General Discussion > Any progress?
Horace, I've been posting regularly in the blogs on this site. Have you been reading them? Evidently not, or you would not have asked the question you did.
Our goal in the beginning, trying to derive the atomic spectra, was ambitious enough, but I never dreamed that we would be able to derive the standard model, in terms of RST theoretical entities.
Yet that is exactly what is happening! By combining the SUDRs and TUDRs into SUDR|TUDR combinations (S|T units), and then combining these into S|T triplets, we are able to identify theoretical photons, electrons, positrons, neutrinos, and three "colors" quarks.
So far, we can combine the quarks into protons and neutrons, and combine these baryons into nucleons of elements, and we can combine quarks and antiquarks into mesons.
Not only this, but we can explain many of the properties of these quantitatively and even the origin of many mysteries that the LST approach raises.
To say the least, we've made a lot of progress. I plan on reporting the results at the next ISUS meeting, but I've been blogging it all along on the New Math and New Physics blogs of this site.
Our goal in the beginning, trying to derive the atomic spectra, was ambitious enough, but I never dreamed that we would be able to derive the standard model, in terms of RST theoretical entities.
Yet that is exactly what is happening! By combining the SUDRs and TUDRs into SUDR|TUDR combinations (S|T units), and then combining these into S|T triplets, we are able to identify theoretical photons, electrons, positrons, neutrinos, and three "colors" quarks.
So far, we can combine the quarks into protons and neutrons, and combine these baryons into nucleons of elements, and we can combine quarks and antiquarks into mesons.
Not only this, but we can explain many of the properties of these quantitatively and even the origin of many mysteries that the LST approach raises.
To say the least, we've made a lot of progress. I plan on reporting the results at the next ISUS meeting, but I've been blogging it all along on the New Math and New Physics blogs of this site.
April 5, 2007 |
Doug
Doug
Doug,
I read your blogs, hoping to learn how you get a sinewave out of the S|T combos, but couldn't find it anywhere.
Many physical experiments (e.g interference patterns) exhibit the sinusiodal shape of photons, but yours are triangular at best.
So forgive me if I am not so impressed with your development, but I will keep track of it...
Reagrds,
Horace
I read your blogs, hoping to learn how you get a sinewave out of the S|T combos, but couldn't find it anywhere.
Many physical experiments (e.g interference patterns) exhibit the sinusiodal shape of photons, but yours are triangular at best.
So forgive me if I am not so impressed with your development, but I will keep track of it...
Reagrds,
Horace
April 10, 2007 |
Horace
Horace
Horace,
We've discussed this on the ISUS forums extensively. Remember that the expansion/contraction of a volume has to be nonlinear, just like the pendulum. When the pedulum is at its highest point and reverses direction, it has slowed to that point non-linearly. When it is reaching the bottom and about to cross the mid-point, it is going at the fastest rate. The contraction to a point and subsequent expansion to a volume has exactly the same dynamics. QED
Doug
We've discussed this on the ISUS forums extensively. Remember that the expansion/contraction of a volume has to be nonlinear, just like the pendulum. When the pedulum is at its highest point and reverses direction, it has slowed to that point non-linearly. When it is reaching the bottom and about to cross the mid-point, it is going at the fastest rate. The contraction to a point and subsequent expansion to a volume has exactly the same dynamics. QED
Doug
April 11, 2007 |
Doug
Doug
Doug,
But the pendulum in your example is rotating around its pivot, hence you get the non-linear trig relationship that is related to geometrical rotation.
There is nothing rotating in your expansion/contraction of a volume. So what makes it non-linear ?
Horace
But the pendulum in your example is rotating around its pivot, hence you get the non-linear trig relationship that is related to geometrical rotation.
There is nothing rotating in your expansion/contraction of a volume. So what makes it non-linear ?
Horace
April 12, 2007 |
Horace
Horace
Let's use an analogy. For instance, a constant stream of air filling a balloon. In the beginning the rate of inflation is high because the volume of the balloon is small compared to the flow-density of the air, but, as the balloon expands, the volume to flow-density ratio increases, so the rate of inflation, or the rate of size increase, slows. The opposite happens when the air is released. The deflation rate increases as the volume decreases.
This is analogous to the swinging pendulum, but it's irrelevant that the pedulum is rotating about a point. It's the non-linear action of the rate of inflation that is compared to the non-linear rate of rotation that is important.
Doug
This is analogous to the swinging pendulum, but it's irrelevant that the pedulum is rotating about a point. It's the non-linear action of the rate of inflation that is compared to the non-linear rate of rotation that is important.
Doug
April 12, 2007 |
Doug
Doug
Doug,
In your baloon analogy you identify two factors: volume and flow-density that form a non-linear relationship.
