Taylor and Wheeler in "Exploring Black Holes" calculate that the spaghettification time, measured from feeling a 1g tidal difference head-to-toe to disintegration at the singularity, is a constant, a little less than one second. For small black holes (3 solar mass) this happens well outside the event horizon. But for large black holes it is inside. But how is one "spaghettified" inside the black hole. The singularity is time-like relative to both your head and feet - so it is not a different distance away and so how does the tidal force arise?

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    $\begingroup$ Hi Brent. For clarification: are you thinking that inside the horizon the Schwarzschild $r$ coordinate is timelike so points at different $r$ are really at different times? If so, this shows only that the Schwarzschild coordinates do not provide a useful way to describe the geometry inside the horizon. The spaghettification force would be measured in the rest frame of the falling observer and in this frame the radial coordinate is spacelike. $\endgroup$ Commented Oct 20, 2017 at 4:30
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    $\begingroup$ Good answer @John Rennie, you should post it as the answer. $\endgroup$
    – Bob Bee
    Commented Oct 20, 2017 at 5:34
  • $\begingroup$ Taylor and Wheeler calculate the spaghettification in the "rain" frame - that is the frame falling in from infinity. So they calculate the acceleration as the second derivative of r (Schwarzschild r) with respect to proper time of the rain. Then they take the derivative to this with respect to r to get the stretching acceleration for a person of height difference dr. So it seems the assumption of the Schwarzschild r is essential to their result. $\endgroup$ Commented Oct 21, 2017 at 5:49
  • $\begingroup$ FWIW, "raindrop" coordinates are also known as Gullstrand–Painlevé coordinates. $\endgroup$
    – PM 2Ring
    Commented Jan 15, 2019 at 5:48
  • $\begingroup$ I take it's constant with respect to the mass of the black hole, and not to the person's height? $\endgroup$ Commented Feb 14, 2020 at 4:27

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Taylor & Wheeler's spaghettification time is valid for the case of "raindrops", a particular motion where the astronaut fell from rest far away from the black hole (as Brent Meeker commented).

As for inside the horizon, maybe it is unhelpful for you to focus on the description of Schwarzschild $t$ and $r$-coordinates swapping roles (as John Rennie commented). Understand that any astronaut anywhere measures 3 dimensions of space and 1 dimension of time, in their local vicinity (the technical term is orthonormal frame or tetrad). The spaghettification is calculated relative to the astronaut's own space and time.

Update: the details involve writing down some vectors. I'll work in Schwarzschild[-Droste] coordinates. The $r$-coordinate vector in this case is $(0,1,0,0)$, which indeed is timelike inside the horizon. Now the raindrop 4-velocity is $$\Big(\frac{1}{1-2M/r},-\sqrt{2M/r},0,0\Big)$$ The radial direction for the raindrop is not $(0,1,0,0)$. Instead we seek a spatial vector orthogonal to the 4-velocity. This is: $$\Big(-\frac{\sqrt{2M/r}}{1-2M/r},1,0,0\Big)$$ which is indeed spatial, so Taylor & Wheeler's calculation is well-grounded.

  • $\begingroup$ @John, so Taylor & Wheeler are wrong to call the $d/dr(d r^2/d tau^2)$ the sphagettification because the $d/dr$ is not a head-to-toe change in acceleration, it is a time-like change within the Schwarzschild radius? $\endgroup$ Commented Feb 9, 2018 at 21:31
  • $\begingroup$ I too was surprised at this, before looking into it. I've added my reasoning above. $\endgroup$ Commented Feb 9, 2018 at 23:26

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