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What is the orbital velocity formula in Schwarzschild metric assuming that the orbit is circular? this formula should also work for relativistic velocities.

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As usual in General Relativity, you have to ask "according to whom?"

There are (at least) three different orbital speeds you could define - (1) the orbital speed as judged by an observer a long way from the central object with $r_o \gg r_s$; (2) an observer at the same radial coordinate as the orbiting object but who is not moving in $\theta, \phi$; (3) an observer travelling in the circular orbit.

The results are: \begin{eqnarray} v_1 & = & \left(\frac{GM}{r}\right)^{1/2}\ ; \tag{1} \\ v_2 & = & \left(\frac{GM}{r}\right)^{1/2} \left(1 - \frac{r_s}{r}\right)^{-1/2}\ ; \tag{2} \\ v_3 & = & \left(\frac{GM}{r}\right)^{1/2} \left(1 - \frac{3r_s}{2r}\right)^{-1/2}\ , \tag{3} \end{eqnarray} where $r_s = 2GM/c^2$.

The first result is somewhat surprising and rather convenient, since it agrees with Newtonian physics. A (probably inelegant) proof is given below.

The third result is derived from $v_3 = r d\phi/d\tau$, where $\tau$ is the proper time and is a consequence of time dilation between the orbiting observer and the distant universe. It is an inferred speed, not a speed that can be measured. The observed relative speed of the orbiting observer and the stationary observer at the same radial coordinate is given by eqn. (2) and can never exceed $c$ (reached at $r = 3r_s/2$, the innermost unstable circular orbit).

Showing that the distant observer sees the same speed as a Newtonian observer

Go back to geodesics in the Schwarzschild metric, which are defined in terms of the two constants of motion that are often labelled $E/mc^2$ and $L/m$, the coordinate specific energy and coordinate specific angular momentum respectively.

In terms of these quantities, the rate of change of radial coordinate with respect to proper time can be written $$ \frac{dr}{d\tau} = \pm c\left[ \left(\frac{E}{mc^2}\right)^2 - \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{m^2r^2c^2}\right) \right]^{1/2}\ , \tag*{(4)} $$ where $$ \left(1 - \frac{r_s}{r}\right)\frac{dt}{d\tau} = \frac{E}{mc^2}\ , $$ $$ r^2 \frac{d\phi}{d\tau} = \frac{L}{m}\ . $$

We can now express the circular velocity as seen by a distant observer, $v_{\rm circ} = r d\phi/dt$ as : $$ v_{\rm circ}^2 = r^2\left(\frac{d\phi}{dt}\right)^2 = r^2\left(\frac{d\phi}{d\tau}\right)^2 \left(\frac{d\tau}{dt}\right)^2 = \frac{L^2}{m^2r^2} \left(1 - \frac{r_s}{r}\right)^2 \left(\frac{mc^2}{E}\right)^2\, . $$ But for a circular orbit $dr/d\tau=0$ and eqn. (4) gives us a relation between $E$ and $L$ -- $$ \left(\frac{E^2}{mc^2}\right)^2 = \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{m^2r^2c^2}\right) \ ; $$ and substituting this into the equation for $v_{\rm circ}^2$, we get $$ v_{\rm circ}^2 = \frac{L^2}{m^2 r^2} \left(1 - \frac{r_s}{r}\right) \left(1 + \frac{L^2}{m^2 r^2 c^2}\right)^{-1} . \tag*{(2)} $$

The last part of the puzzle is to find an expression for $L/m$ in terms of the radial coordinate for a circular orbit. This is done by writing an expression for the effective potential in the Schwarzschild metric and finding where it is a minimum. Thus: $$ \frac{V_{\rm eff}}{mc^2} = -\frac{r_s}{2r} +\frac{L^2}{2m^2r^2c^2} - \frac{r_sL^2}{2m^2 r^3 c^2}\ . $$ Differentiating and equating to zero, we get $$ \left(\frac{L}{m}\right)^2 = \frac{c^2\,r^2\,r_s}{2r - 3r_s}\ . \tag*{(5)} $$

