# Why do things fall into a black hole?

• Singlau
In summary, objects will not fall into the sun because of their inertia, which maintains their orbit. Even if they lose some inertia, they will fall to a lower orbit and not into the sun due to the conservation of angular momentum. The same principle applies to objects near a black hole, where the density of the hole may determine whether the object can orbit or be consumed. In general relativity, if an object has a non-zero initial angular momentum, it can orbit a black hole, but its orbit may or may not be stable. This is due to the equations of motion, which take into account the effects of gravity and the properties of the black hole.
Singlau
Earth will not fall into sun because it has inertia, keeping it in orbit. Even if it suddenly loses some inertia, it falls to a lower orbit and won't fall into the sun, as its inertia increase again due to conservation of angular momentum. So, if a thing fall near a black hole, shouldn't it make an orbit, orbiting so fast to reach equilibrium, instead of being eaten by the massive black hole? Or does it have relation with the density of the hole?

Just using the word "inertia" may be misleading. The point is that whether in classical mechanics or in general relativity, and whether for a black hole(a non-rotating black hole!) or a regular star or planet, if the particle has no initial angular momentum, i.e. if its falling radially, it won't have a stable orbit around the source and will fall right into it.
But even in general relativity and even for a black hole, if the particle has a non-zero initial angular momentum or if the black hole is rotating, the particle may orbit around the black hole and its orbit may or may not be stable.

Let's take a Schwarzschild black hole as an example.

For a Schwarzschild black hole, we have $${\rm d}s^2=-\left(1-\frac{2GM}{r}\right){\rm d}t^2+\left(1-\frac{2GM}{r}\right)^{-1}{\rm d}r^2+r^2{\rm d}\theta^2+r^2{\rm sin}^2\theta{\rm d}\varphi^2$$So the metric tensor ##g_{\mu\nu}## has only diagonal element: $$g_{00}=-\left(1-\frac{2GM}{r}\right),\;g_{11}=\left(1-\frac{2GM}{r}\right)^{-1},\;g_{22}=r^2,\;g_{33}=r^2{\rm sin}^2\theta$$Therefore, we could calculate the connection ##\Gamma^{\mu}_{\nu\lambda}##: $$\Gamma^{0}_{01}=\Gamma^{0}_{10}=\frac{GM}{r\left(r-2GM\right)}$$ $$\Gamma^{1}_{00}=\frac{GM}{r^3}\left(r-2GM\right),\;\Gamma^{1}_{11}=-\frac{GM}{r\left(r-2GM\right)}$$ $$\Gamma^{1}_{22}=-\left(r-2GM\right),\;\Gamma^{1}_{33}=-\left(r-2GM\right){\rm sin}^2\theta$$ $$\Gamma^{2}_{21}=\Gamma^{2}_{12}=\Gamma^{3}_{31}=\Gamma^{3}_{13}=\frac{1}{r}$$ $$\Gamma^{2}_{33}=-{\rm sin}\theta{\rm cos}\theta,\;\Gamma^{3}_{23}=\Gamma^{3}_{32}={\rm cot}\theta$$
Consider the geodesics equation $$\frac{{\rm d}^2x^{\mu}}{{\rm d}p^2}+\Gamma^{\mu}_{\nu\lambda}\frac{{\rm d}x^{\nu}}{{\rm d}p}\frac{{\rm d}x^{\lambda}}{{\rm d}p}=0$$where ##p## is an affine parameter (##p=\tau## if the particle has mass ##m##, where ##\tau## is proper time), we can conclude that$$\frac{{\rm d}^2t}{{\rm d}p^2}+\frac{2GM}{r\left(r-2GM\right)}\frac{{\rm d}r}{{\rm d}p}\frac{{\rm d}t}{{\rm d}p}=0$$ $$\frac{{\rm d}^2r}{{\rm d}p^2}+\frac{GM}{r^3}\left(r-2GM\right)\left(\frac{{\rm d}t}{{\rm d}p}\right)^2-\frac{GM}{r\left(r-2GM\right)}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2-\left(r-2GM\right)\left[\left(\frac{{\rm d}\theta}{{\rm d}p}\right)^2+{\rm sin}^2\theta\left(\frac{{\rm d}\varphi}{{\rm d}p}\right)^2\right]=0$$ $$\frac{{\rm d}^2\theta}{{\rm d}p^2}+\frac{2}{r}\frac{{\rm d}r}{{\rm d}p}\frac{{\rm d}\theta}{{\rm d}p}-{\rm sin}\theta{\rm cos}\theta\left(\frac{{\rm d}\varphi}{{\rm d}p}\right)^2=0$$ $$\frac{{\rm d}^2\varphi}{{\rm d}p^2}+\frac{2}{r}\frac{{\rm d}r}{{\rm d}p}\frac{{\rm d}\varphi}{{\rm d}p}+2{\rm cot}\theta\frac{{\rm d}\theta}{{\rm d}p}\frac{{\rm d}\varphi}{{\rm d}p}=0$$
Since the gravitational field is isotropic, we can define the plane constructed by the particle's position vector ##\vec{r}## and momentum vector ##\vec{p}## when ##t=0## as the equatorial plane, which means that ##\theta=\pi/2,\;\dot{\theta}=0##, and thus we know (from the third geodesics equation) that ##\ddot{\theta}=0##.

