Wikipedia "Time dilation" shows the formula $\sqrt{ 1-v^2-v_e^2-v_r^2.v_e^2/(1-v_e^2) } $
where v is (perhaps) the 3D speed, $v_e$ is the Newtonian escape speed, $v_r$ is the radial speed.
However "Proper time" seems to show the formula $\sqrt{ 1-v_e^2-v_r^2/(1-v_e^2)-v_p^2 }$ where $v_p$ is the speed perpendicular to the gravity vector (i.e. tangential).
(the speeds are all normalized against c and in coordinate time for a distant stationary observer)
The second formula seems to work better. For an infalling object with proper speed $\sqrt(r_s/r)$ and coordinate speed $v_r = (1-r_s/r)\sqrt{(r_s/r)}$ (per @John Rennie), the time dilation drops to zero at the event horizon, $r=r_s$. For an object in circular orbit with proper speed $\sqrt{(r_s/2r)/(1-r_s/r)}$ and coordinate speed $v_p = \sqrt(r_s/2r)$ [Raine & Thomas p.36], the time dilation drops to zero at $r = (3/2)r_s$.
I'd appreciate any help getting this straight.
Asked
Active
Viewed 220 times
1
Roger Wood
- 2,393
-
5Here is a MathJax tutorial. – G. Smith Oct 10 '20 at 17:14
-
1Just enclose all formulas with a dollar sign and put a backlash in front of every "sqrt" for a try. – Gyro Gearloose Oct 10 '20 at 18:08
-
2And a _ for a subscript. – G. Smith Oct 10 '20 at 18:47
1 Answers
3
This turns out to be just algebra. "$v$" is indeed the 3D speed which can be expressed in terms of the radial and tangential (perpendicular to field) speeds: $v^2 = v_r^2 + v_p^2$.
So, starting with the first expression
$\;\;\;\sqrt{ 1-v^2-v_e^2-v_r^2.v_e^2/(1-v_e^2) }$
= $\sqrt{ 1-v_r^2 -v_p^2-v_e^2-v_r^2.v_e^2/(1-v_e^2) }$
= $\sqrt{ 1 - v_p^2-v_e^2-v_r^2-v_r^2.v_e^2/(1-v_e^2) }$
= $\sqrt{ 1 - v_p^2-v_e^2-v_r^2(1+v_e^2/(1-v_e^2))}$
= $\sqrt{ 1 - v_p^2-v_e^2-v_r^2/(1-v_e^2)}$
= $\sqrt{ 1 -v_e^2-v_r^2/(1-v_e^2) - v_p^2}$
which is the second expression
Roger Wood
- 2,393