Free fallen object: convergence Schwarzschild with classic mechanics

I'm trying to verify that, using Schwarzschild metric and using Newton mechanics, the same velocity over time for an object that is fallen to a planet in straight line, until some amount of approximation.

My objective: use Schwarzschild metric to analyze some basic problem (free fallen object speed in this case) and verify the answer comparing it with the well-known Newtonian one.

First step is Newton one. Assume object fallen from infinite directly to the center of the planet, no initial speed. By conservation of kinetic plus potential energy:

$$\frac{1}{2}mv^2 - G\frac{Mm}{r} = 0$$,

$$\frac{dr}{dt} = \sqrt{\frac{2GM}{r}} = c\sqrt{\frac{r_s}{r}}$$

$$t = \frac{2}{3c} \sqrt{\frac{r^3}{r_s}} + \textit{constant}$$

where $$r$$ is the altitude of the object and $$r_s$$ is the Schwarzschild radius ( its expression and remainders I will use can be found here ). Valid when $$r \ge$$ planet radios $$\ge r_s$$.

In Schwarzschild metric, when no angular momentum, we have:

$$\left( 1 - \frac{r_{\rm s}}{r} \right) \frac{dt}{d\tau} = \frac{E}{m c^{2}}$$ and

$$\left( \frac{dr}{d\tau} \right)^{2} = \frac{E^{2}}{m^{2}c^{2}} - \left( 1 - \frac{r_{\rm s}}{r} \right) c^{2}$$

combining these expression we obtain:

$$\frac{dr}{dt} = \frac{dr}{d\tau} \frac{d\tau}{dt} = c \left( 1 - \frac{r_{\rm s}}{r} \right) \sqrt{1 - \left( 1 - \frac{r_{\rm s}}{r} \right) \left( \frac{mc^2}{E} \right) ^2 }$$

I can not see how this expression could approximate to the one in Newton mechanics.

Could be related that in the case of $$E=mc^2$$ and $$h=0$$, wikipedia gives a value of the proper time equal to the $$t(r)$$ found using classic mechanics:

$$\tau = \frac{2}{3c} \sqrt{\frac{r^3}{r_s}} + \textit{constant}$$

• Just thinking out loud here: for $r\rightarrow\infty$, the factor $\left(1 - \frac{r_s}{r}\right)\rightarrow 1$, and the factor $\frac{m^2c^4}{E^2}\rightarrow 1$ so $$c\left(1 - \frac{r_s}{r}\right)\sqrt{1-\left(1 - \frac{r_s}{r}\right)\frac{m^2c^4}{E^2}}\rightarrow c(1)\sqrt{1-1+\frac{r_s}{r}}=c\sqrt{\frac{r_s}{r}}$$ – Alfred Centauri May 20 '20 at 19:07
• @AlfredCentauri: Thanks for your interest in this question. Why $mc^2/E \rightarrow 1$ when $r \rightarrow \infty$ ? – pasaba por aqui May 20 '20 at 19:10
• $E^2=p^2c^2+m^2c^4$ and $p\rightarrow 0$ as $r\rightarrow\infty$ (note that $m$ is the invariant mass here) – Alfred Centauri May 20 '20 at 19:12
• pasaba por aqui, if this is what you're looking for, let me know and I'll write it up as an answer. – Alfred Centauri May 20 '20 at 21:29
• Related: physics.stackexchange.com/q/211930/2451 and links therein. – Qmechanic May 21 '20 at 20:12

Let's consider how to understand $$E$$. It's the name given to the constant of motion arising out of the time-translational symmetry of the Lagrangian (as explained in here). In terms of this constant of motion, the equation of motion for time, as you quote, is given by
\begin{align} \left( 1 - \frac{r_{\rm s}}{r} \right) \frac{dt}{d\tau} &= \frac{E}{m c^{2}} \tag{1} \end{align}
Understand that we'd obtain different trajectories for different values of $$E$$. So, in order to analyze the similarity or otherwise of the GR solution to the Newtonian solution, we need to determine what is the value of $$E$$ that we expect to correspond to the Newtonian case we have considered. Well, since we are interested in a particle who is at rest when $$r/r_s\to\infty$$, let's analyze how different terms in expression $$(1)$$ behave in this limit. $$r/r_s\to\infty$$ is what we already stipulate, so now we analyze what our assumption says about $$dt/d\tau$$. Given that we are stipulating that our particle is performing the radial motion, we can write \begin{align} c^2d\tau^2&=\bigg(1-\frac{r_s}{r}\bigg)c^2dt^2-\bigg(1-\frac{r_s}{r}\bigg)^{-1}dr^2\tag{2} \end{align} Now, our assumption is that when $$r/r_s\to\infty$$ our trajectory is such that $$dr/dt\to 0$$. Thus, we obtain \begin{align} \frac{d\tau}{dt}&\to 1\tag{3} \end{align} Now, using this in conjunction with $$r/r_s\to\infty$$ for expression $$(1)$$, we obtain that $$E=m c^2$$ has to be satisfied for the trajectory of a particle to describe the trajectory of the kind we are interested in. This makes a lot of physical sense. The "potential energy" vanishes far away from the black hole and if the particle is at rest then the total energy is simply the rest energy of the particle, which is what we obtained.
Now, putting $$r/r_s\to\infty, E=mc^2$$ into the final expression of $$dr/dt$$ that you have quoted gives you a result for $$dr/dt$$ in conformity with the Newtonian expectation which is good because we should get such an agreement for slowly moving particles in weak gravity regions.
• Thus, we can say that "initially stalled at $r=\infty$" implies $E=mc^2$ ? – pasaba por aqui May 21 '20 at 20:34
• @pasabaporaqui $r=\infty$ is a meaningless thing to say because $r$ is a dimensionful quantity, but essentially, yes. "initially stalled at such an $r$ that $r/r_s\to\infty$ implies $E=mc^2$ in the given metric" is how I'd put it. – Dvij D.C. May 21 '20 at 20:36