# Paper: A causal mechanism for gravity

## Recommended Posts

Equation 34 gives an example of the gravitational action.

I don't believe the OP is ready for the Langrangian at this time. However the article here provides some details.

##### Share on other sites
On 5/23/2020 at 4:16 AM, Markus Hanke said:

I am honestly not sure if I follow your thought process correctly, since such a notion as “time dilation gradient” does not make much sense to me. But nonetheless, the aforementioned case of an orbit around a rotating mass should be an example. Another scenario that immediately comes to mind would be two parallel beams of light (or any other pp-wave spacetime, for that matter) - if you fire two parallel beams of light in the same direction, there will be no gravitational attraction between them, even though they carry energy. But if you fire the same two beams of light so that they are initially parallel, but travel in opposite directions (i.e. you let emitter and receiver trade places for one of the beams), then they will indeed experience a gravitational attraction.

I'm a visual thinker, so when I say "time dilation gradient" I'm describing a picture which may be using poorly chosen mathematical language. Philosophically, to me, a point mass cannot be affected by gravitation; only masses of non-zero volume can. The volume of a given mass can be viewed as a 3-dimensional time dilation field. In this model, the mass will gravitate in the direction of highest dilation. Two masses moving in the same direction would not detect or consider the other to be experiencing time dilation (beyond the negligible dilation caused by their respective masses). However, if they are moving very quickly in opposite directions, they each calculate (and observe) that the other is experiencing a large amount of time dilation, and there is in fact additional gravitational attraction between the masses.

Producing scenarios where dilation clearly exists but gravitational attraction does not will not suffice. In the dust cloud, and inside Newton's shell, all particles are in uniform time dilation, and therefore experience no gravity. The distant observer is obviously experiencing a different rate of time passage and will, in fact, be drawn toward the cloud or shell, subject to his distance from them. However, that attraction isn't technically due to the time dilation differential between the masses in the shell (for example) and his watch in this model, but rather between the areas of the field comprising his own volume facing the shell and the areas of the field of his own mass further away from the shell. The gradient of the volume.

On 5/23/2020 at 4:16 AM, Markus Hanke said:

However, if we allow the manifold to not be flat, then the situation changes; following the standard prescription for this (refer to any textbook on differential geometry), we must now replace ordinary with covariant derivatives, which do not in general commute. The degree to which they fail to commute is (I use double bars “||” to denote covariant derivatives):

Like I said, I'm a visual thinker. What's the "gradient" of the center of a saddle point? Well...it's zero, or it depends on the direction that you're wanting to move. I think this is saying the same thing that you're saying above. Your solution (and GR's solution) to this is to say that a rank-2 tensor is required for all points; my method for this would be to demand the volume in question be established a priori. Once you know the volume comprising a mass you can determine the direction of gravitational attraction.

On 5/23/2020 at 10:16 AM, studiot said:

I don't know if anyone has considered time dilation at lagrange points and lines where there are at least two such potentials.

Thank-you for the book reference, Studiot, that looks great. After talking things out, I'm starting to wonder if we're talking about 2 sides of the same coin. In order to determine the direction of gravitational attraction for a given point, we need a second point of reference (i.e. a direction) to determine the gradient. I think, mathematically, this is the same thing as demanding a rank-2 tensor a for point if we want to calculate gravity in arbitrary directions.

I really appreciate the back-and-forth on this.

##### Share on other sites
34 minutes ago, rjbeery said:

Thank-you for the book reference, Studiot, that looks great

If you are interested the extract was taken from Professor Frankel's book "The Geometry of Physics"  :  Cambridge University Press

This is a large encyclopedic tome that covers a huge amount of ground from the topology of Kirchoff's laws to the most modern versions of GR.

Back along, Proff Frankel also wrote a much smaller book "Gravitational Curvature an Introduction to Einstein's Theory"  Freeman Publishing.

Here he develops a lot of this, with and without tensors. He also notes and describes the several different fashions in the mathematical presentation of GR.

These may be of interest.

##### Share on other sites
20 hours ago, rjbeery said:

Philosophically, to me, a point mass cannot be affected by gravitation; only masses of non-zero volume can.

Gravitation in GR is geodesic deviation, and thus a geometric property of spacetime; all free-falling test particles experience gravity (they must follow geodesics in spacetime), regardless of whether they have mass or not, and regardless of their internal composition or size.
Remember also that within the Standard Model, all fundamental particles are point-like, i.e. any mass distribution is simply a collection of point particles.

