I do not know about the formalism in string theory, but can

testify that the Einstein Cartan formalism is extremely useful in many models 

of quantum gravity-in particular higher spin theories and Lorentz violating effective models. The fundamental reason for this appears to be that it naturally accommodates the gravitational couplings of fermions, which must involve a torsion component  in addition to the usual Ricci coupling. also EC distinguishes space-time and gauge indices, which is rather convenient.

The question doesn't make sense, since torsion only is relevant when coupling degrees of freedom that carry spin to the metric. In the absence of any such degrees of freedom, choosing to set torsion to zero is just a choice of coordinates. And in their presence torsion is required anyway.

That's why it doesn't make sense distinguishing Einstein-Cartan gravity from general relativity, unless one is interested in the history of physics.

That the equations of motion of the lowest mode of vibration of closed strings-i.e. the classical limit of the theories described that way-do reproduce Einstein's equations is well known:

http://inspirehep.net/record/838

https://lib-extopc.kek.jp/preprints/PDF/1968/6800/6800234.pdf

https://lib-extopc.kek.jp/preprints/PDF/1968/6800/6800238.pdf

There's no meaning to the statement of the Einstein-Cartan formalism being superior to that of general relativity, since they're the same!

Stam: “The question does not make sense...”

When you have intrinsic angular monentum in classical GR that is the continuum limit of a distribution of Kerr masses, the limit includes torsion that is equal to spin density, exactly as in EC; the continuum limit is not GR.

Also, an isolated Kerr mass has “translational holonomy” (TH) around its equator, and torsion = continuum limit of TH/(area inside the loop traversed), which is a definition of torsion in differential geometry. Also, the resulting torsion crudely equals the spin density of the Kerr mass.

So the question does make sense, and I believe is answered in the affirmative.

In order to describe the coupling of particles with non-zero spin to gravity, the formalism is what's known as the Einstein-Cartan formalism. Torsion then cannot br set to zero. That's why Kerr black holes require Einstein-Cartan. However this formalism isn't either in contradiction or alternative to general relativity-it's general coordinate invariance in the presence of spin.

When coupling to particles of spin 0, torsion can be set to zero.

However the distinction between EC and GR is only of historical interest-they're the same thing, when describing spin 0 particles coupled to gravity and gravity in the EC formalism, i.e. in the presence of matter with non-zero spin, does couple to matter only through the metric.

That's what's relevant.

The statements about continuum limits, simply, don't make sense and they're not relevant to the question.

Dear Stam,

The continuum limit may be unnecessary for someone like you who sees EC as a modest extension to GR that enables modeling exchange of intrinsic and orbital angumar momentum (and improves some other things like elminating some singularities from GR).

However, I believe 99% of the gravitational researchers (maybe 99% of researchers over age 40) dont' accept this point of view. For them, I hope the derivation of translatonal holonomy from a single Kerr mass and the continuum limit of distributions of Kerr masses might, just might, ignite a flame insight, or at least respect.

As an example of the problem, in January 2019, I submitted a slightly edited version of "Article Derivation of Einstein-Cartan theory from general relativity

The editor's response was brief: he stands by the original review.

Planck wrote: “A new scientific truth [that alters the elements of a field] does not triumph by persuading its opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it.”

--Max Planck, “Scientific Autobiography and Other Papers,” 1949, p 33-34.Quoted in “The Structure of Scientific Revolutions,” by Thomas Kuhn, 4th edition, 2012, p 150.

I have been on this case for 30 years, and I know what passes for competence out there. I would prefer that someone who sees the light like you would help to get this into the mainstream.

Referee report from Physical Review D (verbatim)

The aim of the manuscript is to derive Einstein-Cartan theory form General Relativity by showing that in some GR configurations involving a distribution of Kerr black holes, there exists torsion using the concept of "translational holonomy." It is not at all clear what "translational holonomy" is supposed to be. The usual holonomy that defines curvature (what the author calls rotational holonomy) has the structure that it does precisely because torsion cancels. Therefore, I do not see how one can "derive" torsion when none existed in the first place. Moreover, the configuration that the author uses to "prove" this is not a stationary configuration as the Kerr masses will feel an attractive force with respect to one another.

My rebuttal

The first sentence of the review correctly states the most important claim of the work.

1) “It is not at all clear what ‘translational holonomy’ is supposed to be.”

The concept of holonomy is defined in references [1] page 87 and [2] page 71.

