Can a Particle Reach Light Speed at r=3M in a Schwarzschild Black Hole Orbit?

[Jorrie]

I understand that light can be in an unstable orbit at Schwarzschild radial coordinate r=3M (geometric units) around a Schwarzschild black hole. I also understand that the local* circular orbital speed of a massive particle around the hole is

$$v_o = r\frac{d\phi}{dt} = \sqrt{\frac{M}{r(1-2M/r)}}$$

which is unstable for r <= 6M. The equation suggests that the local orbital speed equals the speed of light at r = 3M, which (sort of) confirms the orbit of light there.

If this is correct, my question is: if a particle in roughly circular orbit slowly spirals in from 3M < r < 6M, will it only approach the speed of light as it passes r = 3M or can it actually reach it?

*With local I mean as measured by an observer static at radial coordinate r.

 

[George Jones]

I understand that light can be in an unstable orbit at Schwarzschild radial coordinate r=3M (geometric units) around a Schwarzschild black hole. I also understand that the local* circular orbital speed of a massive particle around the hole is

$$v_o = r\frac{d\phi}{dt} = \sqrt{\frac{M}{r(1-2M/r)}}$$

which is unstable for r <= 6M. The equation suggests that the local orbital speed equals the speed of light at r = 3M, which (sort of) confirms the orbit of light there.

If this is correct, my question is: if a particle in roughly circular orbit slowly spirals in from 3M < r < 6M, will it only approach the speed of light as it passes r = 3M or can it actually reach it?

*With local I mean as measured by an observer static at radial coordinate r.

Such a worldline isn't a circle at r = 3M. An observer (static or not) at any event outside (and on the event horizon, and even inside) of a black hole always measures the local speed of a material particle to be less than the local speed of a light.

Your post brings up another interesting point. In order for a particle to spiral down from conditions dr/dt = 0 and dr/dphi =/= 0 to an unstable exactly circular orbit with r near 3M, it would have to start with r *very* large. The closer the r of the unstable circle is to to 6M (below), the closer the starting r is to 6M (above). As r for the unstable circle approaches 3M, the starting r approaches infinity.

 

[pervect]

What makes you think that $$r \frac{d\phi}{dt}$$ is the local circular orbital speed?

You've neglected to account for the time dilation of the static observer relative to the observer at infinity, I think.

You can regard $r d \phi$ as the local distance, because the circumference of a black hole is 2 pi r, but you can't regard dt as the local time. dt is coordinate time, not local time.

 

[Jorrie]

What makes you think that $$r \frac{d\phi}{dt}$$ is the local circular orbital speed?

You've neglected to account for the time dilation of the static observer relative to the observer at infinity, I think.

Oops, yes, I should have written in terms of dphi/dtau:

$$v_o = r\frac{d\phi}{d\tau} = \sqrt{\frac{M}{r(1-2M/r)}}$$

But the rest of formula is correct for the local observer, not so?

For the observer at infinity the orbital speed at r is just [tex]v_{inf} = \sqrt{M/r}[/itex], I think.

 

[pervect]

Such a worldline isn't a circle at r = 3M. An observer (static or not) at any event outside (and on the event horizon, and even inside) of a black hole always measures the local speed of a material particle to be less than the local speed of a light.

Your post brings up another interesting point. In order for a particle to spiral down from conditions dr/dt = 0 and dr/dphi =/= 0 to an unstable exactly circular orbit with r near 3M, it would have to start with r *very* large. The closer the r of the unstable circle is to to 6M (below), the closer the starting r is to 6M (above). As r for the unstable circle approaches 3M, the starting r approaches infinity.

Unless I'm calculating something incorrectly (unfortunately possible), as one approaches r0=3m, the total energy required approaches infinity. I get E=1 (bare escape to infinity) occurring at r0=4m, for r0<4m, the object has to have more than enough energy to escape to infinity.

It takes a little work to get the hang of it, but I find the orbital simulator at http://www.fourmilab.ch/gravitation/orbits/ useful to get some hands-on idea of the orbital behavior.

