And I apologize on behalf of Geoff Mollusc.

I need to make a correction.

L = KE - PE

Newtonian:
L = (1/2)mv^2 - j_b A^b

Then change to the relativistic approach and we know we need to replace (1/2)mv^2 with the Lagrangian for a massive particle which is m/γ. However, (1/2)mv^2 is an increasing function of velocity and m/γ is a decreasing function of velocity. Therefore we need

L = - m/γ - j_b A^b

and not

L = m/γ - j_b A^b

I'm not sure where things stand with this discussion.

If there are further comments on the present posts, I'm happy to discuss.

Otherwise, I can introduce the next topic in the theory.

(Silence does not imply agreement with the present posts. It only implies that you're more interested in seeing the next topic than in discussing the current one.)

So you are abandoning your former claims?

My intention is to take points from the opening post and develop them. Each point will be developed using the same set of assumptions.

The first point that I chose was particle charge. If there is further interest in that topic, I am happy to discuss it. Otherwise, I can choose another point from the opening post and develop it.

I will not later say that no one expressed objections now and therefore they implicitly agreed to whatever was already written. If at some later time someone wants to return to a previous topic then I have no objections. If there are requests to address a particular topic from the opening post (or indeed any question about consequences of the assumptions) then I will do my best to respond.

Key points:
Lorentz transformations of T_{ab}
T_{ab} transformations for v not parallel to M
v not parallel to M gives rest mass to field
rest mass for a field is similar to the weak force



Purpose:

To show how an embedded manifold M has properties similar to the electroweak unification.



Assumptions:

As in the previous posts. M is a 4-manifold embedded in a Minkowski 6-space. There is a field A_v on M. The Lagrangian on M is L = k + F^{ab}F_{ab}.



A few results

M is a 4-manifold embedded in a 6-space. What does the stress-energy tensor T_{ab} look like on M? T_{ab} is a 6-dimensional rank 2 tensor on M. For convenience assume that the tangent space to M is {x_0, x_1, x_2, x_3} at a point p. Then at p, T_{ab} = 0 if a or b is 4 or 5. Now let M move at p so that we get a new stress energy tensor T'_{ab} by a Lorentz transformation of T_{ab}. Let the transforming velocity be in the direction of x_4 with magnitude v. Then T_{ab} transforms such that

T'_{00} = γ^2 T_{00}
T'_{04} = T'_{40} = vγ^2 T_{00}
T'_{44} = (vγ)^2 T_{00}

and is otherwise unchanged. However, if M is in motion at p then the measure dM is not dtdV. Transforming M also transforms dt so dt' = dt/γ. Incorporating that in the measure gives

T'_{00}dt' = γ T_{00}dt
T'_{04}dt' = T'{40}dt' = vγ T_{00}dt
T'_{44}dt' = v^2 γ T_{00}dt.

Now consider if T_{00} has rest mass m and field energy f at p, so T_{00} = m+f at rest. The field energy f transforms in the same way as the rest mass if the velocity of transformation is perpendicular to the manifold, so

T'_{00}dt' = γ T_{00}dt = γ(m+f)dt.

(In these transformations, it is important that the direction of motion is not parallel to the manifold.)

Assume there is a particle with non-trivial topology on M. Then there may be a point p on the particle where the manifold is not flat and the field energy is non-zero. Consider the apparent rest mass of the particle. How does the energy change as a function of particle velocity? The velocities that are possible for the particle are the ones that are parallel to flat spacetime. If those velocities are not parallel to the manifold at p then the field energy at p can contribute to the apparent rest mass.

Far from particles, the manifold is flat and the field transforms masslessly, as in electromagnetism. Close to particles, the manifold is not flat and the field may have apparent rest mass, as in the weak force.

Key points:
electroweak unification
A_v field has 6 dimensions and M has 4
difference in dimension implies gauge groups
flat space has one type of gauge group
non-flat space has a different gauge group



Purpose:

To show how an embedded manifold M has properties similar to the electroweak unification. The description of the A_v field on M can be approximated with gauge groups. The description is an approximation only, not a mathematical equivalence.




Assumptions:

As in the previous posts. M is a 4-manifold embedded in a Minkowski 6-space. There is a field A_v on M. The Lagrangian on M is L = k + F^{ab}F_{ab}.



