New Math, Old Theories

[ImplicitDemands]

12 hours ago, ImplicitDemands said:

9pi area over the integral of its radius is just 4

If it is factored into integral of the area -> A/(integral of r * integral of A) ->9pi/((3/1+1)^(1+1) x (3/2+1)^(2+1)pi) -> 9pi/(3/2)^2 x (3/3)^3 pi -> 9pi/(9/4 x pi) = 4; where r=3

Then you get your 4G/r for:

12 hours ago, Mordred said:

a¨a=−4G3(ρ+3P)+Λ3

 

Edited April 25 by ImplicitDemands

[ImplicitDemands]

On 4/25/2024 at 10:22 AM, ImplicitDemands said:

If it is factored into integral of the area -> A/(integral of r * integral of A) ->9pi/((3/1+1)^(1+1) x (3/2+1)^(2+1)pi) -> 9pi/(3/2)^2 x (3/3)^3 pi -> 9pi/(9/4 x pi) = 4; where r=3

Then you get your 4G/r for:

 

Technically, A/int(r)*int(A) would be how you get the 4 radius for 4/3pi*r^3. You get the 3 radius as the dividend because the radius was 3 before being moved forward by it's own radius along the z plane. Anyway later on it was thought you could fit 12 spheres around the center

https://plus.maths.org/content/newton-and-kissing-problem

But that doesn't work for the factoring of G based on the principle that one can fit only 8 spheres around the central sphere if they were all to be equidistant and all making surface contact. And only 4 of those 8 can be used to measure z. 

Edited April 26 by ImplicitDemands

[ImplicitDemands]

On 4/24/2024 at 4:51 PM, Mordred said:

This is the redshift equation(cosmological) that gets used at all ranges as it takes the evolution of matter, radiation and Lambda.

 

Hz=HoΩm(1+z)3+Ωrad(1+z)4+ΩΛ−−−−−−−−−−−−−−−−−−−−−−−−−√

 

Then Hz is the number in which to factor by the value my equations yield, to be concise about what I'm stating here. It has nothing to do with gravity bending light or even the curvature of space it has to do with the calculus not maximizing surface volume to a concise enough detail, which in turn effects the depth and therefore throws off the redshift measurements. So you really do have to measure it against the calculus if that is the case. 

Pixelate any 2 galaxies in the cosmos from the observatory. 

Edited April 28 by ImplicitDemands

[ImplicitDemands]

The equation for redshift tells you how far away both Galaxy A (nearer) is and Galaxy B (further) is. It tells you how much they have redshifted, and this gives you S(t), which is the distance light would have to travel to redshift to those Hz numbers based on the inverse square law. The problem is that when you redshift by the distance light of Galaxy B has traveled to get to Galaxy A, you find the "redshift discrepancy" between galaxy B's cumulative redshift from galaxy A's position to the observer, and what it should be based on Hz for Galaxy B. So either the distance is changing (and the universe is expanding) or there's some problem with the mathematics being implemented. The same issue could be the culprit, specifically for adjusting the Gravitational Constant, of the three body problem.

Unlike the calculus shown above my method only uses right triangles, this much I can tell you. When you deviate from a 45 degree angle, that's when problems tend to occur.

Edited April 30 by ImplicitDemands

[Mordred]

Well the universe is expanding its why the equations I provided show the expansion with the energy density and pressure relations. 

The basic relation however being Hubbles law. The greater the distance the greater the recessive velocity. Key note this isn't a kinetic based velocity but rather an observer based velocity that depends on separation distance.

\[v_r=H_O d\]

 

 

[ImplicitDemands]

3 hours ago, Mordred said:

Well the universe is expanding its why the equations I provided show the expansion with the energy density and pressure relations. 

The basic relation however being Hubbles law. The greater the distance the greater the recessive velocity. Key note this isn't a kinetic based velocity but rather an observer based velocity that depends on separation distance.

