Mass and curvature (split from Question about matter and space-time)

[Dejan1981]

Why do scientist always say that mass bends space, and not that mass is product of bending of space? To me it makes much more sense that mass and energy is caused by curvaton of space time and not the other way around. They are simultaneous phenomena's and its is impossible to say what was first.

Maybe the mass is manifestation of space curvaton over some level and energy is manifestation of space under some level.

I think that evidence of this hypothesis can be observed in experiment where same amount of mater and antimatter collide. If we presume that mater is a manifestation of space curvaton in one direction and antimatter manifestation of space curvaton in opposite direction than space would act as a two opposite waves colliding. Result would be irregular wave form that would produce mass particles (over some level of space curation) and energy (under some level of space curvaton). This experiment can actually be done.

[Genady]

8 minutes ago, Dejan1981 said:

mass and energy is caused by curvaton of space time

Where are these mass and energy when the spacetime is curved in vacuum?

[Dejan1981]

It doesn't meter where they are. Idea is that we should change the point of view and start from an idea that mater and energy are manifestation of curvaton of space and not that curvaton of space is caused by mater. 

Relationship between mass-energy and spacetime curvature is expressed by the Einstein Field Equations and as every other Equation it works in both ways and it doesn't say what was first mass and energy or the curvaton of space.

 

[Genady]

5 minutes ago, Dejan1981 said:

It doesn't meter where they are.

It was a rhetorical question. The point of that question was that spacetime is often curved in vacuum without mass or energy being anywhere in the vicinity.

[swansont]

42 minutes ago, Dejan1981 said:

It doesn't meter where they are.

Yes, it does. The gravity (i.e. curvature) in some area depends on the mass nearby, but it doesn’t tell you the distribution of that mass. It could be a planet of some size, or a smaller planet with a higher density.

Further, if that mass is rotating, that has an effect. 

[MigL]

2 hours ago, Genady said:

spacetime is often curved in vacuum without mass or energy

Curved space-time has energy.
Gravitational energy gravitates.

Not that I agree with any of the following ...

3 hours ago, Dejan1981 said:

I think that evidence of this hypothesis can be observed in experiment where same amount of mater and antimatter collide. If we presume that mater is a manifestation of space curvaton in one direction and antimatter manifestation of space curvaton in opposite direction than space would act as a two opposite waves colliding. Result would be irregular wave form that would produce mass particles (over some level of space curation) and energy (under some level of space curvaton).

 

[swansont]

4 hours ago, Dejan1981 said:

I think that evidence of this hypothesis can be observed in experiment where same amount of mater and antimatter collide. If we presume that mater is a manifestation of space curvaton in one direction and antimatter manifestation of space curvaton in opposite direction than space would act as a two opposite waves colliding. Result would be irregular wave form that would produce mass particles (over some level of space curation) and energy (under some level of space curvaton). This experiment can actually be done

A related experiment has been done. Antimatter is attracted, not repulsed, by gravity from normal matter.

“antihydrogen atoms, released from magnetic confinement in the ALPHA-g apparatus, behave in a way consistent with gravitational attraction to the Earth. Repulsive ‘antigravity’ is ruled out in this case.”

https://www.nature.com/articles/s41586-023-06527-1

You aren’t going to be able to measure gravitational effects between tiny amounts of matter and antimatter (i.e. particles or atoms)

[Sensei]

6 hours ago, Dejan1981 said:

I think that evidence of this hypothesis can be observed in experiment where same amount of mater and antimatter collide. If we presume that mater is a manifestation of space curvaton in one direction and antimatter manifestation of space curvaton in opposite direction than space would act as a two opposite waves colliding. Result would be irregular wave form that would produce mass particles (over some level of space curation) and energy (under some level of space curvaton). This experiment can actually be done.

When an electron annihilates with a positron, the same amount of energy is before and after the annihilation:

[math]2 m_e c^2=2 h f_c[/math]

During pair-production a particle with a kinetic energy exceeding [math]2 h f_c[/math] (1.022 MeV) is needed to create an electron-positron pair.

https://en.wikipedia.org/wiki/Pair_production

 

f_c - Compton frequency

 

m_e - rest mass of the electron (and positron)

 

Your body emits antimatter all the time. When unstable isotopes of carbon C-14 and potassium P-40 decay, antineutrinos and positrons are produced.

[Markus Hanke]

14 hours ago, Dejan1981 said:

It doesn't meter where they are. Idea is that we should change the point of view and start from an idea that mater and energy are manifestation of curvaton of space and not that curvaton of space is caused by mater.

The source term in the field equations is neither mass (which doesn’t appear at all) nor energy (which isn’t a covariant quantity), but the energy-momentum tensor. This tensor describes the local densities of energy, momentum, stresses and strains, but does not give a unique breakdown or distribution into matter and energy. 

Yes, you can run the field equations backwards and start with a given metric - but that does not give you a unique distribution of matter and energy, just as having a particular energy-momentum tensor does not give a unique metric. In both cases you also need initial and boundary conditions that you must impose.

Physically that just means that the same spacetime geometry can result from entirely different distributions of matter/energy. Also, as has been pointed out, curvature actually requires no local sources - for example, Schwarzschild spacetime is completely empty (T=0 everywhere), but still not Minkowski.

[Dejan1981]

3 hours ago, Markus Hanke said:

The source term in the field equations is neither mass (which doesn’t appear at all) nor energy (which isn’t a covariant quantity), but the energy-momentum tensor. This tensor describes the local densities of energy, momentum, stresses and strains, but does not give a unique breakdown or distribution into matter and energy. 

Yes, you can run the field equations backwards and start with a given metric - but that does not give you a unique distribution of matter and energy, just as having a particular energy-momentum tensor does not give a unique metric. In both cases you also need initial and boundary conditions that you must impose.

Physically that just means that the same spacetime geometry can result from entirely different distributions of matter/energy. Also, as has been pointed out, curvature actually requires no local sources - for example, Schwarzschild spacetime is completely empty (T=0 everywhere), but still not Minkowski.

Gμν (Einstein Tensor): we usually say that it describes how spacetime is curved due to the presence of mass and energy, so its not correct to say that mass and energy are not part of field equations. I think that description how presence of mass and energy is result of spacetime curvature is correct also if we describe this variable. The second explanation is the proper way how I think we should describe this.

Humans first knew about mass and matter (result of how we sense the world) and only much later about curvature of space, so I think this chronology is reason why we presumed that curvature of space is result of existence of mater and not the other way around.

Schwarzschild spacetime is theoretical and it exists on the assumption that the electric charge of the mass, angular momentum of the mass, and universal cosmological constant are all zero. Universe that we live in is not like this. Cosmological constant is not zero for sure because Universe is not static.

[Dejan1981]

13 hours ago, Sensei said:

When an electron annihilates with a positron, the same amount of energy is before and after the annihilation:

2mec2=2hfc

During pair-production a particle with a kinetic energy exceeding 2hfc (1.022 MeV) is needed to create an electron-positron pair.

https://en.wikipedia.org/wiki/Pair_production

 

f_c - Compton frequency

 

m_e - rest mass of the electron (and positron)

 

Your body emits antimatter all the time. When unstable isotopes of carbon C-14 and potassium P-40 decay, antineutrinos and positrons are produced.

Amount of energy is always the same in the system. I didn't question this.

Idea is that when same amount of mater and antimatter collide, and if we hypothesize that mater is product of curvature of space in one direction, and antimatter in opposite, that the space is going to behave as a wave and result will be irregular wave form that will manifest as energy up to the one level of space curvature and as a mater and antimatter over the same level of space curvature.

If mater and antimatter collided in time = 0 and there centers of gravity came to the same place in the space in time = 0 than resulting wave would be zero and energy and mater, antimatter and energy would stop to exist because space would be "flat". This is only mathematical and can never happen in reality. 

[Markus Hanke]

3 hours ago, Dejan1981 said:

Gμν (Einstein Tensor): we usually say that it describes how spacetime is curved due to the presence of mass and energy,

The Einstein tensor describes a certain aspect of overall spacetime curvature, in the sense that it determines the values of a certain combination of components of the Riemann tensor (it fixes 10 out of its 20 independent components). But unlike the Riemann tensor, it is not a complete description of the geometry of spacetime. Thus, the Einstein equations provide only a local constraint on the metric, but don’t determine it uniquely.

3 hours ago, Dejan1981 said:

so its not correct to say that mass and energy are not part of field equations

Physically speaking, mass and energy have of course a gravitational effect, but neither appear as quantities in the source term of the field equations - there’s only the energy-momentum tensor field \(T_{\mu \nu}\). 

3 hours ago, Dejan1981 said:

I think that description how presence of mass and energy is result of spacetime curvature is correct also if we describe this variable.

Here’s the thing with this - the Einstein equations are a purely local statement. So for example, if you wish to know the geometry of spacetime in vacuum outside some central body, the equation you are in fact solving is the vacuum equation

\[G_{\mu \nu}=0\]

which implies 

\[R_{\mu \nu}=0\]

There is in the first instance no reference here to any source term, not even the energy-momentum tensor, because you are locally in a vacuum. During the process of solving these equations, you have to impose boundary conditions, one of which will be that sufficiently far from the central body the gravitational field asymptotically becomes Newtonian; it’s only through that boundary condition that mass makes an appearance at all.

Only in the interior of your central body do you solve the full equations

\[G_{\mu \nu}=\kappa T_{\mu \nu}\]

wherein the energy-momentum tensor describes the overall distribution of energy density, momentum density, stresses, strains, and shear (note that “mass” is again not part of this). By solving this equation along with boundary conditions you can find the metric.

You can work this backwards - you can start with a metric, and calculate the energy-momentum tensor. But here’s the thing: if the tensor comes out as zero, this does not mean that there’s no mass or energy somewhere around, it means only that the metric you started with describes a vacuum spacetime. If it’s not zero, you’re also out of luck, because the energy-momentum tensor alone is not a unique description of a classical system. Knowing its components tells you nothing about what physical form this system actually takes - two physically different systems in terms of internal structure, time evolution etc can in fact have the same energy-momentum tensor. That’s because this tensor is the conserved Noether current associated with time-translation invariance, so what it reflects are a system’s symmetries, but not necessarily or uniquely its physical structure. There’s no unique 1-to-1 correspondence between this tensor and a particular configuration of matter and energy, since all it contains are density distributions. IOW, a given matter-energy configuration will have a unique energy-momentum tensor associated with it, but the reverse is not true - any given energy-momentum tensor can correspond to more than one possible matter-energy configuration. Thus it is not useful to try and define matter-energy by starting with spacetime geometry. Yes, they are of course closely related, but the relationship is not just a trivial equality; there’s many subtleties to consider.

3 hours ago, Dejan1981 said:

Cosmological constant is not zero for sure because Universe is not static.

A zero cosmological constant does not imply a static universe, only that expansion happens at a constant rate.

1 hour ago, Dejan1981 said:

and if we hypothesize that mater is product of curvature of space in one direction, and antimatter in opposite

Matter and antimatter have the same gravitational affects, they are not opposite in terms of curvature.

Edited Wednesday at 11:42 AM by Markus Hanke

[studiot]

Just now, Markus Hanke said:

The Einstein tensor describes a certain aspect of overall spacetime curvature, in the sense that it determines the values of a certain combination of components of the Riemann tensor (it fixes 10 out of its 20 independent components). But unlike the Riemann tensor, it is not a complete description of the geometry of spacetime. Thus, the Einstein equations provide only a local constraint on the metric, but don’t determine it uniquely.

...

etc

Glad you are back and had the time to compose that excellent explanatory post.

