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I've read that the simplified Schwarzschild's metric in the $t$-$z$ space

$$ds^2=\left(1+2\phi\right)dt^2-dz^2$$

could (approximately) reproduce the classical motion of a particle in a constant gravitational field (say near the earth's surface). So we can use the classical weight force potential $\phi=gh=gz$ and the metric becomes

$$ ds^2=\left(1+2gz\right)dt^2-dz^2.\tag{1}$$

I'd like to check that this metric actually leads to a geodesic that, with proper initial conditions, has the classic Newtonian form

$$z=z_0-g t^2$$

Working with the Christoffel symbols using the metric $(1)$ with indexes $\;\{1,2\}\;$ representing the variables $\;\{t, z\}\;$ I got

$\Gamma^1_{12}=\Gamma^1_{21}=\frac{g}{1+2gz}$

$\Gamma^2_{11}=g$

and the geodesic equations become

$\frac{d^2t}{d\lambda^2}+\frac{2g}{1+2gz}\frac{dt}{d\lambda}\frac{dz}{d\lambda}=0$

$\frac{d^2z}{d\lambda^2}+g\left(\frac{dt}{d\lambda}\right)^2=0\;\;$ (previous error corrected)

The second equation seem promising but I can't make sense of the first one, nor get a solution of this system of ODE.

I think that the problem could be connected with the fact that the geodesic equations use an affine parameter ($\lambda$) whilst the $t$ in Newton's free fall motion in not an affine parameter (?).

Is it possible to get classical motion of a particle from these geodesic equations?

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    $\begingroup$ Your metric look like Rindler metric ? en.wikipedia.org/wiki/Rindler_coordinates $\endgroup$
    – Eli
    Aug 16, 2021 at 19:33
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    $\begingroup$ I'm familiar with the approximately Newtonian metric $ds^2 = (1+2\phi)dt^2 - (1-2\phi)(dx^2+dy^2+dz^2)$, but that differs from yours in the spatial part prefactor. I believe there's a whole chapter devoted to it in Schutz's GR text. $\endgroup$
    – Paul T.
    Aug 16, 2021 at 19:57

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The issue seems to be an error in your spatial geodesic equation. This should be a statement of Newton's second law, something like $m\ddot{x} = F = -m\nabla\phi$, where $\phi$ is the gravitational potential. For a uniform gravitational field this should be $\ddot{x} = -g$.

(I find that numerical super- and sub-scripts can get confusing, so I'm going to just use the coordinate letters. I'm also going to use dots to refer to affine derivatives)

spatial geodesic equation

If we write out all of the terms of the sum we get: \begin{align} \ddot{z} + {\Gamma^z}_{\alpha\beta} \dot{x}^\alpha\dot{x}^\beta &= 0 \\ \ddot{z} + {\Gamma^z}_{tt} \dot{t}\dot{t} + {\Gamma^z}_{zz} \dot{z}\dot{z} + 2 {\Gamma^z}_{tz} \dot{t}\dot{z} &= 0 \end{align} In the question you state that only ${\Gamma^z}_{tt}$ is non-vanishing. This leaves $$\ddot{z} + {\Gamma^z}_{tt} \dot{t}\dot{t} = 0$$

(this differs from your spatial geodesic which has $\dot{z}$'s instead of $\dot{t}$'s)

The Christoffel symbol depends on derivatives of the metric. For your metric this reduces to $$ {\Gamma^z}_{tt} = -\frac{g^{zz}}{2}\,g_{tt,z} = +\frac{1}{2}\frac{d}{dz}(1+2\phi) = \frac{d}{dz}\phi.$$ There's our $\nabla\phi$!

If we use proper time as the affine parameter, then $\dot{t}=\frac{dt}{d\tau}=\gamma$, the Lorentz factor. If the particle is moving slowly, then $\gamma\approx 1$.

We can rewrite the spatial geodesic equation as:

$$\ddot{z} \approx -\frac{d}{dz}\phi = -g. $$

An alternate metric

If you used the approximate Newtonian metric I suggested in a comment $$ds^2 = (1+2\phi)dt^2 - (1-2\phi)dz^2, $$ you'd find that the ${\Gamma^z}_{zz}$ Christoffel no longer vanishes. But by the same slow speed approximation as used above $\dot{t}^2 \gg \dot{z}^2$, so you can safely neglect that term.

temporal geodesic equation

The temporal geodesic equation should be a statement about the energy of the system. This is easiest to see by putting the equation in terms of momentum. Multiplying the temporal geodesic by the particle mass, $m$: $$m\ddot{t} + 2m{\Gamma^t}_{tz}\dot{t}\dot{z} = 0$$

$$p^t = m \dot{t} = E \implies m\ddot{t} = \dot{E}$$

If energy is conserved, then $\dot{E} = 0$ at least to our first order approximation. The second term better vanish:

$$\dot{E} + 2m\frac{g}{1+2gz}\dot{t}\dot{z} = 0,$$

but I'm not sure how. We could say $\dot{z}\approx 0$, which doesn't seem fair.

The way I'm currently wrapping my head around it is by going back to sum of Christoffels: $$\ddot{t} + {\Gamma^t}_{tt} \dot{t}\dot{t} + {\Gamma^t}_{zz} \dot{z}\dot{z} + 2 {\Gamma^t}_{tz} \dot{t}\dot{z} = 0.$$

The ${\Gamma^t}_{tt} \dot{t}\dot{t}$ should dominate in the slow speed approximation. We could decide to approximate the other terms to zero at this point. The remaining ${\Gamma^t}_{tt}$ depends on a time derivative of the metric which leads to a term $\frac{d\phi}{dt}$.

With this logic we can say $$\frac{d\phi}{dt}=0 \implies \dot{E}=0.$$ This says energy is conserved in a static gravitational potential, which is true of Newtonian gravity.

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  • $\begingroup$ I think it is (at the end) ...by going back to sum of Christoffels: $\ddot{t}+\Gamma^t_{tt}\dot{t}\dot{t}+\Gamma^t_{zz}\dot{z}\dot{z}+2\Gamma^t_{tz}\dot{t}\dot{z}=0$ $\endgroup$
    – Luca M
    Aug 17, 2021 at 5:58
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    $\begingroup$ @LucaM, good catch! $\endgroup$
    – Paul T.
    Aug 17, 2021 at 15:37

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