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I am wondering if I mix up the notion of proper distance and proper length.

I have two cuves in Schwarzschild space-time describing the flight of two photons (think of it as photons guided in by optical fiber, not geodesics).

$\gamma_1(\lambda) = \big(t_1(\lambda),r_1(\lambda),\frac \pi 2, 0\big )$ and $\gamma_2(\lambda) = \big(t_2(\lambda),R,\frac \pi 2, \varphi_2(\lambda)\big )$ where $r_1(0)=R$ and $\varphi_2(0)=0$.

I now want both photons to fly the same length. With "same length" I mean, that I trimmed the two optical fibers in my lab so that they have the same lenght, measured by a tape.

For this purpose I arbitrarily define the end of the first curve at $\lambda=\Lambda$ and define the corresponding radial coordinate by $r_1(\Lambda)=R+\Delta R$. Now comes the crucial point: Is it correct to say, that the length of the first optical fiber is

$\displaystyle l = \int_0^\Lambda \frac{1}{\sqrt{1-\frac{r_s}{r_1(\lambda)}}} r_1'(\lambda) \; d\lambda = \int_R^{R+\Delta R} \frac{1}{\sqrt{1-\frac{r_s}{r}}} \; dr$ ?

If this is true this would also define a $\Delta \varphi$ for the second optical fiber by

$\displaystyle l \overset{!}{=}\int_0^{\Delta \varphi} R \, d \varphi = R\Delta\varphi$

Is that correct? My problem is, that wikipedia states that I am dealing with proper distance instead of proper length. This confuses me since I want to be sure to have the same length for both optical fibers (measured by a tape).

Qmechanic
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thyme
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1 Answers1

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The distance you seek is in the frame(s) of the optical fibers, which is well-defined because they are timelike (they have mass). I would express the Wikipedia article's points as follows: the proper length is the length of the fiber as measured in its own frame, whereas the proper distance measured by an arbitrary observer is (the proper length) length-contracted by the Lorentz factor $\gamma$. (In curved spacetime, just apply this locally.) We want the fiber's frame(s), so these terms coincide.

Now the photon $\gamma_1$ is radial (so incidentally it is geodesic), so the fiber is radial also. Presuming the fiber is static, its proper length is indeed $\int(1-2M/r)^{-1/2}dr$.

Now for photon $\gamma_2$, the fiber which guides it has fixed $r$ and fixed $\theta=\pi/2$. Does each point along the fiber have fixed $\phi$, that is, is the fiber moving? If moving, we need more details. If not, its proper length is simply $\int r\,d\phi=r\,\Delta\phi$.

Colin MacLaurin
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