A few results

The vector field A_v is a 6-dimensional vector field on M. M has only 4 dimensions. For F_{ab} = A_{b,a}-A_{a,b}, the Lagrangian is L = k+F^{ab}F_{ab}, which is equivalent to electromagnetism if the manifold is flat. F_{ab} is a 6-dimensional anti-symmetric tensor. Anti-symmetric matrices are infinitesimal generators of rotation matrices. F_{ab} is an infinitesimal generator of an element of SO(6). Derivatives A_{a,b} satisfy A_{a,b}v^b = 0 if v^b is perpendicular to the tangent space of M because the field A_v does not change in any direction perpendicular to M. If M is flat in the region being considered then we can assume that it is in the span of {x_0, x_1, x_2, x_3} without loss of generality. Then the components F_{45} and F_{54} have no effect. They correspond to a subgroup SO(2).

If M is not flat then it may not be in the span of {x_0, x_1, x_2, x_3}. Any coordinate axis may be perpendicular to M, except M will never be perpendicular to x_0. M is still 4-dimensional, therefore we need a gauge group. To include the possibility that M is not flat, we can have field components with a gauge group for F_{45} and F_{54} and other field components with a gauge group for F_{ab} for a and b in {1,2,3}. This gives a gauge group SO(3)xSO(2), again a subgroup of SO(6).

Far from particles, the manifold is flat, the gauge group is SO(2)=U(1), and the field is massless (from the previous post) corresponding to electromagnetism. Close to particles, the manifold is not flat, the gauge group is SO(3)xSO(2), and the field can have mass (from the previous post). The group SU(2) surjects onto SO(3) with kernel {I,-I}, therefore SO(3)xSO(2) is locally isomorphic to SU(2)xU(1), the gauge group of the electroweak unification.

To find the charge on non-flat space we can take F_{ab}^{,a}=j_b with j_b the charge current. If the manifold is not flat then F_{ab}^{,a} may have non-zero contribution from any index a in {1,2,3,4,5}. Therefore there can be a contribution to the charge from the components parallel to flat space and the components that are not parallel to flat space. These are components corresponding to the gauge group SO(3) and the gauge group SO(2). Using terms from the electroweak description,

Q = T_z + (1/2)Y_W
with Q the electric charge, T_z the third component of weak isospin, and Y_W the weak hypercharge.

Key points:
strong force
elementary fermions are S¹xP²
S¹xP² can link
linked S¹xP² are quarks
linking implies topological constraints
the links must be close but position is otherwise unconstrained
therefore confinement and asymptotic freedom
modeling the geometry resembles QCD




Purpose:

To show how an embedded manifold M has properties similar to the strong force.




Assumptions:

As in the previous posts. M is a 4-manifold embedded in a Minkowski 6-space. Elementary fermions are S¹xP² attached to the manifold by connected sum.




A few results

A spacelike slice of M can have a S¹xP² attached by connected sum along a torus parallel to the S¹ fiber. A P² can be attached to R². Multiple copies of P² attached to R² can be linked. Fibering over S¹, we can get multiple copies of S¹xP² that are attached to R³ and are linked. Linked S¹xP² cannot be separated in the same way that linked circles cannot be separated in 3 dimensions. If the S¹xP² are close to each other then they exert no force on each other, again in the same way that linked circles in 3 dimensions do not pull each other when they are close.

Interpreting the linked S¹xP² as quarks, they have the properties of confinement and asymptotic freedom. We can apply a mathematical model to the linked quarks. At a time t, there is a spacelike slice of M and each quark has a position in a 5-space. Label some quarks Q, R, S. Then each quark has a 5-position vector. The vectors can be written
Q = (Q_1, Q_2, Q_3, Q_4, Q_5)
R = (R_1, R_2, R_3, R_4, R_5)
S = (S_1, S_2, S_3, S_4, S_5)

We can choose the average position of the quarks to be the origin of their coordinates. Then Q+R+S=0. Now, we want to describe the condition that they can be anywhere relative to each other as long as they are close enough. To do that we can restrict the magnitude of the vectors, for example |Q| = 1. However, this does not allow the quarks to all be close to each other. To do that we can add a non-physical sixth coordinate to make Q', R', and S'. So,
Q' = (Q_1, Q_2, Q_3, Q_4, Q_5, Q_6)
R' = (R_1, R_2, R_3, R_4, R_5, R_6)
S' = (S_1, S_2, S_3, S_4, S_5, S_6)

Now the condition |Q'| = 1 implies |Q| is between 0 and 1. In other words, the quarks can be anywhere relative to each other as long as they are within distance 1 of the center.


