# Intuitively, why do attempts to delay hitting a black hole singularity cause you to reach it faster?

In general relativity, proper time is maximized along geodesics. Inside of a black hole, all future-oriented timelike trajectories end at the singularity. Putting these two facts together, we find that any deviation from geodesic free fall decreases the proper time before one hits the singularity, so as Carroll says, "you may as well sit back and enjoy the ride."

[Edit: As Dale points out, the Schwarzchild singularity does not consist of a single spacetime event, so this argument fails in general: one can in fact extend the proper time experienced by a free faller between the event horizon and the singularity to some extent by firing rockets inward. But this cannot occur appreciably in the limiting case where the free fall begins at rest just outside the horizon, which I'll assume to be the case.]

This is of course very much counter to nonrelativistic intuition. In Newtonian gravitation, if you fire your jetback inward, you slow your inward fall and buy yourself more time. Is there any physical intuition for why this isn't the case inside of a black hole (if you start free falling from rest at the horizon)?

• But this cannot occur if the free fall begins at rest at the horizon, which I'll assume to be the case. This doesn't actually work as a fix. A massive object can't be at rest at the horizon -- we can only consider this as a limiting case. A trajectory with $dr/dt=0$ at the horizon is lightlike. – Ben Crowell Sep 3 '18 at 0:24
• (Those $t$'s look like the Schwarzschild $t$-coordinate, which is not defined at the horizon.) Ben Crowell is certainly correct that massive objects can't be at rest at the horizon, but it's only a limiting case. There is a slight exception for the analytically extended spacetime, at the "bifurcate horizon" as I mention below, although it may be dubious to term this case as "at rest". As for the question, choose coordinates such as Gullstrand-Painleve which are regular at the horizon, and investigate. – Colin MacLaurin Sep 4 '18 at 21:11
• To see the horizon is a null hypersurface, consider the hypersurface $r=\textrm{const}$. This has normal or gradient, expressed as a 1-form: $dr$. In components, this is $(0,1,0,0)$ typically, so has norm-squared $g^{rr}=1-2M/r$, evaluating in say Eddington-Finkelstein or Gullstrand-Painleve coordinates. At $r=2M$ this is zero, so the normal is a null vector, i.e. $r=2M$ is a null hypersurface. – Colin MacLaurin Sep 4 '18 at 21:16
• I want to tweak one of my comments above, which was an unnecessary technicality but it's already written. It is possible for a timelike observer to have $dr/d\tau=0$ at $r=2M$, but only specifically at the bifurcate horizon. However this is not "at rest", as is clarified by a Penrose diagram, or by the fact there is no timelike Killing vector field there. – Colin MacLaurin Oct 5 '18 at 17:13

Actually, it turns out to be incorrect that the optimal strategy is to free fall. There is an optimal strategy for firing your rocket engine which maximizes your proper time from the event horizon to the singularity, and extends it beyond the proper time of a free falling observer.

Here is a paper that discusses the issue and describes strategies for maximizing the proper time to the singularity:

https://arxiv.org/abs/0705.1029v2

Edit: a TL;DR summary of the paper. An infalling rocket can maximize the proper time to the singularity by first making a burn to match the trajectory of a free falling object which started at rest at the horizon. Once the rocket has matched that specific trajectory, then it should turn off engines.

• It is quite amazing that there is a research paper on how to maximize survival time when falling into a black hole. – void_ptr Sep 1 '18 at 20:20
• Is there a simple way to see the flaw in Carroll's argument? – tparker Sep 1 '18 at 22:37
• If you pick two events, then out of all the possible paths connecting those events, the geodesic path is the unique path which maximizes proper time. The requirement is that all those paths connect the same two events. However, by firing the rocket engines it is possible to change the event where you reach the singularity. When changing events the geodesic rule doesn’t apply. – Dale Sep 1 '18 at 22:53
• It is not correct that the change in events in this case is infinitely small. The singularity is a spacelike surface, and the various events composing that surface have finite spacetime intervals between them, not infinitesimal. Firing a rocket engine can make a finite change in the spacetime event where the rocket meets the singularity. – Dale Sep 2 '18 at 12:17
• Can you give a TL;DR of the linked paper? As it stands, it doesn't seem to actually answer the question, instead saying, "there is an answer to this question", although it does make it clear that that question is making an invalid assertion. – Michael Sep 2 '18 at 15:13

My (very limited) intuition for this is that once you cross the event horizon, the singularity is not so much a distant point in space as it is a moment in future time.

In other words, within the event horizon you're firing your rockets not to avoid some point $(x,y,z)$, but rather to avoid next Thursday. From here, I use my intuition about time dilation and the fact that geodesics are trajectories of maximum proper time.