One first has to pick a value of L and M. L=3.462 and m=1 is close to but above the critical value of L of sqrt(12) for m=1. Doing the math or looking at

http://www.fourmilab.ch/gravitation/orbits/figures/roots.gif

indicates that this value of L results in an orbit of r near 6m, and yields slow-decaying spirals of the sort asked for. Also, objects here can only inspiral, they can't outspiral, as E<1.

One uses the simulator by picking L and m, and then clicking on the effective potential diagram (not on the orbital pane!) to "launch" the particle at some particular value of r.

 

[pervect]

Oops, yes, I should have written in terms of dphi/dtau:

$$v_o = r\frac{d\phi}{d\tau} = \sqrt{\frac{M}{r(1-2M/r)}}$$

But the rest of formula is correct for the local observer, not so?

For the observer at infinity the orbital speed at r is just [tex]v_{inf} = \sqrt{M/r}[/itex], I think.

I'm getting something different.

Unfortunately, at the moment I'm getting two different answers by two different routes, and it's getting late.

I think I mispoke when I called r dphi/dtau a velocity, though. It should actually be a celerity (proper distance/ proper time) because tau is the proper time of the orbiting particle.

 

[George Jones]

Unless I'm calculating something incorrectly (unfortunately possible), as one approaches r0=3m, the total energy required approaches infinity. I get E=1 (bare escape to infinity) occurring at r0=4m, for r0<4m, the object has to have more than enough energy to escape to infinity.

Yes, this is right. I even have this as an exercise for some stuff I wrote on black holes (see below)! Duh! I rushed my answer, so as not to miss my bus to work.

It takes a little work to get the hang of it, but I find the orbital simulator at http://www.fourmilab.ch/gravitation/orbits/ useful to get some hands-on idea of the orbital behavior.

I wrote my own Schwarzschild orbital simulator as a Java applet. It doesn't show effective potential or allow the user to play with E and L. It takes as input an r value and spatial velocity (speed and angle with respect to "horizontal") with respect to the orthonormal frame of a static observer, and then ppolts the orbit. I also wrote a set of exercises to illustrate different types of motion.

indicates that this value of L results in an orbit of r near 6m, and yields slow-decaying spirals of the sort asked for..

But this spiral is nowhere near circular at r = 3M. In fact, for this spiral, I get v = sqrt(5/8) = 0.79 [edit: dropped a couple of sqrt's] with respect to a static observer at r = 3M.

 

[George Jones]

Oops, yes, I should have written in terms of dphi/dtau:

$$v_o = r\frac{d\phi}{d\tau} = \sqrt{\frac{M}{r(1-2M/r)}}$$

But the rest of formula is correct for the local observer, not so?

Yes.

Using $r = 6M$ in this gives $v_0 = 1/2$.

To compute the speed at $r = 3M$, use the conserved quantity E twiddle (lose the tiddle from now on) given in pervect's link. From this link, and letting $\tau_p$ and $\tau_s$ be the proper times of the spiraling particle and the static observer at $r =$ something.

$$\begin{equation*} \begin{split} E &= \left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau_p}\\ &= \left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau_s} \frac{d\tau_s}{d\tau_p}\\ &= \left(1 - \frac{2M}{r} \right) \left(1 - \frac{2M}{r} \right)^{-\frac{1}{2}} \gamma, \end{split} \end{equation*}$$

where $\gamma = \left( 1 - v^2 \right)$ and $v$ is the speed of the particle with respect to a static observer.

Now consider the limiting case where $E$ for a particle in orbit at $r = 6M$ equals $E$ for a spiraling particle at $r = 3M$.

$$\begin{equation*} \begin{split} \sqrt{1 - \frac{2M}{6M}} \gamma_0 &= \sqrt{1 - \frac{2M}{3M}} \gamma\\ \frac{2}{3} \gamma_0^2 &= \frac{1}{3} \gamma^2\\ 1 - v_0^2 &= 2 \left(1 - v^2 \right)\\ v &= \sqrt{\frac{1 + v_0^2}{2}} \end{split} \end{equation*}$$

Using $v_0 = 1/2$ gives $v = \sqrt{5/8} = 0.79$.

