Negative Energy quantum symmety as Dark Energy

[caveman1917]

I will now address the first phase of the Icarus2 cosmology, as seen by me.

Just to let you know that i'll be away over the weekend so i'll get to this next week. But in the meantime a quick response.

Icarus2 demonstrated that for a uniform distribution U-- ~ U++ ~ -U+-/2.

Yes that bit is quite obvious actually. Gravitational potential energy is the sum of for every pair of particles.

If we have two distinct sets of particles each of size that means the number of pairs "inside" each set is , so the two will add up to .

Between the two sets the number of pairs is and we have for large . Ergo (the last one has opposite sign). So the total potential energy of the distribution will be zero. Note however that absolute values of potential (energy) are meaningless, the dynamics are decided by their gradients.

However:

For both fluids the mixed potential dominates. This generates a push on both fluids.

You are making the assertion that this generates a push on both fluids, yet you have to show this. Calculating the force directly gives the result that there is no acceleration at all. Using Gauss' law also gives that result. As does using the shell theorem, as well as calculating the net potential (it will be flat). So you cannot simply assert that this "generates a push on both fluids", you'll have to prove this. A computer simulation will not do (specifically because it is not using a uniform distribution).

[utesfan100]

So the total potential energy of the distribution will be zero. Note however that absolute values of potential (energy) are meaningless, the dynamics are decided by their gradients.

You are making the assertion that this generates a push on both fluids, yet you have to show this. Calculating the force directly gives the result that there is no acceleration at all. Using Gauss' law also gives that result. As does using the shell theorem, as well as calculating the net potential (it will be flat). So you cannot simply assert that this "generates a push on both fluids", you'll have to prove this. A computer simulation will not do (specifically because it is not using a uniform distribution).

Interesting. I was mistaken to claim that he is using a fluid limit, as he is modeling the individual particles.

You are correct that what is being shown by the models is that quantum effects produce a non classical early expansion from the initial dense state, resulting in a dispersion into the surrounding vacuum. Positive mass disperses more intensely than negative mass, leading to a state where classical methods can be used.

Phase 3 describes a second quantum effect that briefly compensates for the decelerating acceleration we should observe as the classical gradient flattens due to the expansion.

[utesfan100]

Note however that absolute values of potential (energy) are meaningless, the dynamics are decided by their gradients.

Initially, at the scale of SI units, the potential would be a delta-Dirac function concentrated at the center of the universe. Thus the gradient would be near infinity and the uniform approximation fails and we need to consider quantum effects.

The Icarus 2 model shows that in the low energy limit the quantum effects should push the early universe towards a potential with a concentration of negative mass near the center causing an outward expansion. I can not show that this is valid in the high energy limit of this environment, but don't know of any mechanism that turns this into a contraction.

This pushes the universe into a state where there is a gradient in potential driving an outward acceleration. This will eventually settle to a steady state with decelerating expansion. An additional transient state causes the current, transient, accelerated expansion.

[utesfan100]

I argued that the negative mass of the universe should have settled to a lepton soup. I then argued that gravitational forces should have separated positive and negative masses.

For negative mass charges, like signs still push, but this causes an attraction. Opposite signs still produces a pull, but this causes repulsion.

Further, for electrons and positrons, the electrical forces outweigh gravity significantly. This suggests that on time scales where positive masses have coalesced, the HFM should have separated into positive and negative charge regions. This would introduce huge electric fields throughout the universe that are not observed.

Thus I need a mechanism to counter the collapse of the HFM due to the electric potential to stabilize the HFM. If charges are confined to some region of space, they must be in some form of orbit within that region. This requires an angular acceleration which will cause cyclotron radiation.

By conservation of energy, the positive photon energy of this radiation will take energy from the system. For nearly circular orbits the kinetic energy is roughly double the potential energy of a central inverse square force law set to 0 at infinity. Since the mass is negative, the kinetic energy becomes more negative as we move to tighter and faster orbits.

Thus cyclotron radiation should aid in the collapse, not help prevent it. Without a mechanism, the idea of a uniform HFM is shown not to work.

I am left to suggest that we are on the outer rim of the universe, where there is very little negative mass and where the electric potentials from the central positive and negative charge regions of negative mass cancel to a negligible level.

The steady state for this negative mass medium would be for the positive charges accelerating in the opposite direction from the negative charges. This would happen very early in the universe and generate a very strong magnetic field, amplifying the mechanism described in post #19.

Dark Matter
The ubiquitous presence of this magnetic field provides an answer for the nature of positive mass dark matter as the weight of the magnetic fields of the objects in question.

