If Black holes slowly evaporate over time is there a point where they stop being a black hole?

[Mordred]

That's ok anyone can easily confirm the first equation is the photon propogator its in dozens of textbooks. It a standard form for the Feymann rules. Including the one I provided lmao

1 hour ago, MSC said:

Mordreds better at explaining his math in plain English than you are at refuting it in plain English. If he's made errors then for the sake of the people that understand the math, provide corrections.

Also chill with the personal attacks. Nobody gets graded for anything said here and cutting the attacks out of sincere debate is entirely within your capabilities. 

That's pretty much how I go about debating philosophical and ethical topics. Where I often stand corrected because it's not a precision science like particle physics is but these are pretty uniform academic standards that you don't even have to attend university to infer from just observing debates in different venues. 

I appreciate you taking the time on this topic. Maybe one day I'll understand the math!

however thanks for the support. 

1 hour ago, MSC said:

 

I appreciate you taking the time on this topic. Maybe one day I'll understand the math!

Takes time but it can be accomplished. If you really want to learn I would start with vectors and spinors learn their components and addition rules. Then do so under field treatments. Once you do that GR becomes rather easy. To get a handle on QM and QFT statistical mechanics is the next route as it also uses the vectors/ spinors.  Path integrals uses momentum space hence you need the above as its momentum space/phase space includes probability wavefunctions for all possible paths (path taken is the least action). The amplitude of that wavefunction giving the highest probability ( the weighted sum). Those probabilities are determined using Fourier transformations.  QFT uses the Klein Gordon equation which is Lorentz invariant. However QM uses the Schrodinger equation which is not Lorentz invariant.

This also means the photon propagator can come in different forms depending on the gauge its being applied in. There is also a couple of different forms for Feymann. So you can easily see a solid understanding of vectors/ spinors as standalone and as fields is essential and applies at all levels of physics and in every physics theory.

 Start there the rest will come far easier by default.

should also mention calculus of variations is also applicable for integrals

 

Edited May 6 by Mordred

[Markus Hanke]

17 hours ago, MSC said:

So lets say a stellar mass black hole about 10 times the mass of our sun. Like Gaia-BH1. 3 clocks. One in a safe stable orbit around the black hole, one fixed just before the event horizon, one in freefall.

Let’s further assume that clock (1) is far from the BH, and orbits slowly. This clock will see (2) to be still ticking, but at a much slower rate (compared to itself). It will see (3) to initially fall at increasing speed, while its tick rate gradually slows; as it approaches the horizon its descent will appear to slow more and more, and its tick rate appears slower and slower. Its visual appearance becomes dimmer and redder, and it eventually just fades into invisibility. But it will never be reckoned to have reached the horizon.

(2) sees (1) to be ticking at a much faster rate, and it will visually appear blue-shifted. (3) will appear similar as described above, just at a different rate; it will also never be reckoned to have reached the horizon.

For (3) itself, the time it takes to reach the horizon is finite; the fall time from horizon to singularity is also finite (for a 10 solar mass BH this will be on the order of ~150 microseconds). What tick rates it will see on clocks (1) and (2) will depend on where along its free fall trajectory the clock is - this is a bit of a balancing act between its own position in the gravitational well, and the degree of relative motion between the clocks. However, once it has fallen below the horizon, it should see both (1) and (2) to be ticking faster wrt itself.

17 hours ago, MSC said:

If I drop multiple clocks in free fall, an hour after the other, would the clocks seem to catch up to the first dropped clock, relative to the external stable orbit clock?

After the initial free fall period, they would all be seen to be asymptotically slowing towards the same region of space, while gradually fading from view. But they would never quite catch up.

17 hours ago, MSC said:

See I don't really know how to think about spacetime.

This is not simple, since it’s a mathematical model in four dimensions.