 

vr=HOd

 

 

 

I will admit that there is some truth to the fact that cosmological redshift may not be as much (or at all) connected to the three body problem, nor might it be a problem with the math. But the three body problem is certainly the result of a problem with the mathematics in the way that I have described. As a physicist, I know of other culprits for cosmological redshift that do not really require even calculus to define mathematically, as much as the physics concepts. Although you still need calculus to compare the simplistic math to, it is totally unrelated to the New Math I have been describing so far. It is not new math it is basic arithmetic, the table produced does describe that redshift will increase overtime (tired light). Is my answer related to the curvature of space and how this effects light? Yes and no. Does my solution involve a novel interpretation of quantum physics? Yes. Does my interpretation affirm that the universe is expanding? No. 

The outline I have for it needs revision as I was erroneously operating under the preconceived notion that the system is being guided by events that had occurred in the past instead of events that occur in the future. How egocentric to believe that, just because we have memory then causality itself must be bound by it. 

Edited May 1 by ImplicitDemands

[Mordred]

1 hour ago, ImplicitDemands said:

I will admit that there is some truth to the fact that cosmological redshift may not be as much (or at all) connected to the three body.

  Simply put the way it's done in current modelling is multi body as per a field treatment. One main disadvantage you are having is not knowing just how flexible, interconnected between models physics really is today.

For example using a very high particle count there have been some incredible simulations simply to test our theories and applying the formulas of mainstream physics. One of my favorite involved several supercomputers nearly a year if I recall but it is incredible in its detail.

It tested not only large scale structure formation, it also tested metalicity,(Big Bang nucleosynthesis). Galaxy formation,  etc.

It's really worth watching and then realizing that it's applying the mainstream mathematics.

This is an example of just how capable the way main stream physics does things mathematics truly is.

One further detail there is no restrictions on what mathematical method one uses. You can integrals, derivatives whatever you choose.  Physics will use any mathematical method provided it accurately describes the system or state.

It prioritizes symmetry relations for invariance to all observers, independence of coordinate choice etc for very good reasons. A good way to learn these is gauge group theory. Just to give you some idea of just how detailed our models are mathematically.

Truth of the matter is. If you can mathematically and accurately describe a given system or state etc. The method used isn't incorrect. It becomes a valid method. It may simply not be the most flexible method or may be too restrictive to what it can accurately describe.

Edited May 1 by Mordred

[ImplicitDemands]

When I revised that table it actually reads that the universe is doing something similar to expanding but it is not

Past->Future: Gravity FIELD gets turned off slightly before observation occurs

Future->Past: Anti-gravity FIELD gets turned on when signal was sent which looks to us like a gravity field being turned off as well.

But the process leading to the generation of these fields has not occurred yet. Memory runs off the basis of incorporating a series of other synaptic patterns into the current one. Causality exists outside of the subjective perception of the arrow of time that memory falsely manufactures. It just so happens that in my interpretation the wave-guide is retrograde only. 

 

On 4/30/2024 at 9:51 PM, Mordred said:

It prioritizes symmetry relations for invariance to all observers

If predicting the orbits of three gravitational bodies, my solution to the three body problem would use the central orbiting body (about the z-plane, vector k) as a reference point. From that point, space curves both inward and outward direction as you get closer to and further from the observer, numerically expressing the depth increasing G as you get closer, and decreasing it as you get further. 

Edited May 2 by ImplicitDemands

[Mordred]

What you described above really doesn't make much sense sorry to say. Particularly in how your describing causality in regards to past and future events in regards to gravity. For starters there is no antigravity.  

Also if you have a uniform mass distribution according to Newtons Shell theorem. 

 

[ImplicitDemands]

20 minutes ago, Mordred said:

What you described above really doesn't make much sense sorry to say. Particularly in how your describing causality in regards to past and future events in regards to gravity. For starters there is no antigravity.  

Also if you have a uniform mass distribution according to Newtons Shell theorem. 

 

It is not anti-gravity from our point of view, it is just gravity as far as we perceive it. But our perception of events in the past is rather irrelevant to cause and effect. Simply put what we see as a cause might actually be an effect.

I'd have to show you the numbers, why the pattern produces a redshift from our point of view, why fields get subtracted from our point of view, what matter and energy actually are, why the have wavelengths+frequencies, what those are, etc. 

Edited May 2 by ImplicitDemands

[Mordred]

We look into the past the further away we look. That's well established it's also why our equations use proper time and proper distance  in its equations. The mathematics you've shown so far do not have any time dependency. You haven't got anything equating to a rate of change.