[caveman1917]

You are correct that what is being shown by the models is that quantum effects produce a non classical early expansion from the initial dense state, resulting in a dispersion into the surrounding vacuum. Positive mass disperses more intensely than negative mass, leading to a state where classical methods can be used.

That seems rather wrong since the model uses classical methods all along. It's applying newtonian gravity (and newtonian gravity alone) to a soup of particles at the individual particle level. There are no quantum effects modelled, everything except classical gravity is completely ignored, even though we're working at the particle level.

You'll still need to show that an analysis at the individual particle level would produce the required effects (since an analysis considering a continuous distribution fails to do so). Handwaving to some purported quantum effects that are not even being modelled does not cut it.

Positive mass disperses more intensely than negative mass

Prove this. For example (and this is staying within classical theory) at some point you'll have a dense mixture of positive and negative mass protons and electrons (and perhaps their associated antiparticles). If you're analyzing this at the particle level the main dynamics would be electromagnetic in nature, not gravitational. Gravity dominates over large scales because the electric charges of large regions approximately cancel, leaving mass as the primary source rather than charge. So you will have to show first that the electrodynamics of that mixture produces regions that are approximately neutral and seperates according to the sign of the mass rather than charge or some function of both.

[utesfan100]

That seems rather wrong since the model uses classical methods all along. It's applying newtonian gravity (and newtonian gravity alone) to a soup of particles at the individual particle level. There are no quantum effects modelled, everything except classical gravity is completely ignored, even though we're working at the particle level.

You'll still need to show that an analysis at the individual particle level would produce the required effects (since an analysis considering a continuous distribution fails to do so). Handwaving to some purported quantum effects that are not even being modelled does not cut it.

I am not proporting any uniquely new quantum mechanical effect. Like statistical thermodynamics, I am observing that large scale dynamics emerge from the classical interaction of individual particles on a microscopic level.

Classical gravity at the particle level produces the separation described above at the macroscopic level, as demonstrated by the computer models. The Icarus2 methodology is clearly enumerated.

1) Establish 3 length scales, p << d << r
2) Find some number of points within r, none of which are within d of each other.
3) Place an equal positive and negative mass within p of each point.
4) Model the effects of Newtonian gravity of the particles from an initial rest position.

My claim is that this repeatable computational experiment should always produce an expansion. You argue that it should be a steady state solution. Let us examine a simple geometry of a cube with each vertex within an octant. Let positive unit masses be on the tetrahedron (+++), (+--),(-+-) and (--+) and negative unit masses on the dual tetrahedron (---), (-++), (-+-) and (+--).

A simple balance of the Newtonian forces show that the effects of the opposite sign are stronger than those of the same sign. This repulses the positive masses and attracts the negative masses. In both cases this increases the influence of the negative masses until the negative masses begin expanding due to their own self repulsion.

[utesfan100]

Specifically, using Newtonian gravity, I get the following outward push on all particles, using units where G=1, the edges of the cube=2 and the masses are +/-1.



This push repels the positive masses and attracts the negative masses.

I get that the transition to phase 2 begins once r-/r+ gets below 0.458840. The outward pressure on the positive masses and the momentum of the inward motion of the negative masses will continue to decrease r-/r+.

The steady state of phase 4 occurs when F-/F+=r-/r+, which happens when r-/r+ gets near 0.387740. Because this steady state represents a uniform expansion, this will also be the limiting value of v-/v+. It is expected that the accelerations will be small relative to the velocities at this point, and the forces will drop off inversely with the distance squared. This will produce

This model does not feature the dynamics of phase 3 that causes a transient increase in outward acceleration. The observed increasing acceleration of the expansion then shows that we are still in phase 3.

[Shaula]

A simple balance of the Newtonian forces show that the effects of the opposite sign are stronger than those of the same sign. This repulses the positive masses and attracts the negative masses. In both cases this increases the influence of the negative masses until the negative masses begin expanding due to their own self repulsion.

Please show in detail how this is true for arbitrary distributions of mass.

[utesfan100]

Please show in detail how this is true for arbitrary distributions of mass.

This theory does not aim to cover arbitrary distributions, but rather with the positive and negative masses emerging in pairs, with the points of origin separated by distances larger than the distances between emerging pairs within a bounded region.

I do not have the mathematical skill to cover arbitrary distributions. Indeed, I will presently show that some distributions exist that will collapse and nullify.

By the anti-symmetry of the dynamics, we can find that a second steady state solution exists with r+/r- = F+/F- = v+/v-~ 0.387740. This solution contracts uniformly, and even reaches 0 in finite time. This, however is an unstable equilibrium.

