# Interpreting Schwarzschild coordinates

1. Jan 9, 2009

### Mentz114

Given that the Newtonian 'radius of escape' for light is given by the relation
$$\frac{1}{2}v^2_{esc}=\frac{GM}{R}<\frac{1}{2}c^2$$
from which we get the same value as the Schwarzschild critical radius,
$$R_{crit}=\frac{2GM}{c^2}$$
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 coords without the space-time mixing transformation that got rid of the off-diagonal element. In fact without any transformations at all.

Last edited: Jan 9, 2009
2. Jan 9, 2009

### Philosophaie

$$R_{crit}$$ is fixed for each object (ignoring all the weak gravitational effects from other objects). The Schwarzschild Radius for Earth is about 9mm. The dTdR term means that there is not a pure gravitational field. There must be a strong magnetic or electric field associated with the object in question. If the object in question is the sun the magnetic field filps on a regular basis and there is a question of the solar wind phenomena carrying added magnetic field through the solar system.

Last edited: Jan 9, 2009