# Is it true to say *Space time curvature* $\Leftrightarrow$ *Matter*

Is it true to say Space time curvature and Matter are just the same thing, part of the same coin and that therefore Space time curvature $\Leftrightarrow$ Matter? In other words is Space time curvature just Matter and Matter just Space time curvature?

• This was Wheeler's idea in his geometrodynamics programme. As others have noted below, Clifford conjectured it first, but it's difficult to incorporate, e.g., fermionic matter into such a model. – Alex Nelson Sep 28 '13 at 0:56
• Why is it difficult to include fermionic matter in such a model? – user29727 Sep 28 '13 at 9:04

Your question is interesting from an historical perspective: this is exactly what William Kingdon Clifford conjectured.

But this is not what modern physics thinks is going on. "Stuff" - matter and energy - it's all the same as far as the relevant physics is concerned - does influence the curvature of spacetime. The amount and distribution of "stuff" is measured by a generalized energy density object $T_{\mu\nu}$, called the stress energy tensor and this distribution begets curvature in the spacetime manifold, measured by the Einstein tensor $G_{\mu\nu}$, which curvature in turn defines the paths in spacetime of "free" objects. This description is afforded by the Einstein field equations of general relativity. They look like:

$$G_{\mu\nu} = \frac{8\,\pi\,G}{c^4} T_{\mu\nu}$$

The other things, $G$ and $c$ are constant. For more details, see my answer here.

Can we say that matter is a result of the curvature of space time?

No. Ben's answer has one essential component that ny answer above forgot to put in: although it may look like the matter-energy and curvature are tightly coupled through $G_{\mu\,\nu}\propto T_{\mu\,\nu}$, $G_{\mu\,\nu}$ is not quite the curvature - its something derived from the curvature with its "trace" subtracted away, such the curvature can propagate as waves that become sundered from their matter-energy sources, just as electromagnetic waves can become sundered from the current and charge sources that generate them. It's not only waves either: a "static" curvature field can reach into "empty" space where $T_{\mu\,\nu} = 0$. This is what Ben is driving at when he gave the example of Gravity Probe B in empty space. The relationship between the "curvature field" and matter-energy is very like that between the electromagnetic field and current/charge. One needs the other to get going, but you can see that they are quite separate. Also, I'd like Ben to confirm this, but there is the problem of fermionic matter with things like Wheeler's Geons: things built out of "relativity" objects are going to be tensors, unless some fairly radical changes are made to GR first. Tensors do not transform the same way as fermionic matter fields under rotations and boosts do - you need spinors for that (this is the idea that a spin $1/2$ object takes on a sign swap under a $2\,\pi$ rotation). Whilst you can build tensors from spinors, you can't go the other way (spinors are sometimes said to be "square roots" of tensors, but the analogy to my mind is really rough and not too helpful).

But I'm not talking about gravity, gravitational field.

Actually $G_{\mu\,\nu}$ the Einstein tensor, and this is going to sound utterly crazy so bear with me, doesn't talk about the gravitational field either! It's a wholly geometric object and it describes geometric things like how a test vector changes as it is parallel transported around a closed path and so forth, or how the volume of a test sphere with a given surface area deviates from its Euclidean value. This is the main point of GR - it tells us how the geometry of spacetime looks - in particular it tells us how "coasting" objects will move - along geodesics. In this paradigm, there isn't really a gravitational field: there is simply the geometry of the spacetime manifold. You feel a force on your bottom when you sit on the ground because the Earth's matter-energy curves the space around it such that the geodesics through spacetime would be trajectories where you were accelerating relative to the Earth's surface at $g\,\mathrm{m\,s^{−2}}$: this is the "coasting", Newton's First Law state. But you can't follow these trajectories, instead you ram up against the ground and the ground accelerates you upwards at a rate of $g\,\mathrm{m\,s^{−2}}$ relative to a frame in this coasting state. Although it's a bit weird, in the GR paradigm, to sit on the ground is to be in an endlessly dynamic state where the unbalanced reaction force from the ground endlessly accelerates you upwards relative to what's known as your "momentary co-moving inertial frame". The ground's is the only force on you. The gravitational force in this paradigm is only re-introduced as an inertial force in the D'Alembertain sense if you want to do calculations in the non-inertial reference frame. But, in principle, you don't need the gravitational force idea now - it's just a convenience exactly like the centrifugal force you bring into calculations of a body undergoing circular motion if you want to do calculations in a frame fixed to that body. If you step back and describe the body from an inertial frame, you don't need the centrifugal force anymore - you simply see the nett, unbalanced centripetal force supplied by whatever is tethering the object to move in circular paths (e.g. tension in a rope, electrostatic attraction ...).

