Mordred

My wife is a member of our local Pagan society. I will day I personally like their views. The members I have met tend to be more open minded to the beliefs of others. As for myself I don't particularly follow any form of faith, however that's just me. I do know the local membership has been steadily growing in my locality.

Yes provided you use Maxwell Boltzmann statistics for mixed states or its QFT equivalent. http://arxiv.org/pdf/hep-th/0503203.pdf "Particle Physics and Inflationary Cosmology" by Andrei Linde http://www.wiese.itp.unibe.ch/lectures/universe.pdf:" Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis you can find further details in these two Introductory level articles. Both articles will provide an excellent overview of the major equations used in modern Cosmology.

The answer to that will depend on if the conservation laws of thermodynamics applies to the Universe. Under the LCDM (BB) model the universe is treated as conserved and expansion under thermodynamics is treated as an isentropic and adiabatic expansion. In essence a closed system. So under the LCDM model of the BB the photons become redshifted the density decreases but the total number of photons remain constant. A neat trick results from this mathematically. One can estimate the blackbody temperature at any given value of Z by using the inverse of the scale factor

Interesting thought experiment, I would like to think about this scenario a bit in terms of the stress energy momentum tensor and how it will correlate to the permutation tensor \(H_{ij}\). I will also have to dig into just how symmetric a supernova would be. The answer is yes regardless of how symmetric the core collapse occurs one other factor is that acceleration can also generate GW waves. Though a supernova collapse is never likely to stay symmetric. One simple reason being the typical bulge at the equator of rotation. That's not getting into details such temperature anistropies of the plasma etc. The Shockwaves themselves are sources of GW waves

Not really considering the temperature at the same time is roughly 10^19 GeV which when you convert to Kelvin isn't far off Planck temperature. Using the Bose Einstein statistics that equates to roughly the equivalent to 10^90 photons squeezed into a single Planckian volume. Good luck finding anistropy distribution under those conditions. (Also a symmetric state as all particles are in thermal equilibrium).

If it is its a good one, as a physical wavelength doesn't involve any probability so it's handy being able to readily distinguish between the two types of waveforms. (Physical vs probability). That honestly depends on the system being described for example the Earth due to rotation and Mountains (non uniform mass distribution) can generate GW waves. However the effect is incredibly miniscule. Any non symmetric spinning object can do so. However a symmetric sphere won't regardless of how fast it spins. In the case of inflation you get regions where the expansion rate may vary from other local regions. Though on a global average its roughly uniform. It is those local regions of non uniform expansion rates that can generate GW waves. At \(10^{-43}\) the universe would incredibly uniform in mass/energy density that no GW waves could result. Likely the earliest feasible GW waves would generate via electroweak symmetry breaking and inflation.

One trend I have noticed is the term wavefunction tends to indicate probability functions but the term wavelength tends to refer to physical waves such as Compton and DeBroglie. Do not know if this has become a convention but as an avid reader of numerous physics theories its a trend I have noticed.

We only examine our Observable universe. We can only conjecture based on what we learned from our Observable portion. However that Observable portion also equates to all forms of causality. That is a fundamental distinction. You agree that GR works great for describing what we can observe but that also applies to causality. In cosmology this defines the Cosmic event horizon aka our observable universe. GW waves have wavelengths that far exceed the size of our universe at the earliest stages ie inflation and prior. This has been calculated I recall Bardeen presented a solution for inflation giving the resulting wavelengths at \(10^3\) km for the quarterly wavelengths (needed detector length). Note that's well beyond the volume of the Observable universe at that time. We look for the frozen in effects (that is the literal descriptive oft given ) with regards to traces of relic dynamics due to inflation etc. However those traces are incredibly difficult to detect (relevant wavelengths including redshift due to expansion). I should add gravity waves only result from anisotropic conditions. That list includes mass/energy distribution. The EFE shows this. As you agreed it works well with observational evidence

Not accepting my challenge? Proof yourself mathematically. This is physics derive a QW wave with its wavelength using the mass of the universe. Then run expansion backwards to \(10^{-43} \) seconds. Let me know if you can fit the wavelength inside the observable universe at that time. Go ahead prove me wrong. In order to have a wave you must have a measurable wavelength

There is no assumption it is trivial to perform the calculations something you likely will not have for any of your declarations. You likely don't even know how spacetime via the Einstein field equations pertain mathematically to the GW wave. You cannot have GW waves without sufficient spacetime volume. Go ahead I challenge you to mathematically prove me wrong. Those mathematics is part of our rules and regulations when it comes to rigor and testability. Lol truth be told with your declarations above you likely have never bothered looking at the relevant mathematics of a QW wave. Those same mathematics involving Einsten field equations which you believe are wrong allowed us to predict the existence of GW waves long before we ever measured them.

