What is gravity?

[Genady]

12 hours ago, KJW said:

In another thread, I showed that the flat-space Friedmann–Lemaître–Robertson–Walker (FLRW) metric can be coordinate-transformed to a manifestly conformally flat metric. This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar).

 

 

This is true for homogeneously and isotropically distributed sources. What about other "smeared" sources? The Weyl tensor is not necessarily zero, but the curvature is not away from the source.

[DanMP]

22 hours ago, Genady said:

On 3/7/2024 at 4:46 PM, DanMP said:

You would stil "blame" the affected geometry or some other geometry? Why?

Because Earth or other body you mention are far away and your ship cannot be affected directly by them without "an action at a distance." What it can be affected by directly is the geometry of spacetime

Who said that my ship has to be directly affected by the incoming mass? 

 I wrote:

On 3/6/2024 at 4:35 PM, DanMP said:

3. or because the presence of Earth (more exactly its mass) is changing "the curvature of spacetime" in my proximity? Another question is how exactly the Earth mass does that?

 

22 hours ago, Genady said:

I still don't know what you mean in that question. Does what?

The question was, see above, how the mass can change "the curvature of spacetime" in my proximity?

Edited March 8 by DanMP

[Genady]

16 minutes ago, DanMP said:

Who said that my ship has to be directly affected by the incoming mass? 

The physics is clear then.

[DanMP]

2 minutes ago, Genady said:

The physics is clear then.

Maybe, but your answer is not.

The strike line was unintentional. I try to correct it.

[Genady]

1 minute ago, DanMP said:

Maybe, but your answer is not.

The strike line was unintentional. I try to correct it.

I don't know what is unclear. Clarify.

Restate the unintentionally struck question, please.

[DanMP]

30 minutes ago, Genady said:

Restate the unintentionally struck question, please.

I fixed it:

56 minutes ago, DanMP said:

22 hours ago, Genady said:

I still don't know what you mean in that question. Does what?

The question was, see above, how the mass can change "the curvature of spacetime" in my proximity?

Can you answer this question please?

[Genady]

3 minutes ago, DanMP said:

I fixed it:

Can you answer this question please?

The spacetime geometry obeys certain differential equations. The solution of these equations depends on initial and boundary conditions. The mentioned mass is a part of these conditions.

[DanMP]

6 minutes ago, Genady said:

The spacetime geometry obeys certain differential equations

You are talking about the theoretical/abstract model. I am asking about how the mass actually does that.

[Genady]

2 minutes ago, DanMP said:

actually

What does it mean?

The model tells us how it happens. Some future model might derive this model, i.e., the equations, from more generic principles.

The previous model, aka Newtonian gravity, included a notion of mass acting on the ship. This is an approximation, which is not included in the current best model.

[KJW]

You might be wondering where the formula [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] comes from. I've already provided the clue that it comes from the Killing equation. Here's the derivation:

Let [math]T u^\alpha = T \dfrac{dx^\alpha}{ds}[/math] be a timelike Killing vector field, where [math]u^\alpha = \dfrac{dx^\alpha}{ds}[/math] is a unit tangent vector field of timelike trajectories of stationary observers or other objects in stationary spacetime.

Note that [math]g_{\alpha\beta}\ u^\alpha u^\beta = 1[/math]

The Killing equation: [math]\nabla_\mu (Tu_\nu) + \nabla_\nu (Tu_\mu) = 0 \ \ \ \ \ \ \ \ \ \ \text{(}\nabla_\mu \text{ is the covariant differential operator)}[/math]

[math]\nabla_\mu T\ u_\nu + T\ \nabla_\mu u_\nu + \nabla_\nu T\ u_\mu + T\ \nabla_\nu u_\mu = 0[/math]

[math]u^\nu (\nabla_\mu T\ u_\nu + T\ \nabla_\mu u_\nu + \nabla_\nu T\ u_\mu + T\ \nabla_\nu u_\mu) = 0[/math]

