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

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

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)

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.

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.

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

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.

Well as a physicist I can tell you that you need a lot of preliminary work to make what you have above presentable and complete enough to even have a hope of getting a PH.D on board. The question comes down to how far are you willing to work on your model to develop it ?

Unfortunately in order to compete with Newton Cartan theory as it is today one must show and prove ones theory or methodology is equally complete and accurate. Needless to say what you have is far too rudimentary to provide a reasonable alternative to the Newton Cartan theory. Needless to say a handleful of equations doesn't compare to what is involved in Newton Cartan theory. There is a reason for this, that reason is a physicist will want to be able to determine what will occur at every infinitesimal of every coordinate. This is the completeness of a theories predictability and a measure of its completeness. Needless to say you have years of preliminary work in front of you to approach this degree of predictability.

Which course and how far did you get into it ? This will help me provide a direction and give me an idea on your math skills. For example if you have never worked with tensors before that alone requires a huge study to use properly. Let alone understanding how Newton Cartan applies the Einstein feild equations under the Newton limit with quage group symmetries.

We cross posted while I was getting a link for affine connections see above. Here is the thing, learning tensors requires a large background in preliminaries including differential geometry. These are not the same mathematical objects one is used to as per algebra. Secondly when one is working in Newton Cartan one also must apply the Cartan connections between manifolds. Its not something that is easily or readily explained on a forum however one uses normalized units [latex]c=g=\hbar=1 [/latex] Cartan geometry follows different lemmas than Rheimannian geometry so one must be familiar with the differences. Well here is a 168 page article on Newtan Cartan. You can see fr9m this that the problem is far more complex than merely plugging in values in the above equation. https://www.google.ca/url?sa=t&source=web&rct=j&url=https://www.rug.nl/research/portal/files/34926446/Complete_thesis.pdf&ved=2ahUKEwj9vZykxrDaAhWpsFQKHe3EA4oQFjAHegQIBhAB&usg=AOvVaw2GAhJAD69j8f-S0aGy2o1c 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 ?

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. Here is a brief on affine connections https://en.m.wikipedia.org/wiki/Affine_connection

pretty cool app lol

You will find, if you haven't already done so that the universe temperature is roughly the inverse of the scale factor. There is several methodologies one can use to calculate the temperature at a given Z. One of the easier methods is to use the inverse or Gibbs law however one can also use the Einstein and Fermi-Dirac statistics. It isn't so much a result of energies as its a direct application of the ideal gas laws of a homogeneous and isotropic fluid.

,That isn't based on Koshes snowflake its a tensor, unfortunately the latex doesn't display the superscript and subscript without stacking it on top of each other. However is an antisymmetric tensor representing spatial rotation or torsion of the Yang Mills gauge connection. A and B are the SO(1.4) guage indices. [latex]A^A_B(x)=A_\mu A^A_B dx^\mu[/latex] under the Einstein summation rules for an antisymmetric tensor you will have a superscript indice followed by a subscript indice indicating an antisymmetric tensor. Unfortunately the latex typically stacks the two instead of the superscript followed by the subscript. The formulas I posted are in the coordinate basis of the four momentum/velocity for SO(1.4) the tensor signature is diag (-1,1,1,1,1)(diag=orthogonal) so in the equation in this post A=(0,1,2,3,4) however [latex]\mu=(0,1,2,3)[/latex]. This will correspond to the anti-Desitter spacetime as per certain variations of ADS/CFT. Remember in Newtan Cartan you have embedded fields with different symmetries. In this case we have electromagnetism and gravity. This complicates the Levi_Civita densities to [latex]\epsilon^{\mu\nu\rho\sigma}[/latex] for the SO(1.4) the rolling indices are ABC=0 to 4, for SO(3) i.j,k=(0 to 3), SO(3)=I,jk=(1,2,3) spacetime connection 4d [latex]\mu,\nu,\rho[/latex]=(0 to 3)the Cartan rolling is the tensor [latex]A^{AB}[/latex] with contact vector [latex]V^A[/latex] for the 4d or [latex]V^i[/latex] for the 2d. Torsion is [latex] \tau^A[/latex] for 4d [latex]\tau^i[/latex] for 2d. Does that help or overly confuse lol. PS the Poincare group is SO(1,3)