 

[Jorrie]

But this spiral is nowhere near circular at r = 3M. In fact, for this spiral, I get v = sqrt(5/8) = 0.79 [edit: dropped a couple of sqrt's] with respect to a static observer at r = 3M.

Yea, I was a bit perplexed for a while, because I calculated the v = 0.79 as well.

I suppose one can change the scenario so that the static observer at r=3M gives the particle a transverse boost to close to v=1, but can obviously never achieve v=1, due to the infinite energy required.

I also made these sums before and asked a rather stupid question here, but the discussion was quite interesting, thanks George.

 

[Jorrie]

$$\begin{equation*} \begin{split} \sqrt{1 - \frac{2M}{6M}} \gamma_0 &= \sqrt{1 - \frac{2M}{3M}} \gamma\\ \frac{2}{3} \gamma_0^2 &= \frac{1}{3} \gamma^2\\ 1 - v_0^2 &= 2 \left(1 - v^2 \right)\\ v &= \sqrt{\frac{1 + v_0^2}{2}} \end{split} \end{equation*}$$

Using $v_0 = 1/2$ gives $v = \sqrt{5/8} = 0.79$.

I followed the straightforward route of calculating the specific total energy E/m in the Schwarzschild frame for a circular orbit at r=6M, v=0.408 (v_local=0.5) from:

$$E/m = \frac{1-2M/r}{\sqrt{1-2M/r-v^2}} = 0.943$$

Using this energy, I extracted v for r=3 and got v=0.456 in the Schwarzschild frame. This converts to v_local=0.791 in the local (r=3M) frame.

 

[Chris Hillman]

Jorrie: ditto pervect, you made several errors. I have posted on this topic in the past so you might try to find my discussions.

The single worst error you committed was this: I have repeatedly warned (with hopefully clear and detailed examples) that failure to recognize the existence of multiple distinct operationally significant notions of "distance in the large", and thus (angular) "velocity in the large", even in flat spacetime, inevitably leads to confusion. Your post is only one examples of hundreds we have seen at PF alone.

You might look at [post=1484927]this post[/post], where I gathered links to some of my PF posts dealing with various topics related to black holes.

 

[Jorrie]

The single worst error you committed was this: I have repeatedly warned (with hopefully clear and detailed examples) that failure to recognize the existence of multiple distinct operationally significant notions of "distance in the large", and thus (angular) "velocity in the large", even in flat spacetime, inevitably leads to confusion. Your post is only one examples of hundreds we have seen at PF alone.

Chris, I have read most of what you wrote and will refresh again.

I think the issue in this particular instance is not the Schwarzschild radial parameter r, which is well defined, but the angular or transverse velocity 'at large'. Maybe the mention of r dphi/dtau was superfluous in this post. What is crucial here is the instantaneous transverse (circular orbital) velocity of a particle measured locally by a stationary observer at Schwarzschild radial parameter r in a Schwarzschild vacuum. We seem to have converged here on:

$$v_{o(r)} = \sqrt{\frac{M}{r(1-2M/r)}}$$

Given the definitions, is this the correct equation for it? If it is correct, the energy situation is more or less trivial.

 

[Chris Hillman]

Do you mean the velocity measured by a static observer using his rocket engine to hover at $r=r_0$, as an object in a circular orbit at the same radius whizzes by very close to his position? If so, yes, that would circumvent the "distance in the large" issue, since all such notions of distance agree for sufficiently small distances.

 

[Jorrie]

Do you mean the velocity measured by a static observer using his rocket engine to hover at $r=r_0$, as an object in a circular orbit at the same radius whizzes by very close to his position?

Yep, that, or the observer being inertial and instantaneously static at r_o.

As a side issue, is $v_{o(r)} = r_o d\phi/d\tau$ in such a simple scenario in the Schwarzschild vacuum?

 

[pervect]

Do you mean the velocity measured by a static observer using his rocket engine to hover at $r=r_0$, as an object in a circular orbit at the same radius whizzes by very close to his position? If so, yes, that would circumvent the "distance in the large" issue, since all such notions of distance agree for sufficiently small distances.

Yep, that's exactly what we want to compute.

