Can energy promote a virtual pair?

[tommac]

Can energy promote a virtual pair?

If we had a laser could we inject energy into the VP system to promote it?

[trinitree88]

Can energy promote a virtual pair?

If we had a laser could we inject energy into the VP system to promote it?

tommac. Yes. That's one of the three ways a gamma ray dissipates it's energy supply. They are:
1. Compton scattering off charged particles...mostly electrons, but not limited to.
2. ionization of neutral species, atoms or molecules.
3. pair production. electron/positron pairs mostly, but higher energies can produce more massive pairs, such as muon/antimuon, proton/antiproton, etc.

For electron positron pairs the minimum is a tad over 511 kev or 0.511 Mev per particle, so at an energy of 1.022 Mev, a gamma ray will produce pairs in air. It's why gamma ray bursts were discovered by early military satellites above the atmosphere in the sixties.

[Ken G]

Indeed, one can even say that simple scattering of a laser beam off of dust particles in the air is a kind of "promotion of a virtual pair". The photon you see is a promoted virtual particle, and the recoil of the dust particle is from absorption of the other member of the virtual pair.

[tommac]

tommac. Yes. That's one of the three ways a gamma ray dissipates it's energy supply. They are:
1. Compton scattering off charged particles...mostly electrons, but not limited to.
2. ionization of neutral species, atoms or molecules.
3. pair production. electron/positron pairs mostly, but higher energies can produce more massive pairs, such as muon/antimuon, proton/antiproton, etc.

For electron positron pairs the minimum is a tad over 511 kev or 0.511 Mev per particle, so at an energy of 1.022 Mev, a gamma ray will produce pairs in air. It's why gamma ray bursts were discovered by early military satellites above the atmosphere in the sixties.


Indeed, one can even say that simple scattering of a laser beam off of dust particles in the air is a kind of "promotion of a virtual pair". The photon you see is a promoted virtual particle, and the recoil of the dust particle is from absorption of the other member of the virtual pair.

So what other kinds of energy could be used to promote a virtual pair? Could gravity be used? Could tidal energy be used?

[WayneFrancis]

So what other kinds of energy could be used to promote a virtual pair? Could gravity be used? Could tidal energy be used?

again for the lurkers and again feel free to correct me if I'm wrong Gravity works on existing mass/energy. IE a photon that is already there can have its energy increased as it falls into a gravity well. A particle transfers some of its gravitational potential energy into kinetic energy. The "promotion" of a virtual particle because of gravity requires an event horizon.

Seems tommac is trying to revive Can A Neutron star produce radiation similar to Hawking Radiation? (http://www.bautforum.com/showthread.php/108134-Can-A-Neutron-star-produce-radiation-similar-to-Hawking-Radiation)

[Ken G]

So what other kinds of energy could be used to promote a virtual pair? Could gravity be used? Could tidal energy be used?Gravity and tidal energy are really just mass-energy, associated with rest mass. The only other kind of energy is the energy of motion, and that is frame dependent (including photon energy). So the answer can depend on the reference frame in ways that get extremely difficult, such as with Unruh or Hawking radiation. None of us have the expertise to say with any authority "where the energy comes from" for these, but we know that in the case of Hawking radiation, it ends up coming from the black hole energy. We can say that the net effect of Hawking radiation is to use black hole energy to promote virtual particles to real status, but just how that happens is much harder to put reliably into simple language.

[WayneFrancis]

Gravity and tidal energy are really just mass-energy, associated with rest mass. The only other kind of energy is the energy of motion, and that is frame dependent (including photon energy). So the answer can depend on the reference frame in ways that get extremely difficult, such as with Unruh or Hawking radiation. None of us have the expertise to say with any authority "where the energy comes from" for these, but we know that in the case of Hawking radiation, it ends up coming from the black hole energy. We can say that the net effect of Hawking radiation is to use black hole energy to promote virtual particles to real status, but just how that happens is much harder to put reliably into simple language.

Again Ken says what I was trying to say with much more eloquent words :)

[tommac]

Gravity and tidal energy are really just mass-energy, associated with rest mass. The only other kind of energy is the energy of motion, and that is frame dependent (including photon energy). So the answer can depend on the reference frame in ways that get extremely difficult, such as with Unruh or Hawking radiation. None of us have the expertise to say with any authority "where the energy comes from" for these, but we know that in the case of Hawking radiation, it ends up coming from the black hole energy. We can say that the net effect of Hawking radiation is to use black hole energy to promote virtual particles to real status, but just how that happens is much harder to put reliably into simple language.

Lets forget about Hawking radiation for a second ... what I am trying to confirm here is that it is impossible to create hawking radiation ( or similar ) from a non black hole) .... lets focus for a second on the promotion of a Virtual Particle.

So we are saying that with the energy from a laser we can promote virtual pairs. From previous posts it seems that the amount of energy required by this laser is the exact amount of energy that is equal to the mass of the VPs in the system.

I also believe ( please correct me ) that the mass of the VP pair is limited to the time of their existence as dictated by hup.

OK lets say that we take a long period of time ... say 1 second. If we have E=hv and v = 1 sec ( does that mean that the frequency of the wave thus the period of life of the VP is one second? ) then E=1.05×10−34 J .... please correct me if I am wrong here.

Now lets take ulm/4r^3 where u = standard gravitational parameter. i is the diameters of the VP system and m is the mass. r is the distance to the neutron star.

then could I just combine these two? I think I am slightly off:
using: 2.4×10^−17 m as the size of the VP we can solve for r:

1.05×10−34 J ( N/M ) = (264,000,000,000)(2.4×10^−17 m)(4×10^30 kg)/4(r^3)

(r^3) = (264,000,000,000)(2.4×10^−17 m)(4×10^30 kg)/(4)(1.05×10−34 J)

hmmm ... that seems that at a reasonably far radius that the tidal forces could provide enough energy to promote? However that is not possible right? why? wouldnt quantum tunnelling be needed?

[Strange]

Lets forget about Hawking radiation for a second ... what I am trying to confirm here is that it is impossible to create hawking radiation ( or similar ) from a non black hole)

Confirmed.


.... lets focus for a second on the promotion of a Virtual Particle.

Which has nothing to do with Hawking radiation.

[Ken G]

I also believe ( please correct me ) that the mass of the VP pair is limited to the time of their existence as dictated by hup. That's only if they don't get "promoted," so whatever process will promote them must be effective on that timescale.