I'm by no means a GR expert so if this picture is wrong, corrections are more than welcome :)

• That’s not quite right, because the singularity is not a point in spacetime. It’s an entire hypersurface, and you can extend your life by adjusting where you hit this hypersurface. – knzhou Sep 1 '18 at 20:37
• This is certainly true, but I was looking for a more precise picture of what happens when you fire your jetpack. – tparker Sep 1 '18 at 22:17
• @knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification. – J. Murray Sep 1 '18 at 22:24
• @J. Murray you are correct that the singularity is more like a moment in time, that is why you cannot avoid it. Firing rocket engines will not move you away from Thursday, but it can change your proper time. – Dale Sep 1 '18 at 22:48
• @tparker Well, I don’t know any of those mathematical niceties, but I don’t see how the answer to this concrete computational question would change if we replaced “the singularity” with “the surface $r = \epsilon$”, since the change in the lifetime would then be arbitrarily small. Then there are no mathematical issues. – knzhou Sep 2 '18 at 6:23

Here's a partial answer, although it's still pretty formal. First define $$E := -\left( \frac{2GM}{r} - 1 \right) \frac{dt}{d\tau}, \qquad L := r^2 \frac{d\phi}{d\tau}$$ in the usual Schwarzschild coordinates. If you're free falling, then $E$ and $L$ are constant over your trajectory, but if you can fire your engines then they can change. We can expand out the normalization condition $U \cdot U = -1$ to $$\frac{1}{2} \left( \frac{dr}{d\tau} \right)^2 - \left(\frac{2GM}{r} - 1 \right) \left(1 + \frac{L^2}{r^2} \right) = \frac{1}{2} E^2,$$ which looks somewhat like the statement of conservation of energy (per unit mass) for a nonrelativistic particle with angular momentum $L$.

But there are two weird aspects to this equation:

1. When you expand out the product of the two binomials on the LHS, you get a weird term $-2GM L^2/r^3$ that does not appear in the nonrelativistic case. Unlike the usual centrifugal angular momentum barrier, this is a centripetal angular momentum fictitious "force" that actually sucks the particle inward at small radii. This means that angular momentum is actually your enemy, not your friend, for avoiding the singularity - so you don't want to accelerate in a way that increases its magnitude.

2. The effective total energy $\mathcal{E}$ isn't the physical mechanical energy $E$, but instead $\frac{1}{2} E^2$. In the standard nonrelativistic case, firing your engines to slow your infall decreases your total mechanical energy and helps you delay getting close to the center. (This may seem counterintuitive at first, because we associate highly negative energies with tightly bound orbits and positive energies with unbound orbits, so you might think you would want to increase your energy. But for the purpose of delaying getting close to the center, you actually want to brake and make your energy more negative, at the expense of trapping yourself deeper in the gravity well overall and spending more time near the center once you finally do get there.) But in the Schwarzschild case, $\mathcal{E} = \frac{1}{2} E^2$ means that your effective energy actually depends non-monotonically on your physical energy: if your physical energy $E$ is negative, then making it even more negative actually increases your effective energy $\mathcal{E}$. This means that minimizing your effective energy requires keeping your physical energy at $E = 0$, which indeed corresponds to the optimal geodesic which begins at rest infinitesimally outside the horizon. Any attempt to brake further will overshoot $E = 0$ and send $E$ negative, which will actually increase your effective energy and hurt you.

• the usual convention is to define E as positive (for all observers in the exterior region r>2M), the Lewis & Kwan paper notwithstanding. Also it helps to give E a name: the "Killing energy (per mass)", also interpreted as the "energy (per mass) measured at infinity" – Colin MacLaurin Sep 4 '18 at 20:31
• @ColinMacLaurin Yes, I use the usual sign convention for $E$ in my answer, not Lewis and Kwan's convention. My $E$ is indeed positive for a geodesic that extends outside the horizon - I took out the incorrect statement that it's negative. – tparker Sep 4 '18 at 20:40
• indeed you do... I miscounted minus signs! – Colin MacLaurin Sep 4 '18 at 20:52
• The centrifugal force reversal happens outside the horizon: adsbit.harvard.edu/full/1990MNRAS.245..720A - There are no forces in the radial direction inside, because it is the direction in time. Your analysis suggests that time inside depends on energy, but the total time inside is constant for all bodies and is $r=2M$. The different part is their proper time. The ratio of time to proper time $\dfrac{dr}{d\tau}$ refers to time dilation. While the gravitational potential depends on the time dilation, the dilation simply follows directly from the metric with no need to refer to energy. – safesphere Sep 6 '18 at 4:14
• The quantity $dr/d\tau$ is not called "time-dilation". You might call it the "$r$-component of the 4-velocity", assuming coordinates including Schwarzschild $r$ are being used. In curved spacetime, time-dilation is well-defined for two observers at the same event (same place and time), but not defined between separated observers. One exception is with a "stationary" spacetime, where the time symmetry lets you compare a "gravitational time-dilation" between any two observers at rest. But spacetime is not stationary inside the Schwwarzschild horizon. – Colin MacLaurin Oct 5 '18 at 16:59

In a metric, such as

$$d\tau^2=g_{11}dt^2-g_{22}dr^2$$

the longest interval $d\tau$ between two events obviously is when $dr=0$ simply due to the sign. This is the rest frame with no motion in space and consequently no time dilation due to motion. Any $dr\ne 0$ would result in a motion with a stronger time dilation and therefore decrease the interval or the proper time.