+1

[swansont]

6 hours ago, Dejan1981 said:

If mater and antimatter collided in time = 0 and there centers of gravity came to the same place in the space in time = 0 than resulting wave would be zero and energy and mater, antimatter and energy would stop to exist because space would be "flat". This is only mathematical and can never happen in reality. 

You’re claiming no curvature despite the energy not being zero.

[Dejan1981]

19 hours ago, Markus Hanke said:

The Einstein tensor describes a certain aspect of overall spacetime curvature, in the sense that it determines the values of a certain combination of components of the Riemann tensor (it fixes 10 out of its 20 independent components). But unlike the Riemann tensor, it is not a complete description of the geometry of spacetime. Thus, the Einstein equations provide only a local constraint on the metric, but don’t determine it uniquely.

Physically speaking, mass and energy have of course a gravitational effect, but neither appear as quantities in the source term of the field equations - there’s only the energy-momentum tensor field Tμν . 

Here’s the thing with this - the Einstein equations are a purely local statement. So for example, if you wish to know the geometry of spacetime in vacuum outside some central body, the equation you are in fact solving is the vacuum equation

 

Gμν=0

 

which implies 

 

Rμν=0

 

There is in the first instance no reference here to any source term, not even the energy-momentum tensor, because you are locally in a vacuum. During the process of solving these equations, you have to impose boundary conditions, one of which will be that sufficiently far from the central body the gravitational field asymptotically becomes Newtonian; it’s only through that boundary condition that mass makes an appearance at all.

Only in the interior of your central body do you solve the full equations

 

Gμν=κTμν

 

wherein the energy-momentum tensor describes the overall distribution of energy density, momentum density, stresses, strains, and shear (note that “mass” is again not part of this). By solving this equation along with boundary conditions you can find the metric.

You can work this backwards - you can start with a metric, and calculate the energy-momentum tensor. But here’s the thing: if the tensor comes out as zero, this does not mean that there’s no mass or energy somewhere around, it means only that the metric you started with describes a vacuum spacetime. If it’s not zero, you’re also out of luck, because the energy-momentum tensor alone is not a unique description of a classical system. Knowing its components tells you nothing about what physical form this system actually takes - two physically different systems in terms of internal structure, time evolution etc can in fact have the same energy-momentum tensor. That’s because this tensor is the conserved Noether current associated with time-translation invariance, so what it reflects are a system’s symmetries, but not necessarily or uniquely its physical structure. There’s no unique 1-to-1 correspondence between this tensor and a particular configuration of matter and energy, since all it contains are density distributions. IOW, a given matter-energy configuration will have a unique energy-momentum tensor associated with it, but the reverse is not true - any given energy-momentum tensor can correspond to more than one possible matter-energy configuration. Thus it is not useful to try and define matter-energy by starting with spacetime geometry. Yes, they are of course closely related, but the relationship is not just a trivial equality; there’s many subtleties to consider.

A zero cosmological constant does not imply a static universe, only that expansion happens at a constant rate.

Matter and antimatter have the same gravitational affects, they are not opposite in terms of curvature.

Thank you for taking the time to answer. I found it very interesting and informative.

I agree with your statement about the cosmological constant. My point was that Schwarzschild spacetime doesn't represent the universe that we live in.

About matter and antimatter having the same gravitational affects. I agree with this and I didn't imply that opposite curvature of space has opposite gravitational affect.

To me the most interesting part of your answer is "given matter-energy configuration will have a unique energy-momentum tensor associated with it, but the reverse is not true - any given energy-momentum tensor can correspond to more than one possible matter-energy configuration."

I wasn't talking about grand scale universe where planet will appear because of space curvature. I was talking about quantum scale universe where there is limited number of particles that can exist all of which have unique energy-momentum tensor associated with them. I think that we can't have two type of particles with the same energy-momentum tensor, else we are accepting that types of particles are limitless. I think there is a mechanism that limits the type of particles that can exist.  On the grander scale I agree that two different object can have the same energy-momentum tensor associated with them, but in this case there are other forces that allow this to happen.

