# Why r=0 is inevitable when crossing horizon?

• B
I am puzzled. Please help me out. Thanks.

If we cross the horizon of Schwarzschild black hole horizon, we know that r becomes to timelike.
Why the fututre is direction of decreasing r rather than increasing r?

## Answers and Replies

pervect
Staff Emeritus
Science Advisor
I am puzzled. Please help me out. Thanks.

If we cross the horizon of Schwarzschild black hole horizon, we know that r becomes to timelike.
Why the fututre is direction of decreasing r rather than increasing r?

A time-reversed Schwarzschild black hole is called a white hole. So in a black hole, r decreases to zero, in a white hole the opposite happens.

you mean it is some kinds of defination. Inside we have something like -r=Time?

I find that using light cone direction as the future direction, from diagram we can see that future is towards r decreasing.

Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
This is easier to see in a different coordinate system which is not singular at the horizon such as Kruskal–Szekeres coordinates.

• craigthone
Ibix
Science Advisor
2020 Award
I think @pervect is pointing out that if the singularity is in the past then geodesics point away from the singularity - in other words gravity is repulsive. So we would be surprised if a black hole singularity was in the past.

DrGreg
Science Advisor
Gold Member
This Kruskal–Szekeres diagram from an old post of mine may help: Pink zone I is outside the event horizon.
The dotted purple line is the event horizon.
Blue zone II is inside the event horizon.
The thick blue line is the singularity.
The curved lines are lines of constant ##r## coordinate.
The straight lines are lines of constant ##t## coordinate.
As you approach the event horizon, upwards in the diagram, ##r## is decreasing. It continues to decrease inside the horizon.

• craigthone and vanhees71
Thanks everyone for your kind explainations.

PeterDonis
Mentor
2020 Award
geodesics point away from the singularity

Yes.

in other words gravity is repulsive

No. Gravity in a white hole is still attractive. The geodesics going away from the singularity are decelerating (heuristically--the precise description is more complicated, but it still works out to attractive gravity), which is the time reverse of geodesics accelerating towards a black hole.

Ibix
Science Advisor
2020 Award
Gravity in a white hole is still attractive. The geodesics going away from the singularity are decelerating (heuristically--the precise description is more complicated, but it still works out to attractive gravity), which is the time reverse of geodesics accelerating towards a black hole.
Of course - because in a black hole a particle released from rest undergoes (Schwarzschild) coordinate acceleration. Time reversal would be coordinate deceleration (in Weisschild coordinates...? ). So a white hole emits high speed particles which slow down under the influence of its gravity.

I need to think about this a bit more.

stevendaryl
Staff Emeritus
Science Advisor
The relationship between white holes and black holes can be understood, at least partially, by just looking at the geodesic for radially falling test particles near a black hole.

If we ignore the angular coordinates and assume only radial motion, then the geodesics for a test particle in Schwarzschild coordinates satisfy the equations: (with units such that $c=1$
1. $\frac{m}{2} Q (V^t)^2 - \dfrac{m (V^r)^2}{2Q} = \frac{m}{2}$ where $m$ is the mass of the test particle, $Q = 1 - 2GM/r$, $M$ is the mass of the black hole, $V^t = \frac{dt}{d\tau}$, $V^r = \frac{dr}{d\tau}$. This equation can be obtained from the Schwarzschild metric $ds^2 = Q dt^2 - \frac{1}{Q} dr^2 +$ angular terms by multiplying by $\frac{m}{2}$ and dividing through by $ds^2 = d\tau^2$. Obviously, the factor of $\frac{m}{2}$ isn't doing anything, but I'm putting it in for reasons that will be apparent in a while.
2. $m Q V^t = E$ where $E$ is a constant of the motion. Since the metric does not depend on the coordinate $t$, that implies the existence of such a constant of the motion. We can think of it as the conserved momentum in the time-direction.
We can combine these into a single equation:

$\frac{E^2 - m^2}{2 m} = \frac{m}{2} (V^r)^2 - \frac{GMm}{r}$

where I replaced $Q$ by its definition. Finally, we introduce a new constant, $\mathcal{E} = \frac{E^2 - m^2}{2m}$ to write this in the form:

