Mordred

You can choose to believe what you wish to believe. Both Marcus and I agree the paper isn't a good examination and glosses over essential details. The paper is also very clear on the observer limitation. Quite frankly I have studied far better examinations on BH event and apparent horizons. In particular numerous dissertations on the topic.

I agree with those details being missing. I was thinking along the same lines.

Well DM seeds early large scale structure formation so must symmetry break at an early stage. Right hand neutrinos is considered a candidate for DM. The double slit experiment has no bearing on this. The temperature for DM will depend on its density and degrees of freedom. (Chapter 3 and 4 of the early universe particle physics will discuss temperature contributions of different particle species). I already linked a copy earlier

To be honest I'm not sure myself how they are involving a white hole myself. There is quite a few assumptions expressed in the paper. As stated by myself earlier I wouldn't place too much faith in its accuracy. There is also details with regards to how the Penrose diagrams apply in different regions missing in that paper.

The reality is that you will have observer effects with time, energy etc. There is no philosophy behind that but well tested applications of GR. The paper you linked discusses some of those observer effects...

I never look at the math philosophically. I look at what the math predicts will occur. However I also treat Hawking radiation in the QFT regime. Apparent and event horizons do cause some differences. That paper has a key caveat in that it is only valid for an infalling observer. A BH will not evaporate in a finite time for an observer at infinity. Here is the Arxiv. https://arxiv.org/abs/1510.07157#:~:text=Assuming that the collapsing body,there is no event horizon. The other important detail is that Hawking radiation only occurs if the Blackbody temperature of the horizon exceeds the blackbody temperature of the surrounding universe. So how would you get a back reaction ? ( Temperature varies according to the observer)

I would love to have posted for four lol. Let's try a starter Schwartzchild metric Vacuum solution [latex]T_{ab}=0[/latex] which corresponds to an unaccelerated freefall frame [latex]G_{ab}=dx^adx^b[/latex] if [latex]ds^2> 0[/latex] =spacelike propertime= [latex]\sqrt{ds^2}[/latex] [latex]ds^2<0[/latex] timelike =[latex]\sqrt{-ds^2}[/latex] [latex] ds^2=0[/latex] null=lightcone spherical polar coordinates [latex](x^0,x^1,x^2,x^3)=(\tau,r,\theta,\phi)[/latex] [latex] G_{a,b} =\begin{pmatrix}-1+\frac{2M}{r}& 0 & 0& 0 \\ 0 &1+\frac{2M}{r}^{-1}& 0 & 0 \\0 & 0& r^2 & 0 \\0 & 0 &0& r^2sin^2\theta\end{pmatrix}[/latex] line element [latex]ds^2=-(1-\frac{2M}{r}dt)^2+(1-\frac{2M}{r})^{-1}+dr^2+r^2(d \phi^2 sin^2\phi d\theta^2)[/latex] Stress tensor Dust solution no force acting upon particle (not converted to polar coordinates) [latex] T^{\mu\nu}=\rho_0\mu^\mu\nu^\mu[/latex] [latex]T^{\mu\nu}x=\rho_0(x)\mu^\mu(x)\mu^\nu(x)[/latex] Rho is proper matter density Four velocity [latex]\mu^\mu=\frac{1}{c}\frac{dx^\mu}{d\tau}[/latex] Leads to [latex]ds^2=-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2=-c^2dt^2(1-\frac{v^2}{c^2})^\frac{1}{2}=\frac{1}{\gamma}[/latex] [latex]T^{00}=\rho_0(\frac{dt}{d\tau})^2=\gamma^2\rho_0=\rho[/latex] [latex]\rho[/latex] is mass density in moving frame. [latex]T^{0i}=\rho_0\mu^o\mu^i=\rho^o\frac{1}{c^2}\frac{dx^o}{d\tau}\frac{dx^2}{d\tau}=\gamma^2\rho_0\frac{\nu^i}{c}=\rho\frac{\nu^i}{c}[/latex] [latex]\nu^i=\frac{dx^i}{dt}[/latex] [latex]T^{ik}=\rho_0\frac{1}{c^2}\frac{dx^i}{d\tau}\frac{dx^k}{d\tau}=\gamma^2\rho\frac{\nu^i\nu^k}{c^2}=\rho\frac{\nu^i\nu^k}{c^2}[/latex] Thus [latex]T^{\mu\nu}=\begin{pmatrix}1 & \frac{\nu_x}{c}&\frac{\nu_y}{c} &\frac{\nu_z}{c} \\\frac{\nu_x}{c}& \frac{\nu_x^2}{c} & \frac{\nu_x\nu_y}{c^2}& \frac{\nu_x\nu_z}{c^2}\\ \frac{\nu_y}{c}& \frac{\nu_y\nu_z}{c^2} & \frac{\nu_y^2}{c^2}& \frac{\nu_y\nu_z}{c^2}\\ \frac{\nu_z}{c} &\frac{\nu_z\nu_x}{c^2}&\frac{\nu_z\nu_y}{c^2}&\frac{\nu_z}{c^2}\end{pmatrix}[/latex] Now how much of the above did you understand ? I still haven't included the apparent horizon which is not necessarily the same as the event horizon nor did I include Hawking radiation at this time.

The above makes little sense. Fermions which are antisymmetric relations still obey the conservation laws in application to the mass terms.