Spacetime is not a substance, so resist the temptation to think about it that way. The best I can offer is to think about it as a network of relationships - it tells you how clocks and rulers at different places and times are related to one another. Each nexus within the network corresponds to a physical event (meaning: a specific point in space at a specific instant in time), and has attached to it an object (called the metric tensor) which, if you tell it a direction, will give you the spacetime distance to its closest neighbouring nexus in that direction. So it’s a network of separations between events. In special relativity that separation is the same wherever and whenever you are (the metric tensor is just a matrix of constants), but in general relativity it may explicitly depend on where and when you are. For events that are more widely separated (not neighbouring), you add up all the individual separations between the nexus that are in between, so you perform an integration.

The global properties of that network as a whole influence local relationships, and the form of local relationships puts constraints on how the global network can look like. Also, if something changes locally, the effects of that will propagate outwards and “ripple along” the network. It’s a bit like a spiderweb.

Be careful about the trampoline analogy - it is just a visual plot that tells you how certain length measurements (rulers) are related along a specific coordinate in a specific geometry. It’s not a complete picture of what this spacetime would look like.

11 hours ago, Genady said:

Nothing. The 4D-spacetime curvature becomes infinite.

Technically speaking, it’s the limit of scalar curvature that diverges - the point r=0 isn’t part of the manifold, so no curvature tensors are definable there. This is why the technical definition uses geodesic incompleteness, and not curvature. That just as an aside

Edited May 6 by Markus Hanke

[Mordred]

25 minutes ago, Markus Hanke said:

 

Technically speaking, it’s the limit of scalar curvature that diverges - the point r=0 isn’t part of the manifold, so no curvature tensors are definable there. This is why the technical definition uses geodesic incompleteness, and not curvature. That just as an aside

An important detail as Markus just mentioned R=0, cannot be defined The mathematics breaks down is the short hand descriptive. Its also not part of any finite group. We can't define anything particle related there as well. For the same reasons...if you can't define the spacetime the particles would reside in. Its impossible to define any particle presence 

[Genady]

5 hours ago, Markus Hanke said:

Technically speaking, it’s the limit of scalar curvature that diverges - the point r=0 isn’t part of the manifold, so no curvature tensors are definable there. This is why the technical definition uses geodesic incompleteness, and not curvature. That just as an aside

 

5 hours ago, Mordred said:

An important detail as Markus just mentioned R=0, cannot be defined The mathematics breaks down is the short hand descriptive. Its also not part of any finite group. We can't define anything particle related there as well. For the same reasons...if you can't define the spacetime the particles would reside in. Its impossible to define any particle presence 

Exactly so. A value of a function being undefined somewhere is often part of a function's definition. It does not break math or anything.

A singularity exists, by definition

A notion of a particle "being there" is undefined, by definition  

Edited May 6 by Genady

[Mordred]

Funstrating isn't it. You know somethings there but you can't define it. Any infinite quantity has a finite portion but the finite portion is outside R=0

Edited May 6 by Mordred

[Genady]

1 hour ago, Mordred said:

You know somethings there but you can't define it.

I don't know. Occam tells me, it's nothing.

[Mordred]

 Unfortunately spacetime doesn't curve on its own. One could argue how to describe R=0 till they are blue in the face. Occams nothing is as good as any other way to describe it lol.

[Genady]

That's why I don't argue about it. I don't like my face blue. LOL

[MSC]

On 5/5/2024 at 11:50 PM, Mordred said:

however thanks for the support. 

Anytime. Solidarity within groups of reasonable people feels more urgently needed now more than ever. Without being too OT and alarmist, when dictatorships are at your door, eliticide is shortly behind them. 

On 5/6/2024 at 9:07 AM, Genady said:

don't know. Occam tells me, it's nothing.

I think black holes could just be matter in a transient state constantly in motion toward a convergence in spacetime from our perspective. There has to be an original core made up of the remnant star somewhere but the matter on it's way crosses the event horizon and reaches that core at vastly different rates relative to observer positions located outside of black holes. What pours into the event horizon trickles down to the core along stretching geodesics (did I use that term correctly?) I'm speculating from ignorance but who knows Jack O'Neill effect might give one of y'all an inspiration pump. 

