If something in subatomic particles is neutral direction whether if relative or fixed, it acts as centre of gravity so other subatomic(s) point to it (gravitational attraction)... could that be possible?

"Thought experiment approaches have been suggested as a testing tool for quantum gravity theories.[9][10] In the field of quantum gravity there are several open questions – e.g., it is not known how the spin of elementary particles sources gravity, and thought experiments could provide a pathway to explore possible resolutions to these questions,[11] even in the absence of lab experiments or physical observations."

From link: https://en.wikipedia.org/wiki/Quantum_gravity

I think there’s an implied “if” - since we have no working, tested/confirmed theory of quantum gravity, we don’t know if spin has an effect. It might be present in some proposals, but we don’t know if they are correct.

There are no answers at this point, AFAIK

Particle spin is intrinsic angular momentum, and any momentum is a 'source' of Gravity.

Is that what you meant ?

As said, suggested, implied:

Angular momentum has energy.

Energy sources gravity.

Ergo,

Angular momentum sources gravity.

At the moment, we don't know whether the words "quantum gravity" are akin to what people once used to say: "elastic properties of the luminiferous ether".

It could well be the case that these words finally have to be abandoned.

There is also a spin formalism for general relativity. So, in a way, all of GR is about spin. This is because spinors are more basic objects than space-time 4-vectors (for every 4-vector, or event, there are two spinors representing it).

2 hours ago, MigL said:

Particle spin is intrinsic angular momentum, and any momentum is a 'source' of Gravity.

Is that what you meant ?

Yes, I think so.

4 hours ago, joigus said:

Angular momentum has energy.

Energy sources gravity.

Ergo,

Angular momentum sources gravity.

How much, though? ~10^-16 eV-s of angular momentum vs ~1 MeV of mass for an electron.

(The phrasing did not make it clear to me whether it was claiming a source or the source)

6 hours ago, joigus said:

Angular momentum has energy.

Energy sources gravity.

To be honest, I’m not so sure about this. The energy-momentum that forms the source term in the Einstein equations comes from the Noether current associated with spacetime translations, whereas spin comes from Lorentz invariance. These are different things. It is not in fact possible, AFAIK, to define a unique 4-momentum vector for intrinsic spin, so I don’t see how it could - if taken in isolation - act as a source of gravity.

Or am I missing something?

there have been studies into intrinsic spin couplings to gravity example below

https://arxiv.org/pdf/2502.07604

to date as far as I know there are no measured couplings and the article mentions the key violations that would result including those pertaining to freefall differences. There are numerous papers pertaining to this in the Godel Universe for spin gravity couplings which more often that not employ Einstein Cartan spacetimes.

One of the factors against a rotating universe tight bounds is the lack of any measured spin gravity couplings which experimental data places a very tight bound against any universe rotation.

Mashoom has a paper on it pertaining to Godel universe

https://arxiv.org/pdf/2304.08835

the papers typically employ the magnetic moment of intrinsic spin with their couplings

Edited Saturday at 06:33 AM5 days by Mordred

3 hours ago, Mordred said:

Einstein Cartan spacetimes

Non-zero torsion tensor. I read somewhere (I don't recall where) that the torsion tensor corresponds to spin density (the dimensions match if considered as an extension to the Einstein equation). However, I've also read that the torsion tensor doesn't have a propagating gravitational field, though it is not clear to me precisely what this means.

7 hours ago, Markus Hanke said:

To be honest, I’m not so sure about this. The energy-momentum that forms the source term in the Einstein equations comes from the Noether current associated with spacetime translations, whereas spin comes from Lorentz invariance. These are different things. It is not in fact possible, AFAIK, to define a unique 4-momentum vector for intrinsic spin, so I don’t see how it could - if taken in isolation - act as a source of gravity.

Or am I missing something?

Yeah. To tell you the truth, I found the wikipedia article a tad ambiguous about whether it's "standalone" or it's "contributes to". @swansont had a similar point to make, besides the weakness of an only-spin source. But AFAIK too, you're right. Spinors cannot be represented by 4-vectors.

After looking through the provided papers, it seems clear that none of the authors (related to such idea by the Wikipedia article) mean to say that spin be the one and only source of gravitation. Rather, they set out to find subtle effects of gravity on spin systems safely above the Plankian scale.

6 hours ago, KJW said:

Non-zero torsion tensor. I read somewhere (I don't recall where) that the torsion tensor corresponds to spin density (the dimensions match if considered as an extension to the Einstein equation). However, I've also read that the torsion tensor doesn't have a propagating gravitational field, though it is not clear to me precisely what this means.

