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I'm currently looking at Metric for the Rain Frame in 'Exploring Black Holes' by Taylor & Wheeler (page B-13) and while it's straightforward understanding drrain (which basically equals dr), I'm having a problem getting my head around dtrain. The following is a step-by-step approach but for some reason, the results I get from the metric are negative. (Attached is an extract from the book which shows the metric in full, hopefully this is acceptable).
from the book-
[tex]dt_{rain}=-v_{rel}\gamma dr_{shell}+\gamma dt_{shell}[/tex]
where-
[tex]dr_{shell}=dr\left(1-\frac{2Gm}{rc^2}\right)^{-1/2}[/tex]
[tex]dt_{shell}=dt\left(1-\frac{2Gm}{rc^2}\right)^{1/2}[/tex]
[tex]v_{rel}=-\left(\frac{2Gm}{rc^2}\right)^{1/2}[/tex]
[tex]\gamma \equiv \left(1-\frac{v^2}{c^2}\right)^{-1/2}=\left(1-\frac{2Gm}{rc^2}\right)^{-1/2}[/tex]
Incorporating the above equations into the dtrain equation, I get-
[tex]dt_{rain}=\left(\frac{2Gm}{rc^2}\right)^{1/2} dr\left(1-\frac{2Gm}{rc^2}\right)^{-1} + dt[/tex]
which tallies with equation 14 on page B-13.
Based on a 3 sol mass black hole (Rs=8861.1 m), r=11,000, dr=1 and dt=1, I get the following-
[tex]dt_{rain}=(0.897527 \times 5.142830)+1[/tex]
[tex]dt_{rain}=5.615829[/tex]
The metric for the rain frame (on a radial line only) is-
[tex]d\tau^2=\left(1-\frac{2Gm}{rc^2}\right)dt_{rain}^2 - 2\left(\frac{2Gm}{rc^2}\right)^{1/2}dt_{rain}dr -dr^2[/tex]
The results I get here where dr=1 and dtrain=5.615829 are-
[tex]d\tau^2=(0.194445 \times 5.615829^2)-(1.795054 \times 5.615829)-1[/tex]
[tex]d\tau^2=-4.948399[/tex]
Obviously something is amiss here as dτ2 shouldn't be negative outside the event horizon.
I have a hunch I'm about 95% there but there's a mistake I'm making somewhere. I'd appreciate any feedback regarding getting this right.
Steve
_____________
UPDATE
I've applied the same black hole parameters and radius to standard Schwarzschild metric and I get (more or less) the same answer, dτ2=-4.948385. This is corrected by introducing c2 to dτ2 and dt2 providing an answer for dτ of 0.440959 (in standard Schwarzschild metric) which makes more sense. This tells me I need to incorporate c2 to the rain metric but when applied to both quantities of dtrain, the results are still negative.
______________
from the book-
[tex]dt_{rain}=-v_{rel}\gamma dr_{shell}+\gamma dt_{shell}[/tex]
where-
[tex]dr_{shell}=dr\left(1-\frac{2Gm}{rc^2}\right)^{-1/2}[/tex]
[tex]dt_{shell}=dt\left(1-\frac{2Gm}{rc^2}\right)^{1/2}[/tex]
[tex]v_{rel}=-\left(\frac{2Gm}{rc^2}\right)^{1/2}[/tex]
[tex]\gamma \equiv \left(1-\frac{v^2}{c^2}\right)^{-1/2}=\left(1-\frac{2Gm}{rc^2}\right)^{-1/2}[/tex]
Incorporating the above equations into the dtrain equation, I get-
[tex]dt_{rain}=\left(\frac{2Gm}{rc^2}\right)^{1/2} dr\left(1-\frac{2Gm}{rc^2}\right)^{-1} + dt[/tex]
which tallies with equation 14 on page B-13.
Based on a 3 sol mass black hole (Rs=8861.1 m), r=11,000, dr=1 and dt=1, I get the following-
[tex]dt_{rain}=(0.897527 \times 5.142830)+1[/tex]
[tex]dt_{rain}=5.615829[/tex]
The metric for the rain frame (on a radial line only) is-
[tex]d\tau^2=\left(1-\frac{2Gm}{rc^2}\right)dt_{rain}^2 - 2\left(\frac{2Gm}{rc^2}\right)^{1/2}dt_{rain}dr -dr^2[/tex]
The results I get here where dr=1 and dtrain=5.615829 are-
[tex]d\tau^2=(0.194445 \times 5.615829^2)-(1.795054 \times 5.615829)-1[/tex]
[tex]d\tau^2=-4.948399[/tex]
Obviously something is amiss here as dτ2 shouldn't be negative outside the event horizon.
I have a hunch I'm about 95% there but there's a mistake I'm making somewhere. I'd appreciate any feedback regarding getting this right.
Steve
_____________
UPDATE
I've applied the same black hole parameters and radius to standard Schwarzschild metric and I get (more or less) the same answer, dτ2=-4.948385. This is corrected by introducing c2 to dτ2 and dt2 providing an answer for dτ of 0.440959 (in standard Schwarzschild metric) which makes more sense. This tells me I need to incorporate c2 to the rain metric but when applied to both quantities of dtrain, the results are still negative.
______________
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