Edited May 7 by MSC

[Markus Hanke]

I came across another good video, concerning in-fall into a Kerr BH (mass, angular momentum). This one contains explanations, and also readings on two reference clocks; I thought some readers here might like this:

 

[MSC]

16 hours ago, Markus Hanke said:

I came across another good video, concerning in-fall into a Kerr BH (mass, angular momentum). This one contains explanations, and also readings on two reference clocks; I thought some readers here might like this:

 

Awesome video! Thanks for sharing, but there was one detail about it that bugged me. Go to 1.50 in the video and watch what happens to external time when camera time hits 3.15.20.0000 external time jumps to 99.99.99.9999. What's happening with the clock? 

@Mordred I've been thinking a little. If we have found sterile left spin neutrinos, what is so different about right spin that makes them much harder to detect? Dark matter is more abundant than baryonic matter from what I understand so unless dark matter is a non local phenomenon, why havent we detected it yet and come up with any hard answers about the stuff that makes up so much of the universe?

Also what is meant by sterile in this context?

Another question; kind of OT toward QFT, but could dark matter be the sum of all currently existing virtual particles, making dark energy the vacuum energy that arises from the mass of virtual particles that comes out of it? As I understand it virtual particles aren't massless, have energy and momentum and they tend to share characteristics of their real particle counterparts and don't all have the same lifespans. We can't detect them either so I don't know. Contender or does latice field theory effectively eliminate the need for virtual particles?

[Mordred]

56 minutes ago, MSC said:

 

@Mordred I've been thinking a little. If we have found sterile left spin neutrinos, what is so different about right spin that makes them much harder to detect? Dark matter is more abundant than baryonic matter from what I understand so unless dark matter is a non local phenomenon, why havent we detected it yet and come up with any hard answers about the stuff that makes up so much of the universe?

Also what is meant by sterile in this context?

Ok this can get complex however  the main difference between left hand neutrinos and right hand neutrinos goes beyond simply being opposite in charge..Left hand neutrinos are doublet's while right hand neutrinos are singlets. Those terms directly relates to their respective cross section. In so far as their respective mass terms..

 Now originally it was felt that hand neutrinos would remain massless. (Hence a singlet) ,(also the sterile term) However later finding due to neutrino oscillations strongly indicate that as being incorrect. 

 The Higgs seesaw mechanism along with Majoranni mass terms indicate that the less mass the LHS neutrino the more massive the right neutrino would be.

Now neutrinos being extremely weakly interactive are very difficult to detect. What adds to the problem is the higher mass term it's out of the range of our particle accelerators. We simply cannot produce the amount of energy that would be needed this factor and being weakly interactive are two of the primary factors of why we can't detect them.

Now I realize very few ppl will understand the mathematics but I include them anyways along with the reference articles. (I have it already in my BBN thread on page 2

\[m\overline{\Psi}\Psi=(m\overline{\Psi_l}\Psi_r+\overline{\Psi_r}\Psi)\]

\[\mathcal{L}=(D_\mu\Phi^\dagger)(D_\mu\Phi)-V(\Phi^\dagger\Phi)\]

4 effective degrees of freedom doublet complex scalar field.

with 

\[D_\mu\Phi=(\partial_\mu+igW_\mu-\frac{i}{2}\acute{g}B_\mu)\Phi\]\

\[V(\Phi^\dagger\Phi)=-\mu^2\Phi^\dagger\Phi+\frac{1}{2}\lambda(\Phi^\dagger\Phi)^2,\mu^2>0\]

in Unitary gauge

\[\mathcal{L}=\frac{\lambda}{4}v^4\]

\[+\frac{1}{2}\partial_\mu H \partial^\mu H-\lambda v^2H^2+\frac{\lambda}{\sqrt{2}}vH^3+\frac{\lambda}{8}H^4\]

\[+\frac{1}{4}(v+(\frac{1}{2}H)^2(W_mu^1W_\mu^2W_\mu^3B_\mu)\begin{pmatrix}g^2&0&0&0\\0&g^2&0&0\\0&0&g^2&g\acute{g}\\0&0&\acute{g}g&\acute{g}^2 \end{pmatrix}\begin{pmatrix}W^{1\mu}\\W^{2\mu}\\W^{3\mu}\\B^\mu\end{pmatrix}\]