Now lets take ulm/4r^3 where u = standard gravitational parameter. i is the diameters of the VP system and m is the mass. r is the distance to the neutron star.
That's a tidal acceleration, not an energy. If you just pick numbers out of the air and use formulae that are not relevant either, the results you get won't mean anything. To promote a virtual pair, you need to be able to extract energy from somewhere. Virtual gravitons are no help, so gravity force laws won't do anything. You need some actual mass-energy that couples to the virtual pair somehow, and the presence of an event horizon is somehow important in doing that. I can't tell you why, it has something to do with what Bogolyubov transformations do to the vacuum, and it won't happen in a neutron star, to my knowledge.

[Grey]

Tommac, when you use virtual particles as the mechanism for describing interactions between particles, pretty much every interaction that results in the creation of a particle can be viewed as "promotion" of a virtual particle to "real" status, with the energy of the interaction providing the energy for the new particle. For example, an accelerating charge radiates. In classical EM, the varying electric field (caused by the acceleration) then induces a varying magnetic field, and you get a stable wave propagating outward. If we describe the same situation using quantum electrodynamics, we could say that a charged particle is surrounded by a cloud of virtual photons (in QED, that's what charge "is": the ability to produce and interact with virtual photons). If you then accelerate the charge, you knock it away from the cloud, "orphaning" the virtual photons, and they become real, picking up some of the energy provided by the accelerating force.

Since any interaction can be described this way, you should realize that there's nothing particularly special about the link between Hawking radiation and the "promotion" of virtual particles. It's just one possible type of interaction that can be described that way. The specific mechanism that produces Hawking radiation requires an event horizon, and is related to the Unruh effect. So, sure, there are other ways of making virtual particles into real particles, but they don't have anything to do with Hawking radiation. In QED, that's just what interactions of all types are: particles (real or virtual) interacting with other (real or virtual) particles.

[tommac]

Which has nothing to do with Hawking radiation. Exactly!!!!

[tommac]

That's a tidal acceleration, not an energy.
So can I convert this acceleration to energy?

[tommac]

Tommac, when you use virtual particles as the mechanism for describing interactions between particles, pretty much every interaction that results in the creation of a particle can be viewed as "promotion" of a virtual particle to "real" status, with the energy of the interaction providing the energy for the new particle.

OK ... then you are saying that tidal forces CAN provide enough energy to promote to real status? or NOT ?

[Grey]

OK ... then you are saying that tidal forces CAN provide enough energy to promote to real status? or NOT ?I'm saying that one might be able to construct a situation where gravitational energy can be converted into radiation (the jury is out on whether a charge accelerated purely by gravity will radiate or not), but the tidal force of an object, regardless of its mass or density, won't just spontaneously radiate. Just because there's energy available does not cause that energy to be spontaneously converted into radiation. There has to be some kind of interaction going on that produces the radiation. Massive objects don't spontaneously radiate. Black holes do (well, we think they do, anyway, nobody's ever detected it, nor do we expect to unless we come across a really small black hole) because the introduction of the event horizon has some specific effects on the interaction with the vacuum.

So, to recap: You can't figure out whether an object will emit radiation by figuring out how much energy is available, and then calculating from that value what the most energetic particle it could emit would be. Instead, you have to determine what kinds of interactions are taking place, and determine if those interactions will lead to the emission of radiation.

To summarize the recap: According to our best understanding of neutron stars, they don't emit Hawking radiation or anything similar.

[Ken G]

So can I convert this acceleration to energy?Not to my knowledge, no. What is lacking, as Grey stated, is a specific type of interaction capable of doing it. Does that mean there isn't one? I have no idea, God didn't give me any Cliff notes.

[tommac]

but the tidal force of an object, regardless of its mass or density, won't just spontaneously radiate.

OK ... so wait ... what do you mean by spontaneously radiate?

Simple yes no ... can the tidal effects from a neutron star promote a virtual pair IF the force exerted on the VP system was higher than the energy required to promote? Another question is if a high enough level of tidal force can be produced by a neutron star ... but lets take that as a separate question.

[tommac]

Not to my knowledge, no. What is lacking, as Grey stated, is a specific type of interaction capable of doing it. Does that mean there isn't one? I have no idea, God didn't give me any Cliff notes.

So just to make it clear to me ... you are stating that even if the Neutron star was capable of producing enough energy via its tidal forces where its tidal energy was greater than the energy needed to promote the pair .... it would NOT promote the pair because only a specific type of interaction would be able to do the promotion. Does that correctly describe what you are stating?

[WayneFrancis]

OK ... so wait ... what do you mean by spontaneously radiate?

Simple yes no ... can the tidal effects from a neutron star promote a virtual pair IF the force exerted on the VP system was higher than the energy required to promote? Another question is if a high enough level of tidal force can be produced by a neutron star ... but lets take that as a separate question.

Simple "Yes/No" answers are not good because tommac tends to grab onto them without understanding the bigger picture then tries to apply the answer in a different context that it doesn't actually apply.

Let me bold


I'm saying that one might be able to construct a situation where gravitational energy can be converted into radiation (the jury is out on whether a charge accelerated purely by gravity will radiate or not), but the tidal force of an object, regardless of its mass or density, won't just spontaneously radiate. Just because there's energy available does not cause that energy to be spontaneously converted into radiation. There has to be some kind of interaction going on that produces the radiation. Massive objects don't spontaneously radiate. Black holes do (well, we think they do, anyway, nobody's ever detected it, nor do we expect to unless we come across a really small black hole) because the introduction of the event horizon has some specific effects on the interaction with the vacuum.

So, to recap: You can't figure out whether an object will emit radiation by figuring out how much energy is available, and then calculating from that value what the most energetic particle it could emit would be. Instead, you have to determine what kinds of interactions are taking place, and determine if those interactions will lead to the emission of radiation.

To summarize the recap: According to our best understanding of neutron stars, they don't emit Hawking radiation or anything similar.

[WayneFrancis]

So just to make it clear to me ... you are stating that even if the Neutron star was capable of producing enough energy via its tidal forces where its tidal energy was greater than the energy needed to promote the pair .... it would NOT promote the pair because only a specific type of interaction would be able to do the promotion. Does that correctly describe what you are stating?

Again for the lurkers, because tommac doesn't read anything I write any more. Tidal forces is NOT an energy source. The promotion of a particle in the HR sense REQUIRES an event horizon. A particle falling into a gravity well will emit radiation but this is different then the object causing the gravity well giving up some of its mass to promote the particle.

[Strange]

OK ... so wait ... what do you mean by spontaneously radiate?

Simple yes no ... can the tidal effects from a neutron star promote a virtual pair IF the force exerted on the VP system was higher than the energy required to promote? Another question is if a high enough level of tidal force can be produced by a neutron star ... but lets take that as a separate question.

