As seen in this answer and also in this book chapter 6, page 6-10 expression 22, when you are falling in radially from being at rest at infinity your velocity in coordinate time, as seen from a distant observer, can be described by the expression:

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

If we assume ”$r$” to be the real radial distance and not just the ”r-parameter of Schwarzschild coordinates” this is the velocity $v$. I want to know the expression for the acceleration as a function of ”$r$” and ”$v$” for a small object of mass $m<<M$ moving purely radially inwards or outwards from a compact spherically symmetric mass distribution in the general case.

By taking the derivative with respect to "r" of the expression above I get: $$\frac{dv}{dr}=\frac{0.5}{r}{\sqrt{\frac{r_s}{r}}(1-\frac{3r_s}{r})}c$$

From these two expressions we see that the maximum velocity is $v=\frac{2}{3\sqrt{3}}c$ at $r=3r_s$. From this finding, $\frac{dv}{dt}$ should not be so hard but this will not be a general expression but only the acceleration of something dropped from rest at infinity. In Schwarzschild coordinates the radial velocity of light goes as $v_{light}=c(1-r_s/r)$. At $r=3r_s$ this becomes $v_{light}=2c/3$ so when falling in from being at rest at infinity you reach maximal velocity when $v/v_{light}=1/\sqrt{3}$. Maybe you can say that gravitation becomes repulsive whenever you are travelling towards the center of the gravitational field faster than $1/\sqrt{3}$ times the local speed of light?

Question: What does the general expression for the acceleration of an object in coordinate time look like for an object moving purely radially inwards/outwards from the center of the gravitational field assuming a spherically symmetric non-spinning mass distribution and that the r-parameter of the Schwarzschild coordinates is the real physical distance?

I am looking for an expression for the instantaneous acceleration as a function of "r" and "v".

I know that in the weak fields of our solar system Nasa/JPL are using this expression to calculate Newtonian plus relativistic acceleration:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$

This is based upon a first order expansion of the Schwarzshild solution in isotropic coordinates. For pure radial infall you get:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}-3\frac{v^2}{c^2}\right)\hat{r} $$ which becomes repulsive whenever you are approaching the central mass faster than:


I am hoping for an exact solution for pure radial acceleration in "Schwarzschild coordinate space" and coordinate time similar to the approximate solution for pure radial acceleration you can get from the Nasa equation.

As per the answer by amateurAstro down below the expression should be:


This is consistent with stuffing:




to get (at least this is true for pure radial and pure non-radial motion):

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(\hat{r}-2\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})} -\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})^2}\left((\hat{v}\cdot\hat{r})^2+(1-\frac{2GM}{rc^2}-\frac{v^2}{c^2})|\hat{v}\times\hat{r}|^2\right)\right)$$.

(I am trying to find a velocity-dependent term that together with the classical Newtonian acceleration should reproduce Schwarzshild orbits exactly)

  • $\begingroup$ Can you clarify. The expression I derived below is true for an object with total energy equal to $mc^2$. Are you looking for an expression for a particle with arbitrary energy. And what is the difference between $\bar{v}$ and $v$? $\endgroup$
    – ProfRob
    Commented May 8, 2019 at 7:12
  • $\begingroup$ When you are dropping an object from a radial distance where the radial velocity of light is 0.8c you are going to reach maximum velocity later. If you are "throwing" an object from infinity towards the gravitatinal field you are going to reach the peak velocity earlier. v = |$bar_v$|. Some say that the change of r-coordinate by time is not velocity since "r" is not distance in Schwarzschild coordinates. Yes I am looking for an expression for an arbitrary combination of "r" and "dr/dt". $\endgroup$
    – Agerhell
    Commented May 8, 2019 at 12:52
  • $\begingroup$ Constructing an expression like $$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}(\hat{r}-3\frac{v^2}{c^2(1-\frac{2GM}{rc^2})^2}(\hat{r}\cdot\hat{v})\hat{v}) $$ would have zero acceleration for pure radial infall whenever $v=\frac{c}{\sqrt{3}}(1-\frac{2GM}{rc^2})$, that is $1/\sqrt{3}$ times the local velocity of light which is the point where you have zero acceleration in the drop from infinity case that you described. I would expect the "true" expression to look something like this. $\endgroup$
    – Agerhell
    Commented May 8, 2019 at 13:20
  • $\begingroup$ I'm still not quite clear. Surely any general expression needs a further parameter that specifies the (initial) energy of the object (if it is greater than $mc^2$), or alternatively, a radius from which it is dropped (if it less than $mc^2$)? I have seen such expressions. $\endgroup$
    – ProfRob
    Commented May 8, 2019 at 15:55
  • $\begingroup$ I am looking for an expression for the instantaneous acceleration as a function of "r" and "v". Given the right expression one should be able to calculate from what height an object was dropped from "backwards". $\endgroup$
    – Agerhell
    Commented May 9, 2019 at 5:07

2 Answers 2


As Rob Jeffries suggested in a comment, formulating a solution for arbitrary dr/dt requires another parameter, either the energy per unit mass, E/m, or the radius of the shell where the object was dropped.

