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

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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)?

One possible explanation is that the exhaust from your jetpack necessarily pulls on you gravitationally and therefore makes you fall faster. The harder you fire your engines, the more exhaust you kick out, which pulls on you harder and further speeds your fall. But I don't think this explanation is quite right because (a) you're firing the exhaust radially inward and so into "your future", so it's not obvious that it can causally influence you, and (b) the result to be explained occurs purely within the fixed Schwarzchild metric without needing to incorporate any gravitational back-reaction perturbations from the exhaust.

general-relativity black-holes event-horizon singularities geodesics

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edited Sep 3 at 0:50

asked Sep 1 at 18:44

tparker

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  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 at 0:24
  

  
  
  
  


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  (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 at 21:11
  

  
  
  
  


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  To see the horizon is a null hypersurface, consider the hypersurface $r=textrmconst$. 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 at 21:16
  

  
  
  
  

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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)?

One possible explanation is that the exhaust from your jetpack necessarily pulls on you gravitationally and therefore makes you fall faster. The harder you fire your engines, the more exhaust you kick out, which pulls on you harder and further speeds your fall. But I don't think this explanation is quite right because (a) you're firing the exhaust radially inward and so into "your future", so it's not obvious that it can causally influence you, and (b) the result to be explained occurs purely within the fixed Schwarzchild metric without needing to incorporate any gravitational back-reaction perturbations from the exhaust.

general-relativity black-holes event-horizon singularities geodesics

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edited Sep 3 at 0:50

asked Sep 1 at 18:44

tparker

20.8k42112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 at 0:24
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  (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 at 21:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  To see the horizon is a null hypersurface, consider the hypersurface $r=textrmconst$. 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 at 21:16
  

  
  
  
  

add a comment | 

up vote
18
down vote

favorite

10

up vote
18
down vote

favorite

10

10

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)?

One possible explanation is that the exhaust from your jetpack necessarily pulls on you gravitationally and therefore makes you fall faster. The harder you fire your engines, the more exhaust you kick out, which pulls on you harder and further speeds your fall. But I don't think this explanation is quite right because (a) you're firing the exhaust radially inward and so into "your future", so it's not obvious that it can causally influence you, and (b) the result to be explained occurs purely within the fixed Schwarzchild metric without needing to incorporate any gravitational back-reaction perturbations from the exhaust.

general-relativity black-holes event-horizon singularities geodesics

share|cite|improve this question

edited Sep 3 at 0:50

asked Sep 1 at 18:44

tparker

20.8k42112

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)?

One possible explanation is that the exhaust from your jetpack necessarily pulls on you gravitationally and therefore makes you fall faster. The harder you fire your engines, the more exhaust you kick out, which pulls on you harder and further speeds your fall. But I don't think this explanation is quite right because (a) you're firing the exhaust radially inward and so into "your future", so it's not obvious that it can causally influence you, and (b) the result to be explained occurs purely within the fixed Schwarzchild metric without needing to incorporate any gravitational back-reaction perturbations from the exhaust.

general-relativity black-holes event-horizon singularities geodesics

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edited Sep 3 at 0:50

asked Sep 1 at 18:44

tparker

20.8k42112

share|cite|improve this question

share|cite|improve this question

edited Sep 3 at 0:50

edited Sep 3 at 0:50

edited Sep 3 at 0:50

asked Sep 1 at 18:44

tparker

20.8k42112

asked Sep 1 at 18:44

tparker

20.8k42112

asked Sep 1 at 18:44

tparker

20.8k42112

20.8k42112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 at 0:24
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  (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 at 21:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  To see the horizon is a null hypersurface, consider the hypersurface $r=textrmconst$. 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 at 21:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 at 0:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (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 at 21:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  To see the horizon is a null hypersurface, consider the hypersurface $r=textrmconst$. 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 at 21:16
  

  
  
  
  

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 at 0:24

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 at 0:24

1

1

(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 at 21:11

(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 at 21:11

1

1

To see the horizon is a null hypersurface, consider the hypersurface $r=textrmconst$. 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 at 21:16

To see the horizon is a null hypersurface, consider the hypersurface $r=textrmconst$. 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 at 21:16

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4 Answers
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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.