To summarize all the respones so far: no.

[Strange]

So just to make it clear to me ... you are stating that even if the Neutron star was capable of producing enough energy via its tidal forces where its tidal energy was greater than the energy needed to promote the pair .... it would NOT promote the pair because only a specific type of interaction would be able to do the promotion. Does that correctly describe what you are stating?

Sounds about right.

Also, because you started out talking about it, Hawking radiation is not produced by tidal forces but by the presence of an event horizon.

[ShinAce]

...whatever process will promote them must be effective on that timescale.

End of thread!

A laser hitting a particle is an 'event'. The time of reaction is almost meaningless.

Tidal forces near a neutron star are not instantaneous. You need an event horizon to get past how slow this process can be.

You have forgotten to take time into account.

[tommac]

Tidal forces near a neutron star are not instantaneous. . Huh? Why wouldnt it be instantaneous? It would be as instantaneous as the popping into existance of the VP.

The topology already exists. Please explain why this is not instantaneous!

[ShinAce]

Huh? Why wouldnt it be instantaneous? It would be as instantaneous as the popping into existance of the VP.

The topology already exists. Please explain why this is not instantaneous!

No deal! I'm saying that it takes time for an acceleration to create a certain separation.

Please explain how the tidal acceleration from a neutron star will instantaneously provide enough energy to promote a virtual pair?

[Ken G]

So just to make it clear to me ... you are stating that even if the Neutron star was capable of producing enough energy via its tidal forces where its tidal energy was greater than the energy needed to promote the pair .... it would NOT promote the pair because only a specific type of interaction would be able to do the promotion. Does that correctly describe what you are stating?Basically yes, though I would say it a bit differently. First of all, there is no such thing as "providing energy" via tidal forces alone, you have to tap into some actual energy source, either the rest mass or kinetic energy of some body. That's where the energy needs to come from, and if all you have is a neutron star, then it must come from the neutron star. That only gives you a possibility to promote the virtual particle-- you still need an actual process that can tap into that energy on the required HUP timescale. That is apparently what having an event horizon affords you that you don't get if you don't have an event horizon. I don't know that there is no other process that could do it, I don't even know how an event horizon could do it, but there is no evidence of any such process for a neutron star, and no theory that I know of that predicts there should be.

[tommac]

No deal! I'm saying that it takes time for an acceleration to create a certain separation. Are you confusing WORK with ENERGY?

[Agemegos]

what I am trying to confirm here is that it is impossible to create hawking radiation ( or similar ) from a non black hole)

Confirmed. Hawking radiation is defined to be the radiation that comes from the exterior of the event horizon of a black hole as calculated by Hawking. It is not defined in terms of having been produced by the promotion of a virtual pair. Any process that occurs anywhere other than very close to the event horizon of a black hole is not the subject of Hawking's calculations, and is not covered by the term "Hawking radiation".

[tommac]

Confirmed. Hawking radiation is defined to be the radiation that comes from the exterior of the event horizon of a black hole as calculated by Hawking. It is not defined in terms of having been produced by the promotion of a virtual pair. Any process that occurs anywhere other than very close to the event horizon of a black hole is not the subject of Hawking's calculations, and is not covered by the term "Hawking radiation".

So exactly how close to say a 2 solar mass BH?

[loglo]

Tommac,
No-one knows, not even Hawking.

[Agemegos]

So exactly how close to say a 2 solar mass BH?

I'm not certain, but I think it is distance from the event horizon comparable with the uncertainty in position of the antiparticle to the particle emitted. In my dim understanding the process is analogous to quantum-tunnelling of virtual particles across the event horizon.

[Ken G]

A simple estimate would be to look at the frequency of the photon, maybe it is a gamma ray with frequency 1020 Hz, or maybe it is a visible photon with frequency 1015 Hz, I don't know what one would want it to be. Then say by the HUP that its uncertainty in period could not be more than 10-20 or 10-15 seconds, so its uncertainty in time could not be less than that. At the speed of light, that means we could be uncertain about which side of the EH it is on if it is created a distance less than c times those times, so 3 x 10-12 m or 3 x 10-7 m, which is another way of saying within its wavelength from the EH. Basically, pretty darn close to the EH.

[WayneFrancis]

A simple estimate would be to look at the frequency of the photon, maybe it is a gamma ray with frequency 1020 Hz, or maybe it is a visible photon with frequency 1015 Hz, I don't know what one would want it to be. Then say by the HUP that its uncertainty in period could not be more than 10-20 or 10-15 seconds, so its uncertainty in time could not be less than that. At the speed of light, that means we could be uncertain about which side of the EH it is on if it is created a distance less than c times those times, so 3 x 10-12 m or 3 x 10-7 m, which is another way of saying within its wavelength from the EH. Basically, pretty darn close to the EH.

Let me ask a question. Any photon that is emitted closer then its wave length to an even horizon of a black hole won't be able to actually be seen to an external observer will it?

[Ken G]

Let me ask a question. Any photon that is emitted closer then its wave length to an even horizon of a black hole won't be able to actually be seen to an external observer will it?The point is, there is no meaning to the statement "a photon emitted closer than its wavelength to an event horizon." You don't get to know the photon's location to better than its wavelength, because of the uncertainty principle. And you cannot say that the photon itself knows, in some sense, because it seems to be a fundamental uncertainty. That's one of the "cracks" between GR and QM that Hawking radiation attempts to exploit.

[tommac]

But then why could a laser promote the pair but gravitational tidal forces from anything without an EH not? Are we talking pure energy that is needed to promote OR not?



Basically yes, though I would say it a bit differently. First of all, there is no such thing as "providing energy" via tidal forces alone, you have to tap into some actual energy source, either the rest mass or kinetic energy of some body. That's where the energy needs to come from, and if all you have is a neutron star, then it must come from the neutron star. That only gives you a possibility to promote the virtual particle-- you still need an actual process that can tap into that energy on the required HUP timescale. That is apparently what having an event horizon affords you that you don't get if you don't have an event horizon. I don't know that there is no other process that could do it, I don't even know how an event horizon could do it, but there is no evidence of any such process for a neutron star, and no theory that I know of that predicts there should be.

[ShinAce]

What is 'pure energy'? Or do you mean, 'all forms of energy, including potential energies'?