We want these conditions: |Q'|=|R'|=|S'|=1 and Q'+R'+S'=0. Are these conditions satisfiable? The condition Q'+R'+S'=0 implies Q+R+S=0. This uniquely determines the first five coordinates of the origin of coordinates. Then |Q'|=|R'|=|S'|=1 determines Q_6, R_6, and S_6 up to +/-. However, there is no guarantee that Q_6+R_6+S_6=0. Therefore we are over-constrained by one degree of freedom. If we also allow S_5 to be variable then the conditions are satisfiable.

Now we have a 6-dimensional vector for each quark that describes its possible position in the 5-space at any time. The vectors satisfy |Q'|=|R'|=|S'|=1 and Q'+R'+S'=0. We can convert these 6-dimensional vectors to complex 3-vectors. Denote them with a ''. Then

Q'' = (Q_1 + iQ_2, Q_3 + iQ_4, Q_5 + iQ_6)
R'' = (R_1 + iR_2, R_3 + iR_4, R_5 + iR_6)
S'' = (S_1 + iS_2, S_3 + iS_4, S_5 + iS_6)

And we still have |Q''|=|R''|=|S''|=1 and Q''+R''+S''=0.

Our desired condition is that the position of the quarks is irrelevant as long as they are close enough. The condition of being close enough is automatically satisfied by the unit vector constraint. To make the position irrelevant we can use a gauge group. With complex 3-vectors we can use the gauge group SU(3). The gauge group SU(3) is 8-dimensional. The vectors Q'', R'', S'' have 18 coordinates, 6 constraints from summing to zero, and 3 constraints from being unit vectors. Therefore they have 9 degrees of freedom. The group SU(3) is a 8-dimensional subgroup approximation of the 9-dimensional group that maps between all possible positions.

If we call the vectors Q'', R'', and S'' the color charge, then we have an approximation of QCD.

For mesons, the same description works for two quarks Q'' and R''.

I'm not seeing a Least Publishable Unit (LPU) here.

Could this be a STIPOID (something to interest people only it didn't .. so far)?

I have two papers developing this theory. They are about 25 pages each. The first paper describes the fields, forces, and quantum theory. The second paper describes the particles and topology. So far, everything that I've discussed here is from the first half of the first paper.

Thank you for the responses. I'm having trouble distinguishing whether anyone is reading any of this.

I am. My math skills aren't up to par, but my language and comprehension skills are.

If you aren't a sockpuppet of Mr. Fish, feel free to register at my Joint. We have a few mathematicians and physicists that are probably willing to discuss your papers.

It may take a while to get an adequate response, because I have the most boring forum in the world, according to some others here, anyhow.

Besides, I allow image upload and I have LaTeX available, so you won't have to struggle so much.




Addendum for the peanut gallery: Who asked you?

Key points:
gravity and symmetry breaking
Lagrangian is L = k+F^{ab}F_{ab}
with no electromagnetic field L = k
minimizing Lagrangian in vacuum implies minimizing scalar curvature
therefore the Lagrangian of general relativity in vacuum follows
conservation laws imply stress-energy term
co-dimension 2 implies spontaneous symmetry breaking



Purpose:

To show how an embedded manifold M has properties similar to gravity. Spontaneous symmetry breaking follows, possibly leading to CP violation for particles.




Assumptions:

As in the previous posts. M is a 4-manifold embedded in a Minkowski 6-space. The Lagrangian is L = k+F^{ab}F_{ab} for some constant k.