The radial geometrized Schwarzschild metric inside the event horizon is

$$d\tau^2 = \left(\frac{r_s}{r}-1\right)^{-1} \,dr^2 - \left(\frac{r_s}{r}-1 \right)\,dt^2\tag{1}$$

Where $r$ is the coordinate time and $t$ is a spatial coordinate orthogonal to time and therefore not pointing to the center. As mentioned above, the longest proper time is when $dt=0$ and therefore

$$d\tau^2 = \left(\frac{r_s}{r}-1\right)^{-1} \,dr^2$$

Or

$$d\tau =\dfrac{dr}{\sqrt{\dfrac{r_s}{r}-1}}$$

Solving

$$\tau=-r\sqrt{\dfrac{r_s}{r}-1}-r_s\arctan\left(\sqrt{\dfrac{r_s}{r}-1}\right)+C$$

From $\,r=r_s\,$ to $\,r=0\,$ the longest possible lifetime inside the black hole is

$$\tau=\dfrac{\pi}{2}r_s=\pi M$$

More rigorously, the bound solution of the geodesic equations for the radial metric $(1)$ yields the following geodesics (where $R$ is the radius, from which the fall starts at rest)

$$\tau=\dfrac{R}{2}\sqrt{\dfrac{R}{2M}}\left(\arccos\left(\dfrac{2r}{R}-1\right)+\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right)$$

And

$$t=\sqrt{\dfrac{R}{2M}-1}\cdot\left(\left(\dfrac{R}{2}+2M\right)\cdot\arccos\left(\dfrac{2r}{R}-1\right)+\dfrac{R}{2}\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right)+$$

$$+\, 2M\ln\left(\left|\dfrac{\sqrt{\dfrac{R}{2M}-1}+\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}{\sqrt{\dfrac{R}{2M}-1}-\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}\right|\right)$$

Plotting these functions for the fall from the horizon $r=2M$ confirms no spatial movement $t=0$ (blue line), as well as the maximum proper time $\tau=\pi M$ (green line). Please note that time $r$ on the chart moves from right to left. In comparison, the next plot represents a fall from $r=5M$ showing time $t$ above the horizon diverging to infinity and showing a fast spatial movement along $t$ inside the horizon causing a stronger time dilation that results in a (roughly twice) smaller value of the proper time $\tau$ between the horizon at $r=2M$ and the singularity at $r=0$. The charts show that gravity inside a black hole causes a deceleration of moving bodies $\dfrac{d^2t}{dr^2}\lt 0$ and does not accelerate bodies at rest with the speed of $\dfrac{dt}{dr}=0$.

Using these results, we can now visualize the geometry of a Schwarzschild black hole in a spacetime reduced by one dimension In this diagram, the coordinate $t$ is vertical. Outside the event horizon $t$ represents time; inside the event horizon $t$ represents a spatial dimension that does not point to the singularity. The radial coordinate $r$ is spatial outside the horizon, but represents time inside. Thus the singularity is a line along the spatial dimension of $t$ at the time of $r=0$.

A body falling from the event horizon $A$ has no momentum along the spatial dimension of $t$. Therefore this body is stationary inside and moves only in time along $r$ from $A$ to $B$. Due to the symmetry considerations, this body cannot gain a momentum along the spatial direction of $t$ during the fall. For this reason, a body falling from the event horizon would have the longest possible lifetime inside the horizon, as discussed above. While we call this movement "a free fall", in fact the body remains stationary in space.

A different body in a free fall from infinity or from any point outside would move outside the horizon along the geodesic from $C$ to $D$. Passed the point $D$ time diverges to infinity for an external observer. After crossing the horizon, this body continues moving along the geodesic from $E$ to $F$ (see also the geodesic chart above). Because this body moves in space along the dimension of $t$, the body experiences a time dilation due to motion that shortens its total proper time inside the black hole.

To extend the proper time, the movement along $t$ must be decelerated and stopped, as shown at $G$. After that the body is stationary with no movement in space along $t$ while moving only in time along $r$ from $G$ to $H$. Obviously, provided the deceleration time is negligible, the lifetime of this body is maximized as discussed earlier.

• what software did you use for the charts? – magma Oct 5 '18 at 20:47
• At a glance, why are you ignoring the tangential motion? A generic astronaut falling into a black hole would have nonzero angular momentum. Is it advantageous to remove it or increase it? – A.V.S. Jan 14 at 22:31
• @A.V.S. The tangential motion persists through the horizon thus increasing the speed inside and adding to the time dilation to shorten the proper time. The objective is to stop any spatial movement passed the horizon, including the tangential movement, which we can stop by not having it in the first place. – safesphere Jan 14 at 22:35
• @safesphere: OK and the answer appears correct to me (so +1) although the green line on your “in comparison” plot baffles me – A.V.S. Jan 15 at 5:38
• @A.V.S. Thanks so much! The green line is $\tau(r)$, so the vertical axis for it is the proper taime $\tau$ (I should have indicated this on the plots, sorry). The first plot shows that $\tau$ inside the horizon is $\pi$ and is approx. twice larger than on the second plot. Thus the plots show that the proper time is maximized by starting a free fall from the horizon. – safesphere Jan 15 at 6:54