I believe you may be referring to the antisymmetric affine connections and anti symmetric stress energy tensor vanishing in the Newton limit spacetimes or in spacetimes with zero torsion ( some literature state in the vacuum) but near or in BH regions is non vanishing if I recall this was a means to solve the singularity problem.

However it's been an incredibly long time since I looked at Godel, Gravitoelectromagnetism or Einstein Cartan.

All 3 the above involve intrinsic spin couplings in some fashion or other

Edited Saturday at 04:29 PM5 days by Mordred

15 hours ago, joigus said:

Rather, they set out to find subtle effects of gravity on spin systems safely above the Plankian scale.

That seems more reasonable to me - not that I’m an expert, this is quite a subtle question.

18 hours ago, KJW said:

However, I've also read that the torsion tensor doesn't have a propagating gravitational field, though it is not clear to me precisely what this means.

My own guess - the field equations for torsion in ECT contain no derivatives, and at the same time torsion is completely determined by local matter fields. This implies that torsion vanishes in regions where T=0, and no wave-type equation exists for torsion to “radiate” through vacuum. So it can’t have any propagating degrees of freedom - it’s purely a local phenomenon subject to the local presence of matter.

On 1/23/2026 at 7:21 PM, tylers100 said:

Yes, I think so.

I'm not so sure about that now. Because:

First

I think firstly before trying to understand momentum, motion, etc in regards to gravity, better to try understand the gravity while it is stationary (or neutral direction) and singular before delving into relation, momentum, or motion which is what constitutes complexity.

Singular Object with Gravity

See attached picture. Here is what I see and think:

Newton

What Newton described is gravity (i.e. neutral direction as centre of gravity with toward direction as gravitational attraction towards it).

Einstein

What Einstein described is curvature of spacetime (i.e. loop bi-direction connecting end points of each axis dimension). Kind of similar to the train scene in Matrix 3 when Neo was struck inside a train station for some time, he ran into a train rail underground but only arrived at back where he started. The concept is similar to that.

But uneven distribution of geometry... maybe not so.

Gravity could be symmetric in nature, possibly not uneven distribution of geometry.

In order to understand gravity especially under GR you need to have a good grasp of kinematics. GR uses the 4 momentum and its symmetry relations are freefall states with no force acting upon the object or particle ( which directly applies to the conservation of momentum).

Newton treats gravity as a force acting upon the falling object instead of freefall. GR uses spacetime curvature instead of treating gravity as a force.

Curvature is easily understood if you take 2 or more freefall paths. For example take 2 laser beams in parallel. If the two beams remain parallel spacetime is flat. If the two beams converge it is positively curved. If the beams diverge ( move apart) = negative curvature.

To better understand freefall study the Principle of equivalence.

https://webs.um.es/bussons/EP_lecture.pdf

You can see under the section " Local inertial frames" the freefall paths are approaching one another as the elevator is freefalling toward a center of mass ( positive curvature).

Indeed the equation of GR employ geodesics to describe these paths for photons they are null geodesics and how parallel null geodesics remain parallel converge of diverge are used to describe the curvature terms.

At a more advanced level this is the basis of the Raychaudhuri equations. Which is a good formalism to understand how spacetime geometry affects multiparticle paths with regards to curvature terms.

As mentioned Newton described gravity as a force so the falling objects have the gravitional force acting upon them. Under GR they are in freefall but the spacetime paths become curved. Hence gravity is treated as the result of spacetime curvature .

In terms of geometry the Newton case the geometry is Euclidean and unchanging. This isn't the case under GR. In GR the geometry itself changes resulting in what we describe as gravity.

Edited Sunday at 04:53 PM4 days by Mordred

1 hour ago, Mordred said:

In order to understand gravity especially under GR you need to have a good grasp of kinematics. GR uses the 4 momentum and its symmetry relations are freefall states with no force acting upon the object or particle ( which directly applies to the conservation of momentum).

Newton treats gravity as a force acting upon the falling object instead of freefall. GR uses spacetime curvature instead of treating gravity as a force.

Curvature is easily understood if you take 2 or more freefall paths. For example take 2 laser beams in parallel. If the two beams remain parallel spacetime is flat. If the two beams converge it is positively curved. If the beams diverge ( move apart) = negative curvature.

To better understand freefall study the Principle of equivalence.

https://webs.um.es/bussons/EP_lecture.pdf

You can see under the section " Local inertial frames" the freefall paths are approaching one another as the elevator is freefalling toward a center of mass ( positive curvature).

Indeed the equation of GR employ geodesics to describe these paths for photons they are null geodesics and how parallel null geodesics remain parallel converge of diverge are used to describe the curvature terms.

At a more advanced level this is the basis of the Raychaudhuri equations. Which is a good formalism to understand how spacetime geometry affects multiparticle paths with regards to curvature terms.

