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General relativity states (in simplified vocabulary) that mass curves the space-time, being gravity one of the observable effects of this curvature.

Is there any theory or works starting on the premise: mass (particles) IS a curvature of the space-time ?

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  • $\begingroup$ I think that it is problematic such an identity, due to the regions of spacetime without mass, and having a non vanishing tensor of curvature. $\endgroup$ Feb 4, 2021 at 19:51
  • $\begingroup$ No there is not. See my answer to this post for a simplified explanation of General Relativity. $\endgroup$
    – mmesser314
    Feb 4, 2021 at 20:00
  • $\begingroup$ All energy bends space-time. $\endgroup$
    – jpf
    Feb 4, 2021 at 21:53
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    $\begingroup$ Does this answer your question? Can matter be described as the result of the curvature of space, instead of vice versa? $\endgroup$
    – R. Rankin
    Feb 4, 2021 at 23:47
  • $\begingroup$ As Void's description of Geometrodynamics hints, this is actually a very old idea - in fact it is insinuated by Riemann at the very end of his foundational geometry paper, and was then picked up by William Clifford, then Hermann Weyl, Einstein, and so on. I've been trying to develop it in a loose qualitative way, and I'd love it if you'd look at the outline! It shows promise but it needs a lot of help! $\endgroup$ Feb 11, 2021 at 15:54

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General relativity is that theory. The field equation of general relativity says that mass (more precisely four-momentum or stress-energy) equals Ricci curvature, up to a constant that's merely a unit conversion factor like $c$. There's nothing about one side of the equation causing the other. They're just equal, everywhere, always.

In present physical theories, four-momentum isn't obviously the same as curvature (in special-relativistic field theory, including QFT, it isn't curvature at all), but that's presumably a deficiency of those theories that should be fixed.

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    $\begingroup$ Just a minor technical comment. If you want to relate the Ricci curvature to the stress energy tensor you should re-write the equations using only the Ricci tensor, instead of $G$. This is possible just taking the traces of the two members as you surely know very well... $\endgroup$ Feb 4, 2021 at 20:29
  • $\begingroup$ @ValterMoretti I just meant they're the same degrees of freedom. I deleted the equation altogether because it didn't add much. $\endgroup$
    – benrg
    Feb 5, 2021 at 0:28
  • $\begingroup$ GR doesn't explains inertial mass, nor quantum aspects of particles, nor wave-particle duality, ... . "Just" the gravitational mass. $\endgroup$ Feb 5, 2021 at 17:51
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It is not possible: curvature propagates outside masses, classically gravity propagates from sources. So there is curvature in regions where there is no mass. They are distinct notions though related. The spacetime Riemann curvature is a dynamical notion, it measures the relative acceleration of free falling bodies close to each other.

There is also the problem of giving a precise meaning to the term curvature. I adopted the point of view that curvature is the opposite of flatness. From this perspective the curvature is represented by the so called Riemann tensor, since it vanishes if and only if the spacetime is (locally) flat. However there are other weaker notions like the Ricci curvature which is directly related to the stress energy local content of the universe.

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  • $\begingroup$ Yep, that's the problem with outreach ideas... People think they understand, but they not always do $\endgroup$
    – FGSUZ
    Feb 4, 2021 at 20:03
  • $\begingroup$ Yes, I agree... $\endgroup$ Feb 4, 2021 at 20:05
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    $\begingroup$ "there is curvature in regions where there is no mass" seems based in the assumption that matter is composed of particles deterministics in position and limited in sizes. $\endgroup$ Feb 4, 2021 at 23:17
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    $\begingroup$ "it's not possible" is incorrect Take a look at Wheeler's work on geometrodynamics, he demonstrated that a "ball" (or "geon") of purely gravitational energy could appear to have all of the aspects of a mass from outside, yet have no locally definable mass within it. He did similar intersting things with charge on non-simply connected spacetimes, and he definitely understood General Relativity, he did after all coin the phrase "black hole" and "wormhole" $\endgroup$
    – R. Rankin
    Feb 4, 2021 at 23:56
  • $\begingroup$ This seems to be more an answer to "is curvature matter?" than "is matter curvature?" $\endgroup$ Mar 18, 2021 at 6:50
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The equation $G=T$ (let me put the proportionality constant equal to one for simplicity) is extremely beautiful in its conciseness. But this conciseness can also cause various misconceptions (which indeed appeared among the very developers of the theory).