If r+/r- < 0.387740 the positive masses will be accelerating inward faster than the negative masses. This decreases r+/r-, which goes to 0 in finite time. If we join the collapsed positive masses we find that the negative masses continue to collapse, eventually nullifying after a finite time.

If r+/r- > 0.387740 the negative masses will be accelerating inward faster than the positive masses. This increases r+/r-. At r+/r- = 0.458840 the positive masses begins accelerating outward until r+/r- =1 and we reach the dynamics already described.

Thus r+/r- ~ 0.387740 represents the border between collapse and expansion for a cubical geometry. This allows for a significant deviation from a uniform distribution of r+/r-=1 to produce the desired effect. The conservation of energy suggests that positive and negative mass objects appear in pairs, leading to geometries where the equivalent of r+/r- ~ 1.

In a classical theory the conservation of momentum suggests that separating positive and negative energy is problematic. Given a magnitude of energy and momentum, however, the uncertainty principle gives us a bound on how precisely we can specify the time and location of a particle creation event, allowing for an initial separation. This would lead to pairs whose distance is smaller than the typical distance between emerging pairs. The initial emergence of positive and negative mass pairs is the only place where this theory requires truly quantum mechanical effects.

The classical examination of a single pair shows that the system will accelerate along the vector whose tail is at the negative mass and head is at the positive mass. If our system is bound and the pairs emerge at rest, the result will be that pairs unaffected by nearby pairs will accelerate out of our bounding region, with the positive mass leading. This would bias the distribution towards geometries with r+/r- > 1, suggesting an expansion profile.

The best study of an arbitrary distribution of mass is that presented by Icarus2. He simulated at least three distributions that I consider arbitrary, with each producing the expansion described here.

[utesfan100]

In a classical theory the conservation of momentum suggests that separating positive and negative energy is problematic. Given a magnitude of energy and momentum, however, the uncertainty principle gives us a bound on how precisely we can specify the time and location of a particle creation event, allowing for an initial separation. This would lead to pairs whose distance is smaller than the typical distance between emerging pairs. The initial emergence of positive and negative mass pairs is the only place where this theory requires truly quantum mechanical effects.

It occurred to me that the issue of momentum for an emerging pair could be overcome by generating four particles. Two positive mass particles with opposite momentum paired with two negative mass particles with opposite momentum. This would allow the positive and negative masses to travel in different directions.

Icarus2 showed in his original paper that for a square arrangement similar to the cube described above the same expansion dynamics appear.

[utesfan100]

I misrepresented phase 2 in this statement.

The slowdown in acceleration is not due to the condensing of positive mass into separate regions, but rather to the beginning of outward motion of negative mass. This was explained in my description of phase 1.

I will now discuss, instead, a model of how mass became condensed.

Reviewing the original paper by Icarus 2, I see that this statement is incorrect. The mechanism described in this post is how positive mass becomes condensed, but Icarus2 provided evidence supporting the claim that this condensation decelerates the expansion of positive masses by increasing the total self attraction of the positive masses relative to a uniform distribution.

Since the attraction with negative masses remains nearly identical due to its distributed nature, the negative binding energy of the collapsing masses decreases the total energy of the universe, driving the total energy negative and causing a period of deceleration.

[utesfan100]

The dynamics of Icarus 2's phase 3 are similar to the dynamics of phase 2 described in post #41, but attributed to the condensation of negative mass around positive mass. Essentially, the condensation of positive mass regions produces localized regions with a geometry similar to the case when r+/r- < 0.387740 in post #39.

This increases the total potential energy of the universe, increasing the acceleration of the universe.

This claim is bolstered by models showing that concentrated positive masses, each surrounded by a cloud of negative mass particles, accelerates outward from a rest position.

Error of the Icarus2 model

The validity of summarizing the gravitational potentials as a single value for each over the entire universe can be argued if we are assuming spherical symmetry, as we can for phase 1 and 4. The dynamics of his phase 2 and phase 3, however, clearly violate spherical symmetry.

The changes observed by Icarus 2 produce a binding energy that flows outward with the initial objects, not a new force influencing their motion.

Consider a spherical shell with a radius of 1 and a total mass of 4pi. The self attraction will be equivalent to a mass of 2pi at the center of the sphere.

If we replace this with four spheres with radius r' << r centered on the points of a tetrahedron on the original sphere we would find that the potential energy of the four spheres alone would greatly exceed that of the original sphere, but this does not contribute to the motion of the spheres themselves. It would be better to treat them as 4 separate points with a mass of pi each.

