# What Happens When an Object Accelerating Away from a Black Hole Stops?

• I
• Ivo Draschkow
Since this thread is labeled intermediate, we should be able to use a little math. And the math can be made really simple for the cases under discussion.

A general formula for r' (derivative of Schwarzschild r coordinate by time per very distant observer - Schwarzschild time) for ingoing timelike geodesics is for the exterior coordinate patch (R is Schwarzschild radius):

$$r'=(R/r-1)\sqrt{(R/r-1)/e^2+1}$$

$$r'=(R/r-1)\sqrt{R/r}$$
For free drop from some ##r_0>R##, take ##e^2=1-R/r_0##. For a "thrown" free fall, where the particle, extrapolated back in time, would have some velocity at infinity, take ##e^2=\gamma(v_\infty)^2##.

Without even integrating, it is clear that r' for free fall from infinity becomes zero at infinity and also on approach to the horizont (r=R). The maximum magnitude r' for this case is at r=3R.

For a free faller (from infinity) passing a platform drop, you see r' is nonzero for the free faller at ##r_0##, and that r'=0 at ##r_0## for the drop from this altitude. At lower altitudes, you see that for the platform drop, r' increases in magnitude, then approaches zero at the horizon, but any any altitude from ##r_0## down to R, r' for free fall from infinity is larger in magnitude. Thus, the free faller (from infinity) has always increasing difference in r coordinate from the platform drop, following the coincidence event where it is zero.

While there is a closed form for t=f(r), the integration is quite messy, and doesn't really tell us more than can be deduced from the simple expressions for r'.

Last edited:
Ivo Draschkow

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