[math]\nabla_\mu T\ u^\nu u_\nu + T\ u^\nu \nabla_\mu u_\nu + u^\nu \nabla_\nu T\ u_\mu + T\ u^\nu \nabla_\nu u_\mu = 0[/math]

Considering the individual terms:

[math]\nabla_\mu T\ u^\nu u_\nu = \nabla_\mu T = \dfrac{\partial T}{\partial x^\mu}[/math]

[math]T\ u^\nu \nabla_\mu u_\nu = T\ g_{\nu\sigma} u^\nu \nabla_\mu u^\sigma = T\ \nabla_\mu (g_{\nu\sigma}\ u^\nu u^\sigma) - T\ g_{\nu\sigma} u^\sigma \nabla_\mu u^\nu = -T\ g_{\nu\sigma} u^\nu \nabla_\mu u^\sigma = 0[/math]

[math]u^\nu \nabla_\nu T\ u_\mu = \dfrac{\partial T}{\partial s} u_\mu = 0 \ \ \ \ \ \left(\dfrac{\partial T}{\partial s} = 0 \ \ \text{ by definition} \right)[/math]

[math]T\ u^\nu \nabla_\nu u_\mu = T\ b_\mu[/math]

Therefore:

[math]\dfrac{\partial T}{\partial x^\mu} + T\ b_\mu = 0[/math]

[math]b_\mu = -\dfrac{1}{T} \dfrac{\partial T}{\partial x^\mu}[/math]

(Note that [math]a^\mu = c^2 g^{\mu\nu}\ b_\nu[/math])

[math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math]

 

 

Edited March 8 by KJW

[KJW]

13 hours ago, Markus Hanke said:

Quote

This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar).

Another way to say this is that, in this kind of spacetime and under geodesic motion, shapes (ie angles) are preserved, whereas volumes and surfaces are not.

You and I discussed this in the thread about the cosmological expansion of time as well as space. By coordinate-transforming to the manifestly conformally flat metric, it becomes easier to derive cosmological redshifts due to the simplified geodesic equation for light as a result of the preservation of angles (speeds in spacetime).

 

 

7 hours ago, Genady said:

Quote

In another thread, I showed that the flat-space Friedmann–Lemaître–Robertson–Walker (FLRW) metric can be coordinate-transformed to a manifestly conformally flat metric. This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar).

This is true for homogeneously and isotropically distributed sources. What about other "smeared" sources? The Weyl tensor is not necessarily zero, but the curvature is not away from the source.

A non-zero Weyl tensor can coexist at the same location as a non-zero Einstein tensor. But the energy-momentum tensor corresponds only to the Einstein tensor. I felt the need to distinguish between the curvature that is generated by an energy-momentum source and propagated to remote locations, and the curvature that corresponds to the energy-momentum itself.

It's worth mentioning that inside a spherically symmetric shell of matter, the spacetime is flat. That is, while gravitational curvature propagates from the shell of matter to outside of the shell, it does not propagate from the shell of matter to inside of the shell.

 

 

[KJW]

3 hours ago, DanMP said:

Quote

The spacetime geometry obeys certain differential equations

You are talking about the theoretical/abstract model. I am asking about how the mass actually does that.

There is nothing "abstract" about spacetime geometry. Gravitational time dilation is physically real as is the curvature it implies. It seems to me that many people believe that general relativity is just a theoretical model that describes physics in only an abstract manner typical of theoretical models. But no, geometry is very much a physically concrete notion and the formulae from Ricci calculus very much describes this physically concrete notion. The difficulty arises because the spacetime curvature that describes human-scale physics is quite miniscule, so it becomes very difficult to observe and measure the physically real curvature of spacetime.

There are two possibilities of how mass causes spacetime curvature: 1, mass doesn't cause spacetime curvature, but that mass is how we physically interpret the spacetime curvature that exists fundamentally; 2, mass causes spacetime curvature through a mechanism associated with something like quantum physics, the fundamental forces of nature, or perhaps something entirely different.

 

 

[Markus Hanke]

11 hours ago, KJW said:

You and I discussed this in the thread about the cosmological expansion of time as well as space.