A few results

The Lagrangian on M is L = k+F^{ab}F_{ab} for some constant k. If there is no electromagnetic field then L = k. Then integrating the Lagrangian over the manifold gives ∫k dM. This integral is proportional to the 4-volume of the manifold. We would like to show that minimizing the 4-volume of M implies minimizing the Lagrangian of general relativity. The Lagrangian of general relativity is

∫ R (-g)^(1/2) d^4y
Here R is the scalar curvature and g is the determinant of the metric. The y coordinates in this formula are from a coordinate chart on M. Using our assumptions about M and a coordinate chart, this Lagrangian is well-defined. Now assume an infinitesimal bounded region D on M with a boundary that is fixed in the 6-space and on D the scalar curvature R is nearly constant. Fixing the boundary in the 6-space is not an option for the Lagrangian of general relativity and this distinction is the reason that the Lagrangians are different. The Lagrangians ∫k dM and ∫ R (-g)^(1/2) d^4y both take some value on D. Consider the geodesic curves that begin at the center of D and reach its boundary. The area of the endpoints of those curves must equal the area of the boundary. The area of the endpoints is, to leading order, proportional to 1-Rr^2/(6n). The volume of D is minimized when the area of the endpoints equals the area of the boundary most quickly, as a function of radius. This happens when R is minimized. (Removing bumps reduces volume.) Therefore the Lagrangian ∫k dM implies the Lagrangian ∫ R (-g)^(1/2) d^4y. Therefore Einstein's field equation in the vacuum follows:

R^{ab}-(1/2)g^{ab}R = 0

where g^{ab} is the metric on M, which is inherited from the Minkowski 6-space. However, the equation for the vacuum does not apply where there are knots. Knots do not minimize volume and they do not minimize scalar curvature. If M were truly volume-minimizing then all knots on M would shrink down to points. Instead, quantum properties prevent this from happening. EM fields also do not satisfy volume-minimization because the fields contribute to the Lagrangian as F^{ab}F_{ab} in L=k+F^{ab}F_{ab}. So we would expect that particles and fields would imply that R^{ab}-(1/2)g^{ab}R = 0 is not satisfied. In that case we have

R^{ab}-(1/2)g^{ab}R = S^{ab}
for some rank 2 tensor S. The left side has 0 covariant divergence

(R^{ab}-(1/2)g^{ab}R)_{;a}=0 therefore S^{ab}_{;a}=0.

In the limit that the metric approaches flatness we have S^{ab}_{,a}=0. The tensors with zero covariant divergence are of the form S^{ab}=d(T^{ab}-h^{ab}) where h^{ab} and d are constant as the metric approaches flatness. If there are no particles and fields then S^{ab}=0 and T^{ab}=g^{ab}. Therefore h^{ab}=g^{ab}. Then the equation R^{ab}-(1/2)g^{ab}R = S^{ab} states that the curvature is proportional to the stress-energy tensor T^{ab} minus the stress-energy of gravity and flat spacetime. This matches general relativity where the stress-energy of gravity and flat spacetime do not contribute to S^{ab}.




If there is curvature then the manifold is obviously not flat. For a manifold embedded in a Minkowski spacetime, volume-minimization implies waves. In co-dimension 2 the waves rotate. If a point p at rest is (t,x,y,z,0,0) then when rotating it will be of the form

(t,x,y,z,Bcos(wt),Csin(wt))

The Lagrangian L = k is optimized when the waves have maximum velocity for minimum area. This implies that the waves should be circular. Each particle produces its own waves and those waves interfere. Waves can rotate as

(t,x,y,z, cos(wt), sin(wt)) or
(t,x,y,z, cos(wt), -sin(wt))

The Lagrangian is optimized when all waves are rotating in the same direction. This is spontaneous symmetry breaking and may relate to CP violation. For CP violation the rotation would need to couple to particle geometry.

Key points:
quantum field theory
modify the assumptions for quantum principles
psi-squared probability from particle volume
path integral from recombination




Purpose:

To show how an embedded manifold M can have properties similar to quantum field theory. We need to modify the assumptions of the previous posts to allow M to have quantum properties. If there is a Lagrangian on M that is perfectly optimized, then M is deterministic, which does not match physical observations. Even for a non-deterministic M, accommodating the double-slit experiment requires a sum over histories. We will begin by modifying the assumptions and then provide an outline of a proof to re-create the path integral.