Not from the equations you have so far posted and as how one measures time is relative to the observer you will need a GR treatment.

Edited May 2 by Mordred

[ImplicitDemands]

My revised survey was done on hand this week, it only tables a particle over a distance of approximately 9 planck lengths over about 18 planck times and that is where I saw the pattern of redshift but not enough data to really tell much else. A number was wrong too but that doesn't effect much else other than velocity. 

6 minutes ago, Mordred said:

We look into the past the further away we look. That's well established it's also why our equations use proper time and proper distance  in its equations. The mathematics you've shown so far do not have any time dependency. You haven't got anything equating to a rate of change.

Not from the equations you have so far posted and as how one measures time is relative to the observer you will need a GR treatment.

The mechanics treated this event as occurring over the same duration as if it had happened in positive time (meaning from what I did it was blueshifted). Really it doesn't matter which direction the arrow of time is pointing, the event still translates the same forward and backward but in reverse order. 

The reason cause precedes effect at my interpretation's version of quantum scale is because I could not find the mechanism for photon ejection in our arrow of time. You'd need to understand that matter is held by these fields that originated from events in the future, when those events occur some of these fields will be subtracted/replaced by new events. But the light is actually being ejected in retrograde causality (what we call observation of said signal). 

Edited May 2 by ImplicitDemands

[Mordred]

You still require some term for rate of change as well as some terms regarding force for gravity etc. If your equation is now different than what you have posted so far you should include it. Though as I have already mentioned a uniform mass distribution has zero gravity as per Newtons Shell theorem.

Even how we measure energy also depends on observer just as how one measures redshift depends on the observer.

For that matter how measures volume can sometimes depend on observer a good example being the event horizon of a BH. Different observers will measure the event horizon at different volumes and radius.

SR also teaches us that distance can also be observer dependent hence the length contraction of SR.

I'm sure as an engineer your familiar with signal propagation caused by an EM field. Time dilation can readily relate to this as the coupling constants of the SM produces the mass term.

Mass being resistance to inertia change. 

Just as I'm positive that you understand redshift involves frequencies which requires a rate aka time component.

Edited May 2 by Mordred

[ImplicitDemands]

I'm sorry this topic is titled, "New Math, Old Theories" not "Old Math, New Theories". There is a geometric method not explicitly shown in this thread, how I got the 3.8etc number as opposed to using derivation for area expansion and integration to define the radius of a sphere at one radii into the k vector. I have determined that it is a strong candidate for the solution to the three body problem but admitted however that it might not be the culprit for redshift, since redshift is SELF-inconsistent using calculus. Of course if you were to take this New Theory and apply it to, say, teleporting back in time, you'd need the New Math as well. 

[Mordred]

I don't care what the title is. If your claiming you formula does this or that  it requires the terms that relate to those claims..

Claiming redshift with no time component to describe frequencies is simply wrong old math or new math. Describing past and future gravity terms without anything relating to a force term is just as wrong.

Claiming details concerning different observers without a coordinate system is another example.

So far your equation only shows volume changes you need additional mathematics to do anything beyond that.

On 4/22/2024 at 10:06 AM, ImplicitDemands said:

Differential Calculus

A = pi(r^2) & A' = 2pi(r) x dA/dt; A'/A = (6pi x dA/dt)/(9pi)) -> (6pi x (3+3))/9pi = 2. That's 2r.

This is the equation you posted does it describe anything at all beyond change in area ?

The time component used in that equation would be observer dependent it's not proper time.

proper time using a coordinate system is this for Euclidean geometry (flat spacetime)

\[\Delta\tau=\sqrt{\Delta t^2-\frac{\Delta x^2}{c^2}-\frac{\Delta y^2}{c^2}-\frac{\Delta z^2}{c^2}}\] that's one of its simplest forms. The equation I posted only shows how to convert from coordinate time to proper time for 4d Spacetime using Cartesian coordinates it does nothing else...

to have it do anything beyond that requires additional mathematics its as simple as that

 

Edited May 3 by Mordred

[ImplicitDemands]

1 hour ago, Mordred said:

This is the equation you posted does it describe anything at all beyond change in area ?