What if the centripetal force were gravity, I hear you ask? Well, now in GR, spacetime geodesics begotten by a spherical mass include circular motion paths. So a satellite in a circular (or any other) orbit about a planet is now coasting! There is no force on it. Spacetime is curved, and it's simply following the geodesics.

• Thank you very much for your effort but why can't we say that matter is a result of the curvature of space time (without evoking Einstein field equations)? It make sense for me, just as we say that the Higgs boson is an excitation of the Higgs field. Thank you. Kind regards. – user29727 Sep 28 '13 at 12:37
• You're very welcome: you are clearly probing deeply and I'd really like to compliment you on you're self teaching effort - this is a really fun thing to do. My expertise ends with GR (and that's rusty) so unfortunately I can't follow you into particle physics, which I know very little about. Certainly you can't say "matter is a result of the curvature of space" in GR, so you're talking other theories. Maybe you mean that the language of differential geometry and curvature can be and is used to describe the fibrations of a gauge theory? Certainly there are geometrical analogies there ... – WetSavannaAnimal Sep 28 '13 at 12:55
• ...But I'm not sure what the particle or string theorists would make of the ideas that these are curvatures of space. Maybe you could ask a slightly different question and tag it "particle" and "gauge" to fetch the right expertise for you. – WetSavannaAnimal Sep 28 '13 at 12:56
• @Adobe You may also be aware that many theorists in quantum gravity - if I understand the gist of their musings right - seek to think of gravity as being mediated by a particle, a graviton. This seems to an outsider like me to being as much as saying that the "geometry only" interpretation of GR such as I have described in my answer above is wrong (or probably more accurate to say "incomplete") and that one seeks to "restore" gravity as a force. – WetSavannaAnimal Sep 28 '13 at 13:29
• @Adobe I was looking at your user page and that's a great bibliography the Uni of Chicago has put together to study from. At some stage, you might find Roger Penrose's "Road To Reality" a good thing to read. It's not quite for self learning, but it kind of "ties all the threads" together and lets you know where all the different topics of things like the Uni of Chicago bibliography fit in and how they fit together. – WetSavannaAnimal Sep 29 '13 at 2:20

No, spacetime curvature is not the same as matter.

First of all, measuring two things and stating that the measures are equal doesn't mean that the two things are the same concept. For example, if a polygon has $E$ edges and $V$ vertices, then $E=V$, but that doesn't mean an edge is the same thing as a vertex.

Second, there are many different ways of measuring curvature and many different ways of measuring matter. Only if you take the Einstein tensor as your measure of curvature and the stress-energy tensor as your measure of matter do you get an equality due to the Einstein field equations.

As a specific counterexample, the Gravity Probe B satellite measured some effects due to the curvature of spacetime by the earth. The satellite was in the vacuum of outer space. There was no matter there, but there was curvature.

Another example would be gravitational waves, which can propagate in a vacuum.

• Can we say that matter is a result of the curvature of space time? – user29727 Sep 28 '13 at 8:57
• @WetSavannaAnimalakaRodVance But I'm not talking about gravity, gravitational field. – user29727 Sep 28 '13 at 11:28
• @Adobe: Can we say that matter is a result of the curvature of space time? No, because there are various measures of curvature, and only one measure, the Einstein tensor, relates to matter. In general, the laws of physics are written as equations. Although we often think of them as relationship between a cause on one side of the equation and an effect on the other, this is not really rigorously true. Causality is really a relationship between conditions at one time and conditions at a later time. – user4552 Sep 28 '13 at 13:00

No. A bunch of matter (e.g. earth) generates a gravitational field well outside of itself. A magnet and a magnetic field aren't the same in a similar sense. In general relativity, you can have a curved space even without matter.

• But this gravitational field is caused by the presence of energy? How we have a curved space even without matter? – user29727 Sep 28 '13 at 9:02
• @Adobe: GR admits several solutions of the Einstien equations in vacuum. – Nikolaj-K Sep 28 '13 at 10:06
• Some solutions of Einstein equations are not real in our world – user29727 Sep 28 '13 at 10:26
• @Adobe: All of them. – Nikolaj-K Sep 28 '13 at 11:04
• The solution for the vacuum is $T_{µv} = 0$ – user29727 Sep 28 '13 at 11:26