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edited Sep 2 at 19:25

answered Sep 1 at 19:11

Dale

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  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 at 20:20
  

  
  
  
  


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  Is there a simple way to see the flaw in Carroll's argument?
  – tparker
  Sep 1 at 22:37
  
  

  
  
  
  


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  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 at 22:53
  

  
  
  
  


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  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 at 12:17
  

  
  
  
  


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  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 at 15:13
  

  
  
  
  

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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 :)

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answered Sep 1 at 19:00

J. Murray

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  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 at 20:37
  

  
  
  
  


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  This is certainly true, but I was looking for a more precise picture of what happens when you fire your jetpack.
  – tparker
  Sep 1 at 22:17
  

  
  
  
  


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  @knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification.
  – J. Murray
  Sep 1 at 22:24
  

  
  
  
  


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  @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 at 22:48
  

  
  
  
  


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  @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 at 6:23
  

  
  
  
  

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Here's a partial answer, although it's still pretty formal. First define
$$E := -left( frac2GMr - 1 right) fracdtdtau, qquad L := r^2 fracdphidtau$$
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
$$frac12 left( fracdrdtau right)^2 - left(frac2GMr - 1 right) left(1 + fracL^2r^2 right) = frac12 E^2,$$
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 $mathcalE$ isn't the physical mechanical energy $E$, but instead $frac12 E^2$. 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, $mathcalE = frac12 E^2$ 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 $mathcalE$. 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.

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edited Sep 4 at 20:38

answered Sep 4 at 2:51

tparker

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  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 at 20:31
  

  
  
  
  


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  @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 at 20:40
  

  
  
  
  


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  indeed you do... I miscounted minus signs!
  – Colin MacLaurin
  Sep 4 at 20:52
  

  
  
  
  


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  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 $dfracdrdtau$ 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 at 4:14
  
  

  
  
  
  

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In a metric, such as

$$ dtau^2=g_11dt^2-g_22dr^2 $$

the longest interval $dtau$ 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 $drne 0$ 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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 - left(fracr_sr-1 right),dt^2tag1 $$

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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 $$

Or

$$ dtau =dfracdrsqrtdfracr_sr-1 $$

Solving

$$ tau=-rsqrtdfracr_sr-1-r_sarctanleft(sqrtdfracr_sr-1right)+C $$

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

$$ tau=dfracpi2r_s=pi M $$

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)

$$ tau=dfracR2sqrtdfracR2Mleft(arccosleft(dfrac2rR-1right)+sinleft(arccosleft(dfrac2rR-1right)right)right) $$

And

$$ t=sqrtdfracR2M-1cdotleft(left(dfracR2+2Mright)cdotarccosleft(dfrac2rR-1right)+dfracR2sinleft(arccosleft(dfrac2rR-1right)right)right)+ $$

$$ +, 2Mlnleft(left|dfracsqrtdfracR2M-1+tanleft(dfrac12arccosleft(dfrac2rR-1right)right)sqrtdfracR2M-1-tanleft(dfrac12arccosleft(dfrac2rR-1right)right)right|right) $$

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 $dfracd^2tdr^2lt 0$ and does not accelerate bodies at rest with the speed of $dfracdtdr=0$.

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.

share|cite|improve this answer

edited Sep 6 at 8:00

answered Sep 6 at 0:29

safesphere

6,39111238

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.V.S. I wonder if you'd be so kind as to review my answer above and let me know how I can improve it. I am here to learn and you are the topmost expert in GR, so I would greatly appreciate your help. One error I know about is that the geodesic must be orthogonal to $t$ in the point of $F$ in my improvised chart. I'll try to fix this. In particular, is my understanding correct that this singularity is asymptotically a 1D line $(space,t; time,r=0)$? Asymptotically here is the keyword, so that not to analyze the nature of the singularity itself, but only the spacetime a short interval away.
  – safesphere
  2 days ago
  
  

  
  
  
  

add a comment | 

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4 Answers
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votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
26
down vote

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.

share|cite|improve this answer

edited Sep 2 at 19:25

answered Sep 1 at 19:11

Dale

60718

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  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 at 20:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is there a simple way to see the flaw in Carroll's argument?
  – tparker
  Sep 1 at 22:37
  
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  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 at 22:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 at 12:17
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 at 15:13
  

  
  
  
  

 | 
show 10 more comments

up vote
26
down vote

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.