As mentioned Newton described gravity as a force so the falling objects have the gravitional force acting upon them. Under GR they are in freefall but the spacetime paths become curved. Hence gravity is treated as the result of spacetime curvature .

In terms of geometry the Newton case the geometry is Euclidean and unchanging. This isn't the case under GR. In GR the geometry itself changes resulting in what we describe as gravity.

Geometry and Gravity

Maybe geometry does change but in proportion way, that change alone likely doesn't result in gravity. Geometry arrangement along with being neutral direction whether relative or fixed does. See:

Link Examples

"Figure 2: The field through a cube sliced in half through the faces. We can observe the slight distortion of the field lines between the edges and the center of each face."

...

"Could a moon or satellite, orbit this cubic planet? We notice that there is slightly greater gravitational force of attraction over the corners of the cube, and hence an orbiting satellite would significantly couple with the spin of the cube, refer Fig. 3."

From link: https://ar5iv.labs.arxiv.org/html/1206.3857

(originally link provided on https://www.scienceforums.net/topic/125952-question-about-basics-of-gravity/page/5/)

Point

The point is geometrical arrangement (e.g. of corners) in symmetrical or proportion way while in relative or fixed neutral direction could result in like-hood of gravity.

Edited Sunday at 06:33 PM4 days by tylers100
Clarification.

I wouldn't rely on images to understand gravity. They can often be more misleading.

For example describe how either image shows the equivalence principle between inertia mass and gravitational mass or show how either image describes time dilation when neither image contains a spacetime diagram

Lets put it this way and only you can honestly answer the following. In the time frame since your last post in the your other thread has your understanding of gravity significantly improved ?

In that same time frame would your understanding of gravity improved significantly more if you had instead studied an introductory GR textbook such as Lewis Ryder's General relativity even if you only spend 2 to 3 hours on it a week ?

Edited Sunday at 07:32 PM4 days by Mordred

47 minutes ago, tylers100 said:

Figure 2: The field through a cube sliced in half through the faces. We can observe the slight distortion of the field lines between the edges and the center of each face."

...

"Could a moon or satellite, orbit this cubic planet? We notice that there is slightly greater gravitational force of attraction over the corners of the cube, and hence an orbiting satellite would significantly couple with the spin of the cube, refer Fig. 3."

You originally posted about fundamental particles, which are not cubes.

The effects here are from deviations from a spherical mass distribution.

Lets try a simple example with regards to symmetry relations involved for spacetime.

The Minkowskii metric for example has a specific mathematical statement defining orthogonality which must also be symmetric.

\[\mu \cdot \nu= \nu \cdot \mu\]

This directly applies to vectors more specifically covariant and contravariant vectors. You wont find any image that will teach the above key relationship.

In your first link where is the length contraction as applied under the Lorentz transformations? To give another example

Edited Sunday at 07:47 PM4 days by Mordred

Im going to a phenomena in SR in terms of symmetry that will likely blow your mind and quite frankly that of many other forum members.

If one takes a manifold and In general, manifold is any set that can be continuously parameterised. The number of independent parameters required to specify any point in the set uniquely is the dimension of the manifold, and the parameters themselves are the coordinates of the manifold.

so lets use a simple manifold a sheet of graph paper. Now on that graph paper place two points label one point P and the other Q. The actual label doesn't matter. Now measure the distance between P and Q.

Then roll up the paper into a cylinder. Now this is the tricky part. Following the surface of the paper the distance does not change between P and Q. The geometry is still Euclidean but now in cylindrical coordinates.

So now we introduce two terms " intrinsic geometry and extrinsic geometry. The above case the ds^2 (seperation distance ) is unchanged so it is invariant. The geometry intrinsically is identical to Euclidean flat. Aka the laws of physics is the same regardless of inertial reference frame. There is no intrinsic curvature in this case.

The curvature itself is the extrinsic geometry ( the cylinder viewed from the outside) in the first case think of an ant embedded on its surface.

The above is essential to understand symmetry relations in SR and GR unfortunately when you combine time dilation using the Interval (ct) and apply the Lorentz transformations the example above becomes more complex as the above is a 3 dimensional manifold while spacetime is described by 4 dimensional manifolds.

However there is no limit to the number of ubique parameters that can be used as unique coordinates on a manifold. Aka higher dimensions. ( a parameter can be any set) ie set representing time or charge or temperature etc etc.

The above is something you need the math skills to properly understand and the above is also needed to understand SR ( Minkowskii metric) GR via the field equations including its tensors and gauge gauge groups.

A common term for the above is local vs global geometry.