Translations are not treated the same as rotations in spacetime gauge theory. The root issue is that the geometric structure group (symmetry group) of GR and EC is the semidirect product of the Lorentz group and translations (which has a fixed origin point), not the Poincaré group (which has no preferred origin point). See [3] page 128 for a rather abstract statement of this positon. Therefore we must modify the definition of holonomy for translations.

The definition and construction of translational holonomy is discussed in the manuscript in sections 3.1 and 3.2.

I have provided a brief intuitive description of translational holonomy in section 3 of this document.

2) “The usual holonomy that defines curvature has the structure that it does precisely because torsion cancels.”

This statement is false in Riemannian geometry and in RC. It is a denial that RC can exist.

I believe the referee has in mind the derivation of Riemannian curvature by constructing a piecewise almost–closed loop from four geodesic segments. In this construction, the “failure-to-close” vector is generally nonzero for finite loops. The quantity that vanishes is:

limit[t → 0] (failure-to-close vector)/t^2 = 0.

The manuscript uses only closed loops, in which the “failure-to-close” vector is by definition zero. It use no geodesics.

In RC, my derivations of rotational curvature from rotational holonomy and of translational curvature (affine torsion) from translational holonomy work fine. See manuscript section 3.2. “Definition of curvature in terms of holonomy.”

(manuscript equation 3) limit[t → 0] (rotational holonomy)/t2 = rotational curvature

(manuscript equation 4) limit[t → 0] (translational holonomy)/t2 = translational curvature = torsion

3) “I do not see how one can ‘derive’ torsion when none existed in the first place.”

The disruption of ordinary geometry (singularity?) inside each Kerr mass makes this possible, just as the core of a dislocation in a crystal enables torsion to exist around a dislocation, when there is none in the lattice outside the dislocation.

4) “The configuration that the author uses … is not a stationary configuration…”

This referee’s statement is true. However, non-stationarity does not invalidate the result, any more than the non-stationarity of cosmological solutions in GR, invalidate GR.

The manuscript (constraint m

Once more: The Einstein-Cartan formalism is relevant for setting up and solving the equations of motion of matter, coupled to gravity; not for solving Einstein's equations, where matter enters exclusively through its energy-momentum tensor. It's always possible to express the spacetime metric in terms of the vielbeine and vice versa.

That's why all the issues mentioned are irrelevant and irrelevant for the original question. There's simply no way in which the Einstein-Cartan formalism is an alternative to general relativity-it's the part of general relativity that describes how spinning degrees of freedom (i.e. those that transform non-trivially under rotations) couple to spacetime geometry that's not flat, since it introduces the spin connection. That's all there is to it. So it's relevant for describing the motion of Kerr black holes, as they affect the spacetime geometry, not for describing the motion of spinless particles in the spacetime of a Kerr black hole, for instance.

The description of the spacetime geometry is, still, given by the metric, whatever the properties of matter and matter, even though it's spinning, still contributes to how the metric is affected by it, only through its energy-momentum tensor. Which means that the spacetime geometry can't ``tell'' if it's being affected by scalar fields, gauge fields, spinors or whatever, as long as all these fields lead to the same value of the energy-momentum tensor. The motion of the particles described by such fields, of course, will depend on their spin; that's where the Einstein-Cartan formalism is relevant ( as well as in extensions of general relativity, where fields, beyond the metric, that can't be redefined as being part of matter exclusively, determine the spacetime geometry. The two most well known examples are scalar tensor theories, where additional scalar fields, whose energy-momentum tensor can't describe usual matter, are introduced; and supergravity).

The spacetime geometry is described (in a redundant way) by the metric tensor, in any case and only the invariant quantities make any sense.

``Translational holonomy'' is just a fancy term for describing the fact that, in curved spacetime, translations don't commute, in the case at hand. That's nothing new, by itself, however and it's in all textbooks on general relativity, under the heading ``parallel transport''.

Stam, I am glad we agree that EC = GR where there is no intrinisic angular momentum.

However, I disagree with several of your statements about EC.