Replacing $L^2/m^2$ in eqn. (4) using eqn. (5), \begin{eqnarray} v_{\rm circ}^2 & = & c^2\left(\frac{r_s}{2r- 3r_s}\right) \left(1 - \frac{r_s}{r}\right) \left(1 + \frac{r_s}{2r-3r_s}\right)^{-1} \nonumber \\ & = & c^2 \left(\frac{r_s}{2r- 3r_s}\right) \left(\frac{r - r_s}{r}\right)\left(\frac{2r - 3r_s}{2r -2r_s}\right) \nonumber \\ & = & c^2\, \frac{r_s}{2r} = \frac{GM}{r}\, . \end{eqnarray} Thus the coordinate speed according to a distant observer is $\sqrt{GM/r}$, the same as the Newtonian result!

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$$ v = \sqrt{ \frac{GM}{r}} $$

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  • $\begingroup$ but isn't this from Newtonian mechanics? $\endgroup$
    – Yazdan.M
    Apr 26, 2023 at 19:02
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    $\begingroup$ No, the answer just happens to coincide with the Newtonian answer. $\endgroup$
    – TimRias
    Apr 26, 2023 at 19:04
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    $\begingroup$ No, this is the velocity wrt coordinate time. $\endgroup$
    – TimRias
    Apr 26, 2023 at 19:16
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    $\begingroup$ It works because of the way the $r$ coordinate is defined in the Schwarzschild coordinates. The Schwarzschild $r$ coordinate is not the Newtonian radial distance. $\endgroup$ Apr 26, 2023 at 19:39
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    $\begingroup$ @John Rennie - the circumference is all that matters on a circular orbit, and the r coordinate is C/(2π) which is the same as under Newton, so that the Schwarzschild radial distance measured with stationary rulers is different than under Newton only matters with radial motion, not at circular ones. Here the gravitational time dilatation does the trick. $\endgroup$
    – Yukterez
    Apr 29, 2023 at 2:39
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The velocity relative to a local stationary observer is

$$ \rm v=\frac{\sqrt{G M/r}}{\sqrt{1-r_s/r}}=\sqrt{\frac{G M}{r-r_s}} $$

which is $\rm v=c$ at the photon-sphere at $\rm r=1.5 \ r_s$, while the shapirodelayed velocity is the same as under Newton, see Tim Rias' answer and the reference below:

Cole Miller, ASTR498 wrote: "By a lovely coincidence, in Schwarzschild coordinates the angular velocity observed at infinity is exactly the same as it is in Newtonian physics."

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  • $\begingroup$ Calling it the “ shapirodelayed” velocity is weird since Shapiro delay is a time shift. Redshifted would be more accurate. $\endgroup$
    – TimRias
    Apr 29, 2023 at 7:13
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With velocity you could be asking for one of two things.

(1) The 4-velocity of the observer $\vec{U} = \frac{\partial}{\partial \tau}$.

(2) The velocity measured by another observer $\vec{v} = \frac{\partial x^k}{\partial t}\vec{e}_k$.

(1) Is a geometric object and thus it is perfectly well defined for every observer at every point of the objects worldline. While (2) is not a geometric object and depends completely on your choice of basis and coordinates. Not all choices of coordinates make physical sense, so the only velocity that is physically well defined in the sense of (2) and is measurable is the velocity according to the normal coordinates of the observer. How are these coordinates constructed? Any observer, inertial or not, can claim to be stationary and all measurements done by such observer will be done according to this "stationary" frame of reference. Additionally, the fact that gravitation is locally indistinguishable from inertial special-relativistic movement means that the observer "sees" on his own neighborhood of space time the flat Minkowsky metric. This can be done if at the event the observer performs the measurement we define a basis such that (using -+++ metric convention): $$\vec{e}_t \cdot \vec{e}_t = -1$$ $$\vec{e}_r \cdot \vec{e}_r = 1$$ $$\vec{e}_\theta \cdot \vec{e}_\theta = 1$$ $$\vec{e}_\phi \cdot \vec{e}_\phi = 1$$ This basis can be made into a coordinate system via vector exponentiation, but that is out of the scope of this answer. Regardless, we will refer to this system of coordinates as $\hat{x}^\mu$, notice that the vector basis in general will only be orthonormal at the one event it was defined. Additionally, the requirement that the observer be at rest with respect to this basis is equivalent to saying that its 4-velocity should only have temporal component, thus $\vec{e}_t = \vec{U}$, and since $\vec{U}$ is normalized to -1 $\vec{e}_t$ is automatically normalized.

