone, the dependence is always lurking. Take, for example, heat capacity which seems to defy this rule.

$\displaystyle C=\frac{\text{d}U}{\text{d}T}$

If you look at any values $C$ takes, it is always in terms of $k$. The standard monatomic gas for example has $C=(3/2)k$. If we enforce our “$kT$” principle, we see that

$\displaystyle \frac{\text{d}U}{\text{d}(kT)}=\frac{3}{2}$

so, in fact, any situation that does not follow this principle would be made simpler if it did.

This leads us to believe that perhaps the more relevant physical quantity is $\tilde{T}=kT$. In fact, the only quantity that we can actually measure is $\tilde{T}$ and $T$ is simply the result of multiplying by this seemingly arbitrary constant $k$ which can never be measured in isolation. It is often because we have predetermined $k$ that we can confidently give the value of $T$ but there is no a priori way to determine $T$ without just picking a value for $k$.

In more fundamental theories, we see this even more like, for example, in relativity where we only see the value $\tilde{t}=ct$, not $t$ itself.

This is a nightmare! On the one hand, we have some physical quantities which seem intuitive like $t$ and $T$ but, on the other, we have the ones actually relevant to the theory and laws like $\tilde{t}$ and $\tilde{T}$. How do we resolve this?

## Natural Unit Systems

We propose the following solution to the previous section’s problem: the intuitive quantity and the physical quantity are the same! In other words, $\tilde{T}=T$. What does this mean? Well, this implies that

$\displaystyle k=1$

In other words, it has converted a physical coefficient, $k$, into a stoichiometric one! This simply asserts that instead of separating Kelvin and Joules as measures of two different quantities, let them both be measures of the same thing and treat $k$ as some stochiometric conversion between the two. $T$ and $kT$ are both the temperature because $k=1$!

Note that this doesn’t really change any equations because the coefficients haven’t changed in value. We have simply added additional constraints. We can still keep $k$ in the gas law as we do Avogadro’s number. This just comes with the additional understanding that some units which we previously interpreted as distinct can be naturally converted into each other giving us the option to remove coefficients if we do not deem them necessary to aid computation. $k$ now merely aids computation rather than provide physical insight.

Which quantities are “relevant” is somewhat subjective and these leads to a few unit systems. There are two that are considered the most natural. They are the Planck unit system where

$\displaystyle 1=\hbar=c=k=G=4\pi \epsilon_0 = \frac{\mu_0}{4\pi}$

and the rationalized Planck unit system where

$\displaystyle 1=\hbar=c=k=4\pi G=\epsilon_0 = \mu_0$

The difference is the following. The regular Planck units have equations of the form

$\displaystyle 4\pi Q=\nabla\cdot E \quad\quad E = \frac{Q}{r^2}$

whereas the rationalized realize the form

$\displaystyle Q=\nabla\cdot E \quad\quad E = \frac{Q}{4\pi r^2}$

I personally think the second is more natural. That, combined with a few other reasons, makes me believe rationalized is better but that’s just me.

Note: The reason we differentiate unit system and unit group is because different systems actually interpret the relationship between units differently whereas groups simply choose different sets of units to attend to.

## Physical Interpretation

You may still be unconvinced that such a unit system makes any sense. After all, what is $1$ second of length? Well, it turns out that there is a real physical interpretation for this. If something is $1$ second away in space, it will take $1$ second in time for it to be able to affect you. If something has a mass of $1$ Joule, this means if its mass is converted into energy, it will yield $1$ Joule. All physical quantities can loosely yield such an interpretation!

## Planck Interpretation

When we investigate Planck units further, it quickly becomes clear that its constraints result in all quantities being unitless! In other words, a length can just be $1$ without any additional labeling. What does this mean? Well, we first note that the following quantity $l_P$ is in units of length.

$\displaystyle l_P = \sqrt{\frac{\hbar G}{c^3}} \approx 1.6\cdot 10^{-35} \text{m}$

We know that, under the Planck unit system, $l_P=1$. This means if something has length $L=2$, then that means we can also say it has length $L=2l_P\approx 3.2 \cdot 10^{-35} \text{m}$. There exists a perfectly good translation into the SI unit system! $l_P$ here is called the Planck length and there exists a Planck ____ $Q_P$ for every physical quantity $Q$ such that $Q_P=1$ allowing us to convert from seemingly meaningless values to intuitive ones. The Planck value for the fundamental unit types are given below.

• Planck Length: $l_P \approx 1.616 \cdot 10^{-35} \text{ m}$
• Planck Mass: $m_P \approx 2.176 \cdot 10^{-8} \text{ kg}$
• Planck Time: $t_P \approx 5.391 \cdot 10^{-44} \text{ s}$
• Planck Charge: $q_P \approx 1.875 \cdot 10^{-18} \text{ C}$
• Planck Temperature: $T_P \approx 1.416 \cdot 10^{32} \text{ K}$

Although some of these seem like absurd values, they are in the natural unit system of the universe! For example, after 1 time, the universe was 1 long with a temperature of 1. Amazing! Many more similarly beautiful and simplistic statements can be made.

## Some Nice Equations

With this new system of units, we can profoundly simplify many equations getting to the core of what matters. I have given a few below (I use rationalized Planck).

$\displaystyle E^2=(mc^2)^2+(pc)^2 \rightarrow E^2=m^2+p^2$

$\displaystyle PV=NkT \rightarrow PV=NT$

$\displaystyle G_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu \nu} \rightarrow G_{\mu\nu} = 2T_{\mu\nu}$

$\displaystyle \hat{H}=i\hbar \frac{\partial}{\partial t} \rightarrow \hat{H} = i \frac{\partial}{\partial t}$

I hope I have convinced someone of the beauty of natural units after this (admittedly long) post!

## One thought on “Natural Units”

1. Can you help me on Conformal Field Theory Problem Set 4, question 1? (Please don’t tell anyone that I’m asking for help. I’ve left my email attached.)

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