Mordred

Vacuum can have an energy density ta da lol. That energy density can easily approach infinity keep in mind my original statement had "as close as possible " that allows a QM interpretation on Planck units for cutoff though Gravity has no effective UV cutoff for the mass term. That's a large part of why gravity isn't renormalizable. The IR cutoff is already established. For the record I've had numerous discussions with some mistakes he has made in other articles of his. Sometimes I'm correct other times he is just didn't explain something accurately enough with regards to Victor Toth. Cool character though he's friendly and easily talked to.

How so no one knows what goes o beyond the EH however the equations do lead to the infinite density singularity which everyone agrees is the issue regarding the singularity condition.

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

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.

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 ?

gravitons are still a viable possibility you don't need them to describe a BH or the effects of Hawking radiation on a BH but its also not incorrect to do so. here is the thing about Hawking radiation a virtual particle pair must form outside the event horizon. Due to conservation laws all particles pop into existence as particle pairs primarily but not restricted to conservation of charge. (matter , antimatter for example) which Hawking radiation uses. the matter particle escapes to infinity while the antimatter particle falls in. Its a rather simplistic descriptive but the mass loss is due to being the anti particle of the pair. A photon is its own antiparticle. The difference between them isn't charge but rather its polarity. As a wave it obeys constructive and destructive interference. So anti-photons will annihilate with matter photons. Now this may or may not cause interference with other particles as all particles also have wave and particle like characteristics. However that is moot as the only thing needed is the antiparticle of the pair formed to fall into the BH regardless of what particle is involved. you would get a reduced mass through mass energy equivalence regardless if it is anti photons or some other particle type. As far s I know Hawking never did specify which particle was involved. His original paper simply had particle antiparticle pairs. As photons are generally used with blackbody radiation its the most common treatment. However you also have methods using entropy but entropy in particle physics related to effective degrees of freedom ie spin. charge , flavor, color, energy momentum etc. for example see here "Then Hawking’s black hole emission calculation [9, 10] for free fields gives the expected number of particles of the jth species with charge qj emitted in a wave mode labeled by frequency or energy ω, spheroidal harmonic l, axial quantum number or angular momentum m, and polarization or helicity p as Njωlmp = Γjωlmp{exp[2πκ−1 (ω − mΩ − qjΦ)] ∓ 1} −1 . (5) Here the upper sign (minus above) is for bosons, and the lower sign (plus above) is for fermions, and Γjωlmp is the absorption probability for an incoming wave of the mode being considered." https://arxiv.org/pdf/hep-th/0409024

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. 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

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.

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.

You have to understand that Hawking radiation is a thermodynamic process it's radiation equates to photons as the mediator for the EM spectrum which is used also for blackbody temperature. All equations involving blackbody temperature uses the photon as the mediator for its radiation terms. The other detail to recognize is that any object of any mass can be a blackhole if it's mass is contained in less than Schwarzschild radius. The Smaller the volume of the EH means the rate of Hawking radiation produced increases as the EH shrinks due to mass loss. The smaller the EH the greater the Hawking radiation. The singularity or as close to singularity as possible ie near infinite density would still have sufficient gravity as well as other related forces to maintain an EH even if that EH is smaller than a soccer ball etc.

At that range you wouldn't really need much deflection something as miniscule as a 1 degree defection would likely be sufficient if even that. Likely some form of craft that has sufficient mass to gravitationally causes a change in angle by using its thrusters to simply stay near the asteroid . One could also feasibly save fuel using solar sails to get there. At the mass of the asteroid tethering wouldn't be practical. Though the outgassing method is also viable. Those are the two methods I see as most viable out of the ones I'm aware of.

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.

Just to add to the good answers already posted. One hurdle to overcome is thinking of atoms in accordance to the Bohr model. Which modern physics knows to be incorrect. Instead the atom has a probability cloud with different configurations. All described via the Schrodinger equation. https://www.khanacademy.org/science/physics/quantum-physics/quantum-numbers-and-orbitals/a/the-quantum-mechanical-model-of-the-atom This article from Khan University has a decent coverage.

Theoretically viable it would depend on distance, asteroid size and composition.

Prior to the Higgs field dropping out of thermal equilibrium the universe would certainly have rapid expansion. Simply put all particles are in thermal equilibrium and massless. So their kinetic energy term for momentum certainly overpowered the potential energy terms from any fields present. Once the Higgs field drops out of thermal equilibrium and particles gain mass this definitely helps slow down the expansion. The slow roll stages of inflation corresponds to this. However not all particles drop out equilibrium at the same time even though the Higgs field has. They drop out a different times this can also be seen by the different particle generations of the SM model. For that matter even the Higgs field had stages of how it drops out of equilibrium. Every type of particle that drops out of equilibrium affects the expansion rate. So no Higgs was certainly not the last field to drop out of equilibrium. Any particle type can be treated as a separate field in a multiparticle state.

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.

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. \[v_r=H_O d\]

Your right I didn't bother responding to your logic. As I stated I lost interest. Particularly when you have statements such as information travelling faster than c that you cannot back up with any real physics or mathematics. This includes your holonomic toroid allowing a faster than c wave. This runs counter to well known and understood physics. So any logic based on this is meaningless if you cannot show how that's even possible under mathematics using known physics. Another example is some mysterious toroid travelling at c. It must be something massless to do that. However you can't describe it beyond your verbal claims. I also have no interest in downloading a paper from an outside source when the rules requires that material to presented here.. Who knows you might catch my interest once you start applying some real physics or mathematics. Rather than nothing more substantial than your logic

Sorry to break to you the math is always relevant on physics. You will never convince any professional physicist without that math. As it's your model and conjecture I certainly will not do the work for you. In essence all we have is your claim. With nothing more substantial than a claim. Quite frankly I have already provided clues on what would be needed to prove a hidden variable with regards to the math. The geometry itself is extremely easy. Quite frankly there simply isn't anything of substance beyond your claims. So I have no further interest GL.