The attraction between four points with a mass of pi is less than that of a uniform spherical distribution. This suggests that the concentration of mass due to phase 2 would actually decrease the attraction of the positive masses, increasing the acceleration.

This effect is negligible
In post #29 I suggested that we are on a sphere with a radius 1e15 light years relative to the center of the universe frame. The size of condensed positive masses on the scale of the distances between super clusters would be almost uniform along the surface of this sphere.

Further, these objects are far from being condensed to single points.

While these dynamics do provide a mechanism to produce separated masses, they should not be expected to significantly impact the expansion dynamics of the universe.

[utesfan100]

It would appear that, after the initial mixture has settled in seperate positive/negative regions, the profile would be akin to the one we'd get from a de sitter space where we let the cosmological constant drop like . This is not observed, but perhaps i'm misunderstanding the dynamics you propose.

I do not fully accept that this fairly agrees with the dynamics I have described, however I feel that it is close enough that evidence experimental against this model would constrain this theory tightly.

Particularly, evidence against the cosmological constant dropping like would exclude this model. My limited understanding suggests that determination of its positive sign and estimates of its present magnitude are difficult to observe, much less its time dependence.

[utesfan100]

Can you provide the expansion profile predicted by your positive/negative mass mixture? Preferably in terms of a specific scale factor.

In any case, if you'd derive an actual expression for in your proposal, it would be much easier to check this against observations that in effect constrain that function.

I was thinking a(t) was acceleration function. Are you using this as the unit length parameter? Am I correct in thinking that for Hubble's Law alone ?

Difficulties I will not have time to address before the close of this thread
The velocities, and thus the masses that accelerated them, are relativistic. At the very least the effects of the first order PPN corrections should be considered.

The acceleration profile will cause the flow to be contracted radially. To match observation the net compression will need to be equal in magnitude to the length contraction due to the speed over the visible universe. I have no evidence that such a region exists, much less constitutes a significant region of the universe.

Even relativistically, the local area expansion of the sphere should match the expansion in the equivalent local tangent plane. Presuming that we live in a region where the compression matches length contraction, the expansion will match v(r).

Classical Limit
The physics of a classical inverse square force are well known. Given two spherical shell, one of positive mass 2M and once of negative mass -2M, with r+ > r-, both will accelerate outward as if there is a mass of -1M at the center.

We can define a velocity like the escape velocity that represents the limiting velocity an object at rest at r due to expansion.

We can now define an invariant velocity at infinity, for an object with a given velocity at a given r.

Solving for v(r) gives:

This is a separable differential equation, which Wolfram shows is solvable, generating time as a complicated function of r. If we take v(infinity) >> 2GM/r and conveniently selecting a reference time we can simplify this to:

Solving this for r gives the a(t) you are looking for.

The log term should dominate, giving a near exponential expansion. The perturbative terms should show that this expansion in accelerating slightly.

[caveman1917]

I was thinking a(t) was acceleration function. Are you using this as the unit length parameter?

It's the standard scale factor.

Am I correct in thinking that for Hubble's Law alone ?

No that's for a universe with dominating cosmological constant.

[utesfan100]

No that's for a universe with dominating cosmological constant.

Well, Hubble's law tells us that . This gives us directly that . Thus Hubble's Law is a DeSitter universe.

The cosmological constant is what is used to produce this observed effect in a homogenous fluid solution of GR. This theory models the big bang as a localized explosion in an initially flat universe. There is no cosmological constant, only negative mass within the nearly homogenous spherical shell of the universe our visible universe resides.

The Hubble constant represents the accumulated relative velocity caused by this acceleration since the matter of the universe began expanding, while the central negative mass explains its continued acceleration. This replaces dark energy composing 72% of all energy with an initial total negative energy equal to total positive energy, as expected from the conservation of energy.

[utesfan100]

The steady state for this negative mass medium would be for the positive charges accelerating in the opposite direction from the negative charges. This would happen very early in the universe and generate a very strong magnetic field, amplifying the mechanism described in post #19.

This dynamic would quickly cause the charged negative mass to accelerate along the pole, leaving behind a significantly reduced central negative mass and a surrounding positive mass shell behind. This would lead to the shell collapsing on the negative mass, negating the expansion.

Thus the model fails if charged negative mass particles exist in significant proportions. For the remainder of the thread the negative mass can be assumed to be almost entirely composed of negative mass neutrinos, with just enough leptons to add some magnetic field as they departed.