Something that is often omitted in cursory accounts of the Einstein equations is that they are an incomplete set of equations where matter is present. By themselves they don't determine neither the 4D metric $\gamma$ nor the motion of matter. Additional equations are always necessary: some that tell how the stress-energy-momentum depends on the matter fields, and some that give specific conservation equations for the matter fields – conservations which are not contained in the Einstein equations.

For example, if we have an electromagnetic field, represented by the Faraday tensor $F$, without charges and currents, the complete set of equations is $$ G = T\ , \qquad T=T[F,\gamma]\ , \qquad \mathrm{d}F = 0 \ . $$ The second equation determines how the 4-stress depends on electromagnetic field and metric, and the third expresses the conservation equation of the electromagnetic field (which is completely independent of any metric). Without the last two equations we can't obtain anything from $G=T$. Without $G=T$ we can't obtain anything from the other two equations either.

This is one of the reasons why it's extremely difficult, to say the least, to reduce one particular field to another.


We gain a deeper insight when we rewrite this beautiful equation in a 3+1 space+time form. When we do this various "gauges" are possible. Here I choose one that simplifies the resulting equations a great deal. Here they are (references at the end): $$ \begin{align} \frac{\partial}{\partial t} \pmb{g} &= -2\pmb{K}\pmb{g} \\ \frac{\partial}{\partial t} \pmb{K} &= \pmb{K}\operatorname{tr}\pmb{K} - 2\pmb{K}^2+\pmb{R} -\tau +\frac{1}{2}(\operatorname{tr}\pmb{\tau} - \epsilon) \pmb{I} \\[1em] \epsilon &= \frac{1}{2}[(\operatorname{tr}\pmb{K})^2-\operatorname{tr}(\pmb{K}^2) + \operatorname{tr}\pmb{R}] \\ \pmb{\lambda} &= \nabla\cdot(\pmb{K}-\pmb{I}\operatorname{tr}\pmb{K}) \end{align} \label{einstein}\tag{1} $$

In these equations – completely equivalent to $G=T$ – Latin letters represent fields that embody the spacetime metric and curvature, and Greek letters represent those related to energy-mass, momentum, and stress (force). All the fields in these equations are 3D fields – similarly to how we'd write evolution equations in Newtonian mechanics.

The space-time metric & curvature are represented by the 3D tensor fields $\pmb{g}$ and $\pmb{K}$ – they're basically 3D matrices:

  • $\pmb{g}$ is the 3D metric of 3D space
  • $\pmb{K}$ is the so-called extrinsic curvature, another 3D field that encodes how the 3D metric is embedded in 4D spacetime
  • $\pmb{R}$ is the curvature of 3D space – it can be rewritten in terms of $\pmb{g}$
  • $\nabla$ is the 3D covariant derivative constructed from the 3D metric.

As for the matter fields,

  • $\epsilon$, a scalar field in 3D space, is the energy-mass density
  • $\pmb{\lambda}$, a 3D vector field, is the energy flux, equivalent to momentum density in relativity
  • $\pmb{\tau}$, a 3D tensor field (matrix) is the stress, that is, force per unit area.

Finally, $\pmb{I}$ is just the 3D unit matrix, and $\partial/\partial t$ is a Lie derivative in differential-geometric terms.

All these fields obviously depend on space and time, eg $\pmb{g}(\pmb{x},t)$, and the equations above give their evolution through a sequence of appropriately chosen 3D-space hypersurfaces in spacetime.


