Physics Asked on March 30, 2021
For a blackhole of mass $M$, radius $R$ and Schwarzschild radius $R^*$ (where $R<R^*$), its density $rho$ is defined as $$rho=frac{M}{(R^*)^3}.$$
One reason I read for using the Schwarzschild radius $R^*$ instead of the actual radius $R$ is that no measurements can be made inside of $R^*$ (i.e. for $r<R^*$).
I know that the Schwarzschild metric is $$g_{munu}=text{diag}left[left(-1+{R^*over r}right),left(1-{R^*over r}right)^{-1},r^2,r^2sin^2thetaright],$$
so there is a singular point at $r=R^*$ where $g_{rr}$ is infinite and hence the invariant interval $ds^2$ is infinite.
However, for points inside $R^*$ (i.e. $r<R^*$), the metric and $ds^2$ is finite. Why then is it not possible to make a measurement inside for points inside $R^*$? Is it somehow required to first pass the singular point at $r=R^*$ before a measurement can be made inside $R^*$?
Yes, at the Schwarzschild radius, the metric is so messed up but the fault is in the coordinate system. The metric in Schwarzschild coordinate is faking a singularity. You can call it "A Coordinate Singularity". The Real singularity lies at zero radius (At least in the Schwarzschild coordinate). You can choose a better coordinate system to avoid it (Like the Kruskal-Szekeres Coordinate, And It's easier to understand the cause and effect in this coordinate because the null geodesics are straight 45 degree lines). Being not a true singularity does not mean that the insides of it could be observed, This is because there is a horizon that breaks the causal connection to the insides. And Yes, It is absolutely right that you should make a pass through to make a causal connection with the insides.
Correct answer by Vikash Kotteeswaran on March 30, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP