May 25

How many photons does the universe actually need to work in the way we observe it? 213.205.242.170 (talk) 04:04, 25 May 2019 (UTC)[reply]

[1]--Jayron32 04:37, 25 May 2019 (UTC)[reply]

Or, if you want to put your noggin through the wringer, John Wheeler proposed an answer of "one". He was talking about electrons, not photons, but the same concept applies. --47.146.63.87 (talk) 05:22, 25 May 2019 (UTC)[reply]

Yes I could have asked about electrons insted. THeyre bothe pure balls of energy.--213.205.242.170 (talk) 21:51, 25 May 2019 (UTC)[reply]

Not true. Electrons are matter. They have a

rest mass. Photons are not matter because they do not.--Jayron32 22:50, 26 May 2019 (UTC)[reply

]

"Pure energy" is a term that is essentially meaningless, and one physicists generally don't use, though it tends to show up sometimes in

pop science stuff. It is true that both electrons and photons are both described by current mainstream physics as point particles, with no known internal structure. --47.146.63.87 (talk) 07:43, 27 May 2019 (UTC)[reply

]

"Pure energy" is physically equivalent to "Massless particle" and "moves at speed c", so this is totally meaningful, applies to photon and not to electrons. Gem fr (talk) 21:39, 27 May 2019 (UTC)[reply]

Organians were described as "pure energy".[2] But those were mythical beings. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:57, 27 May 2019 (UTC)[reply]

If you define "pure energy" as synonymous with "a particle with no invariant mass", then sure, it means something. Many non-experts use the term sloppily, though. Also, there's nothing terribly special about a particle with no invariant mass, which raises the question of why it's necessary to use such a term. The term implies there's some special property that particles with no invariant mass have but other particles don't. This is false. Theoretical physicist Matt Strassler states: Photons should not be called `energy’, or `pure energy’, or anything similar. All particles are ripples in fields and have energy; photons are not special in this regard. Photons are stuff; energy is not.. --47.146.63.87 (talk) 04:27, 28 May 2019 (UTC)[reply]

Always moving at speed c IS a "special property that particles with no invariant mass have but other particles don't". Isn't it? In particular, this means photons are stuff you cannot combine into other stuff (like atoms, etc.); said otherwise: they are not stuff, actually. Gem fr (talk) 14:35, 29 May 2019 (UTC)[reply]

Photons only move at c in vacuum. I mean, I suppose you can consider that special, but massive particles can get arbitrarily close to c. I think the core point Strassler is making is that photons aren't magical or uniquely different from all other particles: physicists use the same math. Traveling at c in a vacuum is just a side effect of having zero mass; photons could have mass, and (I believe) all that would happen would be a change in this effect. --47.146.63.87 (talk) 04:24, 30 May 2019 (UTC)[reply]

Photons can be converted into protons or other particles in pair production. (See also two-photon physics.) "Matter"—as the article discusses—is another term that modern physicists often avoid, again, because it tends to give misleading impressions. Particle physicists don't think of things in terms of "matter" and "not-matter". --47.146.63.87 (talk) 07:41, 30 May 2019 (UTC)[reply]

The difference between photons and 'pure energy' is that if you transfer a joule of 'pure energy' as photons, it could be gamma radiation or green light or UVC. It could be polarized left or right or up or down or roughly half and half in some quantum crypto code that has some kind of terribly important information in it. In short, photons are a lot of individual particles of light, which carry energy, but are more than just the energy they carry. Wnt (talk) 15:47, 29 May 2019 (UTC)[reply]

Right, and Strassler is saying that all particles have energy associated with them; photons are not special in this regard. --47.146.63.87 (talk) 04:24, 30 May 2019 (UTC)[reply]

I'm sure that I'm not understanding the concept, so please correct me where I'm wrong. If due to quantum fluctuations within the QED vacuum, positive-mass/negative-mass particle pairs are momentarily created, then it only makes sense for the singularity to lose mass when more negative-mass particles than positive mass particles were to fall in. Is there such a dynamic equilibrium whereby negative-mass particles preferentially, or exclusively fall in? Plasmic Physics (talk) 07:50, 25 May 2019 (UTC)[reply]

no, not "negative-mass". both particles and antiparticle have positive mass (if any mass at all, that is).

The singularity lose mass because it lose energy. because E=mc² so losing energy E is losing mass E/c². And it lose energy because of Hawking radiation.

Remember this is still unproven theory.