share|cite|improve this answer

edited Sep 2 at 19:25

answered Sep 1 at 19:11

Dale

60718

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  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 at 20:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is there a simple way to see the flaw in Carroll's argument?
  – tparker
  Sep 1 at 22:37
  
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  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 at 22:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 at 12:17
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 at 15:13
  

  
  
  
  

 | 
show 10 more comments

up vote
26
down vote

up vote
26
down vote

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.

share|cite|improve this answer

edited Sep 2 at 19:25

answered Sep 1 at 19:11

Dale

60718

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.

share|cite|improve this answer

edited Sep 2 at 19:25

answered Sep 1 at 19:11

Dale

60718

share|cite|improve this answer

share|cite|improve this answer

edited Sep 2 at 19:25

edited Sep 2 at 19:25

edited Sep 2 at 19:25

answered Sep 1 at 19:11

Dale

60718

answered Sep 1 at 19:11

Dale

60718

answered Sep 1 at 19:11

Dale

60718

60718

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  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 at 20:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is there a simple way to see the flaw in Carroll's argument?
  – tparker
  Sep 1 at 22:37
  
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  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 at 22:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 at 12:17
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 at 15:13
  

  
  
  
  

 | 
show 10 more comments

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  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 at 20:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is there a simple way to see the flaw in Carroll's argument?
  – tparker
  Sep 1 at 22:37
  
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  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 at 22:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 at 12:17
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 at 15:13
  

  
  
  
  

12

12

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 at 20:20

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 at 20:20

1

1

Is there a simple way to see the flaw in Carroll's argument?
– tparker
Sep 1 at 22:37

Is there a simple way to see the flaw in Carroll's argument?
– tparker
Sep 1 at 22:37

7

7

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 at 22:53

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 at 22:53

1

1

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 at 12:17

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 at 12:17

3

3

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 at 15:13

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 at 15:13

 | 
show 10 more comments

up vote
6
down vote

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 :)

share|cite|improve this answer

answered Sep 1 at 19:00

J. Murray

5,6532519

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 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 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification.
  – J. Murray
  Sep 1 at 22:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 22:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 6:23
  

  
  
  
  

 | 
show 5 more comments

up vote
6
down vote

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 :)

share|cite|improve this answer

answered Sep 1 at 19:00

J. Murray

5,6532519

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 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 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification.
  – J. Murray
  Sep 1 at 22:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 22:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 6:23
  

  
  
  
  

 | 
show 5 more comments

up vote
6
down vote

up vote
6
down vote

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 :)

share|cite|improve this answer

answered Sep 1 at 19:00

J. Murray

5,6532519

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 :)

share|cite|improve this answer

answered Sep 1 at 19:00

J. Murray

5,6532519

share|cite|improve this answer

share|cite|improve this answer

answered Sep 1 at 19:00

J. Murray

5,6532519

answered Sep 1 at 19:00

J. Murray

5,6532519

answered Sep 1 at 19:00

J. Murray

5,6532519

5,6532519

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 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 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification.
  – J. Murray
  Sep 1 at 22:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 22:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 6:23
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  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 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 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification.
  – J. Murray
  Sep 1 at 22:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 22:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 at 6:23
  

  
  
  
  

3

3

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 at 20:37

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 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 at 22:17

This is certainly true, but I was looking for a more precise picture of what happens when you fire your jetpack.
– tparker
Sep 1 at 22:17

@knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification.
– J. Murray
Sep 1 at 22:24

@knzhou Thanks! That - along with the paper linked by Dale - provides a very clear clarification.
– J. Murray
Sep 1 at 22:24

1

1

@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 at 22:48

@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 at 22:48

1

1

@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 at 6:23

@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 at 6:23

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show 5 more comments

up vote
2
down vote



accepted

Here's a partial answer, although it's still pretty formal. First define
$$E := -left( frac2GMr - 1 right) fracdtdtau, qquad L := r^2 fracdphidtau$$
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
$$frac12 left( fracdrdtau right)^2 - left(frac2GMr - 1 right) left(1 + fracL^2r^2 right) = frac12 E^2,$$
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 $mathcalE$ isn't the physical mechanical energy $E$, but instead $frac12 E^2$. 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, $mathcalE = frac12 E^2$ 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 $mathcalE$. 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.