For others the above is an example of coordinate basis. However the parameters used and subsequent coordinates can be under others "basis". The above should also give a very strong hint of why covariant and contravariant vectors become useful on manifolds 4d and up. (Kronecker delta ) first case 4d needing ( Levi-Cevita ) the above is also useful with regards to Hilbert spaces aka QM.

The above is obviously a 2d manifold mathematical the extrinsic dimensions however requires the z axis to (curl) the 2d plane.

( curl equates to rotational symmetry)

Edited Sunday at 11:37 PM4 days by Mordred

There is an important, albeit not often mentioned, requirement for a coordinate system, without which one real parameter would've sufficed to specify any point in any dimensional manifold.

Edited Sunday at 11:26 PM4 days by Genady

15 minutes ago, Genady said:

There is an important, albeit not often mentioned, requirement for a coordinate system, without which one real parameter would've sufficed to specify any point in any dimensional manifold.

Good point ( pun intended)

On 1/25/2026 at 2:31 PM, Markus Hanke said:

  On 1/24/2026 at 7:59 PM, KJW said:

However, I've also read that the torsion tensor doesn't have a propagating gravitational field, though it is not clear to me precisely what this means.

My own guess - the field equations for torsion in ECT contain no derivatives, and at the same time torsion is completely determined by local matter fields. This implies that torsion vanishes in regions where T=0, and no wave-type equation exists for torsion to “radiate” through vacuum. So it can’t have any propagating degrees of freedom - it’s purely a local phenomenon subject to the local presence of matter.

It occurred to me that pure gravitational curvature would have to be free of the torsion tensor in Einstein-Cartan theory just as pure gravitational curvature is free of the Einstein tensor in general relativity. It's actually not clear to me what the equivalent of the Einstein equation is in Einstein-Cartan theory. However, if the gravitational curvature is sourced from the Einstein tensor, because the Einstein tensor contains the torsion tensor, the torsion tensor ought to act as a source for some of the gravitational curvature even if it is somewhat hidden (note that even the covariant differential operator contains the torsion tensor). One could move all instances of the torsion tensor (including its partial derivatives) over to the source term in the standard Einstein equation of general relativity (assuming integrability conditions are satisfied). I find it interesting that general relativity can deal with rotating objects whereas the torsion tensor seems to be required to deal with (quantum mechanical?) spin.

Note that I'm not considering the torsion tensor as a propagating field, in the same sense that the Einstein tensor appears not to be a propagating field.

Edited Tuesday at 01:47 AM3 days by KJW

Maybe I've got it wrong, but 'intrinsic', to me, means that the separation between two points on a manifold is based solely on the parameters of those points.
IOW, if two points are specified as ( k1, l1, m1 ) and ( k2, l2, m2 ), then the intrinsic separation is along the manifold defined by k, l, and m. The introduction of the parameter n, to the manifold, would imply you can have an extrinsic separation as there is now an 'embedding' dimension, n .

My apologies if I've used incorrect terminology.

Edited Tuesday at 03:09 AM3 days by MigL

1 hour ago, MigL said:

Maybe I've got it wrong, but 'intrinsic', to me, means that the separation between two points on a manifold is based solely on the parameters of those points.
IOW, if two points are specified as ( k1, l1, m1 ) and ( k2, l2, m2 ), then the intrinsic separation is along the manifold defined by k, l, and m. The introduction of the parameter n, to the manifold, would imply you can have an extrinsic separation as there is now an 'embedding' dimension, n .

My apologies if I've used incorrect terminology.

The intrinsic properties of a manifold depend entirely on the manifold itself without any reference to a higher-dimensional embedding manifold (a coordinate transformation is an embedding of a manifold into a manifold of the same dimension).

The distance between two points of a manifold depends on the path between the two points. That is, for the expression:

(ds)² = g_uv dx^u dx^v

ds is not an exact differential (there does not exist a function s(..., x^u, ...) for which ds is the differential). This btw is why there is no absolute time in relativity. If ds were an exact differential, then:

[math]ds = \dfrac{\partial s}{\partial x^u} dx^u[/math]

and therefore:

[math](ds)^2 = \dfrac{\partial s}{\partial x^u} \dfrac{\partial s}{\partial x^v} dx^u dx^v[/math]

[math]g_{uv} = \dfrac{\partial s}{\partial x^u} \dfrac{\partial s}{\partial x^v}[/math]

But the RHS of this expression, as a matrix, has zero determinant, contrary to the requirement that the metric tensor is invertible.

[If the above LaTeX doesn't render, please refresh browser.]

Edited Tuesday at 04:39 AM3 days by KJW

25 minutes ago, KJW said:

This btw is why there is no absolute time in relativity.

Great insight! Never thought about it from this particular angle, though in retrospect it seems obvious +1