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Why do things fall into a black hole?

  1. Dec 30, 2016 #1
    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?
     
  2. jcsd
  3. Dec 30, 2016 #2

    ShayanJ

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    Gold Member

    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.
     
  4. Dec 30, 2016 #3
    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 $$\theta=\frac{\pi}{2}\\\left(1-\frac{2GM}{r}\right)\frac{{\rm d}t}{{\rm d}p}=E\\r^2\frac{{\rm d}\varphi}{{\rm d}p}=L\\\frac{1}{2}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+V\left(r\right)=\frac{1}{2}E^2$$
     
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