In your pendulum analogy you have a pivot, mass, inertia and gravity. You get a trig relationship only because of the pivot. If you did not have the pivot and merely threw the mass straight up, then the competing phenomena (inertia and gravity), would yield a non-linear motion, albeit not based on trig relationsip anymore, but on mere square relationship of accelerated motion.
In expansion/contraction of a RSt volume, you do not identify any such competing factors, so the analogy is lacking.
Nothing in your model imples competing phenomena that would yield non-linear expansion/contraction. Are you proposing a "law" prohibiting discontinuous changes of "direction" ?
Even if you could explain the non-linear relationship somehow, you're still far away from explaining the sinusoidal relationship.
Regards,
Horace
In your baloon analogy you identify two factors: volume and flow-density that form a non-linear relationship.
In your pendulum analogy you have a pivot, mass, inertia and gravity. You get a trig relationship only because of the pivot. If you did not have the pivot and merely threw the mass straight up, then the competing phenomena (inertia and gravity), would yield a non-linear motion, albeit not based on trig relationsip anymore, but on mere square relationship of accelerated motion.
In expansion/contraction of a RSt volume, you do not identify any such competing factors, so the analogy is lacking.
Nothing in your model imples competing phenomena that would yield non-linear expansion/contraction. Are you proposing a "law" prohibiting discontinuous changes of "direction" ?
Even if you could explain the non-linear relationship somehow, you're still far away from explaining the sinusoidal relationship.
Regards,
Horace
April 13, 2007 |
Horace
Horace
Horace,
You are exactly right. However, no one has disproved it, as far as I know, and it's obviously a better approximation than what Larson had. So, until we have someone willing to investigate it more completely, I've been willing to lay it aside in favor of pressing on.
I do have some preliminary thoughts about it, though. For instance, if we take the radius of a unit circle within the unit sphere at two points, 90 degrees apart, these points will converge upon contraction to zero, coinciding at the zero point, and then they will diverge again, upon re-expansion to the unit size, returning to the original locations, separated by 90 degrees.
Thus, the angle theta, between the two radii, is reduced from 90 degrees to 0 degrees, upon contraction, and is increased from 0 degrees to 90 degrees, upon expansion. What we need to determine is the rate of change of the angle. The key parameter is the constant scalar increase. To my mind the increase in the area of the plane of the two radii, as a function of a constant, cannot be linear, because to enlarge a growing area at a constant rate requires an increasing scalar input, not a constant input.
Therefore, if we assume that the scalar input parameter is constant, then the rate of increase of the area, as the plane grows in size, must decrease, as a function of area. Hence, it's straight forward to establish the non-linearity of the function. However, proving that the curve of the function is a sine curve is another matter.
Maybe you could take that on as challenge, you think?
Doug
You are exactly right. However, no one has disproved it, as far as I know, and it's obviously a better approximation than what Larson had. So, until we have someone willing to investigate it more completely, I've been willing to lay it aside in favor of pressing on.
I do have some preliminary thoughts about it, though. For instance, if we take the radius of a unit circle within the unit sphere at two points, 90 degrees apart, these points will converge upon contraction to zero, coinciding at the zero point, and then they will diverge again, upon re-expansion to the unit size, returning to the original locations, separated by 90 degrees.
Thus, the angle theta, between the two radii, is reduced from 90 degrees to 0 degrees, upon contraction, and is increased from 0 degrees to 90 degrees, upon expansion. What we need to determine is the rate of change of the angle. The key parameter is the constant scalar increase. To my mind the increase in the area of the plane of the two radii, as a function of a constant, cannot be linear, because to enlarge a growing area at a constant rate requires an increasing scalar input, not a constant input.
Therefore, if we assume that the scalar input parameter is constant, then the rate of increase of the area, as the plane grows in size, must decrease, as a function of area. Hence, it's straight forward to establish the non-linearity of the function. However, proving that the curve of the function is a sine curve is another matter.
Maybe you could take that on as challenge, you think?
Doug
April 13, 2007 |
Doug
Doug
Doug,
I cannot do it without invoking geometry of some kind, and your reasoning precludes geometry for scalar magnitudes. Doesn't it?
I cannot do it without invoking geometry of some kind, and your reasoning precludes geometry for scalar magnitudes. Doesn't it?
April 16, 2007 |
Horace
Horace
Horace,
No, not really. The length of a line is a scalar magnitude, as is the area of a plane and the volume of a sphere. If the length of a line is increased by a scalar amount, it's length is increased in the direction it is defined. The same is true for an area or a volume. It's the definition of a line that requires a direction property that a scalar doesn't have, but once the line is defined, its length is the scalar part.
The volume of a sphere has no specific direction, only a definite location. Therefore, its magnitude is a scalar value only. However, the unique and interesting thing about the volume is that it contains the lines and the planes of all directions, centered on its location.
