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Derivation of proper time with respect to time

Physics Asked by Peter Anderson on July 23, 2021

I’m confused on how the author got this answer.
He started with this:
$$frac{d(text{proper time})}{dt}=frac{sqrt{dt^2-dx^2}}{dt}
=sqrt{1-v^2}$$

I don’t get how this is solved, specifically where the $1$ came from in $1-v^2$.

One Answer

The author is doing typical "physics mathematics", so it may be considered to be a little sloppy to a pure mathematician. ;)

The $1$ comes from $frac{dt}{dt} = 1$. Perhaps it will make more sense if we do the derivation in a more formal way.

It's common to denote proper time using, $tau$, the Greek letter tau. Note that the author is using natural units where $c$, the speed of light, equals $1$.

Let $$(Deltatau)^2 = (Delta t)^2 - (Delta x)^2$$ Dividing through by $(Delta t)^2$, $$left(frac{Deltatau}{Delta t}right)^2 = left(frac{Delta t}{Delta t}right)^2 - left(frac{Delta x}{Delta t}right)^2$$ or $$left(frac{Deltatau}{Delta t}right)^2 = 1 - left(frac{Delta x}{Delta t}right)^2$$ Taking limits as $Delta t to 0$, $$left(frac{dtau}{dt}right)^2 = 1 - left(frac{dx}{dt}right)^2$$

But $frac{dx}{dt}=v$, so $$left(frac{dtau}{dt}right)^2 = 1 - v^2$$ Taking square roots, $$frac{dtau}{dt} = sqrt{1 - v^2}$$ which is the desired result.


Note that $$gamma = frac{1}{sqrt{1 - v^2}}$$ is the Lorentz factor.

Answered by PM 2Ring on July 23, 2021

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