• Your first paragraph: " The Einstein-Cartan formalism is relevant for setting up and solving the equations of motion of matter, coupled to gravity; not for solving Einstein's equations ..." makes no sense to me.
• Translational holonomy definitely is not equivalent to the fact that certain translations do not commute. I don't know what translations you are treferring to; perhaps integral curves of geodesics. A basis of translations commute if and only if they can be intrepreted as coordinate vectorfields. GR has no torsion, and that hs nothinbg to do with what translations commute.
• Computationally, the metric and torsion can be consisdered as the independent variables of EC. However, the best way to view EC is that the independent variables are the connection coefficients. The flat metric in the MInkowski fibers induces a metric on spacetime. The metric should no be a dynamic variable. I thiknk this is true of GR as well,
• EC eliminates singularities in the Big Bang and in Schwarzschild black holes. This is a big change from GR.
• EC is more than a restatement of GR. I am quite sure that EC will be extremely different from GR in quantum gravity. Spin always plays a much greater role on quantum theories than in the corresponding classial theories. GR cannot allow spin-orbit coupling.

Too many of your statements are fuzzy or false. I don't think it will be productive use of our time to continue discussing this topic.

Dear Richard James Petti

in GR it is the metric which is the independent and dynamic variable. The connection is defined as a dependent variable, in such a way that the torsion is zero. So, one cannot derive from GR that the torsion is non-zero. One can, of course, define different connections, with torsions, and use them as independent variables. but this defines something different, nothing which could be derived from GR. In particular, a theory which gets rid of GR singularities is certainly a different theory, and a different theory cannot be derived from GR.

Your claim that the existence of a singularity in a Kerr BH makes torsion possible even inside GR seems questionable to me. The straightforward counterargument is quite simple - start with a collapsing star, thus, a configuration without singularity. Then, you can always define the time coordinate in such a way that the part behind the horizon will never be reached, but everything outside the horizon is covered. Again, no singularity, straight R^4 topologically for the whole solution. Thus, no torsion visible outside too. So I think the main argument for rejection holds.

Ilja,

It is possible to generate torsion from a space that has no torsion.

It is also possible to generate Riemannian curvature from a flat space has no Riemannian curvature, which is more familiar, so let me show you how to generate curvature from a flat surface.

1. Start with a flat regular polygon with ‘n‘ sides with area ‘A.’ It is composed of n isosceles triangles that meet at the center of the polygon.

2. Remove one of the isosceles triangles. The total angle of the remaining material is 2pi - 2pi/n. That is, the “angular deficit” (compared to 2pi radians at a point in a flat plane) is 2 pi/n.

3. Glue together the two edges where you cut out the triangle. To do this, you must allow the surface to rise up out of the plane to form the tip of a cone.

The slant side of the cone is still flat; that is, its Riemannian curvature is zero. (The appearance that the side of the cone is curved is a property of how the surface is embedded in 3-space; it is not a property of the surface. This kind of curvature is called “ extrinsic curvature.” You can ignore this.)

The area of the resulting surface is less than the original area; the new area is (n-1)/n A. To keep expressions small, denote this area by A1.

At the tip of the cone is an angular deficit of 2pi/n. Curvature of a surface can be defined as the angular deficit di. The curvature function is 2pi/n times a Dirac delta function. (A Dirac delta function is infinitely tall and infinitely narrow, but the integral of this function over a small neighborhood around the tip is 1.) The curvature concentrated at tip of the cone equals the angular deficit of material around the tip.

4. Build a surface that spreads out the curvature to many points on the surface.

To do this, distribute n points fairly evenly over the original flat surface. At each of these points, cut out a small sector of material of angular size 2 pi/n. The resulting surface approximates a surface of constant positive Riemannian curvature of magnitude 2 pi/A1. Riemannian curvature is the integral of the angular deficits over the surface divided by the Area A.

5. Build a surface with smooth curvature that is not concentrated at one point.

If we make ‘n’ larger, the curvature of the surface is 2 pi/A1, but the curvature is spread more evenly over the surface. In the limit as n approaches infinity, the resulting surface is a surface of constant curvature 2pi/A.

This is how to make a positively curved surface out of small (in the limit infinitesimal) pieces of flat surface.

To generate a space with negative Riemannian curvature from a flat space (that is, it has no Riemannian curvature), do everything the same as the procedure to generate positive Riemannian curvature, with one exception:

In step 2, insert an additional isosceles triangle (instead of removing an isosceles triangle). The total angle of the remaining material is 2pi + 2pi/n. That is, the “angular excess” (compared to 2pi radians at a point in a flat plane) is 2 pi/n.

The end result will be a surface that has constant Riemannian curvature –2 pi/A.

If we agree on this, we can move on to how to develop torsion from a space that has no torsion.

