Tomas Vencl
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Thank you. (I used "speed" because it corresponds to Newtonian escape velocity, But I do not need to interpret it as "speed",just ##dr / d\tau## is ok.)PeterDonis said:That quantity is not limited to ##c##; it increases without bound as ##r = 0## is approached. So it cannot be interpreted as a "speed".
If you are fine with that, then I think the answer is that ##dr / d\tau = c## when the true event horizon is crossed, but I have not done the math to confirm that.