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

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


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

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

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

What are my mistakes and what is the right way to think about this problem?

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    $\begingroup$ 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? $\endgroup$ Commented Jun 6, 2021 at 23:40
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    $\begingroup$ 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. $\endgroup$ Commented Jun 7, 2021 at 2:38

1 Answer 1


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.

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    $\begingroup$ (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). $\endgroup$
    – peek-a-boo
    Commented Jun 7, 2021 at 13:53
  • $\begingroup$ 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! $\endgroup$ Commented Feb 1 at 21:50
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    $\begingroup$ @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. $\endgroup$
    – peek-a-boo
    Commented Feb 1 at 22:39
  • $\begingroup$ 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... $\endgroup$ Commented Feb 6 at 16:59
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    $\begingroup$ Yes that’s right. $\endgroup$
    – peek-a-boo
    Commented Feb 6 at 17:32

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