 

[pervect]

For the conserved quantity $E_p$ of a circular orbit of radius $r_0$ (I will also drop the twiddles), I get

$$E_p = \sqrt{\left( 1 - \frac{2\,m}{r_0} \right) \left(1 + \frac{L^2}{r_0^2} \right) }$$

with
$$L^2 = \frac{r_0^2}{\frac{r_0}{m}-3}$$

being the precondition for a circular orbit. This gives$$E_p = \sqrt { \left( 1-2\,{\frac {m}{{ r_0}}} \right) \left( 1+ \frac{1} { \frac { r_0}{m}-3 } \right) } = \frac{1-\frac{2\,m}{r_0}}{\sqrt{1-\frac{3\,m}{r_0}}}$$

Evaluating this at r0=6m gives Ep=.9428, in agreement with Jorrie's specific calculation for that radius.
But I don't quite see how to get general agreement with Jorrie's results, yet.

I find it convenient to rewrite George's result as

$$\gamma = \frac{1}{\sqrt{1-v^2}} = \frac{E_p}{E_s}$$

which can be solved for v

$$v = \sqrt{1-\frac{1}{\gamma^2}} = \sqrt{1-\left( \frac{E_s}{E_p} \right)^2}$$

Here $E_s$ is the energy of a static observer at$r=r_s$, equal to
$$\sqrt{1-\frac{2\,m}{r_s}}$$. Even though the static observer isn't following a geodesic, E_s is constant.

Taking an alternate route, $$r \frac{d\phi}{d\tau}$$ should be equal to the celerity, and this should be given by L/r_s.

Given that
$$cel = \frac{v}{\sqrt{1-v^2}}$$

we can solve for v as a function of celerity
$$v = \frac{cel}{\sqrt{1+cel^2}}$$

We shouldn't expect this route to give the same result as the previous one, though, because it computes only the "orbital" component of the relative velocity, and ignores the components due to the infalling radial velocity.

 

[pervect]

A geometrical motivation for why

$$\gamma = E_p / E_s$$

Consider two 4-velocities u1 and u2, in flat spacetime. Then $u1 \cdot u2 = \gamma$ (for simplicity we adopt a +--- metric signature to avoid minus signs) where $\gamma^2 = 1/(1-v^2)$, v being the magnitude of the relative velocity.

We can basically extend this result to the locally flat space-time. Call the 4-velocity of our orbiting observer up, the 4-velocity of the static observer us. Then we expect that $\gamma = us \cdot up$ in the local tangent space just as it was in flat space-time.

What are the conserved quantites that we call E_s and E_p, the energy per unit mass of a static observer, and the energy per unit mass of the orbiting observer?

They are just

E_s = us_0
E_p = up_0

We have utilized the fact that for a unit mass, the energy-momentum 4-vector P^i is just the 4-velocity u^i, and similarly P_i is just u_i. The conserved energy is therefore P_0 = u_0.

Because the metric is diagonal, and the only non-zero component of us is us_0, we can write:

$$\gamma = us \cdot up = g^{00 }E_s E_p$$

But we also have $g^{00}$ us_0 us_0 = $g^{00}$ E_s E_s = 1, because us is a 4velocity and the norm of a 4-velocity with a +--- metric signature is +1.

Thus $g^{00}$ E_s = 1/E_s, and it immediatly follows that
[tex]\gamma = g^{00}[/itex] E_s E_p = E_p/E_s

 

[Jorrie]

We shouldn't expect this route to give the same result as the previous one, though, because it computes only the "orbital" component of the relative velocity, and ignores the components due to the infalling radial velocity.

Interesting manipulations, but I think both methods give only the transverse component of the local velocity, because they are both based on L. Numerically, the results are the same, it appears.

I suppose the radial velocity component at r=3m might be small in comparison to the transverse component, but I guess it will not stay like that for an in-spiral to the horizon. Would it not approach "all radial" just outside the horizon?

 

[pervect]

The first method actually involves E, not L, so it gives the total velocity, and should be identical to George's method with minor notational differences. The only reason L enters the picture in the first method is to add the contribution of the orbital velocity to the total energy to compute E.

I thought your method was also based on E, but I don't quite follow how you computed E, perhaps that is where the discrepancy arises.