Now we are in a position to derive a general result that relates 4-velocities to measured velocities. Suppose we have an observer with 4-velocity $\vec{U}_{obs}$ and a test particle with 4-velocity $\vec{U}$ both at the same event in spacetime (This is very important). We want to write the 4-velocity of the particle in the observer basis in terms of the normal time derivatives of the normal positions. For this we just use the chain rule: $$\frac{\mathrm{d}\hat{x}^k}{\mathrm{d}\tau} = \frac{\mathrm{d}\hat{x}^k}{\mathrm{d}\hat{t}}\frac{\mathrm{d}\hat{t}}{\mathrm{d}\tau}$$ Now we write $\vec{U}$ in the observer normal basis: $$\vec{U} = \frac{\mathrm{d}\hat{x}^\mu}{\mathrm{d}\tau}\vec{e}_\mu = \frac{\mathrm{d}\hat{t}}{\mathrm{d}\tau}\left(\vec{e}_t + \frac{\mathrm{d}\hat{x}^k}{\mathrm{d}\hat{t}}\vec{e}_k\right) = \frac{\mathrm{d}\hat{t}}{\mathrm{d}\tau}\left(\vec{e}_t + \vec{v}\right)$$ But the 4-velocity must be normalized, so: $$\vec{U} \cdot \vec{U} = \left(\frac{\mathrm{d}\hat{t}}{\mathrm{d}\tau}\right)^2 (-1 + v^2) = -1$$ Since the normal time flows in the same direction as the proper time we take the positive root of the quadratic. $$\frac{\mathrm{d}\hat{t}}{\mathrm{d}\tau} = \frac{1}{\sqrt{1-v^2}} = \gamma(v)$$ So: $$\vec{U} = \gamma(v)\left(\vec{e}_t + \vec{v}\right)$$ Notice this is identical to the special relativistic definition. Now the enlightening part comes when you notice that by construction $\vec{e}_t = \vec{U}_{obs}$ so you get: $$\vec{U} \cdot \vec{U}_{obs} = -\gamma(v)$$ And this formula can be used in any coordinate system.

For example, consider a static observer, this observer has the following 4-velocity: $$\vec{U}_{obs} = \frac{\partial_t}{\sqrt{1 - \frac{r_s}{r}}}$$ The particle in the circular orbit has only temporal and angular components to its 4-velocity and the angular velocity is determined by a conservation law: $$\vec{U} = \frac{\mathrm{d}t}{\mathrm{d}\tau}\partial_t + \frac{l}{r^2}\partial_\phi$$ Normalizing this vector gives: $$\vec{U} = \sqrt{\frac{1 + \frac{l^2}{r^2}}{1-\frac{r_s}{r}}}\partial_t + \frac{l}{r^2}\partial_\phi$$ Then contracting both vectors gives: $$-\sqrt{1 + \frac{l^2}{r^2}} = -\gamma(v)$$ From here you can get $v$ in terms of $l$ and you can get $l$ from the circular orbit condition. $$l^2 = \frac{r^2}{2\left(\frac{r}{r_s} - \frac{3}{2}\right)}$$

$$v^2 = \frac{r_s}{2r}\left(1 - \frac{r_s}{r}\right)^{-1}$$

Now this process only works if the observer is at the same point in spacetime as the particle, now that is not really a surprise since an observer can only affect things locally and measure things locally. You could define a notion of velocity according to normal coordinates in all of spacetime but that bears no relation to direct physical measurements. In real life all measurements of distant objects are done via interaction with light and then you have to do extra work to reconstruct the original dynamics of the particle. But as far as direct measurements go this is pretty much all we can do.

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