That one doesn't, no. And I have already pointed out I did it wrong it was supposed to be:

1 hour ago, Mordred said:

A'/A = (6pi x dA/dt)/(9pi)) -> (6pi x 3)/9pi = 2. That's 2r.

The change in radius relative to observer was

On 4/25/2024 at 10:22 AM, ImplicitDemands said:

A/(integral of r * integral of A) ->9pi/((3/1+1)^(1+1) x (3/2+1)^(2+1)pi) -> 9pi/(3/2)^2 x (3/3)^3 pi -> 9pi/(9/4 x pi) = 4; where r=3

I had also performed that one wrong at first on the previous page. 

Edited May 3 by ImplicitDemands

[Mordred]

12 minutes ago, ImplicitDemands said:

 

I don't have need for this in the three body problem, as it is relative to the observer. 

precisely my point " its relative to the Observer. How do you define one observer from another ? How is it relative ? If I  have an observer a coordinate \(c_1,x_1,y_1,z_1\) living in a gravity well. What effect does it have from an observer moving a 0.99 c etc etc. You have no means of describing one observer from any other observer.

How do I know if you are using strictly Galilean relativity or Special relativity ? am I suppose to read your mind ?

Edited May 3 by Mordred

[ImplicitDemands]

2 minutes ago, Mordred said:

precisely my point " its relative to the Observer. How do you define one observer from another ? How is it relative ? If I  have an observer a coordinate c1,x1,y1,z1 living in a gravity well. What effect does it have from an observer moving a 0.99 c etc etc. You have no means of describing one observer from any other observer.

We have have the gravity as the G term. We don't need all that yet for testing the numbers, if you wanted to get more complicated and try and apply theoretical physics best fix the methods used if they are wrong about the numbers in predicting trajectories for completely different reasons i.e. purely geometrical. 

Best use a system we can test from an Observatory where we don't have to worry too much about it. Mars and its Moons would be perfect for the three body problem. 

[Mordred]

great so I employ full GR for an observer in each case which will get different answers ? Observers affect geometry. length contraction is part of SR. An observer moving at 90 % c won't see a circle. with the equivalence principle inertia has equivalence to gravity with regards to observer effects. Pythagorus theorem  doesn't even work  without conversions to restore Pythagorus theorem. So your triangles wouldn't work correctly.

  You do want your equation to be useful in some cosmology based measurements if the answer is yes then you will need to account for geometry changes. Your going to need to include the effects of curvature and observers in those coordinate changes.

Edited May 3 by Mordred

[ImplicitDemands]

43 minutes ago, Mordred said:

great so I employ full GR for an observer in each case which will get different answers ?

I wouldn't do that either. This a problem with the calculability of depth perception, it is essentially geometrically accurate animation. Earth is the only observer of Mars and its moons. 

7 hours ago, ImplicitDemands said:

If predicting the orbits of three gravitational bodies, my solution to the three body problem would use the central orbiting body (about the z-plane, vector k) as a reference point. From that point, space curves both inward and outward direction as you get closer to and further from the observer, numerically expressing the depth increasing G as you get closer, and decreasing it as you get further. 

On 4/26/2024 at 11:57 AM, ImplicitDemands said:

Technically, A/int(r)*int(A) would be how you get the 4 radius for 4/3pi*r^3. You get the 3 radius as the dividend because the radius was 3 before being moved forward by it's own radius along the z plane. Anyway later on it was thought you could fit 12 spheres around the center

https://plus.maths.org/content/newton-and-kissing-problem

But that doesn't work for the factoring of G based on the principle that one can fit only 8 spheres around the central sphere if they were all to be equidistant and all making surface contact. And only 4 of those 8 can be used to measure z. 

If we were to continue this line of discussion I would have to get back into this. But, alas, the hour is getting late and I must rest. 

Edited May 3 by ImplicitDemands

[Mordred]

So inertial observers.

By the way here's how to do the kissing number problem in four dimensions if your interested. 