[Grey]

But then why could a laser promote the pair but gravitational tidal forces from anything without an EH not? Are we talking pure energy that is needed to promote OR not?What mechanism are you proposing for a laser to "promote" a virtual particle? We talked about some possible arrangements that you could set up to do that at the beginning of the thread. Any situation where you have a laser interacting in such a way as to create new particles could probably be described that way under QED, but a laser beam travelling through empty space won't just start spontaneously emitting other particles any more than a neutron star will. And I'll say again that there's no such thing as "pure energy". Energy isn't a thing by itself. It's a property that particles (or configurations of particles) can have, along with other properties. It's not even an observer-independent property at that.

[Ken G]

But then why could a laser promote the pair but gravitational tidal forces from anything without an EH not? Are we talking pure energy that is needed to promote OR not?We are talking about the need for a mechanism that allows the required conservation laws to hold. With the laser effect you quoted, I was talking about the laser hitting dust particles in the air. The charges in the dust allow for a coupling of the laser to the virtual particles that are being promoted, and allow for conservation of both energy and momentum, because of the mechanism involved in the coupling to the dust. The main thing you need to promote virtual particles is a mechanism that allows for the conservation of all the quantities that need to be conserved. The event horizon permits such a mechanism, because it interferes with the rules for when you need to "pay the piper" on those conservation laws. Dust particles permit such a mechanism, because they couple to light in ways that lets the conservation laws hold. Neutron stars appear to lack any such mechanism, unless there is one we don't know about yet. This is also what Grey is saying.

[tommac]

But what is so special about the EH? Locally for the particles the only thing that is strange about the EH is the tidal forces right? What is the mechanism for the local observers?


We are talking about the need for a mechanism that allows the required conservation laws to hold. With the laser effect you quoted, I was talking about the laser hitting dust particles in the air. The charges in the dust allow for a coupling of the laser to the virtual particles that are being promoted, and allow for conservation of both energy and momentum, because of the mechanism involved in the coupling to the dust. The main thing you need to promote virtual particles is a mechanism that allows for the conservation of all the quantities that need to be conserved. The event horizon permits such a mechanism, because it interferes with the rules for when you need to "pay the piper" on those conservation laws. Dust particles permit such a mechanism, because they couple to light in ways that lets the conservation laws hold. Neutron stars appear to lack any such mechanism, unless there is one we don't know about yet. This is also what Grey is saying.

[Strange]

But what is so special about the EH? Locally for the particles the only thing that is strange about the EH is the tidal forces right? What is the mechanism for the local observers?

In terms of the naive analogy, I guess it is the fact it is one-way barrier. More important is the effect the curvature of spacetime has on different observers' understanding of the energy of the vacuum...

[tommac]

In terms of the naive analogy, I guess it is the fact it is one-way barrier. More important is the effect the curvature of spacetime has on different observers' understanding of the energy of the vacuum...

It is not a one way barrier is it? Only to the distant observer ... locally if I was falling into a black hole I still would be able to touch my toes before, as I was crossing and after having crossed the EH. It is only to an external observer that it is one directional. ( I can not kick my toes out so that an external observer can see me )

[ShinAce]

In terms of the naive analogy, I guess it is the fact it is one-way barrier. More important is the effect the curvature of spacetime has on different observers' understanding of the energy of the vacuum...

I also share this view that the event horizon is special.

If one particle moves an inch away from the event horizon, and the other crosses inwards by an inch, they are no longer two inches apart. The insider thinks it's two inches apart. The outsider does not see the insider. That is a mechanism as far as I'm concerned.

[Strange]

It is not a one way barrier is it? Only to the distant observer ... locally if I was falling into a black hole I still would be able to touch my toes before, as I was crossing and after having crossed the EH. It is only to an external observer that it is one directional. ( I can not kick my toes out so that an external observer can see me )

What a bizarre image :) If your feet were inside the EH and you bent down to touch your toes, you wouldn't be able to unbend again! What goes inside the EH stays inside the EH.

[tommac]

I also share this view that the event horizon is special.

If one particle moves an inch away from the event horizon, and the other crosses inwards by an inch, they are no longer two inches apart. The insider thinks it's two inches apart. The outsider does not see the insider. That is a mechanism as far as I'm concerned.

Huh? Can you explain that? if I fall feet first into a supermassive black hole I may not notice anything different as I cross the event horizon. Only in a smaller black hole will spaghettification happen because of the tidal differences but there is nothing special about the EH from a freefallers perspective ... ( Please correct me if I am wrong here, I am not claiming anything here ... this is just my understanding of mainstream physics )

[tommac]

What a bizarre image :) If your feet were inside the EH and you bent down to touch your toes, you wouldn't be able to unbend again! What goes inside the EH stays inside the EH.

UGGG ... please reread ...

If I am freefalling into a blackhole ( a supermassive one ) if I bend over to touch my toes ... I am fine ... if I "unbend" I am fine ... I am still free falling towards the sigularity ... what I cant do is accelerate back away from and out of the EH ... but since I am freefalling ... and my whole body is accelerating towards the singularity I can have local movement .... it is not like I lose sight of my feet as I cross the EH.

[Strange]

it is not like I lose sight of my feet as I cross the EH.

That is exactly what it is like. That is actually the answer I was going to give to your previous post. Nothing can leave the black hole, including photons. That is why it is called a "black" hole.

[tommac]

The point I am trying to make here is that if the VP is in freefall ( as I assume it would be for a very short time ) then its local experience may not observe anything strange about passing through the EH.

Now to the distant observer there is a barrier. This is where things get tricky, as I have read ... that the local reality may not agree with the reality of the distant observer.

But in any case we are digressing ... I guess the point I am asking for clarrification on is IF there is a local experience for something or someone freefalling past the EH. My understanding of mainstream physics is that there is nothing special for a freefaller ( other than possible signifigant tidal forces ) .




It is not a one way barrier is it? Only to the distant observer ... locally if I was falling into a black hole I still would be able to touch my toes before, as I was crossing and after having crossed the EH. It is only to an external observer that it is one directional. ( I can not kick my toes out so that an external observer can see me )

[ShinAce]

Huh? Can you explain that? if I fall feet first into a supermassive black hole I may not notice anything different as I cross the event horizon. Only in a smaller black hole will spaghettification happen because of the tidal differences but there is nothing special about the EH from a freefallers perspective ... ( Please correct me if I am wrong here, I am not claiming anything here ... this is just my understanding of mainstream physics )

You will notice something when you fire up the engines and try to get back home.

First you ask if the tidal force from a neutron star is enough to promote a virtual pair. The answer seems to be 'no'.

Then you mention supermassive black holes. Are you forgetting that the tidal force at the EH of a supermassive blackhole is likely less than that of any neutron star?

If tidal force alone can promote a virtual pair, then obviously a neutron star would be able to do it.