You haven't really shown anything other than a smattering of wild conjectures...in your OP. You have presented numerous different models and interactions in your OP that it is literally a random collections of ideas. Break each interaction within those your modelling as separate entities under the same field metric. Under fields scalar quantities/temperature are readily modelled. Any quantity that includes a direction of change are naturally vector. Now try to organize each and every coordinate as to what occurs with the chosen interaction. Sounds like calculus to me....try to organize variation of change of any measured value (regardless of what it represents) on a coordinate basis without vector calculus. Good luck on that Now here is the trick Newton Cartan is a torsional related model, good luck modelling torsion or any rotational interaction based theory without use of vectors and calculus. A physics model is pointless unless it describes some form of interaction, a model such as the ideas you have above has many. All not well described under a metric.

This makes absolutely no sense... Why would vector calculus not be applicable for starters ? The purpose of a geodesic equation is to model vectors ( freefall path under relativity) ie via action as one methodology upon particle free fall as per Newton for non relativistic or Einstein Cartan for relativistic. This is the distinction between the two theories. The very purpose of a geodesic So when describing particle or even a waveform under motion why would you not apply vector calculus ?

If you try to rotate one module for work/medical processes then you need the fuel to counterrotate the remaining station as the rotation would gradually affect the station via f=ma and friction. NASA has undoubtedly looked into the possibility of artificial gravity on the station via rotation but probably found the costs and ensuing problems of implementation too problematic to justify compared to simply having crew rotations for medical reasons. Its not the implementation that presents the problem but the cost justification.

You would need to rotate the entire station adding a centrifuge isn't sufficient. Then if you do so the problem is compounded on fuel costs and how difficult it would be to refuel the station.

What you have done is shown how little you understand why vector Calculus is used in physics. It also shows how little you understand the higher dimensions and how they apply under calculus. In Calculus a dimension is simply any independant variable that can change value in an equation without affecting any other variable within that equation. In string theory application it is often different potential regions within the overall global volume ie 3 dimensions for volume then the various field interactions from the fundamental forces. I looked over your post and can see no equation or methodology that reflects Newton Cartan gravity. You have none of the Newton Cartan gravity formulas that I can tell nor have you demonstrated a proper understanding behind the theory itself. Cartan gravity requires a Yang Mills symmetry group [latex] SO(1.4)[/latex] Can you demonstrate the affine connection [latex]\Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\sigma}(\partial_\mu g_{\sigma\nu}+\partial_\nu g_{\mu\sigma}-\partial_{\sigma}g_{\mu\nu})[/latex] unless you can produce the affine connections that will later involve the geodesic equations your theory is meaningless the Guage connection for Cartan gravity being [latex]A_{\mu^A_B}(x)[/latex]

Space by itself void of all forms of particles is simply a volume. One can measure that volume in units of quanta however there is no requirement that the space requires units of quanta as it expands. Such a study has taken place in the past and scientists did test this proposal but found no evidence of space being lumpy implying units of quanta. The tests using photons and neutrinos from distant supernova events showed space is smooth and not lumpy. Time itself is simply rate of change or duration and is a property of change not a thing unto itself.

Belief isn't very scientific. Charge handedness involves vector symmetry. This isn't based on belief but on the mathematical relations of vectors under symmetry. Right handed and left handed charge is rather meaningless as charge polarity itself is an arbitrary choice of which to assign as positive or negative. In point of detail handedness is involved in charges other than the electromagnetic. In all cases it is an assigned vector relation under graph ie an increase in right handed trends towards an increase in positive values. Its only fundamental purpose is its behavior when modelled under graphs in the same manner posotive and negative charge behaves when graphed. Charge is simply a term assigned to attraction. The stronger the attraction the greater the amount of charge.

You need to place [math] on each side in the latter half so I don't activate the command tags replace the % with a / [math] \Delta [%math] [math]\Delta[/math] You will need to refresh the page to see if it worked as well

Where is the fun of learning in that type of answer

Under Cosmology applications, as Swansont mentioned above the definitions vary. The vacuum is a scalar field with no inherent average directional component. The mean energy density itself can be greater or less than the mean ground state ie positive/negative vacuum. The ground state would be some arbitrary baseline state that is set at zero mathematically. Though its true value can be a nonzero energy/mass density.