share|cite|improve this answer

edited Sep 4 at 20:38

answered Sep 4 at 2:51

tparker

20.8k42112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 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 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  indeed you do... I miscounted minus signs!
  – Colin MacLaurin
  Sep 4 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 $dfracdrdtau$ 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 at 4:14
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Here's a partial answer, although it's still pretty formal. First define
$$E := -left( frac2GMr - 1 right) fracdtdtau, qquad L := r^2 fracdphidtau$$
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
$$frac12 left( fracdrdtau right)^2 - left(frac2GMr - 1 right) left(1 + fracL^2r^2 right) = frac12 E^2,$$
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 $mathcalE$ isn't the physical mechanical energy $E$, but instead $frac12 E^2$. 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, $mathcalE = frac12 E^2$ 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 $mathcalE$. 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.

share|cite|improve this answer

edited Sep 4 at 20:38

answered Sep 4 at 2:51

tparker

20.8k42112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 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 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  indeed you do... I miscounted minus signs!
  – Colin MacLaurin
  Sep 4 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 $dfracdrdtau$ 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 at 4:14
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Here's a partial answer, although it's still pretty formal. First define
$$E := -left( frac2GMr - 1 right) fracdtdtau, qquad L := r^2 fracdphidtau$$
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
$$frac12 left( fracdrdtau right)^2 - left(frac2GMr - 1 right) left(1 + fracL^2r^2 right) = frac12 E^2,$$
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 $mathcalE$ isn't the physical mechanical energy $E$, but instead $frac12 E^2$. 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, $mathcalE = frac12 E^2$ 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 $mathcalE$. 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.

share|cite|improve this answer

edited Sep 4 at 20:38

answered Sep 4 at 2:51

tparker

20.8k42112

Here's a partial answer, although it's still pretty formal. First define
$$E := -left( frac2GMr - 1 right) fracdtdtau, qquad L := r^2 fracdphidtau$$
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
$$frac12 left( fracdrdtau right)^2 - left(frac2GMr - 1 right) left(1 + fracL^2r^2 right) = frac12 E^2,$$
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 $mathcalE$ isn't the physical mechanical energy $E$, but instead $frac12 E^2$. 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, $mathcalE = frac12 E^2$ 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 $mathcalE$. 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.

share|cite|improve this answer

edited Sep 4 at 20:38

answered Sep 4 at 2:51

tparker

20.8k42112

share|cite|improve this answer

share|cite|improve this answer

edited Sep 4 at 20:38

edited Sep 4 at 20:38

edited Sep 4 at 20:38

answered Sep 4 at 2:51

tparker

20.8k42112

answered Sep 4 at 2:51

tparker

20.8k42112

answered Sep 4 at 2:51

tparker

20.8k42112

20.8k42112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 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 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  indeed you do... I miscounted minus signs!
  – Colin MacLaurin
  Sep 4 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 $dfracdrdtau$ 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 at 4:14
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 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 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  indeed you do... I miscounted minus signs!
  – Colin MacLaurin
  Sep 4 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 $dfracdrdtau$ 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 at 4:14
  
  

  
  
  
  

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 at 20:31

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 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 at 20:40

@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 at 20:40

indeed you do... I miscounted minus signs!
– Colin MacLaurin
Sep 4 at 20:52

indeed you do... I miscounted minus signs!
– Colin MacLaurin
Sep 4 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 $dfracdrdtau$ 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 at 4:14

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 $dfracdrdtau$ 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 at 4:14

add a comment | 

up vote
0
down vote

In a metric, such as

$$ dtau^2=g_11dt^2-g_22dr^2 $$

the longest interval $dtau$ 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 $drne 0$ 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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 - left(fracr_sr-1 right),dt^2tag1 $$

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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 $$

Or

$$ dtau =dfracdrsqrtdfracr_sr-1 $$

Solving

$$ tau=-rsqrtdfracr_sr-1-r_sarctanleft(sqrtdfracr_sr-1right)+C $$

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

$$ tau=dfracpi2r_s=pi M $$

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)

$$ tau=dfracR2sqrtdfracR2Mleft(arccosleft(dfrac2rR-1right)+sinleft(arccosleft(dfrac2rR-1right)right)right) $$

And

$$ t=sqrtdfracR2M-1cdotleft(left(dfracR2+2Mright)cdotarccosleft(dfrac2rR-1right)+dfracR2sinleft(arccosleft(dfrac2rR-1right)right)right)+ $$

$$ +, 2Mlnleft(left|dfracsqrtdfracR2M-1+tanleft(dfrac12arccosleft(dfrac2rR-1right)right)sqrtdfracR2M-1-tanleft(dfrac12arccosleft(dfrac2rR-1right)right)right|right) $$

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 $dfracd^2tdr^2lt 0$ and does not accelerate bodies at rest with the speed of $dfracdtdr=0$.

