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Exponential function and natural units

Physics Asked by minits on May 2, 2021

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?

2 Answers

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 $0leq v<1$.

Correct answer by Buzz on May 2, 2021

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).

Answered by G. Smith on May 2, 2021

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