Sent from my iPhone

I agree on this. The whole construction gives a non-Euclidean, curved metric, with the curvature somehow artificially localized, but so what.

From point of view of GR the animal you named "angular deficit" is not that important, there is a well-defined metric. And for this metric, there also exists the standard torsion-free connection. Once the curvature is singular in the origin, one can expect that already the Levi-Civita connection is somehow singular there too.

Ilja, I am glad we agree. Here is the next step.

It is possible to generate torsion as the limit of a distribution of discrete dislocations, starting from a space with no dislocations (no torsion).

I will use a perfect discrete cubic 3-deimensional lattice instead of a continuous space, which does not change the basic idea. For convenience, we will name the three principal directions of the cubic lattice as x, y, z; z is the vertical direction.

1. Start with a 3-D cube of lattice that has 2 n + 1 basis units wide in all directions, where n is a positive integer. Its volume is (L n)^3, where L is the length of an edge of the lattice.

2. Draw a line on the top of the lattice in the +x direction from the center to the edge. Cut the lattice with a knife along this line from the top to the bottom of the lattice. The cut creates two new plane boundary surfaces inside the lattice that are parallel to the y-z plane. Remove one layer of vertices from one side of the cut; that is, remove a half-plane of vertices from the cut.

3a. Make an “edge dislocation”: Removed a half-plane of vertices from one face of the cut, and glue together the two free surfaces.

Traverse a square path in the x-y plane that does not surround the point where the cut stops (does not surround the z-axis): take 10 steps in the x-direction, 10 in the y-direction, -10 steps in the x-direction, and -10 steps in the y-direction. This path ends on the same vertex where it started. Any path that does not surround the core of the dislocation (the z-axis) is perfectly normal.

Traverse a path in the x-y plane that surrounds the point where the cut stops (surrounds the z-axis): 10 steps in the x-direction, 10 in the y-direction, –10 steps in the x-direction, and –10 steps in the y-direction. Because you removed a half-plane of vertices, this path is not a closed loop; it ends one vertex away from on the starting point in the direction –x.

The line where the cut ends (the z-axis in the center of the x-y plane) is an edge dislocation. (It is so named because the z-axis consists of an edge of vertices that are not connected to anything on the –x side; it is a free edge.) Materials scientists use the term “Burghers vector” to describe the vector pointing from the starting point to the ending point to the edge dislocation.

A density of edge dislocations corresponds to the trace component of torsion.

3.b Make a “screw dislocation”: Do everything as you did for the edge dislocation, with one exception: Don’t remove any vertices; when you glue together the two free surfaces in the cut, join each vertex to a vertex that is one lattice position up (or down) in the z direction on the other free surface.

Traverse a square path in the x-y plane that does not surround the point where the cut stops (does not surround the z-axis). This path ends on the same vertex where it started. Any path that does not surround the core of the dislocation (the z-axis) is perfectly normal.

Traverse a path in the x-y plane that surrounds the point where the cut stops (surrounds the z-axis): 10 steps in the x-direction, 10 in the y-direction, –10 steps in the x-direction, and –10 steps in the y-direction. This path ends at a vertex that is one lattice spacing up (+z direction) or down (+z direction) from the starting point. (It is like a spiral parking garage ramp: if you drive around the spiral, you move one level up or down in the garage.)

The line where the cut ends (the z-axis in the center of the x-y plane) is a screw dislocation. (It is so named because the shape resembles a screw; a better name might be the spiral dislocation.) Materials scientists use the term “Burghers vector” to describe the vector pointing from the starting point to the ending point to the screw dislocation.

A density of screw dislocations corresponds to the totally antisymmetric component of torsion.

4. Build a surface that spreads out the dislocation defect to many points in the lattice.

To do this, select (n/m)^3 points evenly in the original cubic lattice (m divides n). At each of these points, make an edge or screw dislocation. The resulting lattice approximates a volume with a constant density of dislocations, where the density of dislocations is (1/m)^3 dislocations per basic cell volume (L^3) of the lattice.

5. Build a surface with smooth curvature that is not concentrated at a finite number of points.

To do this, make n extremely large and let m divide n with m

Fine, but what could be the point of the whole construction?

I see an obvious one, the lattice behind the construction defines a connection, and it is probably one with torsion.

But what would be the point of constructing such a connection? Nobody doubts that connections with torsion exist. But the one used in GR, the Levi-Civita connection, is torsion-free. So, one with zero torsion exists too.