While a mechanism has been suggested for removing the quarks of the negative mass standard model, nothing seems likely to remove the negative mass charged leptons. Without such a mechanism, a negative energy dual of the standard model does not reproduce the effects attributed to Dark Energy. This is a major hole in the theory that will not be resolved by the end of the month.

[caveman1917]

Well, Hubble's law tells us that . This gives us directly that . Thus Hubble's Law is a DeSitter universe.

Hubble's constant is itself a function of time, ie and Hubble's law tells us (not )

It's only when you assume that (ie the hubble constant is constant in time) that you get a deSitter universe. This is not necessary for hubble's law though, in general will not be a constant function.

[caveman1917]

There is no cosmological constant, only negative mass within the nearly homogenous spherical shell of the universe our visible universe resides

There are many problems with this.

It requires setting initial conditions by hand such that the universe starts with an interior shell of negative mass and an exterior shell of positive mass. You say that it is the result of some unnamed quantum effects on the particle level starting from a uniform sphere, but you never derived this (and no the computer simulation does not count, it has way too many problems itself, and doesn't even produce that result). Just saying that it is so doesn't make it true, deriving it does. In any case it is highly unlikely to produce that result, because the dynamics will be electrodynamic, not gravitational. The reason gravity dominates over large scales is because charge cancels and mass doesn't, however in your proposal mass also cancels. Electrodynamics works on charge, not mass, so it is highly unlikely to seperate regions on the basis of mass. Note that i said unlikely, not impossible, meaning that you'll have to prove it if you want it to be accepted.

But given that the thread has almost run out of time, and you state that you won't be able to do the derivations in time, let's accept the initial conditions and look what happens. So we have two concentric shells of mass of finite thickness, an interior negative mass one and an exterior positive mass one. You're correct that both would accelerate outwards "together". You're also correct that this would produce a hubble law for observers inside the positive mass shell (if the radius of the shell - the distance to the center - is a couple of orders of magnitude greater than the size of the observable universe for those observers). However that's where it starts to break down.

First, let's look at the acceleration gradient over the positive mass shell. Using the shell theorem (which of course works just as well for negative as positive mass) we see that we can analyze the forces on the positive mass shell as due to a negative mass point source at the center. This means that we have a tidal gradient over the radial direction of the positive mass shell. The interior part will be accelerated outwards faster than the exterior part, so the shell will be getting "squished". Looking at the effects of the positive mass itself, the shell theorem also applies so the interior part doesn't feel any force while the exterior feels an attraction due to the part of the positive mass shell inside its radius. It will also "squish" the shell. So the positive mass shell will expand along the angular directions and contract along the radial direction. So we'd expect to see redshift along two directions in space and a blueshift along the direction towards (and away from) the center. This badly fails observations.

Second, let's look at the stability of the situation. Suppose we have a region in the positive shell with lower density than average. Since negative mass accelerates towards positive mass, this will make the negative mass "under" that region go towards the edges of that region. However positive mass accelerates away from negative mass, so this situation will have the positive mass move away from the edges of that region, effectively making the region even less dense. At some point the density will become so small that you'd effectively have a hole in the shell, yet the dynamics continue. So basically every lower density region will become a hole, and that hole will continue to expand. Note that this is completely different from the accelerating expansion (that was in the shell itself expanding).

At this point your theory requires putting in by hand specific initial conditions (the universe starts as two concentric shells of positive and negative mass) to produce a universe counter to observations. It just doesn't work.

[utesfan100]

You say that it is the result of some unnamed quantum effects on the particle level starting from a uniform sphere, but you never derived this (and no the computer simulation does not count, it has way too many problems itself, and doesn't even produce that result).

My use of the term quantum was unfortunate. I would now call the expansion an emergent property of the stochastic process due to Newtonian gravity.

Placing positive masses at (1, 1, 1), (1, -1, -1), (-1, 1, -1), and (-1, -1, 1) with negative masses at (2.5, 2.5, -2.5), (2.5, -2.5, 2.5), (-2.5, 2.5, 2.5) and (-2.5, -2.5, -2.5) will initially contract, before turning around and accelerating outward. Even for this distribution with an initially high concentration of positive masses we get an expansion profile, which I take as showing that this dynamic is robust.

Changing the 2.5 values to 3 will produce a result that compresses to 0, showing that this dynamic is not absolute.

The three systems studied by Icarus two provide several case studies that generate an expansion profile for an initial distribution I described as follows:

This theory does not aim to cover arbitrary distributions, but rather with the positive and negative masses emerging in pairs, with the points of origin separated by distances larger than the distances between emerging pairs within a bounded region.