I compute L via the fact that d(veff)/dr = 0 for a circular orbit, and given L and r, I compute E from the effective potential. The effective potential is given by the webpage I cited, you can also find it in MTW.

The second method does give only the orbital component of the velocity.

 

[Jorrie]

I thought your method was also based on E, but I don't quite follow how you computed E, perhaps that is where the discrepancy arises.

Yep, I did use the energy method and it gave a result numerically the same as George's and your method. I now understand that the answer is the resultant velocity, including the radial component. I goofed when I said that the road through L (and by implication celerity) will give the same answer; I agree it does not.

I computed E for a circular orbit ($dr/dt=d \theta /dt=0$) in the Schwarzschild frame from MTW's eq. (25.18),

$$\frac{dt}{d\tau} = \frac{E/m}{1-2M/r} = \frac{1}{\sqrt{1-2M/r - r^2(d\phi/dt)^2}}$$

which gave me my

$$E/m = \frac{1-2M/r}{\sqrt{1-2M/r-v_o^2}}$$

However, there is a lot of insight 'embedded' in the method you followed. Tx for the input and the patience.

 

[Jorrie]

Taking an alternate route, $$r \frac{d\phi}{d\tau}$$ should be equal to the celerity, and this should be given by L/r_s.
Given that
$$cel = \frac{v}{\sqrt{1-v^2}}$$

we can solve for v as a function of celerity
$$v = \frac{cel}{\sqrt{1+cel^2}}$$

I'm still slightly confused by the 'alternate route' and what it actually yields. Given that:

$$cel = L/r_s = r \frac{d\phi}{d\tau} = r \frac{d\phi}{dt} \frac{dt}{d\tau}$$

At the static observer, we know the inspiralling particle must have transverse (v_phi) and radial (v_r) velocity components. Shouldn't your next equation have been:

$$cel = \frac{v_{\phi}}{\sqrt{1-v_{\phi}^2-v_r^2}}$$

in order to take the full time dilation (dtau/dt) into account?

If this is correct, how do we now solve for v_phi? I guess the full geodesic equations may be required.

 

[Jorrie]

At the static observer, we know the inspiralling particle must have transverse (v_phi) and radial (v_r) velocity components. Shouldn't your next equation have been:

$$cel = \frac{v_{\phi}}{\sqrt{1-v_{\phi}^2-v_r^2}}$$

in order to take the full time dilation (dtau/dt) into account?

If this is correct, how do we now solve for v_phi? I guess the full geodesic equations may be required.

IF the above is valid, I suppose one may solve it by pervect's previous result:

$$v = \sqrt{1-\frac{1}{\gamma^2}} = \sqrt{1-\left( \frac{E_s}{E_p} \right)^2}$$

where his $v^2 = v_{\phi}^2 + v_r^2$, giving two equations for the two unknowns. But, it depends on whether my interpretation if celerity above is right...

 

[pervect]

The "alternate route" was intended to find only the orbital component of the velocity. To be slightly more formal, this would be the orbital component of the velocity as computed in the frame-field of a static observer.

My derivation was however very informal, and intended to be intuitive. I guess it didn't totally succeed at being "obvious", sometimes the intuitive solutions can take more work to understand than the formal approach. The intuitive approach is also more error-prone than the more formal approach, but if it works, it saves a lot of typing :-).

If you want to see a more formal derivation, you can try http://en.wikipedia.org/w/index.php...ple:_Static_observers_in_Schwarzschild_vacuum, but it may not be the easiest read.

You can probably find something in MTW that might be easier to read. You can think of the orbital velocity as a scalar number being given by a the vector product of some one-form and the total 4-velocity. Similarly, you can think of the radial velocity as a scalar number given by the vector product of a different one form and the 4-velocity. The MTW approach would be to compute these two numbers (orbital velocity and radial velocity) by constructing an orthonormal basis of one forms, and multiplying the 4-velocity of the infalling observer by each one-form to get the appropriate scalar.

While you certainly could find the velocities by solving the geodesic equations, you can also find them by taking advantage of the conserved quantities L and E.

