# Newtonian motion from a simplified Schwarzschild's metric in 1+1D

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?

• Your metric look like Rindler metric ? en.wikipedia.org/wiki/Rindler_coordinates
– Eli
Aug 16, 2021 at 19:33
• 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. Aug 16, 2021 at 19:57

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.

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