The first two paragraphs are accurate enough. It's more accurately described by As a result of expansion particle fields including the Higgs field drop out of thermal equilibrium in accordance to thermodynamic ideal gas laws involving tempersture/ density/ pressure and volume relations. Once the Higgs field drops out of equilibrium particles acquire mass leading to electroweak symmetry breaking. All particles and particle fields has a temperature contribution As for any personal proposals this isn't the section for that. If you choose to pursue personal ideas and a personal hypothesis our rules require that gets done in our Speculation forum. We may not currently know the cause of the cosmological constant. It may be quantum fluctuations due to the Heisenburg Uncertainty principle of the quantum harmonic oscillator which all fields are effected by. Or it may be the Higgs field. There is plenty of research papers suggesting either possibility however nothing is conclusive enough to make any determination between those two possibilities

I beg to differ on this score the FLRW metric is a GR solution and in GR time has dimensionality of length via the Interval (ct). It is that relation that includes length contraction and time dilation. Whether or not its required depends on the spacetime geometry. The simple reason you only really need the spatial component is that observational evidence shows a flat spacetime geometry. That's not some arbitrary choice of the metric. That the findings of all observational evidence. We have very useful methods for seeking spacetime curvature terms at our disposal. One example is distortions curvature causes light paths to bend this leads to distortions. Those distortions are constantly looked for. They can also be useful such as boosting viewing distance by gravitational lensing. That's just one method of detecting spacetime curvature there are others. The point being the metric does factor in the time component simply by being a GR solution. It's simply not needed due to all observational evidence. As far as observer effects, we do indeed need to take those into consideration. The dipole anistrophy due to Earths motion through spacetime in relation to the object we are observing must be factored in. A clear example was the findings of the first Planck dataset that had a dipole anistrophy in its first dataset. That dataset didn't have the correct calibration. That led to all kinds of pop media and scrambling. The next dataset had eliminated that dipole as we then had a better understanding of Earths momentum. As well as other localized effects. There isn't any arbitrary choice made the FLRW metric is quite capable of dealing with curvature. It's simply not needed beyond the weak field limit. You really only need the Minkowsii metric for the weak field limit. In a flat curvature parallel beams will remain parallel. If you have positive curvature those beams will converge. They will diverge for negative curvature. The converging or diverging is detectable and quite apparent in spectography in particular....which makes hydrogen a particularly useful test for distortions in its spectrographic readings. In particular the 21 cm line. That is what spacetime geometry ddescribes. All major findings show miniscule at best curvature best fit of a global geometry is flat. So the FLRW metric follows GR in the appropriate manner described by GR for a flat geometry

https://en.wikipedia.org/wiki/Floquet_theory for A(x) aka Floquet coordinates https://personal.math.ubc.ca/~ward/teaching/m605/every2_floquet1.pdf https://www.cfm.brown.edu/people/dobrush/am34/Mathematica/ch2/floquet.html

Accelerator physics Frenet-Serret Frame/coordinates Hamilton form reference reference 1) https://arxiv.org/pdf/1502.03238 reference 2) Particle accelerator Physics by Helmut Weidemann third edition particle trajectory r(z)=ro(z)+δr(z) define 3 vectors as ux(z) unit vector ⊥ to trajectory uz(Z)=dro(z)dz unit vector || to beam trajectory uy(z)=uz(z)+ux(z) "to form an orthogonal coordinate system moving along the trajectory with a reference particle at r0(z) . In beam dynamics we identify the plane defined by vectorsux and uz(z ) as the horizontal plane and the plane orthogonal to it as the vertical plane, parallel to uy . Change in vectors are determined by curvatures " dUz(z)d(z)=kxUz(z) dUy(z)dz=kyUz(z) k_x and k_y are the curvatures in the horizontal and vertical plane. gives particle trajectory as \[r(x,y,z)=r_o(z)+x(z)U_x(z)+y(z)U_y(z)\] "where\( r_0(z)\) is the location of the coordinate system’s origin (reference particle) and (x,y) are the deviations of a particular particle from \(r0(z)\). The derivative with respect to z is then \[\frac{d}{dz}r(x,y,z)=\frac{dr_o}{dz}+xz\frac{dU_x(z)}{dz}+\frac{dU_y(z)}{dz}+\acute{x}(z)U_x(z)+\acute{y}(z)U_y(z)\] \[dr=U_xdx+U_ydy+U_zhdz\] \[h=1+k_{0x}x+k_{0y}y\] curvilinear coordinate beam dynamic Langrangian \[\mathcal{L}=-mc^2\sqrt{1-\frac{1}{c^2}(\dot{x}^2+\dot{y}^2+h^2\dot{z}^2)}+e(\dot{x}A_x+\dot{y}A_y+h\dot{z}A_z)=-e\phi\] reference 2) 1.8O and 1.81 see floquet coordinates below

Night

Lets put it this way. The SM model including QFT has been so successful that just like the Higgs boson. It was able to predict long before detection over 90 % of the standard model of particles. There is still open questions so it's not complete. However it is simply the best fit for predictability and observational evidence. The VeV is part of that for the Higgs. If it weren't for the VeV range prior to Higgs detection. CERN wouldn't have known what range to look for to calibrate it's detectors.