Now I see bait for a long, winded, reiterative discussion on coordinate singularities. I'm not having it.

To me, it's obvious that the literal disconnection between a pair's realities by an event horizon is physically significant. That is all I have to add.

[Strange]

But in any case we are digressing ... I guess the point I am asking for clarrification on is IF there is a local experience for something or someone freefalling past the EH. My understanding of mainstream physics is that there is nothing special for a freefaller ( other than possible signifigant tidal forces ) .

That is correct. And there would be nothing special about the tidal forces at that point either.

[grant hutchison]

it is not like I lose sight of my feet as I cross the EH.That is exactly what it is like. That is actually the answer I was going to give to your previous post. Nothing can leave the black hole, including photons. That is why it is called a "black" hole.Actually, you don't lose sight of your feet as you fall across the event horizon, and you can also easily touch your toes and straighten up again, if tidal forces are small enough. :)
In Schwarzschild coordinates, photons emitted by your feet just outside the horizon will be moving slowly outwards; photons emitted at the horizon will be hovering at that radial distance; photons emitted inside the horizon will be moving slowly inwards. As an infaller, your eyes will encounter all those photons shortly after they are emitted, because your head follows your feet across the event horizon very rapidly. So you'll see the photons emitted by your feet as they crossed the event horizon, just as soon as your eyes cross the event horizon and intercept those hovering photons. Think of it as the photons remaining stationary in Schwarzschild coordinates, while your head moves inwards at lightspeed: from your point of view, the photons make the journey from your feet to your head at lightspeed. In fact, all the way to the singularity you'll be encountering photons emitted by your feet, so you'll never lose sight of them so long as your body remains intact. Likewise, you can bend down, touch your toes and straighten up again while falling across the event horizon. You cross the horizon so quickly its simply impossible to straighten up fast enough to poke your head back "outside": you'd need to be able to move your head faster than light.

Tommac is forgetting that head and feet are only ever in visual contact when both are on the same side of the event horizon: both have to stay outside, or both have to cross, and outside never receives photons from inside. The situation is not analogous to the parable of the virtual pair.

Grant Hutchison

[tommac]

To me, it's obvious that the literal disconnection between a pair's realities by an event horizon is physically significant. That is all I have to add.

Yes but is it only signifigant to the external observer? Or is it ALSO signifigant to the local infaller ( the pair itself or anything else that is falling along with the pair )?

IF we are saying that it is only signifigant to the external observer ... I actually have less problems understanding the complexities and coming to the realization that wierd stuff happens ...

IF we are saying that the free faller also experiences something "physically signifigant" then I would like to understand what that is.

my understanding of mainstream would suggest that a freefaller in an elevator free falling into the SMBH past the EH would be actually be weightless and would notice no difference unless tidal forces were present. ( Please correct this if I am wrong ... as I am pretty sure this is mainstream ... if it is not please advise )

[tommac]

That is correct. And there would be nothing special about the tidal forces at that point either.

Yes that is what I am getting at. So we are saying ... please correct me if I am misinterpreting your statement ... that hawking radiation is only observed from a distant observer.

[grant hutchison]

Yes but is it only signifigant to the external observer? Or is it ALSO signifigant to the local infaller ...See above. The horizon is significant to everyone.

Grant Hutchison

[grant hutchison]

Yes that is what I am getting at. So we are saying ... please correct me if I am misinterpreting your statement ... that hawking radiation is only observed from a distant observer.Hawking radiation is absent for a free-faller. Hawking radiation is evident for all stationary observers, and all observers accelerating slower (or faster!) than freefall: if you are lowered quasi-statically towards the event horizon, you find yourself immersed in bath of real particles which gets steadily hotter towards the event horizon. Under gravitational redshift, that Unruh radiation becomes Hawking radiation for the distant stationary observer.

Grant Hutchison

[tommac]

Tommac is forgetting that head and feet are only ever in visual contact when both are on the same side of the event horizon: both have to stay outside, or both have to cross, and outside never receives photons from inside. The situation is not analogous to the parable of the virtual pair.

OK ... so then the question is does the Virtual Pair freefall? Otherwise can you explain the not being analogous part? Why isnt it analogous?

[grant hutchison]

OK ... so then the question is does the Virtual Pair freefall? Otherwise can you explain the not being analogous part? Why isnt it analogous?The parable of the virtual pair is told from the viewpoint of a freefalling observer: that's the observer who sees the pair receiving energy from tidal forces.
You suggested that the event horizon isn't a one-way barrier for the freefaller. But the freefaller can touch his toes only because his head, feet and all points in between cross the event horizon while he is performing the bend-and-straighten routine. The event horizon is a one-way barrier for the freefaller. One member of the virtual pair crosses that barrier, and the other doesn't. All of the freefalling observer crosses the barrier.

Grant Hutchison

[tommac]

Hawking radiation is absent for a free-faller. Hawking radiation is evident for all stationary observers, and all observers accelerating slower (or faster!) than freefall: if you are lowered quasi-statically towards the event horizon, you find yourself immersed in bath of real particles which gets steadily hotter towards the event horizon. Under gravitational redshift, that Unruh radiation becomes Hawking radiation for the distant stationary observer.

Grant Hutchison

OK ... so if you are freefalling and a VP appeared in front of you ... you could experimentally see it rejoin ( but possibly not before hitting the Singularity ) but for all other non freefalling observers they would see it radiate.
Is this correct?

[Strange]

Actually, you don't lose sight of your feet as you fall across the event horizon, and you can also easily touch your toes and straighten up again, if tidal forces are small enough. :)

Thanks for the clarification. That's what comes from thinking "classically"; I was imagining tommac drifting slowly through the event horizon doing his calisthenics! Whereas, he would (presumably) be travelling at nearly the speed of light.

[tommac]

OK ... so if you are freefalling and a VP appeared in front of you ... you could experimentally see it rejoin ( but possibly not before hitting the Singularity ) but for all other non freefalling observers they would see it radiate.
Is this correct?

Also is this similar to what happens in undruh ? Or am I totally confused. Where the local observer ( not moving ) sees nothign strange with the VPs BUT for someone that is travelling at near the speed of light they will observe radiation .... sorry to confuse this further .

[grant hutchison]

OK ... so if you are freefalling and a VP appeared in front of you ... you could experimentally see it rejoin ( but possibly not before hitting the Singularity ) but for all other non freefalling observers they would see it radiate.
Is this correct?As a freefaller, you never "see" anything but virtual pairs (and of course you can't see them): there's no Hawking radiation, no Unruh radiation. To observe the Hawking/Unruh radiation you've got to stay outside the event horizon, in which case you see only one real particle.

