# Attemp to encode newtonian gravitation as 3-dimensional space curvature

In lecture 9 of this series of lectures, Professor Frederic Schuller (around time 24:00) is trying to answer the question about the possibility to interpret newtonian gravity as a three-dimensional space curvature in order to write Newton's second law in the form of an auto-parallel equation:

$$m \ddot{x}^i(t) = - m g^i(x(t)) \tag{1}$$

$$g^i(x(t)) \stackrel{?}{=} \Gamma^{i}_{jk}(x(t))\dot{x}^j(t)\dot{x}^k(t) \tag{2}$$

where the latin indices run over $$1,2,3$$.

However, Professor Schuller states that one cannot find "$$\Gamma$$'s" in order to make equation (2) true, because of the fact that the Gamma symbols are functions of position $$x(t)$$ only, and due to the presence of the velocity terms in (2)'s RHS. I have already seen this Phys. SE post, but the answers do not satisfied me.I simply can't understand the Professor's argument.

If I assume the classical example of a particle in a constant radius $$R$$ orbit about a fixed point mass $$M$$ localized in the Origin of $$xy$$-plane such that its trajectory is given by

$$x(t) = (R\cos t, R\sin t,0 ) \tag{3}$$

and

$$g^i (x(t)) = \frac{GM}{||x(t)||^3} x^i(t)= \frac{GM}{R^3} x^i(t) \tag{4}$$

where $$t \in [0,2\pi]$$, for instance.

In this case I can, for example, take the 2nd component of gravitational field:

$$g^2(x(t)) = \frac{GM}{R^2} \sin t\tag{5}$$

and try to write this in the form of eq. (2):

$$g^2(x(t)) \stackrel{?}{=} \Gamma^2_{jk} (x(t)) \dot{x}^j(t) \dot{x}^k(t) \tag{6}$$

I could make

$$\Gamma^2 _{11}(x(t)) = \frac{GM/R^3}{x^2(t)} = \frac{GM}{R^4 \sin t} \tag{7}$$

and make all the other Gammas $$\Gamma ^2 _{jk}$$ vanish, such that

$$\Gamma^2 _{11}(x(t)) \dot{x}^1(t) \dot{x}^1(t) = \frac{GM}{R^2 \sin t} \sin^2 t = \frac{GM}{R^2} \sin t =g^2(x(t))$$

• You proposed a solution for one specific trajectory $x(t)$, but it's not a real solution unless it works for every free-fall trajectory $x(t)$. Can you write down an equation for $\Gamma^i{}_{jk}(x(t))$ that makes (1) work for every free-fall trajectory? Commented Jun 6, 2021 at 23:40
• I didn't noticed that. In this lecture, when the professor shows that newtonian gravity can be encoded as curvature of newtonian spacetime, he was able to write a general formula for the connection coefficients: $\Gamma^i _{00}(x(t))= g^i(x(t))$ and the other $\Gamma$s zero. Commented Jun 7, 2021 at 2:38

Note that the $$\Gamma$$'s are supposed to be functions on (an open subset) of the manifold, or for the sake of argument, they're supposed to be functions $$\Bbb{R}^3\to\Bbb{R}$$ and their definition shouldn't depend on any curves. What you did is take a very particularly defined curve $$x:[0,2\pi]\to\Bbb{R}^3$$ and then you defined $$\Gamma:\text{image}(x)\to\Bbb{R}$$. This is not at all what was required.

The question is:

Do there exist functions $$\Gamma^i_{jk}:\Bbb{R}^3\to\Bbb{R}$$, where $$i,j,k\in\{1,2,3\}$$, such that for all solutions of the equations of motion (i.e for every curve $$x:I\subset\Bbb{R}\to\Bbb{R}^3$$ such that $$\ddot{x}^i=-m\cdot g^i\circ x$$), we have for all $$i\in\{1,2,3\}$$, that $$g^i\circ x =(\Gamma^{i}_{jk}\circ x)\cdot \dot{x}^j\,\dot{x}^k$$.

I hope you realize that this is a much stronger condition than what you produced. The answer to this is no in general.

Of course, if $$g^i$$'s are all zero, then the answer is trivially yes, because we can simply choose all the $$\Gamma$$'s to vanish. But this is of course not the case for the gravitational force, so really, for any non-trivial choices of $$g^i$$, you in general cannot find the $$\Gamma$$'s to make this work, it just doesn't have the right form. We need an extra dimension, so that we can parametrize using that extra coordinate (the $$t$$ coordinate in $$\Bbb{R}^4=\Bbb{R}\times \Bbb{R}^3$$), so that $$\dot{t}=1$$, and then everything works out, as explained later on in the lecture.

Note also that later on in the lecture, the professor chooses $$\Gamma^{\alpha}_{00}=-f^{\alpha}$$ (in his notation) and all other $$\Gamma$$'s to be $$0$$. In this equality, these are functions on the manifold $$M=\Bbb{R}^4$$, they are not functions defined on some particular curve, hence this is a valid resolution.

• (also this is not special to $3,4$ dimensions. If you start with N2L in $n$ dimensions, then by adding an extra dimension you can always encode the $f^{\alpha}$ in terms of $\Gamma$'s). Commented Jun 7, 2021 at 13:53
• Hi peek-a-boo, quick question: since we know that $f=-\nabla \phi$, where $\phi$ is the gravitational potential and $f$ is the gravitational acceleration, how would we write this using the summation convention? Something like $mf^{\alpha} = F^{\alpha} = -\partial_{\alpha}\phi^{\alpha}$? That doesn't really look right to me so I wanted to reach out. Thank you for your time! Commented Feb 1 at 21:50
• @TaylorRendon no. $f=-\text{grad}(\phi)$ means $f^{\alpha}=-g^{\alpha\beta}\partial_{\beta}\phi$, or if you use Cartesian coordinates, then $f^{\alpha}=-\sum_{\beta}\delta^{\alpha\beta}\partial_{\beta}\phi=-\partial_{\alpha}\phi$. Recall $\phi$ is a function, so the exterior derivative $d\phi$ is a covector field. TO get a vector field, you need a metric (or symplectic form in Hamiltonian mechanics… which isn’t the case here) to convert (using the musical isomorphism) the covector field to a vector field, so that’s why we have the $g^{\alpha\beta}\partial_{\beta}\phi$ above. Commented Feb 1 at 22:39
• That makes sense thank you for your response! :) I have one more question if you don't mind: Schuller chooses $\Gamma^{\alpha}_{00} = -f^{\alpha}$ and all other $\Gamma$s to be $0$ and poses the question "is this a coordinate artefact, and so could be transformed away?" The answer is no because you can calculate the Riemann curvature tensor components and find that the only non-vanishing ones are $\text{Riem}^{\alpha}_{0\beta0}=-\partial_{\beta}f^{\alpha}$. Why does calculating the Riemann curvature components answer this question? Is it because the Riemann curvature... Commented Feb 6 at 16:59
• Yes that’s right. Commented Feb 6 at 17:32