After thinking about it some more, I don't think that the 90 degree angle collapses, during contraction, as I was describing above, because the circumference of a plane shrinks in both directions, not just one. Therefore, the angle theta doesn't change, as the plane expands or shrinks. Only the area of the plane segment, defined by theta and the radii, changes.
Nevertheless, intuitively speaking, if we take the area of a given plane, in a volume, at a time t, and increase it by a scalar amount, then, at time t + 1, we increment the area, again, by the same scalar amount, and so on, for n increments, each increase will expand the area of the plane proportionately less, will it not?
Thus, if we plot the percent of increase in area versus the constant scalar increment, what will the curve look like? What if we do the same thing for a line within the volume? What about the volume as a whole? Will the curves for the line, plane and volume be the same, or will they differ?
Given your skill with spreadsheets, Horace, I'll bet you could do this in short order.
Regards,
Doug
April 17, 2007 |
Doug
Doug
Doug,
I looked into it and it immediately became obvious that graphing a volume versus an area (eg. gallons vs. acres) of an linearly vibrating sphere does not yield a sinusoidal realtionship. Neither does the first derivative (rate of change).
So this is a dead end for achieving a sine wave. (square or cubic splines at best).
Since the volume of a sphere in euclidean geometry is given as v=(4/3)*pi*r^3 ,no amount of differentiation or integration will give you a sine or cosine relationship between volume, area or distance of a linearly (triangularly) vibrating sphere.
However, the circumference or area of an intersection of two triangularly vibrating interpenetrating spheres will yield a sine relationship.
I looked into it and it immediately became obvious that graphing a volume versus an area (eg. gallons vs. acres) of an linearly vibrating sphere does not yield a sinusoidal realtionship. Neither does the first derivative (rate of change).
So this is a dead end for achieving a sine wave. (square or cubic splines at best).
Since the volume of a sphere in euclidean geometry is given as v=(4/3)*pi*r^3 ,no amount of differentiation or integration will give you a sine or cosine relationship between volume, area or distance of a linearly (triangularly) vibrating sphere.
However, the circumference or area of an intersection of two triangularly vibrating interpenetrating spheres will yield a sine relationship.
April 20, 2007 |
Horace
Horace
Hmmm, that's interesting, but i'm not sure that we're talking about the same thing yet. Let me try to explain my concept in more detail. For simplicity, let's treat the 1D case first. Since a 1D line, or a 2D plane, on the diameter of a volume, expands/contracts exactly as the volume itself, the dynamics of one is the same as that of the others.
In the case of the line, the expansion extends the origin in two, reciprocal, directions, simultaenuously. At the unit length, the "direction" of the expansion reverses, and the line begins to contract, with each end converging toward the origin. When the line contracts to a point, the "direction" of the contraction again reverses, and the cycle is complete.
Of course, what we are interested in, and what we want to plot, is the time rate of change in the length of the line. We are hoping that the curve will be a sine curve.
We know that if the time rate of change is constant, the curve will be a square wave, not a sine wave. Indeed, when we think of Larson's description of the "direction" reversals, this is the first thing we notice, as several people pointed out to him.
Larson was never able to answer his critics on this point, except to say that, since it had to be sineusodial, it was. However, what happens, when the line we are plotting, increases as the associated 3D volume increases? In other words, if the line is the diameter of the volume, instead of an isolated line, it will extend and contract at the rate that the volume expands and contracts. Thus, what we need to integrate is the scalar increase of the volume, not the line.
Clearly, if at time 0, the origin is a point, with no spatial extent, at time n, the length will have the value given by the volume function, but this function is not linear, as your equation clearly shows. Since we are integrating over a volume, the time rate of change of the diameter cannot be linear.
In fact, since the volume is a function of the cube of the radius, then it follows that the radius is a function of the cube root of the volume.
r^3 = v/((4/3)*pi)
r = (v/((4/3)*pi))^1/3
The time rate of change plot of the diameter of the volume, as a function of the volume, is what we want, but we want the volume to increase over time as a function of a constant, not at a constant rate. In other words, the effect of the constant on the size of the volume gets "diluted," as the volume increases with time.
This should be a simple calculus problem, but never having learned calculus, I can't quite formulate it. Intuitively, I know that at each increment of the clock, the volume increases less than it did in the previous increment. So, the equation would be something like
v = c+(cn),
where c is the integral, and n is the sum of the integration, at time t.
Does this make more sense?
Doug
In the case of the line, the expansion extends the origin in two, reciprocal, directions, simultaenuously. At the unit length, the "direction" of the expansion reverses, and the line begins to contract, with each end converging toward the origin. When the line contracts to a point, the "direction" of the contraction again reverses, and the cycle is complete.
Of course, what we are interested in, and what we want to plot, is the time rate of change in the length of the line. We are hoping that the curve will be a sine curve.
