Non-unifying Geometrized Newton-Cartan Gravity

[Mordred]

These days it takes something incredibly complete and groundshaking to make a physicist take note. Even among physicists who know all the right steps and mathematics its a difficult challenge to get a new model to gain an acceptance for study by other physicists. There is no easy shortcuts, and any idea or model requires a huge bulk of preliminary work. Without that work the best ideas possible simply won't get off the ground.

particularly since GR is so incredibly accurate and diverse in its applications. What I see in your OP isn't nearly ready.

Edited April 10, 2018 by Mordred

[inSe]

 

 Γρμν=1/2gρσ(∂μgσν+∂νgμσ−∂σgμν)

Let's look at everything I picked up on from inspecting this. Because that's the best way to tell how long it will take for me to develop, not where I left off academically

15 hours ago, Mordred said:

, (-1,1,1,1,1)

 

What I was thinking before is that is what's going on with the gamma (n - 1)!. I even say at some point after you gave me this that n is a tiny fraction so that whole left side of the equation is negative. But this part is doing (n-1) (n+1) (n+1) (n+1) (n+1) right?

9 hours ago, inSe said:

I knew the left side of the affinity equation would be negative looking at it and that didn't make sense to me. Now it does, it also makes since now looking at what you just wrote is 2i=A why the affine gauge is a 4d. 

But I'm not familiar with standard form yet. What I don't understand is what to plug in, or where. Esp for the so(3)

 

7 hours ago, Mordred said:

Have you worked with tensors and normalized units ? Also each coordinate infinitisimal of the particle worldline is a seperate calculation. An affine connection is not an affiity equation. 

Doesn't affine stand for affintity? affine map or an affinity (from the Latin, affinis, "connected with") 

7 hours ago, inSe said:

What I do know is n in gamma is going to be a tiny fraction considering this is particle physics, which is how I know the left side of the affinity is the asymmetric. It seems like the affine is the subscript of the gauge?

 

7 hours ago, Mordred said:

Now ask yourself the following question. Does the equations you posted even begin to touch upon Newtan Cartan theory in accordance with the lemmas and axioms under Cartan theory ?

I knew that from the beginning.

 

Now what I failed to realize is that the equation you gave me originally is a fractal: ignorantly saying:

 

"Then with a smooth manifold I have to get a rough koch/hilbert curve manifold before I can even start the geodesic"

But what you gave me was, in fact, the universe, not a koch/hilbert curve, whichwould have been a pointless modulation.

[Mordred]

No I gave you the affine connection. An affine connection is a tangent vector at any given coordinate. In the case of a field it is the tangent vector to each coordinate on a curved or flat geometry. Well that's a bit of an oversimplification.  see this wiki link

https://en.wikipedia.org/wiki/Affine_connection

Edited April 11, 2018 by Mordred

[inSe]

5 minutes ago, Mordred said:

No I gave you the affine connection. An affine connection is a tangent vector at any given coordinate. In the case of a field it is the tangent vector to each coordinate on a curved or flat geometry.

So I was right about gamma, & how it pertained to the four momentum/velocity

[Mordred]

If you claim so I fail to see that in what you have posted thus far. I don't see the coordinate basis of the four momentum in any of the equations you posted. Nor do you even allude to any coordinates that correspond to a field theory.

Edited April 11, 2018 by Mordred

[inSe]

13 minutes ago, Mordred said:

If you claim so I fail to see that in what you have posted thus far

Γρμν

Should produce a NEGATIVE integer, right? 

"(-1,1,1,1,1)"

 this part is representing the Γρ's operation of (n-1) x (n+1)^4 right?

13 minutes ago, Mordred said:

 I don't see the coordinate basis of the four momentum in any of the equations you posted. Nor do you even allude to any coordinates that correspond to a field theory.

We're talking about my awareness, not my knowledge. Am I recognizing what the equation should do when I plug something in, not whether I plugged in the right thing. I haven't plugged anything in yet because as I said earlier I don't know how to turn what's in my OP into what those variables need to be yet. With the proper tools, I can find out quicker than you might think. 

Edited April 11, 2018 by inSe

[Mordred]

No No No that is not what a metric signature is.

a Tensor is a type of metric the signature is the sign convention used in that tensor. Here is an example under the coordinate basis below

[latex]dx^2=(dx^0)^2+(dx^1)^2+(dx^3)^2[/latex]

the metric tensor of GR is

[latex]G_{\mu\nu}=\begin{pmatrix}g_{0,0}&g_{0,1}&g_{0,2}&g_{0,3}\\g_{1,0}&g_{1,1}&g_{1,2}&g_{1,3}\\g_{2,0}&g_{2,1}&g_{2,2}&g_{2,3}\\g_{3,0}&g_{3,1}&g_{3,2}&g_{3,3}\end{pmatrix}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex]

the sign convention diag (-1,1,1,1) is the diagonal components of the above orthogonal tensor

