Given that the Newtonian 'radius of escape' for light is given by the relation(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{1}{2}v^2_{esc}=\frac{GM}{R}<\frac{1}{2}c^2[/tex]

from which we get the same value as the Schwarzschild critical radius,

[tex]R_{crit}=\frac{2GM}{c^2}[/tex]

why can't we identify R with the Schwarzschild coordinate r ?

The deriviation of the Schwarzschild metric starts with the line element in R, T, \theta, \phi and after symmetry constraints there is still a dTdR term. This removed by a transformation between Schwarzschild r,t and T which mixes space and time like a boost. This says that r is not R. This whole procedure looks suspect (to me) because there is no Newtonian line element that has time in it.

We know now that the dRdT coefficient will be zero for a non-rotating source, so we could start from a spherically symmetric metric with only diagonal terms, solve the field equations and arrive at the Schwarzschild coordswithoutthe space-time mixing transformation that got rid of the off-diagonal element. In fact without any transformations at all.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Interpreting Schwarzschild coordinates

**Physics Forums | Science Articles, Homework Help, Discussion**