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.

share|cite|improve this answer

edited Sep 6 at 8:00

answered Sep 6 at 0:29

safesphere

6,39111238

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.V.S. I wonder if you'd be so kind as to review my answer above and let me know how I can improve it. I am here to learn and you are the topmost expert in GR, so I would greatly appreciate your help. One error I know about is that the geodesic must be orthogonal to $t$ in the point of $F$ in my improvised chart. I'll try to fix this. In particular, is my understanding correct that this singularity is asymptotically a 1D line $(space,t; time,r=0)$? Asymptotically here is the keyword, so that not to analyze the nature of the singularity itself, but only the spacetime a short interval away.
  – safesphere
  2 days ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

In a metric, such as

$$ dtau^2=g_11dt^2-g_22dr^2 $$

the longest interval $dtau$ 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 $drne 0$ 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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 - left(fracr_sr-1 right),dt^2tag1 $$

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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 $$

Or

$$ dtau =dfracdrsqrtdfracr_sr-1 $$

Solving

$$ tau=-rsqrtdfracr_sr-1-r_sarctanleft(sqrtdfracr_sr-1right)+C $$

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

$$ tau=dfracpi2r_s=pi M $$

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)

$$ tau=dfracR2sqrtdfracR2Mleft(arccosleft(dfrac2rR-1right)+sinleft(arccosleft(dfrac2rR-1right)right)right) $$

And

$$ t=sqrtdfracR2M-1cdotleft(left(dfracR2+2Mright)cdotarccosleft(dfrac2rR-1right)+dfracR2sinleft(arccosleft(dfrac2rR-1right)right)right)+ $$

$$ +, 2Mlnleft(left|dfracsqrtdfracR2M-1+tanleft(dfrac12arccosleft(dfrac2rR-1right)right)sqrtdfracR2M-1-tanleft(dfrac12arccosleft(dfrac2rR-1right)right)right|right) $$

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 $dfracd^2tdr^2lt 0$ and does not accelerate bodies at rest with the speed of $dfracdtdr=0$.

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.

share|cite|improve this answer

edited Sep 6 at 8:00

answered Sep 6 at 0:29

safesphere

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  @A.V.S. I wonder if you'd be so kind as to review my answer above and let me know how I can improve it. I am here to learn and you are the topmost expert in GR, so I would greatly appreciate your help. One error I know about is that the geodesic must be orthogonal to $t$ in the point of $F$ in my improvised chart. I'll try to fix this. In particular, is my understanding correct that this singularity is asymptotically a 1D line $(space,t; time,r=0)$? Asymptotically here is the keyword, so that not to analyze the nature of the singularity itself, but only the spacetime a short interval away.
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In a metric, such as

$$ dtau^2=g_11dt^2-g_22dr^2 $$

the longest interval $dtau$ 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 $drne 0$ 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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 - left(fracr_sr-1 right),dt^2tag1 $$

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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 $$

Or

$$ dtau =dfracdrsqrtdfracr_sr-1 $$

Solving

$$ tau=-rsqrtdfracr_sr-1-r_sarctanleft(sqrtdfracr_sr-1right)+C $$

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

$$ tau=dfracpi2r_s=pi M $$

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)

$$ tau=dfracR2sqrtdfracR2Mleft(arccosleft(dfrac2rR-1right)+sinleft(arccosleft(dfrac2rR-1right)right)right) $$

And

$$ t=sqrtdfracR2M-1cdotleft(left(dfracR2+2Mright)cdotarccosleft(dfrac2rR-1right)+dfracR2sinleft(arccosleft(dfrac2rR-1right)right)right)+ $$

$$ +, 2Mlnleft(left|dfracsqrtdfracR2M-1+tanleft(dfrac12arccosleft(dfrac2rR-1right)right)sqrtdfracR2M-1-tanleft(dfrac12arccosleft(dfrac2rR-1right)right)right|right) $$

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 $dfracd^2tdr^2lt 0$ and does not accelerate bodies at rest with the speed of $dfracdtdr=0$.