All particles have pointlike characteristics as well as wavelike that is what specifically is described by wave particle duality. In thermal equilibrium however all particles are indistinct from one another. In essence they all would have identical wavefunctions as well as pointlike properties. So you could have gravitons as well as the SM particles however you wouldn't be able to tell them apart. Keep in mind though it's feasible we still have no evidence of a graviton. Gravity is well described without it however via spacetime curvature.

Might help to understand the BB theory starts at \(10^{-43} \) seconds. The theory describes how the Universe expands from that time forward. The theory does not state what created the universe. That is a common misconception. Expansion itself simply follows the thermodynamic laws. So there is nothing unusual there.

Not even close. Not possible,the BB singularity if you run expansion backwards you would hit a volume roughly one Planck length at 10^-43 seconds. At this volume even curvature has no meaning nor does a GW wave. For that matter as there is no curvature term due to the miniscule volume you wouldn't even have gravity.

with water droplets the light enters and leaves the droplet so refraction does apply https://phys.libretexts.org/Bookshelves/College_Physics/Book%3A_College_Physics_1e_(OpenStax)/25%3A_Geometric_Optics/25.05%3A_Dispersion_-_Rainbows_and_Prisms here obviously some light will simply reflect but you get refraction due to light entering and exiting the water droplets. Seriously if a water droplet were a perfect mirror then you would have 100 percent reflection. However a water droplet is semi transparent so naturally refraction will be involved. Anytime light enters a medium you apply refraction. With reflection its simply the surface or internal reflections in both cases Snell's law covers both. Angle of incident= outgoing angle for the reflected case \[\theta_i=\theta_O\] . with refraction obviously you have a much wider range. Quite frankly I can't think of any reason for color separation via reflection as opposed to refraction. I'm fairly positive your well aware of this though Studiot lol. Its still good info for other readers

Snells law of refraction n1sinθ1=n2sinθ2 n_1 is the incident index, n_2 the refracted index, θ1 the incident angle, θ2 the refracted angle. you can get further details here https://www.cis.rit.edu/class/simg232/lab2-dispersion.pdf the article covers the phase velocities and how differences in phase velocities will change the index of refraction however some examples for visible light apply this to red, yellow, violet in regards to a prism such as here https://www.vedantu.com/question-answer/dispersive-power-of-the-prism-and-its-si-class-12-physics-cbse-60d1f94e2f7a4d284fd9ce4a largely just a simplification. also its often referred to as the angular dispersion with regards to the prism

all particles have an antiparticle. Though in some cases some particles are their own antiparticle. photon for example its antiparticle is the antiphoton. The difference is the helicity as the photon doesn't have a charge.

Then you missed the details of Snells law with chromatic dispersion and the subsequent changes of index due to changes in phase velocity of Light through a medium. Are you claiming the following doesn't occur https://en.wikipedia.org/wiki/Dispersion_(optics) https://en.wikipedia.org/wiki/Phase_velocity https://en.wikipedia.org/wiki/Dispersion_relation Obviously some materials are non dispersive, however a prism or water vapours are. The properties of the material and its sometimes dependency of frequency to index of refraction is the determining factor. Some materials are also birefringent though they depend on the polarization. (calcite for example).

I mentioned earlier that the tests for CPT and Lorentz invariance are extensive as well as to high precision. To give a better idea here is a collective of the relevant datatables. Tables D15 and D16 apply to photons. These tables are a bit tricky as there is 44 total parameters 20 for CPT related. These additional parameters involve extensions to describe mathematically the relevant violation. https://arxiv.org/abs/0801.0287 Changing the photon direction is not the same thing as time reversal symmetry involved in the photon/antiphoton symmetry relations. Stimulated emissions involve momentum transfer. Yes you should have Lorentz symmetry in the momentum transfer however the time symmetry in Lorentz has distinctions from CPT symmetry which involve the Dirac equations. Lets do a simple thought experiment take a photon/antiphoton pair. The only difference is the polarity relations and the photon is its own antiparticle. Left hand, right polarizations. They both have the same energy/momentum so the transfer of momentum should be the same when interacting with an atom. We should be clear which time reversal symmetry are we testing for

No the article specifies super conductivity at room temperature under high pressure

My meaning by applying Snells law is to demonstrate in the maths as to how it applies we both agree it does.