20 hours ago, rjbeery said:

However, if they are moving very quickly in opposite directions, they each calculate (and observe) that the other is experiencing a large amount of time dilation, and there is in fact additional gravitational attraction between the masses.

Relative motion is not a source of gravity; the source term in the field equations is the stress-energy-momentum tensor, which, as being a tensor, is covariant under Lorentz transformations. If that were not so, the theory would not be internally self-consistent.

It is important to reiterate that there are two physically distinct types of time dilation - there is kinematic time dilation due to relative motion (which also happens in flat Minkowski spacetime), and there is gravitational time dilation due to curvature of spacetime (which only happens when gravitational sources are present). These two effects can be present simultaneously, but they are nonetheless physically distinct effects.

##### Share on other sites
23 hours ago, rjbeery said:

Philosophically, to me, a point mass cannot be affected by gravitation; only masses of non-zero volume can.

Are you arguing philosophy or physics? Last I checked, gravity was in the realm if science. IOW, what evidence do you have that e.g.an electron doesn’t experience gravity?

And if they don’t experience gravity, do they also not exert gravity?

##### Share on other sites
3 hours ago, Markus Hanke said:

Gravitation in GR is geodesic deviation, and thus a geometric property of spacetime; all free-falling test particles experience gravity (they must follow geodesics in spacetime), regardless of whether they have mass or not, and regardless of their internal composition or size.
Remember also that within the Standard Model, all fundamental particles are point-like, i.e. any mass distribution is simply a collection of point particles.

Yeah my "point particle" objection is philosophical. To me, it's obvious that a point particle is unphysical but that doesn't mean it isn't useful in modeling.

3 hours ago, Markus Hanke said:

Relative motion is not a source of gravity; the source term in the field equations is the stress-energy-momentum tensor, which, as being a tensor, is covariant under Lorentz transformations. If that were not so, the theory would not be internally self-consistent.

This isn't the first time, or even the first forum, where this has been discussed. The covariance of the tensor has also been mentioned in the past, and I understand the thought process -- "if the tensor is zero in one frame then it's zero in all frames." If objects A and B are in relative motion, where does the energy "between" A and B exist? I can't answer this. This enters the realm of philosophy, because we suspect that it exists...but neither at A nor B.

Why would angular momentum cause additional gravity but not linear momentum? Why would photons be affected by other photons traveling in opposite directions...but the same isn't true for massive particles? Those were my initial thoughts, but the fact is that GR does claim that kinetic energy increases gravitational attraction. Not easy to find discussion on the topic, but it does exist.

Quote

According to the general theory of relativity, kinetic energy contributes to gravitational mass. Surprisingly, the observational evidence for this prediction does not seem to be discussed in the literature.

Maybe it would be constructive to explicitly identify topics. I think I've fielded a few objections, and the latest one is whether or not gravitational attraction and time dilation are dependent variables. I don't believe we've produced a counter-example, yet.

19 minutes ago, swansont said:

Are you arguing philosophy or physics? Last I checked, gravity was in the realm if science. IOW, what evidence do you have that e.g.an electron doesn’t experience gravity?

I didn't know we were arguing anything. If my model can't withstand scrutiny then I'll adjust or abandon it. And I do believe that an electron experiences gravity, I just don't believe that it is a point particle (i.e. "EM mass") -- that fact is discussed in the OP, with references, if you're truly interested.

##### Share on other sites
2 minutes ago, rjbeery said:

Yeah my "point particle" objection is philosophical. To me, it's obvious that a point particle is unphysical but that doesn't mean it isn't useful in modeling.

Doesn’t matter if it’s obvious to you. Theory says the electron is a point particle and experiment backs this up.

2 minutes ago, rjbeery said:

Why would angular momentum cause additional gravity but not linear momentum?

Can you transform into a frame where angular momentum goes to zero, the way you can with linear momentum?

##### Share on other sites
38 minutes ago, rjbeery said:

To me, it's obvious that a point particle is unphysical

Science doesn't care about whether things are "obvious" or not. One of the main reasons for the scientific method is to avoid such erroneous "common sense".

39 minutes ago, rjbeery said:

If my model can't withstand scrutiny then I'll adjust or abandon it.

Obviously not true. You have had the same flaws pointed out in great detail by a large number of people over very many years. Yet you cling stubbornly to your baseless "theory".

