If you’re not into reading crackpot theories of physics, I recommend you skip this. If you’re not into reading mathematics - likewise, you should probably skip this, there’s a fair bit. If you really really like mathematics - you maybe should skip this also, but if you don’t, please point out my errors. I’ve found and corrected a few, but I’m reasonably certain I’ve missed others.
That said …
Assume time is a rotation with respect to the polar coordinate r, such that:
dθ/dr = c
where c is the speed of light.
Given a circle of circumference ω at r, with the arclength (distance from θ = 0 on the circle) given by ψ:
dθ/dr = ((dψ/dr)/(dω/dr)) * 2π * c
Then this gives rise to a logarithmic spiral in the polar plane such that, for an additional constant k,
r = k * e^(c * θ)
The time separation between two points (events) in the logarithmic spiral is given by the arclength:
t=((1 + c^2)^(1/2) / c) * (r(θ2) - r(θ1))
The distance separation f, then, is the same value as above, divided by c:
f21 = (1 + c^2)^(1/2) * (r(θ2) - r(θ1))
Which implies that r is not identical to the space separation between these events. If we treat the positions of two particles as two events, and regard one particle as the origin, and the current position of another particle as our event, then the spacial separation (physical distance) between these particles, using θ to represent the total rotation necessary to arrive at the particle, is related by:
f = (1 + c^2)^(1/2) * (r(θ))
or
r(θ) = f / (1 + c^2)^(1/2)
Now, I want an equation for the local force experienced by a particle. Choosing an arbitrary x and y cartesian coordinate system in relation to the polar coordinate system:
x = r * cos(ln(r / k)/c)
y = r * sin(ln(r / k)/c)
Which is ... so frustratingly close to the equation I expect, which should instead look like sin(ln(x))/x.
I could make an unprincipled decision that, since there are two real dimensions (plus time/distance, which are in this approach complex), and this force is distributed instantaneously across the surface area of a sphere, that the equation should be divided by the surface area, for:
x = cos(ln(r / k)/c) / (f * 4π)
y = sin(ln(r / k)/c) / (f * 4π)
Which nets us the equation I'm looking for in one of these two dimensions - whichever has the "correct" orientation, the determination of which is a meaningful question I don't have the answer to yet. (Sine and cosine both yield the behavior I’m looking for, away from the origin.)
However, I sacrifice the idea that this approach is agnostic to the number of real dimensions. It is possible that additional real dimensions somehow cancel out, such that the equation remains consistent, but I don't know how to evaluate that claim.
There is an alternative, however.
See, I have additionally left out some important considerations in the force equations, and in particular I have left out rotation and motion - the rotation ends up being a multiplication by e^(-c*φ), where I -think- φ will be the amount of rotation of the non-origin particle over the course of the rotation represented by θ. However, the chirality of the logarithmic spiral, and its rotational direction, are expected to correspond to whether or not the particle is matter or antimatter, and in these cases, the behavior can start to look significantly different.
Rotating the spiral, for reference, changes the equation such that, for a rotation of φ, we get the following equation:
r = k * e^(c * θ) * e^(-c * φ)
r = k * e^(c * (θ - φ))
Which results in the following changes to our equations, with some hope that I put all the parentheses in the correct locations:
x = k * e^(c * (θ - φ)) * cos(ln((k * e^(c * (θ - φ)) / k)/c))
y = k * e^(c * (θ - φ)) * sin(ln((k * e^(c * (θ - φ)) / k)/c))
Which is - a bit messier, and it’s somewhat ambiguous how to interpret the rotations involved, as θ and φ are, in a static inertial frame (that is, the particles are not moving with respect to one another), identical; they’ve both rotated the same amount. Granted, if we are interpreting x and y as curvature - this is exactly right; they are identical in that case. However, in that case, I’d expect the value to be 0, which it - well, it seems to be for y, assuming my math is correct, suggesting sine is the correct function to use. There’s something like a bootstrapping issue here, if you haven’t noticed: This implies a particle won’t observe any curvature relative to the origin particle, unless it is already observing curvature relative to the origin particle, which is necessary to invalidate the shared inertial reference frame. However, this nicely implies that a particle doesn’t experience its own curvature. This is actually pretty important; earlier versions of this nonsense didn’t have this property, and trying to figure out distances to use for curvature became a recursive nightmare I never managed to resolve.