Given that we know L for a circular orbit by my equation for L vs r0 for a circular orbit (which I derived from the webpage http://www.fourmilab.ch/gravitation/orbits/ by taking d(veff)/dr=0), we can compute $d\phi/dtau$ by the relationship

dphi/dtau = L/r^2

Of course you might want to double-check that what I posted was actually right...

You can use another relationship on the webpage to find dr/dtau, specifically

(dr/dtau)^2 + Veff^2 = E^2

where I won't type out the full equation for Veff (that's on the webpage, I think you can find it).

However, to convert (dr/dtau) to the radial velocity in the frame-field of the statonary observer, you'll need to construct the ONB of one-forms I was talking about.

With dr/dtau, dphi/dtau, and in addition

dt/dtau = E/(1-2m/r)

you can compute all the components of the 4-velocity from E and L, so the only remaining task is to find the ONB of one forms I was talking about.

I'll probably write them down in another post, but right now I have to go. Sorry for neglecting this thread, I've been a bit busy recently.

 

[pervect]

OK, let's write down the ONB of one-forms, and use them to show how to compute the radial and orbital components of the velocity in the frame of the static observer.

Symbolically, dr, dphi, and dt are (or at least can be interpreted as) one-forms, because they can be thought of as taking a vector as an input, and producing a scalar as an output. See for instance MTW, pg 53-59, for more about one-forms. See also pg 60 and 232 for more discussion of "basis one-forms".

MTW suggest visulaizing one-forms as stacks of plates, and the associated scalar as the number of plates pierced by a given vector.

Note that MTW adopts the convention of using boldface for a differential when it is intended that the differential be interpreted as a one-form (which is something I generally don't do).

While, dr,dt, and dphi are one-forms, they aren't orthonormal. We need to construct a new basis

d(rhat), d(phihat), d(that)

that IS orthonormal. Since we are assuming (or at least I am assuming) that we are in the equatorial plane, theta=pi/2, and we can drop it because dtheta=0 always.

Then we can write:

d(that) = sqrt(1-2m/r) dt
d(rhat) = (1/sqrt(1-2m/r)) dr
d(phihat) = r sin(theta) dphi = r dphi (since theta=pi/2)

How do we know this is an ONB? Because the Schwarzschild metric can be written as

ds^2 = -d(that)^2 + d(rhat)^2 + d(phihat)^2

which has unity coefficents

The orbital component of the velocity will be

d(phihat)/d(that) = (r / sqrt(1-2m/r)) dphi/dt = (r/sqrt(1-2m/r) (dphi / dtau) / (dt/dtau)

The radial component of the velocity will be

d(rhat)/d(that) = (1/(1-2m/r)) dr/dt = (1/(1-2m/r)) (dr/dtau) / (dt/dtau)

Hopefully that will get you the right answer, or at least started. Note that the 4-velocity will give you all the (d/dtau)

I should probably work out some actual numbers but I will wait to see if that's needed. Hopefully, you can work out the numbers, and show that they actually make sense - for instance, compute the radial and orbital velocites, and show that the total velocity is given by

vtotal^2 = vradial^2 + vorbital^2

If this doesn't work out, there's a problem somewhere, either with what I wrote, or with your calculation, or with both...

 

[Jorrie]

Hopefully that will get you the right answer, or at least started. Note that the 4-velocity will give you all the (d/dtau)

I should probably work out some actual numbers but I will wait to see if that's needed. Hopefully, you can work out the numbers, and show that they actually make sense - for instance, compute the radial and orbital velocites, and show that the total velocity is given by

vtotal^2 = vradial^2 + vorbital^2

If this doesn't work out, there's a problem somewhere, either with what I wrote, or with your calculation, or with both...

Tx pervect. I understand the process now, so I'll stick in numbers and see if it works out as expected.

 

[Jorrie]

Local transverse and radial velocities

Tx pervect. I understand the process now, so I'll stick in numbers and see if it works out as expected.

