I encountered a mental block in my understanding of natural units. According to natural units, $c = 1$, which would imply that if we wanted to convert velocity from SI units ($m/s$) to natural units (dimensionless), we would use the relation:

$$c = 1 \implies 3 \times 10^8 m/s = 1$$


$$1 \frac ms = \frac 13 \times 10^{-8}$$ (in natural units)

Now, if in a given problem, the velocity is $v_{SI} = 220 \times 10^3 m/s$ (same as $220 km/s$), then the corresponding value in natural units should be:

$$v_{nat} = 220 \times 10^{3} \times \frac 13 \times 10^{-8} = 0.000000733 = 7.33 \times 10^{-4}$$

If I am asked to square the velocity ($v_{SI}^{2}$ or $v_{nat}^{2}$), then I seem to arrive at a problem.

$$v_{SI}^2 = 4.84 \times 10^{10} (m/s)^2$$

Converting to natural units, $$\frac{v_{SI}^2}{c^2} = 5.38 \times 10^{-7}$$

which I also obtain if I square the velocity in natural units,

$$v_{nat}^{2} = 5.38 \times 10^{-7}$$

However, I find this counter intuitive, because while working entirely in conventional units (SI, CGS, MKS, etc.), the square of a velocity ($v_{SI}^{2} = 4.84 \times 10^{10} (m/s)^2$) would yield a larger value of the velocity, whereas in the natural units, it yields a significantly smaller value. Am I doing something wrong, or have I got the intuition wrong? Is this an intended effect of utilizing natural units, and if so, how does this justify effects we observe?

  • 1
    $\begingroup$ what do you mean by "the square of a velocity would yield a larger value of the velocity"? do you mean that $v^2>v$? note that such an expression would be meaningless. $\endgroup$ Commented Aug 25, 2017 at 12:54
  • $\begingroup$ Try computing $v^2$ in SI units for something going 1 centimeter per second. $\endgroup$ Commented Aug 25, 2017 at 13:08
  • $\begingroup$ I understand your points @AccidentalFourierTransform and David Hammen. It's just that the velocities we deal with in SI units often take value >1 and the intuition built up tells you that generally, squaring one value would yield a larger value. Natural units set the natural limit as the upper limit, making all velocities such that squaring the number yields a smaller number. I am not comparing $v^2$ and $v$, just stating where my intuition went wrong :) $\endgroup$
    – smashwhat
    Commented Aug 25, 2017 at 14:20
  • $\begingroup$ Also, to be really honest, I did understand where I was going wrong by the time I typed out the whole question. But, I had spent so much time thinking about it in my head, that I decided to post the question anyway, since 1) there wasn't any such question here already, and 2) I expected the community to post better responses than I could (which they have now) :) I hope this is not against some SE principle :P $\endgroup$
    – smashwhat
    Commented Aug 25, 2017 at 14:27

3 Answers 3


I kind of understand why you find this unintuitive. It may be easier if we visualize it using areas (in SI units for the moment).

We get an analogous - as you call it - unintuitive result if we express the side length of a square in meters, say $10\,\mathrm{m}$. The area of the square is $100\,\mathrm{m}^2$, a number that is larger than the side length. If we express the side in kilometers, we have $0.01\,\mathrm{km}$. Squaring this gives an area of $0.0001\,\mathrm{km}^2$, a number that is smaller than the side length.

But, what this actually means is that the reference area, namely $\mathrm{m}^2$ and $\mathrm{km}^2$ have a much bigger ratio than the reference lengths $\mathrm{m}$ and $\mathrm{km}$. And since all we do is express the areas in multiples of these reference areas and the lengths as multiples as the reference lengths, it is obvious that one area can yield a smaller number than the reference length, and the other a larger number than the reference length.

For your question in particular, one time you express the velocity in multiples of $\mathrm{m}/\mathrm{s}$ and the other in multiples of $c$. For these, the same argument as above holds.

  • 1
    $\begingroup$ Here's another example. An American measures the side of a square as $2$ feet, calculates its area as $4$ square feet and says: "the area is bigger than the width". Someone from almost any other country measures the same square and finds that its width is $0.6m$, calculates the area as $0.36m^2$ and says: "the area is smaller than the width". $\endgroup$
    – badjohn
    Commented Sep 6, 2017 at 15:30

The problem is when you compare $v^2$ to $v$. These both have different dimensions, so you can't meaningfully make this comparison.

Let's leave dimensions alone for a moment, and consider just a mathematical form of $y=x^2$. What we're interested the relationship between $y$ and $x$, not the actual numbers. If we have $x$ from $-1$ to $1$, then $y\leq |x|$, with $y\geq |x|$ outside of this region. But the shape of the graph doesn't change, that is the relationship between $y$ and $x$ is the same at all points. So whether $y$ is smaller than $x$ or not, doesn't bother us, as the same mathematical form remains true.


The reason for natural units is that time and space are more closely related than is apparent from classical physics. Space and time are combined into a space-time. Interval in space-time is measured by

$$\Delta s^2 = \Delta x^2 - c^2 \Delta t^2 $$

This is much like classical physical, where horizontal distance is measured in miles and altitude in feet. It turns out that horizontal distance and altitude are closely related. You measure distance in miles by

$$\Delta d^2 = \Delta h^2 + (1/5280)^2\Delta a^2$$

You can do a similar calculation in feet.

$$\Delta d^2 = (5280)^2\Delta h^2 + \Delta a^2$$

Or better, convert altitude to miles. The distance calculation gets simpler.

$$\Delta d^2 = \Delta h^2 + \Delta a^2$$

For space-time, this is means natural units. Measure time in years and distance in light-years. Or seconds and light seconds. With these units

$$\Delta s^2 = \Delta x^2 - \Delta t^2 $$

Velocity is a ratio of distance to time. $$v = \Delta x / \Delta t $$ is simpler than $$v = \Delta x / c\Delta t $$

It is like measuring slope. If you choose the right units, you can use

$$slope = \frac{\Delta a}{\Delta h}$$ instead of $$slope = \frac{\Delta a}{5280\Delta h}$$

  • $\begingroup$ This is a nice answer for a different question. The OP isn't asking why we use natural units. $\endgroup$
    – PM 2Ring
    Commented Aug 25, 2017 at 20:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.