Additionally, it is much harder to understand the graph when you use a slightly modified version of these equations in which I substitute x for θ - φ. If any readers happen to care to try that exercise, just remember that x there is a rotation, not a distance, and that distance scales exponentially with rotation - each full rotation is crossing exponentially more distance than the rotation before it. The nice periodic behavior is expected there, as we’ve effectively eliminated the exponential relationship.
A fascinating property of logarithmic spirals, is that such a rotation is also equivalent to a rescaling - that is, it is equivalent to making the spiral - which is, in an important sense, the particle - smaller (depending on the direction of rotation, I am assuming a particular direction of rotation). The solution to the problem above, in which we need to divide by 4πr^2 to arrive at the “correct” equation, thus may lay in one of these two equations. Or maybe not; I don’t really see how it could. Perhaps the scaling involved also needs to be applied to the mass of the origin particle, which has yet to be included, in which case we would need to divide by e^(c*φ):
x = (k * e^(c * (θ - φ)) * cos(ln((k * e^(c * (θ - φ)) / k)/c))) / e^(c*φ)
y = (k * e^(c * (θ - φ)) * sin(ln((k * e^(c * (θ - φ)) / k)/c))) / e^(c*φ)
Which seems like it might be a plausible solution; I think I expect φ to be larger than θ - φ. But I’m very thoroughly out of my depth here.
The comment on the reference frames above should also imply another consideration: The original equations are in a static inertial frame; the particles are not in relative motion. Even in the next set of equations, attempting to account for this, still make an assumption of a static inertial frame with respect to the value “c”, which remember originates in the speed of a particle through time. That is, the value of "c" above, in particular, will "change" if the particles do not share an inertial frame of reference (I think) - mind, this isn’t actually a change in the value of the speed of light, but rather a change to the (relative) rotational velocity of a particle; a particle “experiences less time” as it approaches the speed of light in space. I expect this change to give rise to motion, rather than motion giving rise to the change in c, once fully considered.
However, I kind of expected this to just fall out of the equations in the form of a complex component somewhere whenever the value of c “changes”. We can plausibly get there if k is negative, and we swap out the natural log for the complex natural log - which might be the correct thing to do, given that we’re working in a complex plane. But, well - I’m out of my depth here.
I expect that the changes to the particles over time, with respect to their chirality (both of spiral and rotation - antimatter is traveling “backwards in time”, which matches behavior in more conventional physics, but may make more sense in this context, in which “time” has more to do with the progression of “change” than the progression of “history”), will give rise to the electromagnetic force; if we consider the motion of a particle whose distance is held fixed as the original spiral rotates (such as, say, yourself, in relation to the planet you’re standing on), it is rotating in a closed circle around the origin. Insofar as I understand the Kaluza-Klein theory, which granted isn’t much beyond a superficial understanding (in particular I don’t know how well the theory holds up in 2+1 dimensions, as opposed to 3+1+1 dimensions - the actual theory posits five dimensions, one of which is time, and one of which is a compact closed dimension), this should create the effect of electrical charge in and of itself in a relativistic framework. (I expect we can inject the above considerations into a proper relativistic framework by interpreting the equations of force above as a mass-energy distribution equation in GR.)
However, if motion does “change the value of c”, which remember was introduced in the beginning to refer to the motion of a particle through time - then motion should create a (relative) change in charge. I would expect that this is canceled out somewhere else, as, as far as I know, the Kaluza-Klein theory, while unfashionable, is consistent with evidence, and I’d expect motion in the closed dimension there to be likewise affected - but if it isn’t, this is a serious problem with this entire framework, as, as I understand it, charge is pretty well confirmed by evidence to be Lorentz-invariant. Perhaps the solution lies in the idea that distance and time are the same dimension, here? Maybe not. I’m still working on actually understanding the mathematics involved; my grasp on both partial derivatives and tensor notation are weak.
So, if anybody has reached this far - I’m kind of curious why, but thank you, anyways, for humoring my crackpot nonsense. I hope you enjoyed reading it, anyways.