Grant Hutchison

[ShinAce]

IF we are saying that the free faller also experiences something "physically signifigant" then I would like to understand what that is.


Waving goodbye to your significant other and never seeing a reply would constitute significance to me. You cannot communicate with the outside world, even though you can see it. That is significant, and physical.

[grant hutchison]

Also is this similar to what happens in undruh ? Or am I totally confused. Where the local observer ( not moving ) sees nothign strange with the VPs BUT for someone that is travelling at near the speed of light they will observe radiation .... sorry to confuse this further .Unruh radiation comes from acceleration, not speed. The accelerating observer in flat space finds himself immersed in a bath of real particles. Someone trying to hover above an event horizon is similarly accelerating, and finds a similar bath of real particles. Factor in the effects of gravitational redshift, and you find that the hot particle soup experienced by an observer hovering just outside the horizon looks like Hawking radiation to a stationary observer at infinity.

Grant Hutchison

[tommac]

Waving goodbye to your significant other and never seeing a reply would constitute significance to me. You cannot communicate with the outside world, even though you can see it. That is significant, and physical.

Please reread grants post above. You can wave to your signifigant other and your signifigant other can see then wave back to you as long as you both are in free fall.

[grant hutchison]

Please reread grants post above. You can wave to your signifigant other and your signifigant other can see then wave back to you as long as you both are in free fall.You're missing the important bit: so long as you are both on the same side of the event horizon. If your significant other is outside the horizon, he can't see you waving from inside the horizon, unless he also crosses the horizon.

Grant Hutchison

[tommac]

You're missing the important bit: so long as you are both on the same side of the event horizon. If your significant other is outside the horizon, he can't see you waving from inside the horizon, unless he also crosses the horizon.

Grant Hutchison

Assuming he is hovering ... if both are free falling ( so both will eventually be through the EH ) then you are not limited to that. The point is that the light will not leave the EH ... but if I am outside the EH and fall through, I can catch up to the light that was shining towards me. You say so in your post above ( but in way better terms ).

[tommac]

Sorry : didnt see this last bit:

unless he also crosses the horizon.

Which I am assuming as he is a freefaller.

[ShinAce]

I did read Grant's post. Twice actually. I thought to myself, where's the cherry to be picked? I did not find one. I see that you have.

My post and Grant's post are not conflicting.

edit: If you were right, we have an easy way to probe beyond the event horizon. Let Alice fall in and radio back. Let Bob fall quickly toward the event horizon but then turn around before crossing it. If you're guess is right, Bob will hear Alice's signal and have 'inside' information. No such inside information is possible nor has it been suggested by anyone in the thread, other than yourself. The burden of proof is yours. As usual, you are free to cherry pick this post as well.

[grant hutchison]

Which I am assuming as he is a freefaller.But both must cross the horizon if they are to stay in contact, in which case both lose the ability to communicate with the outside world. That's a significant event, just as ShinAce told you.

Grant Hutchison

[tommac]

But both must cross the horizon if they are to stay in contact, in which case both lose the ability to communicate with the outside world. That's a significant event, just as ShinAce told you.

Grant Hutchison

hmmmm ... lets say I tie a rope to a flashlight and set it to blink every second and dangle it towards the black hole from my spaceship then I jump out of my spaceship with the flashlight below me. we are both free falling towards the black hole.

Lets assume a super massive black hole so to avoid signifigant tidal forces.

Will I notice anything different about the flashlight at any point around the EH? Would I be able to tell that I crossed the EH? I am assuming the answer is no ... I realize that my freefall moves me into the light and that for light that was released at the EH I would need to move up to the EH in order to see that light but I notice nothing strange about my relationship with the flashlight. It keeps blinking at the same rate and remains the same distance away from me I believe even the rope stays equally taught.

Lets say I had an experiment that could detect Virtual pairs existance and could detect if radiation happened. I was able to contain this entire experiment in an elevator the elevator was in a free fall into a BH.

my understanding is that inside of the elevator virtual pairs would be created at the exact same rate as in empty space and nothing would radiate this would be true outside of the EH, passing the EH and inside of the EH.

[tommac]

If you were right, we have an easy way to probe beyond the event horizon. Let Alice fall in and radio back. Let Bob fall quickly toward the event horizon but then turn around before crossing it.

Yes but you are missing the point that before Alice could receive the radio signal she would also be inside the Event horizon. I believe there is some ladder proof that shows this ...

If alice was hovering she would not be freefalling and would not be able to catch up with the signal.

If she is free falling she will catch up to the signal ... but as grant noted if the signal was released from the EH then Alice will have to wait until she is at the EH to receive that signal. But she notices nothing special about her relationship with Bob ... she crosses the EH and doesnt realize it everything for her is the same before, at the time of crossing and after crossing the EH.

[grant hutchison]

And if I drive with a paper bag over my head, I won't see oncoming traffic.
Pointing out ways in which you can fail to detect the event horizon, by limiting to your view, doesn't make make the event horizon disappear, or render it irrelevant for processes, like Hawking radiation, which span the event horizon.

Grant Hutchison

[tommac]

But both must cross the horizon if they are to stay in contact, in which case both lose the ability to communicate with the outside world. That's a significant event, just as ShinAce told you.

Grant Hutchison

What is "the event" though, if neither party observes anything? I realize that is their point of no return ... but they may not be made aware of that and nothing special happens to them as they pass that point. it is only a signifigant event to an external observer.

[tommac]

And if I drive with a paper bag over my head, I won't see oncoming traffic.
Pointing out ways in which you can fail to detect the event horizon, by limiting to your view, doesn't make make the event horizon disappear, or render it irrelevant for processes, like Hawking radiation, which span the event horizon.

Agreed. BUT you also said that the free fallers dont detect any hawking or undruh radiation. I was also reading in a book that hawking had a discussion about this same issue when they were discussing information loss. The deal was that to the external observer hawking radiation happens but for the free faller it doesnt. I think it was that if someone was able to take the bit of lost information and then jump back into the EH with it then you would have 2 copys of the same bit of information or soemthing like that I need to look up the details ... but they were able to work around the issue by saying that the freefaller would hit the singularity before he could detect the copy therefore it is not a possible scenerio in any time frame.

[grant hutchison]

What is "the event" though, if neither party observes anything?The event is crossing the event horizon. Being unaware of an event does not render it non-existent or insignificant.

Agreed. BUT you also said that the free fallers dont detect any hawking or undruh radiation.Unruh. Yes. Relevance?