##### Share on other sites
5 hours ago, swansont said:

Can you transform into a frame where angular momentum goes to zero, the way you can with linear momentum?

This is now a moot point. GR predicts that linear velocity is in fact a source of gravity.

##### Share on other sites
25 minutes ago, rjbeery said:

This is now a moot point. GR predicts that linear velocity is in fact a source of gravity.

Since when provide a citation from. A peer reviewed source.

##### Share on other sites
14 hours ago, rjbeery said:

If objects A and B are in relative motion, where does the energy "between" A and B exist?

Kinetic energy is an observer-dependent quantity, so it is best understood as a relationship between the two reference frames in spacetime. It is in itself not a source of gravity.

14 hours ago, rjbeery said:

Why would angular momentum cause additional gravity but not linear momentum?

Neither one of these are in themselves sources of gravity. What enters into the field equations as part of the energy-momentum tensor are momentum density and momentum flux. These are neither linear nor angular (the distinction is just a convention anyway). If there is any kind of momentum present in a gravitational source, then it will contribute to one or both of the aforementioned quantities, but the way it does so is not always trivial; in fact, finding the energy-momentum tensor for a given distribution of matter-energy can be a very difficult task, particularly if the distribution is not static or stationary. If the kinetic energy is evenly (statistically) spread out over the entire distribution, then you can sometimes simplify things by letting it enter as a contribution to another component of the tensor, the energy density.

14 hours ago, rjbeery said:

Those were my initial thoughts, but the fact is that GR does claim that kinetic energy increases gravitational attraction. Not easy to find discussion on the topic, but it does exist.

This is just the last case I mentioned above - refer to equation (16) in that paper. The kinetic energy becomes a contribution to the energy density term of the tensor. Physically this means you are describing a different system (one that has a higher temperature as compared to a reference system), not the same system in motion.

14 hours ago, rjbeery said:

I don't believe we've produced a counter-example, yet.

That’s because you haven’t produced a model yet, you have thus far only described an idea you have had, and how you yourself understand that idea. The next step from here would be for you to actually write down a model - i.e. a field equation for the time dilation field you are proposing -, and then see what kind of predictions that model yields, and how they compare against experiment and observation.
Remember, it is always good to have ideas, but it is for yourself to investigate the scientific value of that idea - you can’t just assume your idea is “right”, and then ask for others to show you wrong.

14 hours ago, rjbeery said:

If my model can't withstand scrutiny then I'll adjust or abandon it.

Yes, that is the right approach

##### Share on other sites
4 hours ago, Markus Hanke said:

If there is any kind of momentum present in a gravitational source, then it will contribute to one or both of the aforementioned quantities, but the way it does so is not always trivial; in fact, finding the energy-momentum tensor for a given distribution of matter-energy can be a very difficult task, particularly if the distribution is not static or stationary.

It's confusing to me that you claim momentum isn't a source of gravity, but that it "contributes to energy flux" which is a source of gravity. If massive objects, A and B, are moving in opposite directions at a substantial speed and pass near each other, will the gravitational attraction between them be more than what is calculated by their rest masses?

##### Share on other sites

For the specific system where the kinetic energy is evenly 'spread out' over the system, then there is a contribution to the energy density.
It contributes to ED because it is describing a system that has a higher temperature than the reference system.
Not motion of the system.

Seemed simple enough to follow...

##### Share on other sites

I think part of your confusion lies in that the Einstein field equations including the stress energy momentum tensor.

Don't just describe spacetime curvature (gravity) but also describes how particles move in spacetime. The trick is the metric and Ricci tensor can both modify the stress energy tensor and vise versa.

Remember the expression I gave mass tells spacetime how to curve spacetime tells matter/ particles how to move. The Einstein field equations cover both statements.

##### Share on other sites
40 minutes ago, MigL said:

Seemed simple enough to follow...

If massive objects, A and B, are moving in opposite directions at a substantial speed and pass near each other, will the gravitational attraction between them be more than what is calculated by their rest masses?

##### Share on other sites
Posted (edited)
38 minutes ago, Mordred said:

I think part of your confusion lies in that the Einstein field equations including the stress energy momentum tensor.

Don't just describe spacetime curvature (gravity) but also describes how particles move in spacetime. The trick is the metric and Ricci tensor can both modify the stress energy tensor and vise versa.

Remember the expression I gave mass tells spacetime how to curve spacetime tells matter/ particles how to move. The Einstein field equations cover both statements.

