Radial motion in the Schwarzshild spacetime

[grav-universe]

Simple, because you can't make a valid metric out of two differentials calculated for two DIFFERENT conditions. This is explained by several different posters on page 1.

It is the same condition, or rather the same invariant, as calculated by three different observers: dτ² as measured by the freefaller, dt_s² - dr_s² as measured by a stationary observer that coincides with the freefaller, and (1-2μ/r) dt² - (1- 2μ/r)^-1 dr² as inferred by a distant observer using Schwarzschild coordinates.

 

[xox]

It is the same condition,

No, it isn't:

$dr_s$ is calculated from the metric by setting the condition $dt=0$ in $ds^2=\frac{dr^2}{1-2 \mu/r}-(1-2 \mu/r)dt^2$
$dt_s$ is calculated from the metric by setting a DIFFERENT condition, $dr=0$, into $ds^2=(1-2 \mu/r)dt^2-\frac{dr^2}{1-2 \mu/r}$

Three of us pointed out this kind of error on the first page, please read it, no point in repeating the same mistake over and over.

 

[grav-universe]

No, it isn't:

$dr_s$ is calculated from the metric by setting the condition $dt=0$ in $ds^2=\frac{dr^2}{1-2 \mu/r}-(1-2 \mu/r)dt^2$
$dt_s$ is calculated from the metric by setting a DIFFERENT condition, $dr=0$, into $ds^2=(1-2 \mu/r)dt^2-\frac{dr^2}{1-2 \mu/r}$

Three of us pointed out this kind of error on the first page, please read it, no point in repeating the same mistake over and over.

Well right, time and distance are two different measurements, but when combined, they give the same metric for the overall spacetime.

 

[xox]

Well right, time and distance are two different measurements, but when combined, they give the same metric for the overall spacetime.

You CAN'T "combine" bits of two different solutions obtained for two different conditions, this is the mistake that you keep repeating.