Rather pointless article. It didn't apply a single QCD formula. The formulas it does apply are well known FLRW metric formulas involving the equations of state of Lambda and DM. We cross posted the second article isn't much better. Treating Lambda as a force or as negative pressure isn't particularly accurate as lambda doesn't have a potential gradient. So there would not be any net force in a particular direction. Pressure under GR has flux. Neither of these articles particularly help you defend your position with regards to the OP.

Text books are about the next best thing to formal training. I never pay attention to anything YouTube unless I can quarantee the poster is a well accredited physicist in the field of his or her expertise. ( The field of physics is highly diverse. I can quarantee someone like Swansont far beats my skills in his specialty. While I have my own specialty (cosmology)). So research on a topic should never be blind faith. If you cannot find numerous support on a theory by different professional opinions then be wary. Lol though I give credits to Studiot for applied engineering physics, Marcus for relativity and Janus for astrophysics. The information in this thread does not meet any criteria to question the second law in thermodynamics in any cosmology related studies I am familiar with including QFT related applications. You are absolutely correct to question the above. So +1 for that.

Well let's try an example if you have two particles in 3d space it would take 6 dimension values to define their location. However you can reduce the degrees of freedom between the two particles to 5 via the constraint relation for two particles at a fixed distance. This is the part that gets tricky as most examples are rigid bodies. However in GR there aren't rigid bodies. These little sticky points is where I would need advise. Chemistry for example involves a different descriptive for degrees of freedom for a state than I would describe for a particle. Studiot would describe a more engineering aspect. Which involves more constraints due to application design so the effective degrees of freedom can be reduced due to the additional constraints.

I understood that from the beginning. You will always have a delay of signal between two events regardless of valid (v<c) observer. A v>c will see time reversal and a v=c observer is invalid. Surprisingly enough GR does handle v>c observers but it suffers the same singularity conditions for v=c.

Night its a good question that is often confused. Your not the first to hit this stumbling block. So +1 For the question and attempt to understand. Any v=c reference frame is invalid because of garbage answers. Ie the object being everywhere in the universe at the same instsnce. An obvious impossibility. A photon for example doesn't exist the entirety of the universe simultaneous.

Any observer less than c is valid. However a reference frame v=c isn't. Any reference frame less than c will measure c as being constant regardless of the observer velocity. So you will have a separation distance between two events but never instantaneous. The v=c does lead to mathematical singularities. Another reason for being invalid as a reference frame ie a particle moving at near c does not form a black hole. Another common misconception.

As stated there isn't a valid reference frame for a v=c observer. The photon reference frame leads to nonsense conclusions hence invalid.

Lol I prefer answers that can be proven via experimentation or mathematics. Any time it comes to opinion without recourse from belief then its not provable. At least in my opinion lol. ( however that is a distraction from thread topic which involves commonly misunderstood terms such as dimension or degrees of freedom)

A V=c is not a valid reference frame as the separation distance is \(ds^2=0\) which would imply the photon exists everywhere in the universe which is obviously nonsense. The only valid reference frame is a v<c observer of the photon path. Transfer of information of any kind is limited to c. Where you are getting confused is that the v=c isn't a valid reference frame. You must use the separation distance of an observer less than c. In essence there is no instantaneous communication between any two events in any valid reference frames an observer with v=c isn't a vsid reference frame

Change your $ sign to [math] for latex. Your dang close to latexing on this site. Obviously you close with a [/math] instruction Example [math]\theta[/math] Also here here for simpler format \(\theta\) Last one was done in same format of the first post here https://www.scienceforums.net/topic/108127-typesetting-equations-with-latex-updated/

How can you get thrust from a gravity wave when the oscillations are fluctuating from a background spacetime geometry in contraction and expansion. Ie the x and y axis for the H+ or h× axis ? There isn't an inherent single direction but a quadrupole change. This has differences compared to a water like Dipolar behavior.

The difficulty with dimensions as opposed to degrees of freedom is a good topic to examine. There are factors that can reduce the needed degrees of freedom. I'm still looking into a simple way to highlight this detail. I have no issues with pinned threads in philosophy but I am the worse philosopher lol. So I cannot contribute in that department.

I truly have a difficult time making sense of your post. So let's deal with what I can make sense of. A CMB signal such as photons which is the mediator boson for the electromagnetic field. Can be recieved as static on our radio frequencies. This is due to Cosmological redshift. If a particle travels at c then the seperation distance is [math]ds^0[/math] however this shows that a reference frame of v=c isn't a valid reference frame. A different observer ( not v=c can) recognize the constant speed of light value. If you have some other thought in mind you will need to clarify.

Symmetric is the term your looking or in this case also commutative. (Under constant velocity) This is shown as the inner product of the Minkowskii group is symmetric via [math]\mu \cdot \nu=\nu \cdot \mu[/math] The equations are linearized however that doesn't necessarily describe reversible functions.

Your image cut off the top equation. Here is a hint the purpose to square the expectation value is to return a positive norm. All probability functions and energy densities are positive norm.

I have a copy of his second edition. It's a decent textbook. Griffith has a section on Dirac notation.

+1 to both of you. It's nice to see this caliber of thread on this forum. @Lizwi have you looked at Dirac notation yet ? You will find this notation helpful to understand https://www.google.com/url?sa=t&source=web&rct=j&url=https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_04.pdf&ved=2ahUKEwj16sf4sazqAhVTvZ4KHc8WCzsQFjAaegQIAxAB&usg=AOvVaw28oEJ_F36vW2fsOmfOJg2B This will help you in your studies.