Assumptions, first approach:

M is a 4-manifold embedded in a Minkowski 6-space. The function L = k+F^{ab}F_{ab} when integrated over M is some constant, ∫L dM = constant.




Results of first approach

If the integral ∫L dM is constant then the manifold will tend towards the integral-minimizing manifold because of probability. Consider, for example, a random path of fixed length from point B to C. It is possible for the path to go in a straight line from B to some point D and from D to C such that the path is stretched as tight as it can be. It is possible, but it is unlikely. This is likewise true for each sub-part of the path. If we allow the path to have unbounded curvature then the path will be arbitrarily close to the straight line from B to C with probability 1. Likewise, for any random manifold with a fixed "action" (in this case the action is not optimized, only fixed) the manifold will be arbitrarily close to the optimal manifold with probability 1. Therefore L serves the same purpose as the Lagrangian though it is not actually the Lagrangian.

The manifold is no longer deterministic, but the probability of a non-optimizing result is 0. Also, the double slit experiment would not produce a sum-over-histories result. We therefore continue to modify the assumptions.




Assumptions, second approach:

Assume a set of histories that determine the probability of any observational result. The set has finite measure. Each history, or path, is a 4-manifold embedded in a Minkowski 6-space. The function L = k+F^{ab}F_{ab} when integrated over each path is some constant, ∫L dM = constant. The histories branch and they interact by recombining when they are "close enough" for some definition of close enough.




Results of second approach

The set of paths saturate the most likely paths that are arbitrarily close to the optimal manifold. When those paths are saturated, less likely paths have non-zero probability. Therefore there is genuine non-determinism.

If two paths are topologically distinct then they cannot recombine. However, if they have the same topology and they are close to each other then they can recombine. Let a point p have coordinates in each of two paths, then the recombination would just be the average of those two coordinates: x_v(p1) and x_v(p2) in the individual paths recombine to (1/2)[x_v(p1)+x_v(p2)].

Now a rotating knot that has phase angle θ_1 in path 1 and phase angle θ_2 in path 2 would have a point p with these coordinates in each of the two paths:

(t, x, y, z, bsin(θ_1), bcos(θ_1))
(t, x, y, z, bsin(θ_2), bcos(θ_2))

and the average of the two would be

(t, x, y, z, (1/2)[bsin(θ_1)+bsin(θ_2)], (1/2)[bcos(θ_1)+bcos(θ_2)])

One could express the last two coordinates using a single complex number to make e^{iθ_1} and e^{iθ_2} and then the average is (1/2)[e^{iθ_1} + e^{iθ_2}].

We constrain the manifold such that it cannot pass through itself (by definition of embedding). A knot on the manifold constrains the set of paths. To compare, consider a one-dimensional path that is constrained to pass through a circle of radius r. The size of the radius determines the measure of the set of paths consistent with the requirement. Likewise, the size of a knot on the manifold determines the measure of the set of paths. In general, the probability is proportional to the "interior volume" of the knot. There is no space that is strictly inside of the knot but a linear scaling of the knot geometry scales the interior volume by an equal amount.

If two paths recombine then the recombination affects the points on those paths by the averaging, as above. The points on a knot are likewise affected by averaging. Those points have coordinates of the form

(t, x, y, z, bsin(θ_1), bcos(θ_1))

and averaging produces

(t, x, y, z, (1/2)[bsin(θ_1)+bsin(θ_2)], (1/2)[bcos(θ_1)+bcos(θ_2)])

Which can be expressed as (1/2)[e^{iθ_1} + e^{iθ_2}]. Two of the knot dimensions are affected by the recombination. Therefore two dimensions of the knot's interior volume are affected by the recombination. Therefore the probability is affected as the square of the averaged amplitude which is

|(1/2)[e^{iθ_1} + e^{iθ_2}]|^2

If we only need to consider two paths then this reproduces the result from the double-slit experiment.

If a subset of all paths has a knot at a particular location then those knots recombine and their phases add as above. The paths that have no knot at that location do not recombine because they are topologically different at that location. The total probability of observing a particle at that location is therefore the measure of the paths that recombine divided by the measure of the total set of paths. Assigning complex amplitudes to each of the contributing paths gives the description familiar from quantum mechanics.

The recombination happens at every time. From here the reasoning is the same as for the path integral.