We know that if the time rate of change is constant, the curve will be a square wave, not a sine wave. Indeed, when we think of Larson's description of the "direction" reversals, this is the first thing we notice, as several people pointed out to him.
Larson was never able to answer his critics on this point, except to say that, since it had to be sineusodial, it was. However, what happens, when the line we are plotting, increases as the associated 3D volume increases? In other words, if the line is the diameter of the volume, instead of an isolated line, it will extend and contract at the rate that the volume expands and contracts. Thus, what we need to integrate is the scalar increase of the volume, not the line.
Clearly, if at time 0, the origin is a point, with no spatial extent, at time n, the length will have the value given by the volume function, but this function is not linear, as your equation clearly shows. Since we are integrating over a volume, the time rate of change of the diameter cannot be linear.
In fact, since the volume is a function of the cube of the radius, then it follows that the radius is a function of the cube root of the volume.
r^3 = v/((4/3)*pi)
r = (v/((4/3)*pi))^1/3
The time rate of change plot of the diameter of the volume, as a function of the volume, is what we want, but we want the volume to increase over time as a function of a constant, not at a constant rate. In other words, the effect of the constant on the size of the volume gets "diluted," as the volume increases with time.
This should be a simple calculus problem, but never having learned calculus, I can't quite formulate it. Intuitively, I know that at each increment of the clock, the volume increases less than it did in the previous increment. So, the equation would be something like
v = c+(cn),
where c is the integral, and n is the sum of the integration, at time t.
Does this make more sense?
Doug
April 21, 2007 |
Doug
Doug
Doug,
The derivative of a polynomial of the general form: x^(1/3) is always: 1/(3*((x^2)^(1/3)))
In the case of: v/((4/3)*pi))^(1/3), just substitute: v/((4/3)*pi) for "x" in the above formula.
Differentiation or integrating a polynomial NEVER yields a sine or cosine
I can't really decipher this sentence: "we want the volume to increase over time as a function of a constant, not at a constant rate..."
Do you mean "constant flow" as in "gallons per minute" ?
If yes - then indeed the radius of the volume will increase as : r=(v/((4/3)*pi))^(1/3), which is slower and slower just like the cube root
The derivative (rate of change) of a cube root is 1/(3*((x^2)^(1/3))) which indeed is a decreasing function...
The derivative of a polynomial of the general form: x^(1/3) is always: 1/(3*((x^2)^(1/3)))
In the case of: v/((4/3)*pi))^(1/3), just substitute: v/((4/3)*pi) for "x" in the above formula.
Differentiation or integrating a polynomial NEVER yields a sine or cosine
I can't really decipher this sentence: "we want the volume to increase over time as a function of a constant, not at a constant rate..."
Do you mean "constant flow" as in "gallons per minute" ?
If yes - then indeed the radius of the volume will increase as : r=(v/((4/3)*pi))^(1/3), which is slower and slower just like the cube root
The derivative (rate of change) of a cube root is 1/(3*((x^2)^(1/3))) which indeed is a decreasing function...
April 21, 2007 |
Horace
Horace
Horace,
Yes, that is what I mean. If the volume increased at a constant rate, the flow would have to increase to keep the volume increase constant, but I am talking about a volume increasing from a constant flow.
So, where does that leave us?
Yes, that is what I mean. If the volume increased at a constant rate, the flow would have to increase to keep the volume increase constant, but I am talking about a volume increasing from a constant flow.
So, where does that leave us?
April 21, 2007 |
Doug
Doug
It leaves us with a x^(2/3) parabola, and in case of periodicaly reversing flow - with a spline made out of these parabolas.
I would upload a graph of these functions, but you can't find a way to upload attachments to your site, and I cannot find the "Members Area" on the ISUS site anymore.
Horace
I would upload a graph of these functions, but you can't find a way to upload attachments to your site, and I cannot find the "Members Area" on the ISUS site anymore.
Horace
April 22, 2007 |
Horace
Horace
Horace,
You can email it to me at douglasbundy@hotmail.com. I've complained about the limited capabilities of this host's forum software, but it falls on deaf ears.
Doug
You can email it to me at douglasbundy@hotmail.com. I've complained about the limited capabilities of this host's forum software, but it falls on deaf ears.
Doug
April 22, 2007 |
Doug
Doug
Dane
April 24, 2007 |
Horace
Horace
Err typo... and no way to edit - aaaarrrrrrrgggghhh
Done
Done
April 24, 2007 |
Horace
Horace
I didn't get it. Hotmail probably blocked it. Please try again, but this time send it to douglasbundy@yahoo.com.
Thanks,
Doug
Thanks,
Doug
April 25, 2007 |
Doug
Doug
Sent it again
April 28, 2007 |
Horace
Horace
Have youmade any progrss recently?
Any new ideas ?
Horace