A shell observer at $r_0$, sees an object passing by at proper radial velocity $v_0$, with corresponding energy/momentum as measured in the local Minkowski frame (using units with c = G = 1) $$m^2=E_0^2-p_0^2=E_0^2-\gamma _0^2 m^2 v_0^2$$ $$\frac{E_0^2}{m^2}=\gamma _0^2 v_0^2+1=\frac{v_0^2}{1-v_0^2}+1=\frac{1}{1-v_0^2}=\gamma _0^2$$ where the subscripts indicate local shell coordinates. In Schwarzschild coordinates, energy per unit mass is constant and defined as $$\frac{E}{m}={ \left(1-\frac{2 M}{r}\right)}\frac{dt}{d\tau}$$ From equations 11 and 12 in section 6.3 of the Taylor, Wheeler, Bertschinger book (TWB) linked in the original question $$\frac{E_0}{m}=\left(1-\frac{2M}{r_0}\right)^{-1/2}\frac{dt}{d\tau}$$ $$\frac{E}{m}=\sqrt{1-\frac{2M}{r_0}}\frac{E_0}{m}=\gamma _0 \sqrt{1-\frac{2 M}{r_0}}=\sqrt{\frac{1-\frac{2M}{r_0}}{1-v_0^2}}$$ The equation for direct radial inward motion is obtained from TWB section 8.4, equation 17, by setting the angular momentum to zero and taking the negative root . $$ \left(\frac{dr}{d\tau}\right)^2=(E/m)^2-\left(1-\frac{2M}{r}\right)$$ $$\frac{dr}{dt}=\frac{dr}{d\tau}\frac{d\tau}{dt}= \frac{1-\frac{2 M}{r}}{E/m}\frac{dr}{d\tau}$$ $$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{1-\frac{1-\frac{2M}{r}}{(E/m)^2}}$$ Differentiating, and a bit of algebra yields the following result for the acceleration. $$\frac{d^2 r}{dt^2}=-\frac{2M}{r^2}\left(1-\frac{2 M}{r}\right) \left(1-\frac{3}{2}\frac{1-\frac{2 M}{r}}{(E/m)^2}\right)$$ These can be rewritten in terms of the Schwarzschild radius $r_s=2M$ and proper velocity, $v_0$, at $r_0$. $$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-\frac{1-\frac{r_s}{r}}{(E/m)^2}}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right) \left(1-\frac{3}{2}\frac{1-\frac{r_s}{r}}{(E/m)^2}\right) =-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right) \left(1-\frac{3}{2}(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}\right)$$

For known $dr/dt$ at a radius $r$, $E/m$ can be determined by inverting the previous expression for $dr/dt$. Then substitute $E/m$ in the equation for acceleration above.


The alternate approach, dropping an object from rest at radius $r_0$, uses equations 37 and 38 from section 6.7 of TWB, resulting in

$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right) \sqrt{\frac{\frac{r_s}{r}-\frac{r_s}{r_0}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=\frac{r_s\left(1-\frac{r_s}{r}\right)} {2 r^2\left(1-\frac{r_s}{r_0}\right)} \left(1+\frac{2r_s}{r_0}-\frac{3r_s}{r}\right)$$

The shell radius $r_0$ necessary to obtain a desired velocity at shell $r_1$ can be found using $$\frac{E_0}{E_1}=\sqrt{\frac{1-\frac{r_s}{r_1}}{1-\frac{r_s}{r_0}}} =\frac{\gamma_0}{\gamma_1}$$ where $\gamma_0=1$ for an object dropped from rest at $r_0$.

As an example, a plot is shown below using $v_1=1/2$ at $r_1=2r_s$. For these values $\gamma_1=2/\sqrt{3}$, $E/m=\sqrt{2/3}$ and $r_0=3r_s$. For reference, the green and red curves are Schwarzchild coordinate velocity and acceleration for an object dropped from rest at infinity, $E/m=1$. The dashed blue and brown curves are for an object dropped from rest at $r_0$, and the solid blue and brown curves for an object launched from $r_1$ with a Schwarzschild velocity of $$\left(\frac{dr}{dt}\right)_{r_1} =-\left(1-\frac{r_s}{r_1}\right)v_1 =\frac{1}{4}$$ Note that the dashed curves are overlaid by the solid ones from $r_1$ to $r_s$, since the initial conditions were chosen to result in the same velocity at $r_1$. enter image description here

  • $\begingroup$ Now you could rewrite the fourth full line expression from below to get an expression for $r_0$ as a function of $r$ and $dr/dt$.? If you repace $r_0$ in the third full line expression from below below I gues you get an expression for dv/dt as a function of $r$ and $dr/dt$ only, not caring about the "initial altitude of dropping"? $\endgroup$
    – Agerhell
    Commented May 20, 2019 at 11:47
  • 1
    $\begingroup$ @Agerhell, I do not think that this suggestion will work. Note that $r_0$ in that equation is a constant, chosen so that $dr/dt$ is a particular value at some smaller radius where the object is dropped from rest. It is not a function of $r$ and $dr/dt$. If you want acceleration for a known $dr/dt$ at radius $r$, the expression for $dr/dt$ from the first derivation above can be inverted to find $E/m$. Then this result can be used in the acceleration equation. I edited my answer to add this approach, based on your comment. $\endgroup$ Commented May 20, 2019 at 17:07
  • $\begingroup$ Following you instructions I managed to find exactly what I was looking for. Thanks! If you want to you could edit in the final expression into you answer to make it look even better. $\endgroup$
    – Agerhell
    Commented May 24, 2019 at 15:19

Edit: This is only the freefall from infinity case; not the general case. More needs to be added.

Acceleration can be written as $$\frac{d^2 r}{dt^2} = v \frac{dv}{dr} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{2r^2}\right){(1-\frac{3r_s}{r})}c^2 $$

Acceleration is negative (i.e. $v$ becomes more negative) whilst $r>3r_s$, changes sign to a "deceleration" when $r<3r_s$, reaches a minimum and then increases to zero at $r = r_s$ (but never gets there since $v \rightarrow 0$).


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