The electromagnetic field with its mass and momentum creates gravity, which changes the direction of movement of electromagnetic waves.

In this case, gravity is an additional source of gravity or anti-gravity. I'm leaning toward the latter.

Edited by SergUpstart

##### Share on other sites
20 minutes ago, rjbeery said:

If you consider the TWO masses as parts of the system, then the energy of the system is increased, and so the contribution to the energy density of the system is increased.
The system, as a whole, will have a larger gravitational potential, but I don't think the 'attraction' between the individual members of the system will change.
( I could be wrong, but if Markus says the calculations are not trivial, don't expect me to do them )

25 minutes ago, SergUpstart said:

In this case, gravity is an additional source of gravity or anti-gravity.

Gravity is an additional source of gravity, simply because gravity is self-coupling, due to the non-linearity of the EFEs.
IOW gravity gravitates. BUT there is no such thing asanti-gravity.

##### Share on other sites
6 minutes ago, MigL said:

Gravity is an additional source of gravity, simply because gravity is self-coupling, due to the non-linearity of the EFEs.
IOW gravity gravitates. BUT there is no such thing asanti-gravity.

Let's write Newton's law of universal gravitation in the differential form

div g = - G*ro

The divergence Operator shows where the field has a drain and where it has a source. If the divergence is positive at a given point in space, it means that the source of the field is there, and if it is negative, then the flow of the field is there.From the above equation, it follows that the gravitational field has only drains. Where are the origins? in infinity? But then gravity must propagate at an infinite speed, which is not true. So where is the origin of the gravitational field?

##### Share on other sites
Posted (edited)
2 hours ago, SergUpstart said:

Let's write Newton's law of universal gravitation in the differential form

div g = - G*ro

The divergence Operator shows where the field has a drain and where it has a source. If the divergence is positive at a given point in space, it means that the source of the field is there, and if it is negative, then the flow of the field is there.From the above equation, it follows that the gravitational field has only drains. Where are the origins? in infinity? But then gravity must propagate at an infinite speed, which is not true. So where is the origin of the gravitational field?

The Div operator is a vector. If you have a uniform matter field with no anistropies then you have no curvature and no gravity.

Strange as this may be to understand but the term gravity is rather misleading.

Marcus mentioned tidal forces. So let's add some detail.

Take that uniform distribution. Now in that uniform distribution place two massive particles in free fall with an initial velocity or in this case momentum can be used ( momentum is the rest mass multiplied by the velocity) the two terms are not equal.

The paths of those two particles do not change nor do they accelerate due to gravity. The field is uniform in distribution.

Now take a region with spacetime curvature such as a planet.

Drop the same two objects. The paths will converge toward the center of mass.

So to understand the origin of gravity one can only answer the local anistropy regions of the mass distribution. The very term gravity is replaced by spacetime curvature.

With EM the potential difference (voltage) induces current flow. The resistance restricts the current.

Mass is resistance to inertia change or resistance to acceleration. So you can see the similarity. Photons do not couple with any field they interact with so have no invariant (rest mass). Though they do not couple they still interact with other particles in a region. We describe this interaction via the path of least action which a good study source would be Feymann path integrals. (The Feymann path integrals are also curved paths). Though gravity is not involved in the latter case.

Another way to look at a uniform field of mass. (Which can be gained through all other fields in a multi particle system)

Is take a multiparticle system uniformly distributed. You can arbitrarily choose any test particle in that field a the centre of mass then apply the shell theorem. However as any particle can be chosen with no difference you effectively have a scalar field. Gravity is at minimal treatment a vector field. (Attraction)

Edited by Mordred

##### Share on other sites
2 hours ago, MigL said:

If you consider the TWO masses as parts of the system, then the energy of the system is increased, and so the contribution to the energy density of the system is increased.
The system, as a whole, will have a larger gravitational potential, but I don't think the 'attraction' between the individual members of the system will change.
( I could be wrong, but if Markus says the calculations are not trivial, don't expect me to do them )

Yeah I'm not asking for calculations, just a qualitative description. It's a pretty simple question: does GR predict that momentum energy of a mass increases gravitational attraction towards that mass?

##### Share on other sites
Posted (edited)
1 hour ago, Mordred said:

The Div operator is a vector.

You probably just forgot to write the word operator after the word vector.