What do you think about idea that there is a border of space curvature where if it's crossed it manifest it's self as a matter, and under that limit as an energy? Is it possible that light exist on that border and that's why it behaves like a wave and like a particle?

[Markus Hanke]

21 minutes ago, Dejan1981 said:

Thank you for taking the time to answer. I found it very interesting and informative.

No probs, it’s my pleasure 👍

21 minutes ago, Dejan1981 said:

My point was that Schwarzschild spacetime doesn't represent the universe that we live in.

And of course you are right in that, it’s a solution with idealised boundary conditions. But my point was simply to show that the Einstein equations admit vacuum solutions - ie that the local (and even global) absence of sources does not automatically imply flatness of spacetime.

24 minutes ago, Dejan1981 said:

I was talking about quantum scale universe where there is limited number of particles that can exist all of which have unique energy-momentum tensor associated with them.

GR is a purely classical theory, and the energy-momentum tensor can (to the best of my knowledge) not be promoted to any kind of meaningful quantum operator. IOW, general relativity can only handle classical systems.

27 minutes ago, Dejan1981 said:

but in this case there are other forces that allow this to happen.

There are no special mechanisms needed - as previously mentioned, this tensor arises from Noether’s theorem, and thus it describes some of the symmetries of a system, but not directly its internal structure or evolution. So it’s not surprising that you can have different systems with the same EMT.

31 minutes ago, Dejan1981 said:

What do you think about idea that there is a border of space curvature where if it's crossed it manifest it's self as a matter, and under that limit as an energy? Is it possible that light exist on that border and that's why it behaves like a wave and like a particle?

Again, GR is a purely classical theory, it has nothing to say about quantum systems. Also, you have to remember that “mass” and “energy” are Newtonian concepts, which don’t always straightforwardly translate to GR, due its non-linear nature. So this question doesn’t seem very meaningful to me.

Consider this: rest mass yields curvature. But so does angular momentum, electric charge, stress, strain, shear, and energy density in general (EM fields etc). But the effects of these don’t linearly superimpose, which is to say you can’t take a spacetime geometry and say “this curvature is because of mass, that curvature because of angular momentum,…”.

Also remember that some Newtonian concepts don’t have a gravitational effect at all, for example kinetic energy - you can’t turn something into a black hole simply by making it go very fast with respect to yourself, so one must be very careful when talking about energy as a source of gravity.

And then you have non-static and non-stationary gravity. In GR, you can have gravitational radiation that propagates, you can have isolated wave fronts, gravitational solitons, all kinds of things.

So you see, there’s a lot of subtleties to consider.

19 hours ago, studiot said:

Glad you are back and had the time to compose that excellent explanatory post.

Been around this whole time, but as a silent reader - we’ve recently had a lot of threads on politics, which I have very little to say about. I try and stick with GR

[Dejan1981]

28 minutes ago, Markus Hanke said:

Consider this: rest mass yields curvature. But so does angular momentum, electric charge, stress, strain, shear, and energy density in general (EM fields etc). But the effects of these don’t linearly superimpose, which is to say you can’t take a spacetime geometry and say “this curvature is because of mass, that curvature because of angular momentum,…”.

I'm not sure that this is completely true. The solution for a charged neutron star is similar to the Reissner-Nordström metric, meaning the total curvature is the sum of gravitational curvature + electric field curvature. In GR 𝑇𝜇𝜈 includes Mass-energy of the neutron star (dominant) and Energy of the electric field, which contributes additional curvature.

16 hours ago, swansont said:

You’re claiming no curvature despite the energy not being zero.

This is purely theoretical scenario and it can never happen in reality. Idea is that space behaves like a wave and that mater is manifestation of curvaton of space in one direction and antimatter in opposite direction. Two opposite wave will nullify them self's. If mass and energy are result of space curvature there would be no mass and energy left as an manifestation of space curvature. Let' say that we have a ideal body that consist of a ball of mater and the same mass of antimatter surrounding it as a sphere having the same center of gravity. Mater and antimatter are separated with a vacuum and exist in a vacuum. Would this body be able to exist if mater is manifestation of curvaton of space in one direction and antimatter in opposite direction?