Grant Hutchison

[Grey]

Lets say I had an experiment that could detect Virtual pairs existance...Completely separate from the points grant is making, let me point out that it's not possible to detect a virtual particle, not even in principle. That's part of the point.

[tommac]

Being unaware of an event does not render it non-existent or insignificant.

How can it be an event if it is not measurable?

[tommac]

Completely separate from the points grant is making, let me point out that it's not possible to detect a virtual particle, not even in principle. That's part of the point.

Then how do we know that they are there? Isnt there something about two plates close together or something?

[grant hutchison]

How can it be an event if it is not measurable?It is measurable. The infallers can easily discern, with arbitrary accuracy, the precise moment at which they cross the event horizon, using measurement and observation of the outside world. That crossing is an "event" for them, one in which they'll be interested, because it marks the point of absolute no-return for them, and the point at which they can no longer signal the outside world.

Grant Hutchison

[Grey]

Then how do we know that they are there? Isnt there something about two plates close together or something?Predictions made using quantum electrodynamics, modeling all interactions of exchanges of virtual particles, are matched by experiment to about one part in a trillion. QED predicts slightly different values for some measurable quantities that are different (by really small amounts) from prior predictions without using a virtual particle model, and experiment verifies those results. But it's worthwhile noting that there are essentially infinitely many virtual particles of all types everywhere at all times in that model. Figuring out how to make calculations with infinitely many particles took a bit of mathematical trickery, taking advantage of the fact that the more complicated a particular interactions is, the less likely it is to happen. But the same theory that makes those predictions also tells us how virtual particles behave, and one of those features is that you can't detect them directly. A virtual particle cannot possibly trigger a signal in a particle detector, for example.

[tommac]

It is measurable. The infallers can easily discern, with arbitrary accuracy, the precise moment at which they cross the event horizon, using measurement and observation of the outside world. That crossing is an "event" for them, one in which they'll be interested, because it marks the point of absolute no-return for them, and the point at which they can no longer signal the outside world.

Grant Hutchison

Not to be a buster here ... but how could they tell if they were in an elevator. I know I am nit picking and yes I know that there is a significance of crossing that line BUT I dont get the event part. Using equivalence wouldnt the free faller not be able to detect any acceleration and be totally weightless? in a similar fashion to floating in outer space?

[tommac]

Fair enough ...


Predictions made using quantum electrodynamics, modeling all interactions of exchanges of virtual particles, are matched by experiment to about one part in a trillion. QED predicts slightly different values for some measurable quantities that are different (by really small amounts) from prior predictions without using a virtual particle model, and experiment verifies those results. But it's worthwhile noting that there are essentially infinitely many virtual particles of all types everywhere at all times in that model. Figuring out how to make calculations with infinitely many particles took a bit of mathematical trickery, taking advantage of the fact that the more complicated a particular interactions is, the less likely it is to happen. But the same theory that makes those predictions also tells us how virtual particles behave, and one of those features is that you can't detect them directly. A virtual particle cannot possibly trigger a signal in a particle detector, for example.

[tusenfem]

Not to be a buster here ... but how could they tell if they were in an elevator. I know I am nit picking and yes I know that there is a significance of crossing that line BUT I dont get the event part. Using equivalence wouldnt the free faller not be able to detect any acceleration and be totally weightless? in a similar fashion to floating in outer space?


tommac can you please stop this kind questioning, this is like a child claiming "If I close my eyes I cannot see you so you cannot see me either."

Rather clear answers are given to your questions. However, you don't seem to like the answers and therefore you start arguing in a very nerving way (but what if , but then what if, and the but what if then ....) to hopefully get the answer that you do like, so it fits into whatever new off-the-chart view of black hole event horizons you have.

This is getting to be infractionable.

I must commend grant for being so patient and answereing all these "but what if" questions.

[grant hutchison]

Not to be a buster here ... but how could they tell if they were in an elevator. I know I am nit picking and yes I know that there is a significance of crossing that line BUT I dont get the event part. Using equivalence wouldnt the free faller not be able to detect any acceleration and be totally weightless? in a similar fashion to floating in outer space?They can't tell, if they're in an elevator. They can't tell, if they have bags over their heads. They can't tell, if they're stupid. They can't tell, if they're asleep. They can't tell, if they're not paying attention.
So what? Do you think that restricting their ability to observe is going to change the fundamental nature of spacetime?

Grant Hutchison

[tommac]

To which question?


tommac can you please stop this kind questioning, this is like a child claiming "If I close my eyes I cannot see you so you cannot see me either."

Rather clear answers are given to your questions. However, you don't seem to like the answers and therefore you start arguing in a very nerving way (but what if , but then what if, and the but what if then ....) to hopefully get the answer that you do like, so it fits into whatever new off-the-chart view of black hole event horizons you have.

This is getting to be infractionable.

I must commend grant for being so patient and answereing all these "but what if" questions.

[tommac]

Do you think that restricting their ability to observe is going to change the fundamental nature of spacetime?

No but due to equivalence there is nothing special that is happening for the local observer ... the event really matters to the local observer if he is trying to escape. The point here is that for the local free faller there is nothing special about crossing the event horizon so why would AND you are saying that he cant detect any radiation. So my point is that there is a difference of opinion between the freefaller and the distant observer about the going ons at the EH. This is backed up in many books.

[tommac]

tommac can you please stop this kind questioning, this is like a child claiming "If I close my eyes I cannot see you so you cannot see me either."


No what I am asking is DOES IT HAPPEN? and I think Grant has already answered no it doesnt happen. Just looking to make sure that I am understanding him correctly.

[tommac]

They can't tell, if they're in an elevator. They can't tell, if they have bags over their heads. They can't tell, if they're stupid. They can't tell, if they're asleep. They can't tell, if they're not paying attention.
So what? Do you think that restricting their ability to observe is going to change the fundamental nature of spacetime?

Grant Hutchison

If they cant tell then does it matter?

[tommac]

Lets say we accept grants definition of an event ( although I dont think it is clear ) ( AND I need to say Grant is one of the very few people on this board whos word I take as fact so dont mean this as a knock to grant ) ... then what does the meaning of an undetectable event have to do with anything? That is the point ... there is no way to detect that something happened BUT the fact that it happened is enough to qualify as an event to distiguish enough and to act to create radiation ... I am missing something ... but what?

[tommac]

See above. The horizon is significant to everyone.

Grant Hutchison

Agreed. But why is :


To me, it's obvious that the literal disconnection between a pair's realities by an event horizon is physically significant relevant?

[tommac]

This is exactly my point ... what does the free faller experience?