Which corresponds to

 

[latex]\frac{dx^\alpha}{dy^{\mu}}=\frac{dx^\beta}{dy^{\nu}}=\begin{pmatrix}\frac{dx^0}{dy^0}&\frac{dx^1}{dy^0}&\frac{dx^2}{dy^0}&\frac{dx^3}{dy^0}\\\frac{dx^0}{dy^1}&\frac{dx^1}{dy^1}&\frac{dx^2}{dy^1}&\frac{dx^3}{dy^1}\\\frac{dx^0}{dy^2}&\frac{dx^1}{dy^2}&\frac{dx^2}{dy^2}&\frac{dx^3}{dy^2}\\\frac{dx^0}{dy^3}&\frac{dx^1}{dy^3}&\frac{dx^2}{dy^3}&\frac{dx^3}{dy^3}\end{pmatrix}[/latex]

under your partial derivatives at each infinitesimal coordinate. You must apply matrix mathematics and follow the rules of tensor calculus to use the formulas I provided.

Edited April 11, 2018 by Mordred

[inSe]

2 minutes ago, Mordred said:

No No No that is not what a metric signature is.

a Tensor is a type of metric the signature is the sign convention used in that tensor. Here is an example under the coordinate basis below

dx2=(dx0)2+(dx1)2+(dx3)2

the metric tensor of GR is

Gμν=⎛⎝⎜⎜⎜⎜g0,0g1,0g2,0g3,0g0,1g1,1g2,1g3,1g0,2g1,2g2,2g3,2g0,3g1,3g2,3g3,3⎞⎠⎟⎟⎟⎟=⎛⎝⎜⎜⎜−1000010000100001⎞⎠⎟⎟⎟

the sign convention diag (-1,1,1,1) is the diagonal components of the above orthogonal tensor

Which corresponds to

 

dxαdyμ=dxβdyν=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜dx0dy0dx0dy1dx0dy2dx0dy3dx1dy0dx1dy1dx1dy2dx1dy3dx2dy0dx2dy1dx2dy2dx2dy3dx3dy0dx3dy1dx3dy2dx3dy3⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟

under your partial derivatives at each infinitesimal coordinate.

So I don't even know what this thing's doing really. Damn. 

[Mordred]

Takes a large amount of study to understand how tensors are used in field theories. Think of them as an organization of vectors almost like a datatable. The row and columm applied at each infinitesimal will be denoted by the subscripts and superscripts. For example

[latex]G_{\mu\nu}=G_{1,2}[/latex] indicates use the first row and second column of the above tensor. However don't feel bad most ppl don't know what a tensor is. Here this may help

https://www.ese.wustl.edu/~nehorai/Porat_A_Gentle_Introduction_to_Tensors_2014.pdf

This is one of the easier to understand articles teaching tensors.  (at least that I have found outside of textbooks)

Edited April 11, 2018 by Mordred

[inSe]

37 minutes ago, inSe said:

I don't know how to turn what's in my OP into what those variables need to be yet.

But I do know, & you know, that whatever rudimentary equations I did know in my OP, was somehow enough to match the measurements of BH proton evaporation rates & the rate at which pair-particle spins got synchronized in the Afshar experiment & other experiments like it assuming this synchronization is not instantaneous 

So whatever revelation that's described by those equations could be described by a geodesic. No matter the difference in that which can be defined by a geodesic (everything) & that which can be defined by those equations (the oscillation frequency of a BH with the mass of a proton=the oscillation frequency of a proton or the velocity of gravity of an electron plus the velocity of an electron equals the velocity of action at a distance). 

Edited April 11, 2018 by inSe

[Mordred]

Why would you think that? Have you ever looked at the mean lifetime of a proton?

[inSe]

2 minutes ago, Mordred said:

Why would you think that? Have you ever looked at the mean lifetime of a proton?

You're assuming the lifetime of what we think a proton is involves only one proton. What we're looking at when we measure a proton could literally be a billion different objects forming & vanishing in one location. 

Edited April 11, 2018 by inSe

[Mordred]

no that is not how particle decay work. In order for a particle to decay it must follow numerous conservation rules including conservation of mass, charge,baryon number,lepton number, spin isospin and parity. There is only so many allowable decays in particle physics (confirmed by experiment). Proton decay has never been observed and its mean lifetime is of the order 10^34 years as a lower bounds. (far older than the age of the universe itself lol

Edited April 11, 2018 by Mordred

[inSe]

How did time dilation for "λmax of the proton’s quasar is the proton’s normal λmax but to the negative power of the proton’s length divided by twice the Schwarzschild radius" literally equal to the entropy available in a quasar around a black hole with the mass of a proton match with the evaporation rate of a proton that size perfectly in my OP?

