Can matter be described as the result of the curvature of space, rather than the curvature of space being the result of matter, and energy being the cause of the curvature of space?

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    $\begingroup$ One difficulty that has not yet been addressed in the answers is that matter has to obey quantum mechanics, but it is not yet clear how to get the curvature of spacetime to obey quantum mechanics. $\endgroup$ Commented Dec 5, 2018 at 14:59
  • $\begingroup$ speculation: Maybe describing something as "matter" or "spacetime curvature" is ultimately a matter of taste, like say a particle or wave representation. $\endgroup$
    – R. Rankin
    Commented Dec 6, 2018 at 10:01
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    $\begingroup$ One historical tidbit: Einstein & Rosen (1935) represented an electron as Schwarzschild's curved spacetime. However in retrospect, we reinterpret their work as a "wormhole" solution with nothing to do with an electron. $\endgroup$ Commented Dec 12, 2018 at 0:18

2 Answers 2


Maybe one day.

This idea, at least it's mathematical genesis seems to have begun with Riemann and later Clifford. In 1870 Clifford (a very good mathematician), building upon Riemann gave a lecture stating:

1) That small portions of space are in fact analogous to little hills on a surface which is on the average flat namely that the ordinary laws of geometry are not valid.

2) That this property of being curved or distorted is continually being passed on from one portion of space to another after the matter of a wave.

3) That this variation of curvature of space is what really happens in that phenomena we call the motion of matter, whether ponderable or etherial.

4)That in the physical world nothing else takes place but this variation, subject (possibly) to the law of continuity.

This was the type of thinking that Led Einstein to consider space and time as dynamic entities and develop General Relativity(using Riemann's then developed geometry). Many others have sought to describe more of the universe than gravity through this type of program.

As of now, it's a no; however some progress has been made.

Because Electromagnetism curves spacetime like matter does, it has been shown (Rainich, Misner, Wheeler) that just the "footprints" left on spacetime by electromagnetic fields are enough to reconstruct the fields themselves (up do a "duality" rotation). This of course only holds for classical electrodynamics. This was called "Geometrodynamics" by Wheeler (who was famous for coining other phrases like blackhole and wormhole as well).

Wheeler showed analytically that a properly constructed ball of gravitational and electromagnetic radiations would possess the properties of a massive object (he called this a geon) and he applied similar topological ideas such that electric charges can appear to exist when there is in fact none (a trick of spacetime geometry).

To go further, one would need to be able to describe the gauge fields of the standard model in terms of geometry and properly quantize it. Whether these things can be found withing the topology of general relativity's spacetime, well that jury is still out.

The equations involved here are highly nonlinear, especially when you have the feedback effect of describing say the electromagnetic field through geometry which in turn effects the geometry again.

Anyway, when I read your question I am literally reading a book in front of my face:

"The Geometrodynamics of Gauge Fields. On the geometry of Yang-Mills fields and Gravitational gauge theories" Eckehard W. Mielke

I'll update this when I'm done maybe, it's an excellent book by the way.

PS: I had the pleasure of seeing Kip Thorne Lecture this (last?) year and he seems a large proponent of geometrodynamics. As one of the founders of LIGO (we're FINALLY detecting gravitational waves!!! still can't believe it) there's a potential to test such theories in the foreseeable future.


The answer is yes !

In General Relativity, there's a not very well known theorem called the Campbell-Magaard theorem that states that any metric in 4D spacetime (including matter) could be represented (or embeded) as a metric in empty 5D spacetime (Ricci flat spacetime, wich implies pure geometry) :



There are several simple solutions to Einstein's equation in 5D vacuum ($R_{AB}^{(5)} = 0$) that represent matter in 4D ($R_{\mu \nu}^{(4)} \ne 0$). This could be interpreted as matter made of pure geometry, in higher dimensions spacetimes.

Here's a very simple example. Consider the following metric in 5D spacetime ($\theta$ is a cyclic coordinate in the fifth dimension) : \begin{equation}\tag{1} ds_{(5)}^2 = dt^2 - a^2(t) (dx^2 + dy^2 + dz^2) - b^2(t) \, d\theta^2. \end{equation} Substitute this metric into the 5D Einstein's equation without any matter : \begin{equation}\tag{2} R_{AB}^{(5)} = 0. \end{equation} Then you get as a non-trivial solution these two scale factors : \begin{align}\tag{3} a(t) &= \alpha \, t^{1/2}, & b(t) &= \beta \, t^{- 1/2}. \end{align} This is the same as pure radiation in an homogeneous 4D spacetime : \begin{equation}\tag{4} ds_{(4)}^2 = dt^2 - a^2(t) (dx^2 + dy^2 + dz^2). \end{equation} With \begin{equation}\tag{5} R_{\mu \nu}^{(4)} - \frac{1}{2} \, g_{\mu \nu}^{(4)} \, R^{(4)} = -\, \kappa \, T_{\mu \nu}^{(4)}, \end{equation} and $T_{\mu \nu}^{(4)}$ describing a perfect fluid of incoherent radiation ($p_{rad} = \frac{1}{3} \, \rho_{rad}$):

\begin{equation}\tag{6} T_{\mu \nu}^{(4)} = (\rho_{rad} + p_{rad}) \, u_{\mu}^{(4)} \, u_{\nu}^{(4)} - g_{\mu \nu}^{(4)} \, p_{rad}. \end{equation}

This is the subject of the induced matter hypothesis, and is extremely fascinating! For more on the Campbell-Magaard theorem and the induced-matter theory :




  • $\begingroup$ How are $\rho_{rad}$ and $p_{rad}$ related to the geometric parameters $a(t)$ and $b(t)$? $\endgroup$
    – lurscher
    Commented Dec 5, 2018 at 14:25
  • $\begingroup$ @lurscher : $\rho_{rad} \propto a(t)^{- 4}$, which is the standard relation between energy density of radiation and the scale factor, in FLRW cosmologies. $\endgroup$
    – Cham
    Commented Dec 5, 2018 at 14:40
  • $\begingroup$ @Cham Is this a special 4d case of the Nash embedding theorem? $\endgroup$
    – R. Rankin
    Commented Dec 6, 2018 at 4:15
  • $\begingroup$ @R.Rankin, I don't know the Nash embedding theorem. Have you some references? $\endgroup$
    – Cham
    Commented Dec 6, 2018 at 12:12
  • $\begingroup$ Appears, Nash embedding is a case of whitney embedding. Here's a great stackexchange answer, but there are papers on them everywhere if your looking: math.stackexchange.com/questions/236285/… $\endgroup$
    – R. Rankin
    Commented Dec 6, 2018 at 12:30

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