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.

share|cite|improve this answer

edited Sep 6 at 8:00

answered Sep 6 at 0:29

safesphere

6,39111238

In a metric, such as

$$ dtau^2=g_11dt^2-g_22dr^2 $$

the longest interval $dtau$ 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 $drne 0$ 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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 - left(fracr_sr-1 right),dt^2tag1 $$

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

$$ dtau^2 = left(fracr_sr-1right)^-1 ,dr^2 $$

Or

$$ dtau =dfracdrsqrtdfracr_sr-1 $$

Solving

$$ tau=-rsqrtdfracr_sr-1-r_sarctanleft(sqrtdfracr_sr-1right)+C $$

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

$$ tau=dfracpi2r_s=pi M $$

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)

$$ tau=dfracR2sqrtdfracR2Mleft(arccosleft(dfrac2rR-1right)+sinleft(arccosleft(dfrac2rR-1right)right)right) $$

And

$$ t=sqrtdfracR2M-1cdotleft(left(dfracR2+2Mright)cdotarccosleft(dfrac2rR-1right)+dfracR2sinleft(arccosleft(dfrac2rR-1right)right)right)+ $$

$$ +, 2Mlnleft(left|dfracsqrtdfracR2M-1+tanleft(dfrac12arccosleft(dfrac2rR-1right)right)sqrtdfracR2M-1-tanleft(dfrac12arccosleft(dfrac2rR-1right)right)right|right) $$

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 $dfracd^2tdr^2lt 0$ and does not accelerate bodies at rest with the speed of $dfracdtdr=0$.

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.

share|cite|improve this answer

edited Sep 6 at 8:00

answered Sep 6 at 0:29

safesphere

6,39111238

share|cite|improve this answer

share|cite|improve this answer

edited Sep 6 at 8:00

edited Sep 6 at 8:00

edited Sep 6 at 8:00

answered Sep 6 at 0:29

safesphere

6,39111238

answered Sep 6 at 0:29

safesphere

6,39111238

answered Sep 6 at 0:29

safesphere

6,39111238

6,39111238

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.V.S. I wonder if you'd be so kind as to review my answer above and let me know how I can improve it. I am here to learn and you are the topmost expert in GR, so I would greatly appreciate your help. One error I know about is that the geodesic must be orthogonal to $t$ in the point of $F$ in my improvised chart. I'll try to fix this. In particular, is my understanding correct that this singularity is asymptotically a 1D line $(space,t; time,r=0)$? Asymptotically here is the keyword, so that not to analyze the nature of the singularity itself, but only the spacetime a short interval away.
  – safesphere
  2 days ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.V.S. I wonder if you'd be so kind as to review my answer above and let me know how I can improve it. I am here to learn and you are the topmost expert in GR, so I would greatly appreciate your help. One error I know about is that the geodesic must be orthogonal to $t$ in the point of $F$ in my improvised chart. I'll try to fix this. In particular, is my understanding correct that this singularity is asymptotically a 1D line $(space,t; time,r=0)$? Asymptotically here is the keyword, so that not to analyze the nature of the singularity itself, but only the spacetime a short interval away.
  – safesphere
  2 days ago
  
  

  
  
  
  

@A.V.S. I wonder if you'd be so kind as to review my answer above and let me know how I can improve it. I am here to learn and you are the topmost expert in GR, so I would greatly appreciate your help. One error I know about is that the geodesic must be orthogonal to $t$ in the point of $F$ in my improvised chart. I'll try to fix this. In particular, is my understanding correct that this singularity is asymptotically a 1D line $(space,t; time,r=0)$? Asymptotically here is the keyword, so that not to analyze the nature of the singularity itself, but only the spacetime a short interval away.
– safesphere
2 days ago

@A.V.S. I wonder if you'd be so kind as to review my answer above and let me know how I can improve it. I am here to learn and you are the topmost expert in GR, so I would greatly appreciate your help. One error I know about is that the geodesic must be orthogonal to $t$ in the point of $F$ in my improvised chart. I'll try to fix this. In particular, is my understanding correct that this singularity is asymptotically a 1D line $(space,t; time,r=0)$? Asymptotically here is the keyword, so that not to analyze the nature of the singularity itself, but only the spacetime a short interval away.
– safesphere
2 days ago

add a comment | 

 

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