For photon beams you would want the current densities of the polarity related wavefunctions for both both quantum and macroscale regimes. LOL you also need these relations for your readouts on your test equipment truth be told. We do push the umbrella at looking for CPT at respectable energy levels and are always pushing for test methodologies at higher. In point of detail its also looked for in our highest energy level test equipment such as our particle accelerators. Lots of ongoing research is still looking for new methods to test for it of course.

While I still feel its highly unlikely to get any CPT violations with the device. Its a nice change to discuss some serious physics. A couple of details to consider. CPT of photons tie into U(1) gauge symmetry. However it also ties in rather closely to Lorentz invariants. In answer to the question "has macroscopic tests been performed for CPT violations" then the answer in light of Lorentz symmetry and its connections with CPT would become Yes we have. This article is an example of tests performed on the International space station utilizing atomic clocks. https://arxiv.org/abs/hep-ph/0306190 this is the reference 14 of the above paper https://arxiv.org/pdf/hep-ph/0306190.pdf I chose this one simply due to its uniqueness. More commonly known methods involve synchrotrons, Penning traps, etc Here for example is the Zeus detector results. https://arxiv.org/pdf/2212.12750.pdf A synchrotron test https://arxiv.org/abs/0905.4346 You will likely find the information handy if anything it may grant some further ideas on how to modify your planned detector. As you can now include Lorentz invariants. Mathematically these papers use SME which which is the standard model extension to include violating terms. this article has a decent listing of the Langrangian for the major ones https://arxiv.org/pdf/hep-ph/0506054.pdf hope this helps, if anything its informative to other readers as well. Granted that also opens a few doors in what to look for with your experiment. This article will help get a handle on polarizations which you will need with your setup https://scholar.harvard.edu/files/schwartz/files/lecture14-polarization.pdf As it is informative I will add some mathematical relations that you may or may not choose to use but are of use. Useful relations Parity Many textbooks describe this as a mirror reflection its not quite accurate but its a useful analogy. Its more useful to use an inversion of coordinates for parity difference described below reflection of the x-y plane \[(x,y,z)\longrightarrow(z,-y,z)\] inversion (combines the reflection with a 180 degree rotation) the reason this is easier to work with is you don't need to choose a mirror plane. \[(x,y,z)\longrightarrow (-x,-y,-z)\] \(\mathbb{P}:(t,x,y,z)\longrightarrow(t,-x-y-z)\Rightarrow \mathbb{P}(\vec{V}=-\vec{V}\) gives \[\mathbb{P}((\vec{V})\cdot(\vec{W})=\mathbb{P}(\vec{V})\cdot(\vec{W})=(-\vec{V})-\cdot(\vec{W})=\vec{V}\cdot\vec{W})\] \[\mathbb{P}((\vec{V})\times(\vec{W})=\mathbb{P}(\vec{V})\times(\vec{W})=(-\vec{V})\times(\vec{W})=\vec{V}\times\vec{W})\] above denotes changes from left to right hand coordinate systems. note two types of vectors those that reverse signs under parity (vector V and those that don't pseudovector A gives \(\mathbb{P}:\vec{V}\times\vec{A}=-\vec{V}\times\vec{A}=-\vec{V}\times\vec{A}=\) a vector \(\mathbb{P}:\vec{V}\cdot\vec{A}=-\vec{V}\cdot\vec{A}=-\vec{V}\cdot\vec{A}=\) a pseudovector with twice applied parity \(\mathbb{P}^2=\mathbb(I)\) eugenvalues of parity plus or minus 1. scalars and pseodovectors have eugenvalue +1, while pseudoscalars and vectors have eugenvalue -1. with GR parity has the sign convention (+1,-1,-1,-1) \[(\Lambda_\mathbb{P})^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}\] under Lorentz transformation \[g_{\alpha\beta}(\Lambda_\mathbb{P})^\alpha_\nu (\Lambda_\mathbb{P})^\beta_\mu=g_{\mu\nu}\] by examining the photon in a hydrogen atom we can find the parity of a photon. this is too lengthy so you can get the gist here https://en.wikipedia.org/wiki/Spherical_harmonics of course QED rules further include the above. The photon for energy conversation has parity -1

Is the optical isolator you have there applying the Faraday rotator ? If so your going to need to account for the 45 degree polarity shift.

Lol your last statement shows you have little knowledge on how a particle physicists conducts his tests to determine particle properties.

What makes you believe only Einstein concluded c is invariant to all observers ? No physics work goes unchallenged that never happens. Every physics theory gets examined and tested by others. That is an essential part of the scientific methodology.