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First of all, I should say that I understand if this is put on hold for being unclear... but I'll try my best to make it as clear as it can get.

For all the time I spent learning Newtonian Gravity (school etc), I've associated it with the fact that the gravitational field "dissipates" with the area it covers based on the distance it is from a given point. This seemed reasonable to me, if we think of gravity as a field generated by a particle. And this seems consonant with the fact that it depends on the inverse of the square of distance:

$$\displaystyle |F|=G \frac{Mm}{||\textbf{x}-\textbf{x}_0||^2}. $$

I don't know if this is a standard interpretation, my first question is therefore:

  1. Is it?

Okay, going on with the reasoning:

We then have that:

$$\displaystyle F=-G \frac{Mm}{||\textbf{x}-\textbf{x}_0||^3}(\textbf{x}-\textbf{x}_0)$$

By following my reasoning, imagining if we were in a two dimensional space (euclidean space, as in Newton's POV), we would get:

$$\displaystyle F=-G \frac{Mm}{||\textbf{x}-\textbf{x}_0||^2}(\textbf{x}-\textbf{x}_0)$$

and, in one spatial dimension:

$$\displaystyle F=-G \frac{Mm}{||\textbf{x}-\textbf{x}_0||}(\textbf{x}-\textbf{x}_0)$$

Therefore, the force of gravity would be "constant". My second question, then, is:

  1. Does this make physical sense?

Now, I don't know Einstein's General Relativity Theory, but I know differential geometry, and I know that gravity is associated to the curvature tensor. So, if we understand gravity this way, then we can simply give a $1$-manifold some metric, and gravity will be the result of this metric. But I don't know if the metric must be given by something specific, or if it can be arbitrary. If it is arbitrary, this creates much more freedom to gravity: It wouldn't be so restricted as to necessarily imply that the "field" should be constant.

Therefore, my last questions are:

  1. What would be gravity according to Einstein's General Relativity Theory in a $1$-manifold?

  2. Would it be VERY different from the Newtonian's perception?

  3. If not, is there any support for any (or none) of those interpretations to be physically valid?

Qmechanic
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Aloizio Macedo
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1 Answers1

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General relativity in just one dimension will always be flat, as all 1D manifolds are diffeomorphic to flat space :

\begin{equation} ds^2 = -f(t) dt^2 \end{equation}

As you can perform the variable change

\begin{equation} \frac{dt'}{dt} = \frac{1}{\sqrt{f(t)}} \rightarrow t' = \int \sqrt{f(t)} dt \end{equation}

Giving you

\begin{equation} ds^2 = - dt'^2 \end{equation}

that is for one dimension of time and zero of space. If you have one dimension of time and one dimension of space (1+1 D), then you can apply a variation of the Gauss Bonnet theorem, which states that in two dimensions, the following formula applies :

\begin{equation} \frac{1}{2} \int_\mathcal{M} R d^2x + \int_{\partial \mathcal{M}} k_g dS + \sum_i \theta_i = 2 \pi \chi(\mathcal{M}) \end{equation}

With $R$ the curvature, $k_g$ the geodesic curvature on the boundary of the manifold, $\theta$ the exterior angle at non-smooth points of the boundary, and $\chi$ is the Euler number of the manifold. Since spacetimes are usually considered without boundaries, you obtain roughly

\begin{equation} \int_\mathcal{M} R d^2x \propto \chi(\mathcal{M}) \end{equation}

Meaning that the Einstein action will be a constant, hence its variation will always be 0, meaning that

\begin{equation} T_{\mu\nu} = 0 \end{equation}

Or, if you allow a cosmological constant,

\begin{equation} T_{\mu\nu} = \Lambda g_{\mu\nu} \end{equation}

You can also throw in various other terms to get a result, but in basic GR, spacetime is basically static in two dimensions.

A used toy model is obtained by the following cheat, using the action

\begin{equation} S = \int d^2 x \sqrt{-g} [\varphi R + \frac{1}{2} (\partial \varphi)^2 + \Lambda + \mathcal{L}_M] \end{equation}

Which gives the equation $R - \Lambda = T$, $T$ the trace of the stress energy tensor, the so called planar general relativity.

Slereah
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