Gem fr (talk) 13:05, 25 May 2019 (UTC)[reply]

Remember that YOU are still an unproven theory. You are still consistent and explainable though. --Jayron32 15:33, 25 May 2019 (UTC)[reply]

when any particle comes from infinity to event horizon it was not there yet , it is bringing mass when come nearer , this may not mean evaporation of singularity ,but when anti particle creates we may be able to think the new born particle's mass , saving quantum information and obeying conservative of mass law , caches itself from black hole body .--Akbarmohammadzade (talk) 04:58, 29 May 2019 (UTC)[reply]

explainable, not so much (lots of hidden variables), but observable for sure, while we just got our first image of a black hole. So we are still some way to observing Hawking radation and a vanishing black hole. Gem fr (talk) 20:20, 25 May 2019 (UTC)[reply]

Wait a minute, how can both particles have positive mass/energy when net mass/energy cannot be created or destroyed? One particle must cancel the other for the law of conservation to be upheld. It is not enough for simply the charges to cancel. Plasmic Physics (talk) 22:40, 25 May 2019 (UTC)[reply]

again, E=mc². So conversion from energy to mass allows for creation of mass, at the expense of some energy. This has been observed quite a lot (for instance by CERN) Gem fr (talk) 12:03, 26 May 2019 (UTC)[reply]

Yes, I know this. Mass-energy equivalence does not address my question above. In fact, it is an axiom upon which my original question was constructed upon. Plasmic Physics (talk) 11:52, 31 May 2019 (UTC)[reply]

Moreover, if the singularity absorbs a particle with positive mass/energy, shouldn't it's mass increase instead of decrease? Plasmic Physics (talk) 02:54, 26 May 2019 (UTC)[reply]

Not if the original energy comes from it. Also, I smell some confusion: Hawking radiation comes from Apparent horizon, not the singularity itself Gem fr (talk) 12:03, 26 May 2019 (UTC)[reply]

As far as I'm aware, they energy comes from the QED vacuum fluctuations, not from the black hole - virtual pair production is a process that is independent of gravitational anomalies. I am quite aware of the distinction between an apparent horizon and a singularity, I was just mistaken in thinking the horizon and singularity was non-identical insofar as time-dependent processes as measured by the external observer, are concerned. Plasmic Physics (talk) 11:52, 31 May 2019 (UTC)[reply]

Time and energy are

Virtual particles normally exist for a short enough time that their existence does not conflict with conservation of energy. CodeTalker (talk) 03:47, 26 May 2019 (UTC)[reply

]

OK, but what about the question of my adenumdum? Plasmic Physics (talk) 04:35, 26 May 2019 (UTC)[reply]

This is approaching the limits of my understanding, so perhaps someone with more knowledge can chime in, but there are a couple of points to consider. When the virtual particle escapes the black hole, it is no longer slipping under the "short enough time" loophole in the law of conservation of energy, so to speak. It is now a real particle, and the energy embodied in that particle must have come from somewhere. The only place it can come from is the black hole's mass. Also, you are objecting that a virtual particle falling into the black hole should increase its mass. But the virtual particle isn't a real particle -- it didn't get created because there was enough energy to embody in a particle somewhere. Since it's virtual, it's not carrying energy/mass. Finally, I think the "virtual pair where one escapes and one falls in" model can't be pushed too far. It's not exactly what's really happening; it's an approximate model to help one understand the event without resorting to quantum mechanics calculations. CodeTalker (talk) 16:07, 26 May 2019 (UTC)[reply]

I remember seeing the same explanation in virtually the same words in something I read. But my mind boggles at a notion of a universe that "notices" that the bookkeeping has been cheated and "finds" mass to deduct. It seems like a very sloppy idea. What is stranger is that the Unruh effect dictates that the virtual particles seen by the accelerating observer are not there for the free-falling observer, i.e. for the person falling in the hole there is no Hawking radiation. I feel like there has to be a saner way to interpret this, using the same math. Wnt (talk) 21:36, 26 May 2019 (UTC)[reply]

A similar phenomenon actually occurs in classical physics. Consider the case of an accelerating charged particle, which is predicted by Maxwell's equations to emit electromagnetic waves. Now ask, is a charged particle at rest with respect to the surface of the Earth accelerating? Classical mechanics says no, but general relativity says yes. Electromagnetic waves are not observed to emanate from one. Now ask, is a charged particle free-falling to the Earth's surface accelerating? Classical mechanics says yes, but general relativity says no. Electromagnetic waves are observed. So did we disprove general relativity? No. Maxwell's laws are formulated for an inertial observer. In the case of an accelerating observer, inertial charged particles are expected to emit electromagnetic waves, but co-accelerating particles are not. So wtf? Well, a stationary charged particle is associated with a static electric field. To an inertial observer who is not in the same reference frame, that field is distorted, but still stationary relative to the particle. To an accelerating observer, that field is changing, and has a wave-like component. Someguy1221 (talk) 22:07, 26 May 2019 (UTC)[reply]

Fascinating point. How about observers at the Equator and North Pole? They feel acceleration in perpendicular directions, so they can scarcely be "co-accelerating", can they? But they remain at the same position relative to one another and, I assume, see no electromagnetic wave emission... Wnt (talk) 02:22, 27 May 2019 (UTC)[reply]