Right hand neutrino singlet needs charge conjugate for Majorana mass term (singlet requirement)

\[\Psi^c=C\overline{\Psi}^T\]

charge conjugate spinor

\[C=i\gamma^2\gamma^0\] 

Chirality

\[P_L\Psi_R^C=\Psi_R\]

mass term requires

\[\overline\Psi^C\Psi\] grants gauge invariance for singlets only.

\[\mathcal{L}_{v.mass}=hv_{ij}\overline{I}_{Li}V_{Rj}\Phi+\frac{1}{2}M_{ij}\overline{V_{ri}}V_{rj}+h.c\]

Higgs expectation value turns the Higgs coupling matrix into the Dirac mass matrix. Majorana mass matrix eugenvalues can be much higher than the Dirac mass.

diagonal of

\[\Psi^L,\Psi_R\] leads to three light modes v_i with mass matrix

\[m_v=-MD^{-1}M_D^T\]

MajorN mass in typical GUT 

\[M\propto10^{15},,GeV\]

further details on Majorana mass matrix

https://arxiv.org/pdf/1307.0988.pdf

https://arxiv.org/pdf/hep-ph/9702253.pdf

Edited May 9 by Mordred

[MSC]

1 hour ago, Mordred said:

Now originally it was felt that hand neutrinos would remain massless. (Hence a singlet) ,(also the sterile term) However later finding due to neutrino oscillations strongly indicate that as being incorrect. 

 The Higgs seesaw mechanism along with Majoranni mass terms indicate that the less mass the LHS neutrino the more massive the right neutrino would be.

So does that make RHS neutrinos WIMPs or when you say more massive than the LHS do you just maen relatively more massive but still not WIMPs?

Are the states of the LHS and RHS correlating toward them being an entangled pair or does that just not come into it at all?

1 hour ago, Mordred said:

Now neutrinos being extremely weakly interactive are very difficult to detect. What adds to the problem is the higher mass term it's out of the range of our particle accelerators. We simply cannot produce the amount of energy that would be needed this factor and being weakly interactive are two of the primary factors of why we can't detect them.

Are some of the bigger colliders that are allegedly in the works going to be capable of detecting them?

 

[Mordred]

15 minutes ago, MSC said:

So does that make RHS neutrinos WIMPs or when you say more massive than the LHS do you just maen relatively more massive but still not WIMPs?

Are the states of the LHS and RHS correlating toward them being an entangled pair or does that just not come into it at all?

Are some of the bigger colliders that are allegedly in the works going to be capable of detecting them?

 

All neutrinos are weakly interactive [WIMP] the M for massive simply denotes it has an invariant mass aka rest mass.

We hope they can but we only have theorized cross sections to know where to look in terms of mass.

[Markus Hanke]

6 hours ago, MSC said:

Go to 1.50 in the video and watch what happens to external time when camera time hits 3.15.20.0000 external time jumps to 99.99.99.9999. What's happening with the clock? 

What you’re referring to is the instant that the camera crosses the event horizon.

The reading on the bottom clock (“External time”) refers to what an external observer who is at rest far away from the BH would see. But the problem is that from such an observer’s point of view, the falling camera never reaches the horizon at all; just before it gets there, it will appear to fall slower and slower, and will visually appear redder and dimmer, until it fades into invisibility. There is no instant on the distant observer’s clock at which the camera is reckoned to have crossed the horizon - the entire region of the horizon and below cannot be mapped into the external observer’s coordinate system at all, because from his point of view such a region cannot be accessed (and vice versa, the falling camera cannot get back out either). Thus, once the horizon is reached, the times on the falling camera’s clock no longer correspond to any times on the external clock, even though the camera continues to fall and continues to accumulate time normally on its own clock. There simply is no longer any meaningful notion of simultaneity at all, whether relativistic or not.

So you see, time on its own in GR is a purely local concept, specific to a specific observer. It may not be shared by others. In order to ensure agreement between observers, you must now use covariant quantities instead.