Due to the property of Killing vector field, we can get two conserved quantities: ##E\equiv \left(1-2GM/r\right){\rm d}t/{\rm d}p={\rm constant}## and ##L\equiv r^2{\rm sin}^2\theta{\rm d}\varphi/{\rm d}p=r^2{\rm d}\varphi/{\rm d}p={\rm constant}##

Besides, we know that ##-g_{\mu\nu}u^{\mu}u^{\nu}=\epsilon##. For photons we have ##\epsilon=0##, and for particles with mass ##m##, ##\epsilon=1##. Therefore, we can conclude that $$-\left(1-\frac{2GM}{r}\right)\left(\frac{{\rm d}t}{{\rm d}p}\right)^2+\left(1-\frac{2GM}{r}\right)^{-1}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+r^2\left(\frac{{\rm d}\varphi}{{\rm d}p}\right)^2=-\epsilon$$replace ##\left(1-2GM/r\right){\rm d}t/{\rm d}p## with ##E## and ##r^2{\rm d}\varphi/{\rm d}p## with ##L##, we know that $$-E^2+\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+\left(1-\frac{2GM}{r}\right)\left(\frac{L^2}{r^2}+\epsilon\right)=0$$ which means $$\frac{1}{2}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+V\left(r\right)=\frac{1}{2}E^2$$where ##V\left(r\right)=\epsilon/2-GM\epsilon/r+L^2/\left(2r^2\right)-GML^2/r^3##

In conclusion, we have the equations of motion as follow $$\theta=\frac{\pi}{2}\\\left(1-\frac{2GM}{r}\right)\frac{{\rm d}t}{{\rm d}p}=E\\r^2\frac{{\rm d}\varphi}{{\rm d}p}=L\\\frac{1}{2}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+V\left(r\right)=\frac{1}{2}E^2$$

## 1. Why do things get sucked into a black hole?

Things get sucked into a black hole because of its extremely strong gravitational pull. This pull is so strong that even light cannot escape, which is why black holes are known as "black" - they do not emit any visible light.

## 2. How are black holes formed?

Black holes are formed when a massive star collapses upon itself, creating a singularity - a point of infinite density and zero volume. This singularity is surrounded by an event horizon, which is the point of no return for anything that enters it.

## 3. What happens to objects that fall into a black hole?

Objects that fall into a black hole are stretched and torn apart by the immense tidal forces near the event horizon. As they get closer to the singularity, they are crushed into an infinitely small point, known as the central singularity.

## 4. Can anything escape from a black hole?

According to current scientific understanding, nothing can escape from a black hole once it has passed the event horizon. This includes light, matter, and even information.

## 5. Is there a way to escape from a black hole?

Currently, there is no known way to escape from a black hole once an object has passed the event horizon. However, there are theories that suggest alternate universes or tunnels through space-time may exist within black holes, but these are purely speculative at this time.

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