If I have some angular frequency e.g. $\omega [\text{rad } \text{s}^{-1}]$, I can easily express this as an energy as $E = \hbar \omega [\text{ J}]$.

Now suppose I am working in Natural (Planck) units where $\hbar = c = G = k_e =k_B = 1$. In this case $E = \omega$.

I have two sources of confusion:

  1. What are the units of energy now?
  2. What about if instead of energy I am dealing with a length? i.e. How would I express 10 m in natural units?

For my second point, I understand that we could choose a unit e.g. kg to express a length in. In this case would we just convert by a factor of $c^2 /G$?

Thanks for any help/guidance.

  • $\begingroup$ What's natural units? $\endgroup$ Jun 3, 2017 at 18:22

4 Answers 4


If you set $\hbar = c = G = 1$ then any physical quantity can be declared equivalent to any other physical quantity and you can transform an expression that says that this is so in any other unit system, e.g. SI units by putting in the right factors involving $\hbar$, $c$, and $G$. This is most easily done by constructing quantities with the dimensions of length, time and mass using $\hbar$, $c$, and $G$, these expressions are known as the Planck length, Planck time and Planck mass, see here for details.

Then given an equation in $\hbar = c = G = 1$ units, you can then find the correct SI units equivalent by dividing all the variables in the equation by the Planck unit of the variable and multiplying the end result by the Planck unit of the desired physical quantity. Because you start out with $\hbar = c = G = 1$, you are not changing anything to the equation, but because it's now also dimensionally correct in SI units, it's is the correct SI equation when you substitute the SI values for $\hbar$, $c$, and $G$.

Example, if you equate a mass $M$ to a length $L$ in $\hbar = c = G = 1$ units:

$$M = L$$

then we are free to put in arbitrary powers of $\hbar$, $c$, and $G$ because they are equal to 1 anyway. We can then divide the l.h.s. by the Planck mass and the right hand side by the Planck length, this yields:

$$M \sqrt{\frac{G}{\hbar c}} = L\sqrt{\frac{c^3}{\hbar G}}$$

We can write this as:

$$L=\frac{M G}{c^2}$$

which is dimensionally correct in SI units.

You can play this game with any arbitrary expression, e.g.

$$M^3 \exp(M/L) = L^5$$

can be transformed into a dimensionally correct SI expression with almost no effort (I leave this as an exercise for the OP).

  • $\begingroup$ Nice simple explanation! $\endgroup$
    – daydreamer
    Oct 16, 2020 at 21:47

You are free to chose any one unit to use (in various powers) for expressing values in de-dimensionalized systems. Gertain chooses to use seconds, more than a few introductory general relativity books use length, but in my field (particle physics) we tend to use electron-volts (i.e. energy) as the basis for all measurements.

In that case energy is expressed as, well, energy. As is mass. Both length and time are expressed as inverse energy (you can get that from the equation you are considering and then noting that velocity has to be dimensionless).


First of all, radians are dimensionless so you should not write: $\omega[rad\ s^{-1}]$.

Dimensional analysis

The units of energy in natural units the dimensions of energy are $s^{-1}$. This makes sense since $E=\omega$.

The units distance in natural units are $s$. This makes sense since wavelength = $c *$ travel time. But $c$ = 1 so distance and time have the same units.

Some intuition behind natural units

To interpret these natural units let us look at this simple example : $v = 0.5 c$, in natural units this would be $v=0.5$. This means that the velocity is half of the light speed.

Similarly, $E=0.7 s$ should be interpreted as: the energy is 0.7 times $\hbar*s$

Edit to clarify OP's question in the comments

It should be noted that we cannot set an infinite amount of units to 1 since this would result in severe inconsistencies in the units. One example is G that is NOT set to 1 in natural units.

Let us assume G = 1. We find that $r_s = GM/c^2$ so that $[r_s] = m = s = [M] = [E] = s^{-1}$ ?!

However if $[G] = m^3 *kg*s^2 = ms$ (since $kg = [E] = 1/s$ and $m=s$) we find that $r_S = GM/c^2$ so that $[r_S] = m = s = G [E] = ms*s^{-1} = m$ so that there is no problem...

So, don't set all units to 1. For example G should not be set to 1 ! (bottom line, if you find inconsistencies you have set to many units to 1)

  • $\begingroup$ Hmm ok, I understand the first part of your answer re energy, but not the second. For example, a common expression for the radius of a black hole is $r = G M /c^2$. Converting this to natural units seems to put $r$ in units of mass, no? $\endgroup$ Jun 3, 2017 at 13:49
  • $\begingroup$ @user1887919 Or mass in units of distance, which is what my intro general relativity text did. The mass of the Earth is a bit over a centimeter. $\endgroup$ Jun 3, 2017 at 15:23
  • $\begingroup$ @user1887919 I have edited my answer to include this question. $\endgroup$
    – gertian
    Jun 3, 2017 at 16:20
  • $\begingroup$ But in Planck units $G$ is set to 1? $\endgroup$ Jun 3, 2017 at 17:59

Answer to your 1st question is that in natural units we use energy as only unit to describe other quantities. Energy can be taken in term of eV mostly, so now we can use energy units eV to describe other quantities like mass, distance etc. For reference you can see Peskin (An introduction to Quantum Field Theory).

For your second question.

$$0.511 M eV = (3.862*10^{-11}cm)^{-1}$$ From this relation we can derive $$(1M eV)^{-1}= 1.973*10^{-13} m.$$


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