The cube scenarios add to the robustness of this emergent property.

I will agree that a more rigorous proof may be needed for a final proof.

No one has provided ANY distribution fitting this description that fails to exhibit this emergent property.

In any case it is highly unlikely to produce that result, because the dynamics will be electrodynamic, not gravitational. The reason gravity dominates over large scales is because charge cancels and mass doesn't, however in your proposal mass also cancels. Electrodynamics works on charge, not mass, so it is highly unlikely to separate regions on the basis of mass.

I have already conceded (post #47) that the negative mass must not be charged, and conceded that the lack of a mechanism to explain the lack of charged negative mass leptons is a major hole in this theory.

First, let's look at the acceleration gradient over the positive mass shell. Using the shell theorem (which of course works just as well for negative as positive mass) we see that we can analyze the forces on the positive mass shell as due to a negative mass point source at the center. This means that we have a tidal gradient over the radial direction of the positive mass shell. The interior part will be accelerated outwards faster than the exterior part, so the shell will be getting "squished". Looking at the effects of the positive mass itself, the shell theorem also applies so the interior part doesn't feel any force while the exterior feels an attraction due to the part of the positive mass shell inside its radius. It will also "squish" the shell. So the positive mass shell will expand along the angular directions and contract along the radial direction. So we'd expect to see redshift along two directions in space and a blueshift along the direction towards (and away from) the center. This badly fails observations.

Since I also argue (post #30) that we are moving relativistically relative to the center of the universe frame at z > 100,000 the impact of length contraction should also come into play. In post #44 I argued that the uniformity of our visible universe requires the compression you describe above must match the length contraction due to our motion from the center of the universe. Transforming to our local frame will then nullify the red shift and blue shifts you speak of.

The anthropic principle suggests that we should not expect to be in an unusual corner of the universe. Which is why I stated this explanation is incomplete.

I have no evidence that such a region exists, much less constitutes a significant region of the universe.

Second, let's look at the stability of the situation. Suppose we have a region in the positive shell with lower density than average. Since negative mass accelerates towards positive mass, this will make the negative mass "under" that region go towards the edges of that region. However positive mass accelerates away from negative mass, so this situation will have the positive mass move away from the edges of that region, effectively making the region even less dense. At some point the density will become so small that you'd effectively have a hole in the shell, yet the dynamics continue. So basically every lower density region will become a hole, and that hole will continue to expand. Note that this is completely different from the accelerating expansion (that was in the shell itself expanding).

This is the essence of phase 2 and phase 3 of the original Icarus2 dynamics.

Phase 2 is the observation that the contraction of positive masses is stronger than the negative masses in a positive mass shell.
Phase 3 is that eventually even the negative masses crowd around the positive mass regions.

A corollary is that this will leave large voids in the shell.

I argued in post #42 that Icarus 2 was in error in his original modeling of this dynamic, and that this effect generates a negligible outward acceleration based on the geometry required for this model, even on my own terms.

The primary effect of this dynamic is to explain why we see super clusters separated by huge voids.

At this point your theory requires putting in by hand specific initial conditions (the universe starts as two concentric shells of positive and negative mass) to produce a universe counter to observations. It just doesn't work.

I will agree that this theory is far from its final form. This model should require three parameters to describe what is explained by dark energy.

1) E, the total positive (and the equivalent amount of negative) energy of the universe.
2) z, the time dilation of our speed from the center of the universe.
3) r, the distance to the center of the universe.

This replaces:
1) The energy density of our universe
2) The current Hubble expansion rate
3) The proportion of the energy that is dark energy

The physical composition of Dark Energy has yet to be explained to me, nor why we should have so much of it.

The conservation of energy suggests that from an empty starting point any generated positive mass should have generated an equal negative mass. Negative mass neutrinos have not been excluded as a possibility, though charged negative mass particles must be a negligible proportion of the negative mass.

All things being equal, it is more compelling for me to assume that we are some empirically derived distance from the center of the universe than to suppose the existence of some unknown medium composing 72% of all energy.

I concede that all things are not yet equal.

[utesfan100]

Hubble's constant is itself a function of time, ie and Hubble's law tells us (not )

It's only when you assume that (ie the hubble constant is constant in time) that you get a deSitter universe. This is not necessary for hubble's law though, in general will not be a constant function.

Indeed. Given the expansion emergent property of gravity, H(t) must be increasing in this theory.

[utesfan100]

Specifically, using Newtonian gravity, I get the following outward push on all particles, using units where G=1, the edges of the cube=2 and the masses are +/-1.