[Markus Hanke]

30 minutes ago, Dejan1981 said:

The solution for a charged neutron star is similar to the Reissner-Nordström metric, meaning the total curvature is the sum of gravitational curvature + electric field curvature.

No, the metric is the unique solution that arises from the field equations for an electrically charged isolated spherically symmetric non-rotating body in an asymptotically flat spacetime. You can’t uniquely decompose this into two separate metrics that somehow add together, at least not if you want those metrics to themselves be valid solutions to the equations. What you can do though is consider the Reissner-Nordström metric as a special case of the Kerr-Newman metric.

In GR, if you take two valid solutions to the field equations and add them together, the result will in general not be a new valid solution; and valid solutions are not generally decomposable into the sum of other valid solutions. That’s because the field equations are non-linear.

[studiot]

Just now, Dejan1981 said:

I'm not sure that this is completely true. The solution for a charged neutron star is similar to the Reissner-Nordström metric, meaning the total curvature is the sum of gravitational curvature + electric field curvature. In GR 𝑇𝜇𝜈 includes Mass-energy of the neutron star (dominant) and Energy of the electric field, which contributes additional curvature.

Firstly you seem to me to be describing curvature here and in previous posts in this thread using too few dimensions.

When you get to space or spacetime there is no such thing as 'opposite curvaure' .

 

Secondly you might like to review a simpler system, that of the difference between moment of inertia and product of inertia and the inertia tensor.

This offers a simpler explanation as to why the are many solutions a given moment of inertia, but the product is unique.

[Markus Hanke]

@Dejan1981 I’d just like to add that there is in fact a way to “split” curvature, though not along the lines of what you suggested.

Imagine you have a very small volume - like a very small ball - that is in free fall towards a larger body. One might ask what happens to this ball, and this can be analysed in terms of volume and shape. As it turns out, if the ball free-falls in a vacuum, its total volume remains conserved, and only its shape distorts over time as it approaches the central body. If however the free fall happens inside an energy-momentum distribution (eg in a very strong electromagnetic field), both its shape and its volume will change.

The rate at which a test volume in free fall begins to change is encoded in the Ricci tensor Rμν , which, as you might recall, appears directly in the Einstein equations. This is what the vacuum equations physically mean:

 

Rμν=0

 

means that a small volume in free fall through vacuum is preserved. The shape of the ball is encoded in another object, the Weyl tensor Wμνϵδ , which does not vanish, and which does not appear in the field equations. If the free fall happens in something other than vacuum, the Ricci tensor never vanishes:

 

Rμν=κTμν−κ12gμνT

 

while the Weyl tensor may or may not vanish.

EDIT: The last term in the above should have a factor ½, not 12. For some reason I’m unable to edit the LaTeX, and the rendering looks off too. No idea why.

The point is this: the full description of the entire spacetime geometry is given by the Riemann tensor, and it so happens that this tensor can always be covariantly decomposed into a linear combination of the Ricci tensor and its trace, as well as the Weyl tensor. This is called the Ricci decomposition. 

What I wrote above is the geometric interpretation, and you can also give it a more physical one - the Ricci tensor represents curvature due to local sources, whereas the Weyl tensor represents curvature due to distant sources, for example gravitational radiation.

So if you want to decompose curvature, this would be a well-defined and observer-independent way to do it.

Edited yesterday at 11:48 AM by Markus Hanke

[swansont]

4 hours ago, Dejan1981 said:

This is purely theoretical scenario and it can never happen in reality. Idea is that space behaves like a wave and that mater is manifestation of curvaton of space in one direction and antimatter in opposite direction. Two opposite wave will nullify them self's. If mass and energy are result of space curvature there would be no mass and energy left as an manifestation of space curvature.

Matter + antimatter has energy, both before and after annihilation (since energy is conserved), so saying there’s no energy left is incorrect.