Indeed. Just wanted to mention this again quickly in case other readers find it helpful to have some geometric intuition what the various aspects of curvature - Riemann, Weyl, Ricci - actually mean.

[DanMP]

17 hours ago, Genady said:

The model tells us how it happens. Some future model might derive this model, i.e., the equations, from more generic principles.

OK, thank you for all your answers and for your patience. 

 

13 hours ago, KJW said:

There are two possibilities of how mass causes spacetime curvature: 1, mass doesn't cause spacetime curvature, but that mass is how we physically interpret the spacetime curvature that exists fundamentally; 2, mass causes spacetime curvature through a mechanism associated with something like quantum physics, the fundamental forces of nature, or perhaps something entirely different.

Thank you! I prefer the second one. 

[Eise]

On 3/8/2024 at 8:53 PM, KJW said:

There are two possibilities of how mass causes spacetime curvature: 1, mass doesn't cause spacetime curvature, but that mass is how we physically interpret the spacetime curvature that exists fundamentally; 2, mass causes spacetime curvature through a mechanism associated with something like quantum physics, the fundamental forces of nature, or perhaps something entirely different.

Hmm. I don't think naming it a cause, and even a mechanism, is a good way expressing the relationship between mass and spacetime curvature. In my opinion that would mean that physics would be able to describe the mechanism, and that implies new laws of nature. I think that we recognise mass because of its curvature (or inertia).

Nobody asks for how a charge 'causes' an electrical field. So why should one do it for a gravitational field, even if we now know that this field is a geometrical curvature of spacetime?

So if these are the only two possibilities, I opt for option 1.

[swansont]

1 hour ago, Eise said:

Nobody asks for how a charge 'causes' an electrical field. So why should one do it for a gravitational field, even if we now know that this field is a geometrical curvature of spacetime?

how it causes it is a different issue than recognizing that it does happen. 

[DanMP]

2 hours ago, Eise said:

Nobody asks for how a charge 'causes' an electrical field.

I would ask. There is no need to, but I would like to know. And this is the beauty of physics: there still are questions to ask and things to investigate/discover.

2 hours ago, Eise said:

In my opinion that would mean that physics would be able to describe the mechanism, and that implies new laws of nature

and/or new theories ...

Since we know there is a dark matter, we should be opened to such possibilities.

[Moontanman]

7 minutes ago, DanMP said:

I would ask. There is no need to, but I would like to know. And this is the beauty of physics: there still are questions to ask and things to investigate/discover.

I know I am ignorant on this topic but I would be surprised to find out, that if no one knows how a charge creates an electrical field, that no one is looking into the problem.  

[Bufofrog]

22 minutes ago, Moontanman said:

that no one is looking into the problem.

There is no problem.  Why charge creates a field is not a physics problem.

[Moontanman]

3 minutes ago, Bufofrog said:

There is no problem.  Why charge creates a field is not a physics problem.

The question was how not why. 

[Bufofrog]

19 minutes ago, Moontanman said:

The question was how not why.

Still not a problem.  I don't think there's an answer to that question.

It is the same as asking how does a positive charge attract a negative charge.  That type of question is more of a why than a how IMO.

Edited March 12 by Bufofrog

[Moontanman]

10 minutes ago, Bufofrog said:

Still not a problem.  I don't think there's an answer to that question.

It is the same as asking how does a positive charge attract a negative charge.  That type of question is more of a why than a how IMO.

The difference between how and why is how our technological civilization was formed, how is always important, why is often nothing but a subjective desire.  

[Bufofrog]

6 minutes ago, Moontanman said:

The difference between how and why is how our technological civilization was formed, how is always important, why is often nothing but a subjective desire.

OK.

[Genady]

1 hour ago, DanMP said:

I would ask.

Does the QFT provide a sufficient answer?

 

1 hour ago, Moontanman said:

how a charge creates an electrical field

It does not. Not in QED, the best current theory.

[Bufofrog]

1 hour ago, Genady said:

It does not. Not in QED, the best current theory.

Of course... [Slaps forehead]