Deriving radial velocity as observed from infinity

[stevebd1]

In Wheeler and Taylor's 'Exploring Black Holes', on pages 3-12 and 3-13, the bookkeeper measure of radial velocity (i.e. radial velocity as measured from infinity) is derived. Basically the equation for 'Energy in Schwarzschild geometry' is established-

$$\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}=1$$

The book states-

'From the energy equation and the Schwarzschild metric, we can find an expression for $dr/dt$, the rate of the change of the r-coordinate with far-away time t for a stone starting from rest at a very great distance. To obtain this derivative, square terms on either side of the right-hand equality, multiply through by $d\tau^2$, and equate it to the Schwarzschild metric equation for $d\tau^2$ in the case of radial fall $(d\phi=0)$:

[tex]\left(1-\frac{2M}{r}\right)^2dt^2=d\tau^2=\left(1-\frac{2M}{r}\right)dt^2-\frac{dr^2}{\left(1-\frac{2M}{r}\right)}[/itex]

Divide through by $dt^2$, solve for $dr/dt$, and take the square root to obtain

$$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\left(\frac{2M}{r}\right)^{1/2}$$I'd appreciate if someone could show the process of derivation between the second and third equation.

 

[tiny-tim]

[tex]\left(1-\frac{2M}{r}\right)^2dt^2=d\tau^2=\left(1-\frac{2M}{r}\right)dt^2-\frac{dr^2}{\left(1-\frac{2M}{r}\right)}[/itex]

Divide through by $dt^2$, solve for $dr/dt$, and take the square root to obtain

$$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\left(\frac{2M}{r}\right)^{1/2}$$


I'd appreciate if someone could show the process of derivation between the second and third equation.

Hi stevebd1!

Ignore the dtau² in the middle, and the coefficient of dt² becomes (1 - 2M/r)(1 - 2M/r - 1)

 

[stevebd1]

Hi tiny-tim

Thanks for the response. I'm probably missing something elementary here but could you shed some light on how you arrived at that coefficient for dt²?

 

[tiny-tim]

Hi tiny-tim

Thanks for the response. I'm probably missing something elementary here but could you shed some light on how you arrived at that coefficient for dt²?

Yup … in

$$\left(1-\frac{2M}{r}\right)^2dt^2=\,\cdots\,=\left(1-\frac{2M}{r}\right)dt^2-\frac{dr^2}{\left(1-\frac{2M}{r}\right)}$$

rearrange to $$\left[\left(1-\frac{2M}{r}\right)^2\ -\ \left(1-\frac{2M}{r}\right)\right]dt^2\ =\ -\frac{dr^2}{\left(1-\frac{2M}{r}\right)}$$

which is $$\left[\left(1-\frac{2M}{r}\right)\left(1-\frac{2M}{r}\right\ -\ 1)\right]dt^2\ =\ -\frac{dr^2}{\left(1-\frac{2M}{r}\right)}$$