5 minutes ago, Mordred said:

no that is not how particle decay work. In order for a particle to decay it must follow numerous conservation rules including conservation of mass, charge,baryon number,lepton number, spin isospin and parity. There is only so many allowable decays in particle physics (confirmed by experiment). Proton decay has never been observed and its mean lifetime is of the order 10^34 years as a lower bounds. (far older than the age of the universe itself lol

New BHs smaller than planck (which any black hole with the mass of a proton would be like 10^-54 meters refer to OP) just keep forming around the general area for 10^34 years

[Mordred]

Utter nonsense A BH evaperates via Hawking radiation which has nothing to do with protons. It is a function of thermodynamics that if the surface of a BH blackbody temperature is higher than the blackbody temperature of the universe the BH can then evaporate. (Currently no know BH"s meet this criteria

[inSe]

I'm saying none of those concepts were described by a geodesic, or even a metric - but they don't to be in order to be right. They just can't define much at all without being described in a metric

1 minute ago, Mordred said:

Utter nonsense A BH evaperates via Hawking radiation which has nothing to do with protons. It is a function of thermodynamics that if the surface of a BH blackbody temperature is higher than the blackbody temperature of the universe the BH can then evaporate. (Currently no know BH"s meet this criteria

Not there, we're talking about matter, energy, anything that represents an increased thermodynamic state as squishing a three-dimensional euclidean geometrized brane together. Any increase in entropy represents a thinning of that brane. Which is why bhs evaporate quicker under high thermal pressure, or last longer when in a lower temperature 

[Mordred]

Nothing you have described matches the theories you have mentioned. I could have picked apart your OP to death, I however chose a more friendly approach to guide you into learning where the errors lie in. ( its often a better approach to teaching complex concepts) posters tend to get too defensive by the former approach and won't learn when on the defensive

8 minutes ago, inSe said:

 

Not there, we're talking about matter, energy, anything that represents an increased thermodynamic state as squishing a three-dimensional euclidean geometrized brane together. Any increase in entropy represents a thinning of that brane. Which is why bhs evaporate quicker under high thermal pressure, or last longer when in a lower temperature 

This is incorrect. If the temperature of the EH (blackbody is higher than the Blackbody of the universe evaporation can occur if the other way around the BH will absorb the surrounding temperature adding this to its mass. This process does not require branes which is a mathematical object.

Edited April 11, 2018 by Mordred

[inSe]

6 minutes ago, Mordred said:

This is incorrect. If the temperature of the EH (blackbody is higher than the Blackbody of the universe evaporation can occur if the other way around the BH will absorb the surrounding temperature adding this to its mass

Not if the EH is the surface of a sphere of three negative euclidean dimensions. The process you're referring to can be viewed as micro-BH absorption. Which is why doesn't necessarily occur.  Did you read my OP? 

10 minutes ago, Mordred said:

This process does not require branes which is a mathematical object.

Math is an uncovering of nature, not some esoteric invention. We like to think we invented it. We just gave it symbols. 

Edited April 11, 2018 by inSe

[Mordred]

Yeah and just what is a negative Euclidean dimension pray tell? Please feel free to post the coordinate basis of a negative Euclid plane

Edited April 11, 2018 by Mordred

[inSe]

1 minute ago, Mordred said:

Yeah and just what is a negative Euclidean dimension pray tell?

A negative density medium.

Edited April 11, 2018 by inSe

[Mordred]

Density is not a Euclid plane but a thermodynamic condition. Perhaps you may have greater success in gaining interest in your idea if you take the time to properly express key words and theories in the proper context and recognized definitions of said buzzwords

Edited April 11, 2018 by Mordred

[inSe]

8 minutes ago, Mordred said:

Please feel free to post the coordinate basis of a negative Euclid plane

Same as positive. It would be values that coincide with increases in mass. If positive & negative areas cancel out, you can imagine why. Constantly remove a distance, & you're constantly accelerating by the length of the distance removed. 

Edited April 11, 2018 by inSe

[Mordred]

How can a 2 dimensional object such as aa plane have either a positive or negative density ? That requires a volume which is 3 dimensional ( were not talking about probability densities in this case

Edited April 11, 2018 by Mordred

[inSe]

Just now, Mordred said:

How can a 2 dimensional object such as aa plane have either a positive or negative density ? That requires a volume which is 3 dimensional

I've never said plane, I've been using the word brane the whole time

[Mordred]

6 minutes ago, inSe said:

Same as positive. It would be values that coincide with increases in mass. If positive & negative areas cancel out, you can imagine why. Constantly remove a distance, & you're constantly accelerating by the length of the distance removed. 

Then why did you answer in this post in this manner. OK same question applies to a brane. What is the minimal dimension of said brane ? A point particle is a brane of zero dimensions