In vector calculus, divergence is a vector operator that operates on a vector field, producing a scalar field giving the quantity of the vector field's source at each point. More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point.

From the point of view of physics (and in a strict sense and in the sense of intuitive physical image of a mathematical operation) the divergence of a vector field is a measure of the extent to which a given point of space (or rather a sufficiently small neighborhood of a point) is a source or a drain of this field:

div F>0 — point field is the source;
div F<0 — the field point is a drain;
div F=0-there are no drains and sources, or they compensate for each other

It is  quote from the Russian-language version of Wikipedia, the English-language version does not have this

Edited by SergUpstart

##### Share on other sites
Posted (edited)

Div operator is a short hand but you got the point. +1

An off topic side note the mass distribution is also how a matter only universe can expand. Which is a very tricky concept to understand. One would think a matter only universe would collapse. If you think about my last post one can see that as anistropies develop ie LSS and galaxy formation the density of the void regions decrease. In essence local gravitational anistropies aid expansion.

However that's off topic...and involves the term and formula for critical density...

Edited by Mordred

##### Share on other sites
Posted (edited)
23 hours ago, rjbeery said:

It's confusing to me that you claim momentum isn't a source of gravity, but that it "contributes to energy flux" which is a source of gravity.

It’s momentum flux, not energy flux.

19 hours ago, rjbeery said:

If massive objects, A and B, are moving in opposite directions at a substantial speed and pass near each other, will the gravitational attraction between them be more than what is calculated by their rest masses?

What you are describing here is a relativistic 2-body problem, for which there is no closed analytical solution to the field equations; you can only treat this case via numerical methods. I don’t know what exactly happens here in terms of GR; I have never done this simulation myself.

However, if we slightly change the scenario, then I can give you a definitive answer: let’s say there is only one (spherically symmetric) gravitating body plus an observer whose own gravitational influence is negligible. Spacetime around this mass is simply the Schwarzschild metric. If we now introduce relativistic motion (i.e. mass and observer move at nearly the speed of light with respect to one another), how will that change the gravity exerted by the mass? The appropriate solution to the Einstein equations for this case is called the Aichlburg-Sexl Ultraboost - at first glance this metric looks very different from the Schwarzschild metric, however, closer inspection reveals that these two metrics are actually just diffeomorphisms of each other. In other words, we are dealing with the same physical spacetime, it’s just that events in it are labelled differently. All curvature invariants are the same (this can be explicitly shown, though it is tedious) between these two solutions. Thus, relative motion does not increase gravity; you are still in the same spacetime with the same geometry, it is just “seen” differently (roughly analogous to how different inertial frames in SR are related by a simple rotation of the coordinate system about some hyperbolic angle in spacetime). If this weren’t so, you could construct unresolvable paradoxes just by introducing relative motion, and the model would not be internally self-consistent.

I should also remind you that, if we are looking at the vacuum outside the mass, the energy-momentum tensor is always zero there. It is only non-vanishing in the interior of the mass distribution. Therefore, whether there is relative motion or not, you are actually solving the same equation: $$R_{\mu \nu}=0$$; the only thing that changes are initial and boundary conditions.

Edited by Markus Hanke

##### Share on other sites
1 hour ago, Markus Hanke said:

What you are describing here is a relativistic 2-body problem, for which there is no closed analytical solution to the field equations; you can only treat this case via numerical methods.

Honestly, that's pretty surprising -- two massive bodies passing each other at great speeds seems to be a pretty fundamental physics problem.

Let me ask you a question, Markus. Do you believe that time dilation and the so-called "gravitational attraction" (i.e. spacetime curvature) are independent variables in GR?

##### Share on other sites
1 hour ago, rjbeery said:

Honestly, that's pretty surprising -- two massive bodies passing each other at great speeds seems to be a pretty fundamental physics problem.

Fundamental doesn't mean easily solvable.

1 hour ago, rjbeery said:

Let me ask you a question, Markus. Do you believe that time dilation and the so-called "gravitational attraction" (i.e. spacetime curvature) are independent variables in GR?

It depends how you define independent. And, perhaps, how you define "gravitational attraction" as you are equating it to spacetime curvature. They are clearly related because they have the same cause.

But the usual meaning of gravitational attraction is the "force of gravity" (the acceleration due to gravity, g here on Earth). In which case, there is direct relationship between the two. You can have two places with the same value of g but different time dilation. (Or the same time dilation but very different values of g.)

## Create an account

Register a new account