It is well known that if a black hole fulfills certain conditions, it's tidal forces will rip incoming objects apart even before they cross the event horizon. So at some point, the curvature of space seems to be so intense that your feet are attracted by orders of magnitude higher than your head, leading to your painful death.

I was curious about the analogous effect in the temporal direction, that is how high will the gravitational time dilation between your head and feet get at some point before reaching the event horizon. Using the Schwarzschild-metric and the geodesic equation, I can try to answer this question and ultimately want to plug in some numerical values for a person who is 1.90m tall just for fun. However, I doubt that I interpret the radius coordinate in the right way, and would like to know your opinion on how to best approach this problem. I will show you my approach first:
From the geodesic equation of a radial free fall into a schwarzschild black hole follow the two equations:

  1. $ \frac{dt}{d\tau}(1-\frac{r_s}r)=F$

  2. $(\frac{dr}{d\tau})^2+(1-\frac{r_s}r)=F²$

Where I set c=1, F is a constant of integration and other equations were eliminated because of $d\phi=d\theta=0$ as well as $\theta=\frac{\pi}2$
$dt$ is the time interval measured by an observer who is "infinitely" far away. $d\tau$ is the time interval measured by the person falling inwards. I could now write down equation 1) two times for my head and feet. Where r is the radial coordinate of my feet, and $r_{head}$ the one for my head So:

$ \frac{dt}{d\tau_{head}}(1-\frac{r_s}{r_{head}})=F$

$ \frac{dt}{d\tau_{feet}}(1-\frac{r_s}{r})=F$

I could now eliminate $dt$ and F, thus obtaining:


Now comes my question:
In flat spacetime, I could say that $r_{head}=r+1.90$. But as has been discussed in many threads, because of the curved spacetime the radial coordinate in the Schwarzschild-metric cannot be interpreted as the physical distance from the center/singularity, which means I shouldn't be able to use this simple expression. And because I'm particularly interested in the extreme regime of a black hole, it should deviate extremly.

When trying to obtain some actual numerical values from problems regarding black holes, how can I properly deal with the radius coordinate, specifically in this example?

Of course it might be that my approach written above is just wrong. I could maybe simply look at the line element, set $dr=d\phi=d\theta=0$ and then write down the resulting relationship between $dt$ and $d\tau$ two times, for head and feet in the same way as above. In any case the question on how to deal with the radial coordinate is still relevant.


2 Answers 2


You've described the situation as though there are two effects: a tidal force that spaghettifies the person's body, and a time dilation effect that makes time run at different rates at their head and toes. This is actually not a good way to conceptualize it. If we drop two test particles into the black hole successively along the same radial line, separated by a distance $h$, then we can say that the distance between them continues to be exactly $h$, if we define simultaneity appropriately, or we can say that the distance between them becomes bigger than $h$, if we define simultaneity based on the proper time of each particle. In the former case there is time dilation but no stretching, while in the latter there is stretching but no time dilation. But these are just two different descriptions.

Regardless of which of these interpretations you take, we have

$$\frac{\Delta \tau}{\tau} = \frac{ah}{c^2},$$

where $\tau$ is the proper time, $\Delta \tau$ is the difference in proper time between the head and the foot when the proper distance equals $h$, and $a$ is the relative acceleration of the head relative to the feet.

The Newtonian tidal force in the radial direction is $-2GM/r^3$. I would have thought that this would be modified by relativistic corrections, and it is modified in the Kerr spacetime (for a rotating black hole) if you take $r$ to be the Kerr $r$ coordinate. However, it turns out that for a free-fall trajectory, when you set the spin of the black hole to zero and let $r$ go over to the Schwarzschild $r$, you get exactly the Newtonian expression for the tidal force, at all radii down to the singularity. For a calculation of all this, see Lima Junior et al., https://arxiv.org/abs/2003.09506, eq. (35). I think there must be some easier way to prove that the Newtonian expression is exact in the Schwarzschild case, but I have no idea what it is.

As an example, the result is that for a solar-mass black hole, when you're inside the horizon at $r=R_s/2$, with $h=1.9$ m, we have $\Delta \tau/\tau=1.6\times10^{-6}$, i.e., the effect is still quite small, whereas the mechanical strain is huge enough to have killed you already.

In your calculation, you're using expressions for the time dilation in a black hole. The problem is that these expressions are only valid for a static observer. We're talking about a free-falling observer. (And once you get inside the horizon, there are no static observers.) This is why your expression gives weird results like negative numbers when you're inside the horizon. Even if you're outside the horizon, it doesn't make sense to define or calculate time dilation in this way for a free-falling observer.

  • $\begingroup$ Thank you. Can you explain me how to derive the first equation or send me a link where I can see it? In my approach, I tried to solve the problem analogously to the problem where you calculate the time dilation between GPS satellites and earth residents. There, you also set up the time dilation formula for both systems and eliminate dt in the same way as above: physics.stackexchange.com/questions/421255/… As in my problem, a GPS satellite is also in free-fall, why can they compute it that way and I cant? $\endgroup$
    – MegAmaNeo1
    Commented Dec 11, 2020 at 9:00

For a stellar-mass (or larger) black hole, with a Schwarzschild radius measured in kilometers (or more), the difference between $\Delta r$ in Schwarzschild coordinates and the proper distance between your head and feet is fairly insignificant, as long as you are outside the horizon. So you can take $\Delta r$ to be 1.9 meters and be wrong by only one part in a thousand or so.

The proper distance does not “deviate extremely” from $\Delta r$ unless you are outside a hole whose Schwarzschild radius is the same scale as your height, or smaller. We know of no such black holes.


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