This push repels the positive masses and attracts the negative masses.

I get that the transition to phase 2 begins once r-/r+ gets below 0.458840. The outward pressure on the positive masses and the momentum of the inward motion of the negative masses will continue to decrease r-/r+.

The steady state of phase 4 occurs when F-/F+=r-/r+, which happens when r-/r+ gets near 0.387740. Because this steady state represents a uniform expansion, this will also be the limiting value of v-/v+. It is expected that the accelerations will be small relative to the velocities at this point, and the forces will drop off inversely with the distance squared. This will produce

This model does not feature the dynamics of phase 3 that causes a transient increase in outward acceleration. The observed increasing acceleration of the expansion then shows that we are still in phase 3.

Simulation has shown that dynamics, and those of post #39, are accurate, but the calculated transition values are in error. The transition in the direction of the internal masses occurs at r-/r+=0.783010, or r+/r-=1.277123. (Not 0.458839)

The steady state solution occurs at r-/r+=0.609582, or r+/r-=1.640469. (Not 0.387740)

The values of 2.5 in post #50 put the dynamics clearly into the contract to a point region. A value of 1.63 will exhibit the described dynamics. This reduces the robustness argument, but not fatally.

Simulations of a model with r+=1 and r-=1.63 initially at rest show:
Positive masses begin accelerating out at r+=0.29, r-=0.37.
Negative masses pass positive masses at r+=r-=0.25.
Negative masses begin accelerating out at r+=0.23, r-=0.18.
Negative masses reach minimum distance at r+=0.24, r-=0.07.
Negative masses begin accelerating inward again at r+=0.55, r-=0.43. Negative velocity is much larger than positive velocity here.
Negative velocity slows to below the positive velocity at r+=1.70, r-=1.63.

The intense acceleration near the turn around point produces a large outward velocity that pushes the system into a nearly cubic state with the negative masses accelerating inward, but unable to overcome the outward velocity within the time allowed by the simulator.

[utesfan100]

CM1a: Please rigorously derive the result that it "accelerates rapidly outwards".

The outline of a proof would be as follows:

For any bounded distribution of masses we can define a center of momentum frame where is a minimum. Relative to this frame we can further define an origin where is a minimum.

Hard Step: Then show that is increasing when and , and further that these inequalities are stable.

Since is constant, this requires the average radius to increase without bound.

This would rigorously prove the existence of the emergent expansion due to stochastic Newtonian gravity with negative mass. The proof of the hard step might even make this dynamic publishable.

[caveman1917]

The outline of a proof would be as follows:

For any bounded distribution of masses we can define a center of momentum frame where is a minimum.

If they start at rest, won't every frame be a center of momentum frame?

The proof of the hard step might even make this dynamic publishable.

Well that's the reason i've been asking you to do these derivations. So that your argument would be of publishable quality. There's really no point arguing with vague statements as to what it "should be". It's a lot more productive to take it slowly step by step and rigorously derive the results at each step. Yes it will take a lot of time, and quite a few papers, but that just happens to be the hard work part of coming up with a replacement for a mainstream theory. Rushing through the foundations with only vague ideas to get to the end result will only result in the entire thing falling down at the first breeze of someone poking a hole in your foundations.

[utesfan100]

If they start at rest, won't every frame be a center of momentum frame?

Start at rest in which frame?

Note that is positive or 0 for both positive and negative masses. When we transform from the rest frame of the masses to another frame, all masses will now have a velocity and the sum will be positive. Thus the rest frame, with this sum=0, is minimum, and thus the center of momentum frame.

Well that's the reason i've been asking you to do these derivations. So that your argument would be of publishable quality. There's really no point arguing with vague statements as to what it "should be". It's a lot more productive to take it slowly step by step and rigorously derive the results at each step. Yes it will take a lot of time, and quite a few papers, but that just happens to be the hard work part of coming up with a replacement for a mainstream theory. Rushing through the foundations with only vague ideas to get to the end result will only result in the entire thing falling down at the first breeze of someone poking a hole in your foundations.

My purpose in presenting ideas here is to sanity test ideas to see if they are worth developing further.

Sometimes it is possible to have an interesting idea that produces some interesting initial results, but simple considerations that were neglected rule out the idea. Often a peer can highlight these considerations, removing the need to do the rigorous derivations for a failed theory.

Indeed, several branches of this theory have been pruned by the considerations presented in this very thread.

[caveman1917]

Start at rest in which frame?

With respect to eachother.

Note that is positive or 0 for both positive and negative masses.