Putting everything that you wrote above together did give the correct answers. The two local velocity components of the in-spiralling object, as measured by the observer static at r_s, simplified to:

local orbital (transverse) speed: (edited to conform to your previous notation and without 'twiddles')

$$v_{orb} = \frac{\sqrt{1-2M/r_s}}{r_s} \frac{L}{E_p} = \frac{E_s}{E_p} \frac{L}{r_s}$$

local radial speed:

$$v_{rad} = \frac{\sqrt{E_p^2-V_s^2}}{E_p}$$

For the case that we computed here, (spiralling in from circular at r=6M to r_s=3M) the results are: $v_{orb} = 0.707$ and $v_{rad} = -0.3535$, exactly half of the orbital magnitude. The resultant total velocity is $v_{tot} = 0.79$, as computed previously. Energy and angular momentum are conserved, as expected. It is interesting to note that as the particle (and the static observer's position) approaches the horizon (r_s -> 2M), Veff -> 0 and so does v_orb, while v_rad -> 1.

(Edit: taken out note for typo.)

Thanks for the very helpful inputs.

 

[Jorrie]

While, dr,dt, and dphi are one-forms, they aren't orthonormal. We need to construct a new basis

d(rhat), d(phihat), d(that)

that IS orthonormal. Since we are assuming (or at least I am assuming) that we are in the equatorial plane, theta=pi/2, and we can drop it because dtheta=0 always.

Then we can write:

d(that) = sqrt(1-2m/r) dt
d(rhat) = (1/sqrt(1-2m/r)) dr
d(phihat) = r sin(theta) dphi = r dphi (since theta=pi/2)

How do we know this is an ONB? Because the Schwarzschild metric can be written as

ds^2 = -d(that)^2 + d(rhat)^2 + d(phihat)^2

which has unity coefficients

The orbital component of the velocity will be

d(phihat)/d(that) = (r / sqrt(1-2m/r)) dphi/dt = (r/sqrt(1-2m/r) (dphi / dtau) / (dt/dtau)

The radial component of the velocity will be

d(rhat)/d(that) = (1/(1-2m/r)) dr/dt = (1/(1-2m/r)) (dr/dtau) / (dt/dtau)

I have noticed a slightly 'easier' path to these results by recognizing how a local velocity transforms between Schwarzschild coordinates and local coordinates (an observer static at radial parameter r). For simplicity, work with the radial and orbital (transverse) velocity components separately. What pervect wrote above naturally leads to the following conversion equations:

For transverse (orbital) velocity components:

[1] $$v_{t(r)} = \frac{d\hat \phi}{d\hat t} = \frac{rd\phi/dt}{\sqrt{1-2m/r}} = \frac{v'_{t(r)}}{\sqrt{g_{tt}}}$$

where $g_{tt} = 1-2m/r$, $v_{t(r)}$ is the transverse velocity measured by a local static observer at radial coordinate r and and $v'_{t(r)}$ is the transverse Schwarzschild coordinate velocity component at r, as 'measured' by a static distant ('infinity') observer. Likewise for the radial velocity component.

[2] $$v_{r(r)} = \frac{d\hat r}{d\hat t} = \frac{dr/dt}{1-2m/r}= \frac{v'_{r(r)}}{g_{tt}}$$

It is easy to obtain the Schwarzschild coordinate velocities from a given E and L (see pervect's post #23 above), where

$$v'_{t(r)} = r\frac{d\phi}{dt} = r\frac{d\phi}{d\tau} \frac{d\tau}{dt} = \frac{{g_{tt}}}{r} \frac{ L}{ E}$$

$$v'_{r(r)} = \frac{dr}{dt} = \frac{dr}{d\tau} \frac{d\tau}{dt} = \frac{g_{tt}\sqrt{ E^2- V_{eff}^2}}{ E}$$

These Schwarzschild coordinate velocities can now be transformed to the frame of the local static observer at r by the transformation equations [1] and [2], giving:

$$v_{t(r)} = \frac{v'_{t(r)}}\sqrt{g_{tt}}} = \frac{\sqrt{g_{tt}}}{r} \frac{ L}{ E}$$

$$v_{r(r)} = \frac{v'_{r(r)}}{g_{tt}} = \frac{\sqrt{ E^2- V_{eff}^2}}{ E}$$

as obtained before. This is obviously just an interesting 'shortcut' when the transformation equations are taken to be true and they follow rigorously from the analysis that pervect has done. IMO, it is however educational to note the existence of these transformations.