Rate of change of position coordinate with respect to proper time for a particle moving in the Schwarzschild spacetime Consider the Schwarzschild metric as follows:
$ds^2=\Big(1-\frac{2M}{r}\Big)dt^2- \Big(1-\frac{2M}{r}\Big)^{-1}dr^2-r^2 d\theta^2-r^2\sin^2\theta d\phi^2$
We can write that the Lagrangian is as follows:
$$ L=\Big(1-\frac{2M}{r}\Big)\dot{t^2}- \Big(1-\frac{2M}{r}\Big)^{-1}\dot{r^2}-r^2 \dot{\theta}-r^2\sin^2\theta  \dot{\phi^2}$$
By solving the Eular Lagrange in time , theta and phi we can rewrite as follows:
$$\frac{1}{2}\dot{r^2} +\frac{1}{2}(1-\frac{2M}{r})(\frac{l^2}{r^2 }-L) =\frac{1}{2}E^2 \space\space\space(1)$$
whereby $l = r^2 \dot{\phi}$ and $E = \dot{t}(1-\frac{2M}{r})$
Now consider how a massive particle that is initially at rest at r coordinate position 6M.This of course means that at t = 0, r = 6M.
My question is how would I go about working out $ \dot{r} = \frac{dr}{d\tau}$ where $\tau$ is the proper time.
I know that for this example $ \dot{r}  = -\frac{1}{\sqrt3} \sqrt{\frac{6M}{r}-1}$ however I am struggling to show it.
So far I have managed to note that a large particle means that L = -1.
I have also noted that when t = 0 $\dot{r} =\dot{\phi} = \dot{\theta}=0 $ allowing us to deduce $E = \sqrt{\frac{2}{3}}$
Combing all of this I tried rearranging (1) to isolate $\dot r$ however the term $(\frac{l^2}{r^2 }-L) $ seems to be giving me problems as I am unsure how to deduce l in a form that allows me to finally show that $ \dot{r}  = -\frac{1}{\sqrt3} \sqrt{\frac{6M}{r}-1}$ as my notes state I should
 A: The big idea here is that the particle starts from rest and so it will fall radially inwards with no component in either the $\theta$ or $\phi$ directions. You could prove this by finding the geodesic equation for those two coordinates and showing that if $\dot{\theta}, \dot{\phi}, \ddot{\theta}, \text{and}\  \ddot{\phi}$ all begin at 0, then they remain 0 and so one only has to worry about the $\dot{r}$ and $\dot{t}$ components.
Like you have already shown, there is a constant of the motion $E$ and because of this you don't have to worry about the Euler-Lagrange equations because the motion is uniquely determined by the normalisation of the 4-velocity, $U^\mu U_\mu=-1$ along timelike geodesics. In other words just set your Lagrangian (but with $\dot{\theta}$ and $\dot{\phi}$ set to 0) equal to -1, eliminate $\dot{t}$ with your constant of the motion $E$, and rearrange to solve.
$$
U^\mu U_\mu = \left(1-\frac{2m}{r}\right)\dot{t}^2 - \left(1-\frac{2m}{r}\right)^{-1} \dot{r}^2 = -1
$$
Now we eliminate $\dot{t}$ with our expression for $E$:
$$
U^\mu U_\mu = -\left(1-\frac{2m}{r}\right)^{-1}E^2 - \left(1-\frac{2m}{r}\right)^{-1} \dot{r}^2 = -1 \\
\dot{r}^2+E^2 = 1-\frac{2m}{r}
$$
Like you said, we have $\dot{r}=0$ initially so we can determine the constant $E$. For generality, say we begin falling from a radius $r_0$, then:
$$
E^2 = \frac{2m}{r_0}-1\\
\dot{r}^2 = 2m\left(\frac{1}{r}-\frac{1}{r_0}\right)
$$
Then substituting $r_0 = 6m$ and taking the negative square root since we're falling in should give you the answer from your notes.
This is a great example of how we can exploit symmetry to make our lives a little easier.
A: Wonderful answer by Alex G!  This provides a nice plot of velocity versus position.

Follow up question: If I put in a value for M, say 1500 meters so that this black hole is roughly the mass of our sun and initially r = 9000 meters, is it fair to calculate a time of fall from this dr/dt?  I get .000064 seconds of proper time by breaking this up into 700 time steps and manually integrating.
But then can I figure out the time as might be witnessed by a viewer on Earth by increasing each time step by time dilation based on velocity and gravity (I am replacing our sun with this black hole and using a sort of solar system reference frame)?  I end up with 0.00011 seconds.
My final time step does have the gravity-related dilation factor of well over 1000 but the proper time for that time step is 1.6e-13, that is, my manual integration is trending towards a zero*infinity value.
I ask this because I sometimes come across the statement that "nothing falls into a black hole" because the time to do so, on Earth, would be infinite.  I would appreciate learning that that notion is not correct.
