Exponential function and natural units

Question

The argument of the exponential function has to be dimensionless. By switching to natural units, velocity (for example) becomes dimensionless. Surely, I cannot take the exponential of a velocity now and give it a physical sense, right? However, how do I resolve this contradiction?

Buzz · Accepted Answer

If you are working in natural units and write an expression like $exp(v)$, that really means $exp(v/c)$. Having rescaled the velocity $v$ by the speed of light $c$, in these units, what we write as $v$ is actually the ratio $v/c$ that is often known as $beta$ is discussions of relativity.  Note that $beta$, being a ratio of two speeds, is a pure number.
You can write expressions like this one for the Lorentz factor,
$$gamma=frac{1}{sqrt{1-v^{2}}},$$
and that's fine, because in these dimensionless units, $v$ is a quantity that lies in the range $0leq v<1$.

G. Smith · Answer

Surely, I cannot take the exponential of a velocity now and give it a physical sense, right?

Your assumption is incorrect.
In hyperbolic motion under constant proper acceleration $alpha$, the distance traveled after proper time $tau$ is $x=alpha^{-1}(cosh{alphatau}-1)$ in natural units where $c=1$. Hyperbolic cosines can be written in terms of exponentials, and $alphatau$ is a dimensionless velocity-like quantity (an acceleration multiplied by a time).

Exponential function and natural units

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