I do believe you would observe electromagnetic waves, and this is easier to see from the perspective of the observer at the north pole. Ignoring the rest of the Earth, the particle on the equator is essentially making a slow orbit around the particle at the north pole, though offset by a few thousand miles. Of course, charged particles following a circular trajectory emit electromagnetic waves, and incidentally is why the idea of electrons orbiting a nucleus was so bizarre to physicists prior to the development of quantum mechanics. Comparing again to Unruh radiation, what I find so peculiar about this type of classical phenomenon is that while all observers record the same interactions, perspectives differ as to who is actually emitting the radiation. I believe the focus on the electric field itself, and how it can change from static to dynamic simply from a change in reference frame can also make the quantum version seem less outrageous, even without understanding the math. The quantum

vacuum state is of course occupied by quantum fields. If a boring old electric field can interact with inertial particles differently than it does with accelerating particles, it's suddenly not so weird that quantum fields might do the same thing. And if a particle is just an excitation in a quantum field, maybe it's also not so weird that even the existence of a particle could be a matter of perspective. Someguy1221 (talk) 10:19, 27 May 2019 (UTC)[reply

]

Is this the paradox of radiation of charged particles in a gravitational field? (Happened to stumble upon that from equivalence principle.) --47.146.63.87 (talk) 07:49, 28 May 2019 (UTC)[reply]

Ditto, why should mass be deducted from the black hole to square the imbalance? The black hole isn't intrinsically connected to the escaping particle, because their existence is independent. Only when the captured particle is actually assimilated by the singularity, does it become connected. Secondly, why does the escaping particle become real, but not the captured one? This is all assuming that the virtual-to-real particle-pair transition model is accurate. If this is only an approximation of what is happening, what is actually happening? Lay it on me, I studied cosmology and quantum physics, including the likes of the Schrodinger equation. Plasmic Physics (talk) 06:06, 27 May 2019 (UTC)[reply]

An observer outside the event horizon of a black hole can't "see" anything beyond the event horizon. You can't "see" how much of the black hole's mass is in the singularity and how much is still infalling. For an outside observer, as soon as something passes the event horizon, its mass is added to the black hole's mass. In the case of Hawking radiation, the particle that falls in has negative energy according to a distant observer. (Basing this off the article.) Remember, mass and energy are the same thing, and in particle physics are routinely interchanged. Note that you can also extract mass–energy from a spinning black hole outside its event horizon, in the ergosphere. --47.146.63.87 (talk) 08:36, 27 May 2019 (UTC)[reply]

This make a whole lot of sense actually. So, I was sort of right by considering the particle pair as being negative-mass/positive mass, except that the captured particle only attains hypothetical negative mass, at the instant of capture at that, due to the relativistic nature of a black hole. Furthermore due to the above stated no-hair theorem, the conversion of the particle to negative-mass is not actually required, since the assimilation of mass/energy by a black hole across the horizon is instantaneous from the outside perspective. Is this correct? Except, now we've got a contradiction - what about the internal perspective? Shouldn't the internal observer measure capture particles as still having positive mass causing the singularity to gain mass/energy? Plasmic Physics (talk) 09:56, 27 May 2019 (UTC)[reply]

According to general relativty, a local observer actually can't tell when they've crossed the event horizon. I'm not an expert and could definitely be wrong about this, but I think that means there is no contradiction, because the local observer can't distinguish between a particle "emitted" at the event horizon as part of Hawking radiation, a particle that fell in from outside the event horizon, or a particle that popped up spontaneously inside the event horizon from vacuum fluctuations. Of course, part of the issue is we need quantum gravity to say what "really" goes on at the event horizon at the quantum scale. --47.146.63.87 (talk) 05:45, 28 May 2019 (UTC)[reply]

If you ask me, that would make the case for a contradiction even stronger, although not the same contradiction. As the local observer can't distinguish between events occurring either side of the event horizon, then they must by definition, observe no Hawking radiation. Plasmic Physics (talk) 11:31, 31 May 2019 (UTC)[reply]

This isn't my field, but I've read on occasion that

quantum tunneling. Some papers (unknown quality) include [3] [4] The second suggests there is a close connection between the two ideas -- I wonder if the classic textbook illustration of quantum tunneling through a barrier could be done with a "pair of virtual particles" appearing inside the barrier and leaving to opposite sides??? I looked that up and came up with this paper just now. My experience in physics though is that anything I can speculate, I'll find a source for, which people will 95% of the time tell me is bogus. :) Wnt (talk) 21:29, 25 May 2019 (UTC)[reply

]

Our virtual particle article does say "Quantum tunnelling may be considered a manifestation of virtual particle exchanges." CodeTalker (talk) 17:48, 26 May 2019 (UTC)[reply]

On the subject of virtual particles, I found this post by physicist Matt Strassler helpful. --47.146.63.87 (talk) 05:45, 28 May 2019 (UTC)[reply]