Trying Again With Crackpot Nonsense
Alright, let’s start over.
I think the piece I’m going to focus on, to begin with, will be the scaling factor, e^(-k*φ). This should approximate 1/d, where d is distance from the sun, for the scale of the solar system; we’ll permit some error, for these purposes, because I’m going to use θ instead. So I expect to get approximately correct values for Venus, Earth, Mars, Jupiter, and Saturn; it’s “okay” if Uranus and Neptune are off a bit, but ideally, they should also be correct.
Let’s also rederive our equations a bit.
The arclength of a logarithmic spiral is (k^2 +1) ^ 1/2 / k * (r2 - r1) - since we’re measuring from the origin, that is, r1 = 0, and assuming the arclength is the distance d we’re concerned with, this simplifies to:
d = r * (k^2 +1) ^ 1/2 / k
r = k * d / ((k^2 +1) ^ 1/2)
In our original equation, we divided out this k, then c, to do unit conversions. We’re not doing this here, because the assumptions that went into that attempt are no longer considered valid, at least not yet.
The fundamental equation for theta in our logarithmic spiral is:
r = a * e^(k * θ)
So:
θ = ln(r / a) / k
Plugging this back into our scaling equation, and substituting θ for an approximation of φ, we get:
e^(-k * ln(r / a) / k)
Which simplifies to:
e^(-ln(r / a))
Via exponential identities, this nets us:
a / r
Which, since we’re looking for an equation which approximately 1/d, is … well, surprisingly good.
1 / d ≈ a / r
Given d = r * (k^2 +1) ^ 1/2 / k, this means this equivalency holds for sufficiently large values of k; we’re not too worried about the value of a, as it is a constant.
Alright, our first sanity check passes. In the interest of showing my process, incidentally, I kept dropping that minus sign when I first wrote this out, and got r / a instead. I only noticed because my spreadsheet was more accurate, and r / a contradicted the mathematical results there, which suggested the value should be inversely proportional to r, rather than directly proportional to r.
And c is looking promising as a value for k. In terms of the scaling factor for our planets - I don’t feel the need to validate them specifically, at least not at this stage.
I didn’t really expect this, incidentally, and this sort of thing, iteratively, is why I’m still doing this instead of looking for a new way of approaching the mathematics: Things keep working out better than I expect. I was seriously expecting to have to play with the equations at length, and tinker with the constants, to get something that “looked” like 1/d.
Alright; the next question is whether or not our equation can net us our second 1 / d, to get us to the inverse square law. Neglecting the scaling equation:
y = a * sin(θ) * e^(k * θ)
θ = ln(r / a) / k
y = a * sin(ln(r / a) / k) * e ^ (k * ln(r / a) / k)
y = a * sin(ln(r / a) / k) * e ^ (ln(r / a))
y = r * sin(ln(r / a) / k)
We begin to see an issue, here. We need y to approximate 1 / d, as before. That’s not what we are looking at. What about the relativistic version?
y = a * sin(θ - φ) * e^(k * θ - k * φ)
This could, conceivably, work. Our scaling factor, which we’ve already determined, is e^(-k * φ), which we’ve already established can be equivalent to 1 / d. Via identity:
y = a * sin(θ - φ) * e^(k * θ) * e^(-k * φ)
So we can derive the inverse square law, here. The challenge turns to this portion of the equation:
a * sin(θ - φ) * e^(k * θ)
In a pure approach, we should expect this to approximately yield g, Newton’s gravitational constant, for the range of distances under consideration. That exponent, however, might turn out to be a problem for us; it really depends on what θ and φ mean. For y = e^x, as x approaches 0, y approaches 1; so when θ ≈ φ, the exponent part of the equation drops out. But the sine function approaches 0, erasing the entire equation. Pulling the a out of our a/r - twice - we get:
a^3 * sin(θ - φ) * e^(k * θ) = 6.6743 * 10^-11
And here I’m back to the point I’ve been stuck many times before - there are simply too many unknown parameters. Maybe k is c; that’s one constant, I still have another. Maybe the value of a is nicely small, such that between it and sin(θ - φ), we can counterbalance the tendency of e^(k * θ) to get insanely large. But also - I’m kind of bad at intuitive understand of rotation, and having two rotations, while sensible from a relativistic perspective, just makes it that much harder for me to think about this.
Let’s try anyways.
Using our previous definition of φ = θ * 0.99999997879961263496368333568363, we get:
a^3 * sin(θ * 2.12003873346 * 10^-8) * e^(k * θ) = 6.6743 * 10^-11
Substituting for θ, changing our parenthesis for visual clarity,
a^3 * sin[ln(r / a) * 2.12003873346 * 10^-8 / k] * e^[ln(r / a)] = 6.6743 * 10^-11
And, finally, substituting in d for r:
a^3 * sin[ln(k * d / (a * (k^2 +1) ^ 1/2)) * 2.12003873346 * 10^-8 / k] * e^[ln(k * d / (a * (k^2 +1) ^ 1/2))] = 6.6743 * 10^-11
Given a d of 1.48 * 10 ^ 11, and supposing k = c:
a^3 * sin[ln(299792458 * 1.48 * 10 ^ 11 / (a * (299792458^2 +1) ^ 1/2)) * 2.12003873346 * 10^-8 / 299792458] * e^[ln(299792458 * 1.48 * 10 ^ 11 / (a * (299792458^2 +1) ^ 1/2))] = 6.6743 * 10^-11
… well. I don’t see a quick and easy way to solve for a. Excuse me for a moment while I go plug random numbers into a spreadsheet.
So a ≈ 15.12, if k=c, and if all my other nonsense here is correct.
How well does this hold up for, say, Venus?
Not very well. Small deviations in distance, as expected, lead to exponential changes in our values.
It looks like we need a more proper accounting for, and examination of, the differences between θ and φ, if this approach is to work at all.