Several things may be noticed about these equations:

  1. We see that the first pair of equations tell how the metric & curvature propagate through time.
  2. The second equation can also be rewritten in terms of the 3D stress only: $$ \frac{\partial}{\partial t} \pmb{K} = \pmb{K}\operatorname{tr}\pmb{K} - 2\pmb{K}^2+\pmb{R} -\frac{1}{4}(\operatorname{tr}\pmb{K})^2 + \frac{1}{4}\operatorname{tr}(\pmb{K}^2) -\frac{1}{4}\operatorname{tr}\pmb{R} -\pmb{\tau} + \frac{1}{2}\operatorname{tr}\pmb{\tau} $$
  3. The propagation of metric & curvature is determined by themselves ("self-interaction") and also by the energy density and the stress (or the stress only, if we like). This is the reason why metric & curvature can propagate also where there's no matter.
  4. The second pair of equations don't determine any evolution, but constrain the values of energy density and momentum and the values of curvature on every 3D space hypersurface.

And somw very important remarks must be added:

  • The system of equations $\eqref{einstein}$ above – which is just $G=T$cannot be solved as it is, because it's underdetermined: way more variables than equations.
  • The energy-mass density, energy flux, and stress of matter depend on the specific matter field, usually from its conserved charges and currents (typical examples are a baryon field or the electromagnetic field).

So in order to really determine the evolution of all the fields in $\eqref{einstein}$ we need additional equations (and initial and boundary conditions for the full system). The latter are the constitutive equations of matter, which determine its specific properties and behaviour (for example if it's a gas, liquid, solid, elastic, viscous, some other kind of field, and so on). Typically they are functional equations of this form: $$ \begin{align} \epsilon &= \epsilon[\nu, \pmb{\phi}, \theta, \pmb{g}, \pmb{K}] \\ \pmb{\lambda} &= \pmb{\lambda}[\nu, \pmb{\phi}, \theta, \pmb{g}, \pmb{K}] \\ \pmb{\tau} &= \pmb{\tau}[\nu, \pmb{\phi}, \theta, \pmb{g}, \pmb{K}] \end{align} \label{const}\tag{2} $$ where $\nu$ and $\pmb{\phi}$ are further 3D fields that usually represent charges and currents specific to the matter field – in the example above, baryon density & flux; but they might be electromagnetic fields. These further fields have additional charge-conservation equations. Finally, $\theta$ is the temperature.


From this slightly broader point of view we see that it's difficult to say "such-and-such field is just the manifestation of such-and-such other field". For example, looking at the second pair of equations in $\eqref{einstein}$ one might be tempted to say "energy density and energy flux are just 'manifestations' of the curvature". But the necessity of the constitutive equations $\eqref{const}$ shows that such an interpretation is not possible: those fields have specific properties that are not determined by metric & curvature alone.


For the full derivation of the equations above (which are routinely used in numerical relativity), the analysis of their Newtonian limit, and for the discussion of the constitutive equations in specific cases, I recommend the brilliant book by Gourgoulhon:

see especially chapters 5 (equations $\eqref{einstein}$ are in section 5.3.2) and 6.

Other texts which derive and discuss the same results are for example

Also check out

for an in-depth discussion of matter's specific description and equations.

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Some aspects of matter-energy are captured in geometry

Let us first consider what the question is even about. How can matter be identified with curvature? It would mean that one should be able to read off all the properties of matter only from space-time quantities such as the Riemann curvature tensor $R^\mu{}_{\nu\kappa\lambda}$ and it's derivatives.

A part of that can be, indeed, done in Einstein gravity, where we have \begin{align} R^{\mu\nu} - \frac{1}{2} R g^{\mu\nu} = 8 \pi G T^{\mu\nu}\,, \end{align} where $R^{\mu\nu}, R$ are the Ricci tensor and Ricci scalar respectively, which are contractions of the Riemann curvature tensor. This means that we can completely read off all components of the momentum-energy tensor $T^{\mu\nu}$ from the Riemann tensor. As Valter Moretti points out, there are other degrees of freedom of curvature (such as freely propagating gravitational waves) that are not related to matter, but this isn't a big issue, there is simply "matter" curvature and "non-matter" curvature. We know how much matter and energy is there at every point, with how much momentum, and what are its internal stresses. But is that enough?

