[Physics] How to express an energy in natural units

absolute-unitsdimensional analysisunit conversionunits

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.

Best Answer

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).

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