Radial motion in the Schwarzshild spacetime

[George Jones]

No, it doesn't. It is a pure coincidence that dr_s/dt_s gives the right answer, this was pointed out to you repeatedly by different posters. The correct derivation is considerably more involved than dividing dr_s by dt_s. The first page of this thread is occupied by posts that show your approach to be unfounded.

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

I have posted that there is more going on than meets the eye, but it is more than a coincidence the division $dr_\mathrm{S}/dt_\mathrm{S}$ gives the physical velocity. I"m not sure, but WannabeNewton might be gnashing his teeth over what seems to be a particularly egregious example of physicists' handwaving mathematics (often useful!). Actually, the division is mathematically rigourous(!), and does give the answer in a somewhat intuitive way.

the 4-velocity of a radially freely falling observer, who starts at rest at spatial infinity, written relative to the unphysical coordinate frame $\{\partial_t, \partial_r\}$ is simply $u = (1 - 2M/r)^{-1}\partial_t - (2M/r)^{1/2}\partial_r$

The trick is take what Theactualbman defined seriously as differential forms, and to evaluate these differential forms using Ed's 4-velocity. Then, the division makes mathematical sense. There is a slight abuse of notation in using "d" on the right of Theactualbman's definitions, and in suppressing the notation for the evaluation using Ed's 4-velocity $u$.

$$\frac{dr_\mathrm{S} \left(u\right)}{dt_\mathrm{S} \left(u\right)} = -\sqrt{ \frac{2M}{r}}$$

 

[grav-universe]

Hello,

The following has been confusing my friends and I, I want to make sure I have this clear as it is fairly elementary. (note set c = 1)

Ed is falling radially into a black hole, the Schwarzschild metric is:

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

his proper time is dτ² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

If Simon is stationary and is coincident with Ed at some radius then he measures the proper distances and times given by:

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

I think that's all right. But then is Simon's metric dt_s² - dr_s² ≈ ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr² or are these really meant to be different bases?

Many thanks.

SR is valid locally, so right, Simon measures the metric as just

ds² = dt_s² - dr_s²

Using Schwarzschild coordinates, the relation between the local stationary observer and that of a distant observer at infinity is

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

as you wrote, so the metric according to what the distant observer infers becomes

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

as you have also. The Schwarzschild metric is formulated according to what the distant observer infers. And yes,

dr_s / dt_s

is the locally measured velocity.

 

[xox]

Simon measures the metric as just

ds² = dt_s² - dr_s²

Several of us spent a full page (the first page) showing that this is not true. There is no justification for claiming the above. While this is trivially true:

Using Schwarzschild coordinates, the relation between the local stationary observer and that of a distant observer at infinity is

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

...this isn't:

so the metric according to what the distant observer infers becomes

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

as you have also.

 

[grav-universe]

Several of us spent a full page (the first page) showing that this is not true. There is no justification for claiming the above.

While this is trivially true:

...this isn't:

It is correct. Why wouldn't it be?

 

[xox]

It is correct. Why wouldn't it be?

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.

 

[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.