 

THE KISSING NUMBER IN FOUR DIMENSIONS Oleg R. Musin

"In three dimensions the problem was finally solved only in 1953 by Sch¨utte and van der Waerden. In this paper we present a solution of a long-standing problem about the kissing number in four dimensions. Namely, the equality k(4) = 24 is proved. The proof is based on a modification of Delsarte’s method."

https://arxiv.org/pdf/math/0309430

you wish to use vectors well you have those relations here. Its rather detailed.  as you linked Newton and the kissing number above thought you might find it handy with regards to your spheres

 

Edited May 3 by Mordred

[ImplicitDemands]

I will say that when you go from a radius of 3 to a radius of 4, that this is accurate even with my maths if you look at the object straight on. However, if you were to do it this way when calculating depth you'd need to realize that if you fitted the 7, 11, or 12 other spheres around that surface volume (9pi) you would need to adjust your xy and yx planes because now all the other spheres need to be placed differently giving different perceived radii to stack further iterations upon. I don't think calculus takes this into consideration. 

I do know that when it's 8 spheres around the sphere of origin the way I do it if you took the two spheres on the bottom and right and had them where you have the two spheres in front it would look exactly the same as it did before. No change to the xy and yx planes. 

Edited May 3 by ImplicitDemands

[ImplicitDemands]

On 4/25/2024 at 10:22 AM, ImplicitDemands said:

If it is factored into integral of the area -> A/(integral of r * integral of A) ->9pi/((3/1+1)^(1+1) x (3/2+1)^(2+1)pi) -> 9pi/(3/2)^2 x (3/3)^3 pi -> 9pi/(9/4 x pi) = 4; where r=3

Which yields accurate changes in depth perception over the z direction. But what about x and y? The objects along the horizontal and vertical plane if lined up at the same point in the z line doesn't matter because those objects are further away too but because they are at the same point in the z line the changes in depth for x and y can't be shown this way even though they are further from the observer looking straight on. It is a dimensional problem. Of course knowing this I'd figured out a wholly different way to get an accurate calculation of all dimension, it having to do with the post above.

This is the problem Newton would have had with the motion of more than two gravitational orbits not a physics problem, really it would be a problem for more than one object if it weren't for our ability to find the Lagrangian coordinate between the two. Due to the fact that redshift is based on the doppler effect it couldn't be a culprit for redshift but like I said my actual physics based dataset for the actual mechanics of light doesn't say enough to rule out tired light as a cause. It says there is an increase redshift over greater distances due to some assumptions about what light is and what forces at play. Obviously you'd need a larger dataset to measure against cosmological redshift. It even explains why it takes so long for light to be emitted from inside the sun very nicely and why neutrinos are exempt. It even says something about particle wave duality at close distances. It says a different color has nothing to do with the doppler effect at all. The light's wavelength isn't increasing because it is coming from an object that is moving away, or because it has some negative motion, which doesn't make sense to begin with you can't change the velocity of light, that's what relativity is about. 

Edited May 17 by ImplicitDemands

[ImplicitDemands]

On 5/16/2024 at 8:20 PM, ImplicitDemands said:

The objects along the horizontal and vertical plane if lined up at the same point in the z line doesn't matter because those objects are further away too but because they are at the same point in the z line the changes in depth for x and y can't be shown this way even though they are further from the observer looking straight on.

Quick clarification, this is difficult to explain. If you stack spheres on top of one another in the z line from one central point, but there are spheres left, right, top, and bottom around it about the x and y lines. The spheres you stack on top of those peripheral spheres (x and y) in the z line will be smaller from your point of view because they are not coming from the center. Not only that, if you were to use the shape of a cone with the pointy end facing toward an observer to factor in for how the dimensions of the circles around the central circle shrink, the circles will overlap the central circle! The viewing angle of the observer's perspective of gravitational bodies warps the dimensions they move along in a way calculus cannot geometrically fix. Of course if it is one gravitational body the central radius is all you need, and with two we can use where the Lagrangian coordinate aught to be to fix it. With 3 onward this fix becomes exceedingly arduous and approximated. 

You see with all the geometry we have we've never really found the solution to that particular problem, we speak of higher dimensions but we never really learned how to get a proper math for 3 dimensions. I have, I do know how that math works:

 

On 5/2/2024 at 10:48 PM, ImplicitDemands said:

I do know that when it's 8 spheres around the sphere of origin the way I do it if you took the two spheres on the bottom and right and had them where you have the two spheres in front it would look exactly the same as it did before. No change to the xy and yx planes. 

 

Edited May 19 by ImplicitDemands