Right, i missed that you were squaring it. This however begs a different question: How do you square a vector? Or more importantly, why would you want to square a vector? (this also goes with that other vector you're squaring in the next expression).

Momentum is a vector quantity, a center of momentum frame is one in which the norm of the vector sum of all constituent momentum vectors is minimized. Negative mass or not, the norm will always be positive. The momentum vector for a negative mass particle will simply point "the other way" from a positive mass particle, but its norm will be the same.

Given this, with an equal distribution of positive and negative mass, every frame in constant velocity wrt the rest frame will have zero momentum and will thus be a center of momentum frame, and my previous question stands. Not because there are "negative" and "positive" ones that "cancel" so you have to square it, but because the momentum vector for the negative particle points in the other direction than that for the positive particle, with equal length, and thus their vector sum gives the null vector with norm zero.

[utesfan100]

With respect to eachother.



Right, i missed that you were squaring it. This however begs a different question: How do you square a vector? Or more importantly, why would you want to square a vector? (this also goes with that other vector you're squaring in the next expression).

Momentum is a vector quantity, a center of momentum frame is one in which the norm of the vector sum of all constituent momentum vectors is minimized. Negative mass or not, the norm will always be positive. The momentum vector for a negative mass particle will simply point "the other way" from a positive mass particle, but its norm will be the same.

Given this, with an equal distribution of positive and negative mass, every frame in constant velocity wrt the rest frame will have zero momentum and will thus be a center of momentum frame, and my previous question stands. Not because there are "negative" and "positive" ones that "cancel" so you have to square it, but because the momentum vector for the negative particle points in the other direction than that for the positive particle, with equal length, and thus their vector sum gives the null vector with norm zero.

With positive mass, the center of momentum frame is usually defined as the frame where the norm of the sum of momentum is 0.

As you observe, when there is an equal amount of positive and negative mass the momentum will be 0 in every frame if it is 0 in any frame. If the negative mass equals the positive mass and the center of momentum frame of the positive mass differs from the center of momentum frame of the negative mass, there is no frame with a net 0 momentum. This requires a different definition of the center of momentum frame.

An alternate definition of the center of momentum frame is that the sum of the norms of the momentum is a minimum. When all mass is positive this is identical to the first definition. The square of a vector, the dot product with itself, is a convenient norm. This definition is well behaved when negative mass is included.

[caveman1917]

When all mass is positive this is identical to the first definition.

Hardly, the norm of the sum is not identical to the sum of the norms. Take two unit mass particles travelling along the x-axis with unit velocity but opposite direction. This frame is clearly the center of momentum frame, indeed . However the sum of the norms is . This holds for every frame moving left or right wrt the center of momentum frame with a speed of at most 1, for example taking the frame of the leftmost particle:

In any case, if you'd want to use the norm you should use the norm (ie ) not the inner product (ie ) since an inner product space is a stronger condition than a normed vector space. Whilst not technically wrong, it's a bit inelegant

[utesfan100]

Hardly, the norm of the sum is not identical to the sum of the norms. Take two unit mass particles travelling along the x-axis with unit velocity but opposite direction. This frame is clearly the center of momentum frame, indeed . However the sum of the norms is . This holds for every frame moving left or right wrt the center of momentum frame with a speed of at most 1, for example taking the frame of the leftmost particle:

In any case, if you'd want to use the norm you should use the norm (ie ) not the inner product (ie ) since an inner product space is a stronger condition than a normed vector space. Whilst not technically wrong, it's a bit inelegant

The square of the Euclidean norm satisfies the mathematical definition of a norm. This is a more general concept than the absolute value.

http://en.wikipedia.org/wiki/Norm_%28mathematics%29

The norm of the sum is clearly different than the sum of the norms. This is also why the center of momentum frame is where the norm of the sum is 0, but only a minimum for the sum of the norm. When using a Euclidean norm we get the singularity you observe. Squaring this norm fixes this singularity.

For the test case you pick, the obvious center of momentum frame has a sum of norm of 1^2+1^2=2. From the frame of one of the masses we have a sum of norms 2^2+0^2=4. In general, a frame moving in the same line as the momentum of the particles has a value , which has a minimum of 2 at a=0. But this is the center of momentum frame.

[utesfan100]

The square of the Euclidean norm satisfies the mathematical definition of a norm. This is a more general concept than the absolute value.

On second thought, examining the case of an obtuse triangle shows that the square of the Euclidean norm fails the triangle identity. Thus, what I was mistakenly calling a norm it is not a norm.

In my previous posts, when I used the word norm, I meant the square of the Euclidean norm.