Skip to main content
added 37 characters in body
Source Link
J. Murray
  • 71.4k
  • 5
  • 98
  • 203

In this context, "interior" does not mean inside the event horizon of a black hole. Rather, one imagines computing the metric for a spacetime which features the presence of a star, modeled as a spherically-symmetric fluid. The exterior solution corresponds to the (vacuum) region outside of the star, and the interior solution corresponds to the (non-vacuum) region within the star itself.

If the entire mass of the star is compressed to a radius less than the Schwarzschild radius $R_s = 2GM/c^2$, then the star undergoes gravitational collapse and a singularity forms. However, this radius is far smaller than the radius of a typical star (the Schwarzschild radius of the sun is $R_s \sim 10$ km while its actual radius is $R_{\odot }\sim 10^5$ km), so no collapse occurs and no singularity exists.

As a non-relativistic analogy, note that if you compute the gravitational potentialfield strength of a spherical mass distribution with uniform density, total mass $M$, and radius $R$ in a Newtonian framework, you obtain

$$\phi(r) = \begin{cases} \frac{GM}{R} \frac{r}{R} & r< R \\ \frac{GM}{r} & r\geq R\end{cases}$$$$\mathbf g(r) = \begin{cases} \frac{GM}{R^2} \frac{r}{R} \hat r & r< R \\ \frac{GM}{r^2}\hat r & r\geq R\end{cases}$$

Outside of the star, the gravitational potentialfield strength goes like $\sim 1/r$$\sim 1/r^2$ and therefore increases with decreasing $r$. However, once you get inside the star, the gravitational potentialfield strength decreases smoothly back to $0$ at the origin and there is no singularity.

In this context, "interior" does not mean inside the event horizon of a black hole. Rather, one imagines computing the metric for a spacetime which features the presence of a star, modeled as a spherically-symmetric fluid. The exterior solution corresponds to the (vacuum) region outside of the star, and the interior solution corresponds to the (non-vacuum) region within the star itself.

If the entire mass of the star is compressed to a radius less than the Schwarzschild radius $R_s = 2GM/c^2$, then the star undergoes gravitational collapse and a singularity forms. However, this radius is far smaller than the radius of a typical star (the Schwarzschild radius of the sun is $R_s \sim 10$ km while its actual radius is $R_{\odot }\sim 10^5$ km), so no collapse occurs and no singularity exists.

As a non-relativistic analogy, note that if you compute the gravitational potential of a spherical mass distribution with uniform density, total mass $M$, and radius $R$ in a Newtonian framework, you obtain

$$\phi(r) = \begin{cases} \frac{GM}{R} \frac{r}{R} & r< R \\ \frac{GM}{r} & r\geq R\end{cases}$$

Outside of the star, the gravitational potential goes like $\sim 1/r$ and therefore increases with decreasing $r$. However, once you get inside the star, the gravitational potential decreases smoothly back to $0$ at the origin and there is no singularity.

In this context, "interior" does not mean inside the event horizon of a black hole. Rather, one imagines computing the metric for a spacetime which features the presence of a star, modeled as a spherically-symmetric fluid. The exterior solution corresponds to the (vacuum) region outside of the star, and the interior solution corresponds to the (non-vacuum) region within the star itself.

If the entire mass of the star is compressed to a radius less than the Schwarzschild radius $R_s = 2GM/c^2$, then the star undergoes gravitational collapse and a singularity forms. However, this radius is far smaller than the radius of a typical star (the Schwarzschild radius of the sun is $R_s \sim 10$ km while its actual radius is $R_{\odot }\sim 10^5$ km), so no collapse occurs and no singularity exists.

As a non-relativistic analogy, note that if you compute the gravitational field strength of a spherical mass distribution with uniform density, total mass $M$, and radius $R$ in a Newtonian framework, you obtain

$$\mathbf g(r) = \begin{cases} \frac{GM}{R^2} \frac{r}{R} \hat r & r< R \\ \frac{GM}{r^2}\hat r & r\geq R\end{cases}$$

Outside of the star, the gravitational field strength goes like $\sim 1/r^2$ and therefore increases with decreasing $r$. However, once you get inside the star, the gravitational field strength decreases smoothly back to $0$ at the origin and there is no singularity.

Source Link
J. Murray
  • 71.4k
  • 5
  • 98
  • 203

In this context, "interior" does not mean inside the event horizon of a black hole. Rather, one imagines computing the metric for a spacetime which features the presence of a star, modeled as a spherically-symmetric fluid. The exterior solution corresponds to the (vacuum) region outside of the star, and the interior solution corresponds to the (non-vacuum) region within the star itself.

If the entire mass of the star is compressed to a radius less than the Schwarzschild radius $R_s = 2GM/c^2$, then the star undergoes gravitational collapse and a singularity forms. However, this radius is far smaller than the radius of a typical star (the Schwarzschild radius of the sun is $R_s \sim 10$ km while its actual radius is $R_{\odot }\sim 10^5$ km), so no collapse occurs and no singularity exists.

As a non-relativistic analogy, note that if you compute the gravitational potential of a spherical mass distribution with uniform density, total mass $M$, and radius $R$ in a Newtonian framework, you obtain

$$\phi(r) = \begin{cases} \frac{GM}{R} \frac{r}{R} & r< R \\ \frac{GM}{r} & r\geq R\end{cases}$$

Outside of the star, the gravitational potential goes like $\sim 1/r$ and therefore increases with decreasing $r$. However, once you get inside the star, the gravitational potential decreases smoothly back to $0$ at the origin and there is no singularity.