$\mathcal{E} = \frac{m}{2} v^2 - \frac{GMm}{r}$

where I dropped the superscript on $V^r$. The interesting thing about this equation is that it is EXACTLY the Newtonian energy equation for the motion of a test particle near a mass $M$, except that the "radial velocity" is with respect to proper time, $\tau$, rather than coordinate time. This equation is a little complicated to solve exactly, but knowing how Newtonian gravity works, we can immediately get a qualitative description of the motion (assuming that we extrapolate proper time maximally into both the past and future):
• For $\mathcal{E} > 0$ and $v > 0$, the test particle will escape to infinity, rather than fall into the black hole. So let's assume that $\mathcal{E} < 0$.
• If the test particle is ever momentarily at rest ($v=0$) above the event horizon at $r=2GM$, then $|\mathcal{E}| < \frac{m}{2}$. Let's assume that is the case.
• At a finite "time" $\tau_1$ in the past, the test particle was rising from the singularity at $r=0$, with the velocity $v$ ever-decreasing as it gets higher.
• At some "time" $\tau_2 > \tau_1$, the test particle rose above the event horizon at $r=2GM$.
• At some "time" $\tau_3 > \tau_2$, the test particle reaches its maximum "height" $r_{max} = \frac{GMm}{|\mathcal{E}|}$ above the event horizon.
• At some "time" $\tau_4 > \tau_3$, the test particle falls back below the event horizon.
• Finally, at some finite time $\tau_5 > \tau_4$, the particle falls into the singularity at $r=0$
So the qualitative behavior of the test particle, as a function of proper time, is pretty unsurprising: Initially the particle is rising, but eventually gravity turns it around, and it starts falling, and it keeps falling faster and faster until it hits the singularity at $r=0$. But this seems to contradict what we know about black holes, that once a test particle is below the event horizon, it can never rise above it again. So how to resolve this apparent contradiction?

Well, the relationship between proper time $\tau$ and coordinate time $t$ is singular at the event horizon, which can be seen from the equation:

$m Q V^t = E$ or $V^t = \frac{E}{mQ}$

At the event horizon, $Q \equiv 1-\frac{2GM}{r} \Rightarrow 0$, so $V^t \Rightarrow \infty$. Without explicitly solving for $t$ in terms of $\tau$, we can reason that
• As $\tau \Rightarrow \tau_2$, the moment when the test particle rose from the event horizon, then $t \Rightarrow -\infty$.
• As $\tau \Rightarrow \tau_4$, the moment when the test particle falls below the event horizon, then $t \Rightarrow +\infty$.
So the events with $\tau < \tau_2$ took place in the infinite past, and the events with $\tau > \tau_4$ will take place in the infinite future, as reckoned using Schwarzschild coordinates. A realistic black hole is one that hasn't always existed, but was created by the collapse of a star. So for such a realistic black hole, the region of spacetime $t \Rightarrow -\infty$ doesn't exist (or at least, is not described by the Schwarzschild geometry). So for realistic black holes, the event of a test particle rising above the event horizon never happens.

In our idealized, eternal black hole, however, there is a region of spacetime in which the test particle is below the horizon, and rising. (The time period $\tau < \tau_2$). This region of spacetime is the white hole region. From a certain point of view, this doesn't seem to be a region defined by anything in the geometry, but is defined by kinematic properties of test particles: In this region of spacetime $V^r \equiv \frac{d r}{d \tau} > 0$.

But that brings up a puzzle: This way of picking out the white hole region relies on a kinematic property of the test particle, the sign of $V^r$. If you have a bunch of different test particles, then what guarantees that they all single out the same region of spacetime? Why can't $V^r$ be positive for one test particle, and negative for another test particle? Here's what I think is the answer. Proper time $\tau$ is just an affine parameter of a geodesic. But if $\tau$ is such a parameter, then so is $-\tau$. So we are free to choose a parametrization such that $V^r > 0$ for all particles in one region of spacetime (the region below the horizon, in the infinite past, according to $t$), and $V^r < 0$ for all particles in another region (the region below the horizon, in the infinite future). This makes, for an eternal black hole, it completely arbitrary which region we consider the black hole, and which region we consider the white hole. But in a realistic black hole, only one of these regions will exist, so we can say that only the black hole region exists. This amounts to choosing the Schwarzschild time coordinate $t$ and the proper time parameter $\tau$ so that both increase as a particle falls toward the event horizon from above.

This actually brings up yet another puzzle: From the point of view of Schwarzschild geometry, the sign of the $t$ coordinate is an arbitrary choice, so we can choose it so that $t$ increases as an object falls toward the event horizon. But for thermodynamics, the sign of $t$ is not arbitrary; the third law says that entropy increases with increasing $t$. So there is a question of why the arbitrary modeling choice, that $t$ increases as you fall into a black hole, matches the nonarbitrary thermodynamic arrow of time.

• m4r35n357