The thing is that we still need to specify the dynamics of the matter, and the Einstein equations are simply not enough. They do imply that the stress-energy tensor $T^{\mu\nu}$ is conserved, $T^{\mu\nu}{}_{;\nu} = 0$ as an integrability condition, but we need more to describe the dynamics that we actually observe. The problem is that $T^{\mu\nu}$ does not carry any information about the charges of the matter. Here by charges I mean things such as baryon number, lepton number, and electromagnetic charge.


EM charge is not in $T^{\mu\nu}$

Let me give you an example. Consider a cloud of charged dust such that its stress-energy tensor is \begin{align} T^{\mu\nu}_{\rm dust} = \rho u^\mu u^\nu\,, \end{align}
where $\rho$ is its mass density and $u^\mu$ its four-velocity field. Now let me give you the stress-energy tensor of the EM field \begin{align} T^{\mu\nu}_{\rm EM} = \frac{1}{\mu_0}\left(F^{\mu \kappa} F^\nu{}_\kappa - \frac{1}{4} g^{\mu\nu} F^{\kappa \lambda} F_{\kappa \lambda} \right) \,, \end{align}
where $F^{\mu\nu}$ is the Faraday tensor and $\mu_0$ is vacuum magnetic permeability. Notice that there is no mention of the charge of the dust cloud. Now the conservation of the total stress-energy tensor $T^{\mu\nu}_{\rm dust} + T^{\mu\nu}_{\rm EM}$ is not enough to determine its evolution uniquely (you can verify this yourself). We need to specify the Maxwell equation $F_{\mu\nu;\kappa} + F_{\kappa\mu;\nu} + F_{\nu\kappa;\mu} = 0$ and the source equation \begin{align} F^{\mu\nu}{}_{;\nu} = \mu_0 j^\mu = \mu_0 \tilde{q} \rho u^\mu\,, \end{align} where $\tilde{q}$ is the charge per unit mass of the cloud. That is, to specify the evolution of the system of the electromagnetic field and a charged dust cloud, we need to know a number $\tilde{q}$ and that is nowhere to be found in the stress-energy. This means that there is also no way to find it in the geometry!

In other words, the stress-energy tensor is like a black and white photo. It captures a lot, but some important aspects are lost!


Geometrodynamics to the rescue...?

The last section was a purely classical discussion, so what about a more fundamental quantum approach? In quantum field theory we learn that the conserved charges correspond to conservation laws in gauge field theories, the basic building block of the Standard model of particle physics. Even there we see that the stress-energy tensor cannot capture everything about the composition of matter that is superposed in it.

There are some famous examples of "charge without charge" by Misner and Wheeler. They constructed pseudo-charge by connecting two space-time regions with a wormhole with an electromagnetic field. They were then able to find an EM field such that even though there was apparently a non-zero divergence of $\vec{E}$ on one end of the wormhole, the electric field lines were actually diverging from the other space-time region through the wormhole and no charges were involved. As a result, if you were unaware of the wormhole, you would conclude that the two throats of the wormhole were actually two oppositely charged particles!

The efforts of mainly Misner and Wheeler on this matter are summarized in the Wiki article Geometrodynamics. Skipping other issues, the main problem here is that Misner & Wheeler's constructions actually only ever involved a single gauge field $A_\mu$ corresponding to the Abelian gauge group $U(1)$. However, in the Standard model we have a bunch of independent gauge fields that simply cannot be all read off from the stress-energy tensor. This remains true even in Grand Unified Theories where a single gauge group emerges, since it still has many independent gauge fields (more than the Standard model!). In other words, the dynamics we observe is too complex to be simply captured in a $T^{\mu\nu}$.

Of course, one could speculate that there is a plethora of topological defects in space-time that would somehow mimick all these flavours of the gauge fields, but this is far from easy, and as far as I know, has never been done. That is, there is no theory that would actually work for a non-negligible part of the phenomena we observe and that would model matter purely as space-time curvature.

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