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Consider a tiny device which locally measures the Karlhede's Invariant and blinks a bright pulse of light whenever that measurement reads a change of sign in the quantity. Consider a spatially extended object such as a rod which has two of these devices attached to the ends of the rod, say A and B. Now align the rod radially with respect to a Schwarzschild black hole so that the end B is closer to the black hole than end A. Now drop it.

Assume that the black hole is massive enough so that a single reference frame can be attached to the entire rod and the times measured at both ends are the proper times of the comoving observer at the instant of measurement.

At some point during the free-fall, the tiny light bulbs at the ends A and B flash pulses of light (because the Karlhede's Invariant flips sign as one crosses the event horizon). What can we say about the time difference between these flashes according to the comoving observer?

  1. B flashes before A,
  2. A and B flash simultaneously, or
  3. A flashes before B.

Please provide some calculation, not just intuition (because by intuition, I would go with the first choice). Thank you.

Nanashi No Gombe
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1 Answers1

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The proper time along a radial free-fall path in Schwarzschild metric is given by $1/3{\sqrt(2/GM)}({\sqrt(r_A^3)}-{\sqrt(r_B^3})$ where $r_A > r_B$. With $r_B = 2M$ one obtains the free-fall proper time from $r_A$ to the event horizon.

This is not yet sufficient however because we need to know $r_A$ expressed by the proper length of the rod. The spacelike form of the metric leads to $dr = d\sigma{\sqrt(1-2M/r)}$. Integration yields a quite lengthy formula which is available in web. Now we are able to calculate the elapsed proper time for the upper end of the rod to reach the event horizon given $r_B = 2M$

The caveat here is that due to unavoidable tidal forces not all points along the rod are in free fall. This true only for a particular point, the point which is force free. Only this point describes a geodesic. See the unanswered question
Schwarzschild metric: where is the force free point on a radial rod?. We can neglect this aspect though in case of a short rod and a large black hole.

timm
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  • The spacelike form of the metric, setting dϕ=0, since we are only interested in the radial part becomes $dσ^2=dt^2(1−2M/r)+dr^2/(1−2M/r)$. To calculate the proper distance of the rod we have to freeze time, so $dt=0$, which leads to $dr=dσ(√1−2M/r)$. – timm Aug 05 '18 at 16:37