Unruh radiation comes from acceleration, not speed. The accelerating observer in flat space finds himself immersed in a bath of real particles. Someone trying to hover above an event horizon is similarly accelerating, and finds a similar bath of real particles. Factor in the effects of gravitational redshift, and you find that the hot particle soup experienced by an observer hovering just outside the horizon looks like Hawking radiation to a stationary observer at infinity.

Grant Hutchison

[tommac]

As a freefaller, you never "see" anything but virtual pairs (and of course you can't see them): there's no Hawking radiation, no Unruh radiation. To observe the Hawking/Unruh radiation you've got to stay outside the event horizon, in which case you see only one real particle.

Grant Hutchison

Yes ... so this is the reason I am also questioning the significance of crossing the EH ... You say here there is no Hawking Radiation and no Unruh ... And you agree that in an elevator you would not be able to detect crossing the event horizon ... So I think we are in agreement. Not sure about what we are in disagreement about.

[tommac]

Let me take a different approach to my questioning ...

What is the minimum space that mass can fit into?

[tommac]

OK ...now it all makes sense. As the definition of an event is: (t, P(x,y,z)) then yes I agree freefalling past the EH is an event.

[WayneFrancis]

That is exactly what it is like. That is actually the answer I was going to give to your previous post. Nothing can leave the black hole, including photons. That is why it is called a "black" hole.

What tommac fails to see when he is "unbending" is he isn't climbing out of the gravity well while falling into a black hole.. In his falling frame his unbending still has his head falling towards the singularity where here on the surface of the Earth when you "unbend" you are lifting your head out of the gravity well just a little bit.

[WayneFrancis]

The point I am trying to make here is that if the VP is in freefall ( as I assume it would be for a very short time ) then its local experience may not observe anything strange about passing through the EH.

Now to the distant observer there is a barrier. This is where things get tricky, as I have read ... that the local reality may not agree with the reality of the distant observer.

But in any case we are digressing ... I guess the point I am asking for clarrification on is IF there is a local experience for something or someone freefalling past the EH. My understanding of mainstream physics is that there is nothing special for a freefaller ( other than possible signifigant tidal forces ) .

Which should be a clue to tommac on why the tidal forces have NOTHING to do with HR.

HR can happen very close to the EH of any black hole. At the EH of a super massive black hole where there is very little tidal forces and the EH of a VERY small black hole where the tidal forces are huge.

[WayneFrancis]

Actually, you don't lose sight of your feet as you fall across the event horizon, and you can also easily touch your toes and straighten up again, if tidal forces are small enough. :)
In Schwarzschild coordinates, photons emitted by your feet just outside the horizon will be moving slowly outwards; photons emitted at the horizon will be hovering at that radial distance; photons emitted inside the horizon will be moving slowly inwards. As an infaller, your eyes will encounter all those photons shortly after they are emitted, because your head follows your feet across the event horizon very rapidly. So you'll see the photons emitted by your feet as they crossed the event horizon, just as soon as your eyes cross the event horizon and intercept those hovering photons. Think of it as the photons remaining stationary in Schwarzschild coordinates, while your head moves inwards at lightspeed: from your point of view, the photons make the journey from your feet to your head at lightspeed. In fact, all the way to the singularity you'll be encountering photons emitted by your feet, so you'll never lose sight of them so long as your body remains intact. Likewise, you can bend down, touch your toes and straighten up again while falling across the event horizon. You cross the horizon so quickly its simply impossible to straighten up fast enough to poke your head back "outside": you'd need to be able to move your head faster than light.

Tommac is forgetting that head and feet are only ever in visual contact when both are on the same side of the event horizon: both have to stay outside, or both have to cross, and outside never receives photons from inside. The situation is not analogous to the parable of the virtual pair.

Grant Hutchison

as always, you've said it very eloquently.

[WayneFrancis]

Yes but is it only signifigant to the external observer? Or is it ALSO signifigant to the local infaller ( the pair itself or anything else that is falling along with the pair )?

IF we are saying that it is only signifigant to the external observer ... I actually have less problems understanding the complexities and coming to the realization that wierd stuff happens ...

IF we are saying that the free faller also experiences something "physically signifigant" then I would like to understand what that is.

my understanding of mainstream would suggest that a freefaller in an elevator free falling into the SMBH past the EH would be actually be weightless and would notice no difference unless tidal forces were present. ( Please correct this if I am wrong ... as I am pretty sure this is mainstream ... if it is not please advise )

Correct which is why tidal forces have nothing to do with promotion of virtual pairs thus asking if tidal forces on a neutron star can promote one of the pair is nonsensical.

A very loose analogy is like asking if a green plastic plate can boil water like a red microwave oven does because the green plastic plastic plate has a colour too. Boiling water has nothing to do with the colour of the object. Promotion of of a VP has nothing to do with the tidal forces a VP experiences but the fact that there is an EH which can be used by the pair to separate them.

[WayneFrancis]

Please reread grants post above. You can wave to your signifigant other and your signifigant other can see then wave back to you as long as you both are in free fall.

Yes but with virtual particles one of the pair is NOT in free fall. One of the pair has escape velocity in a direction away from the black hole.

[WayneFrancis]

Assuming he is hovering ... if both are free falling ( so both will eventually be through the EH ) then you are not limited to that. The point is that the light will not leave the EH ... but if I am outside the EH and fall through, I can catch up to the light that was shining towards me. You say so in your post above ( but in way better terms ).

Light can not hover. Light can only got at light speed. So one photon falls in the other photon has a light cone where it escapes the the black hole.

Can someone draw the light cones for him so he understands?

[WayneFrancis]

Let me take a different approach to my questioning ...

What is the minimum space that mass can fit into?

This isn't even moving the goal post this is a completely different game.

[grant hutchison]

Light can not hover. Light can only got at light speed. So one photon falls in the other photon has a light cone where it escapes the the black hole.

Can someone draw the light cones for him so he understands?Tommac is just repeating what I said, but missing out the coordinate choice. In Schwarzschild coordinates, light emitted outwards at the event horizon does "hover" (its radial coordinate remains fixed), while light emitted inwards moves rapidly towards the singularity. (Hence the parable about "space" moving inwards at lightspeed at the event horizon, with outward directed photons swimming against this current.) The lightcone at the event horizon is tilted right over so that its outward-facing slope is parallel to the time axis.
A freefaller moves inwards with this apparent spacial flow, and so can see a sheaf of photons emitted in the vicinity of the event horizon as his head passes through them. Thus, he has a view of his feet all the way through the event horizon.

Grant Hutchison

[grant hutchison]

Yes ... so this is the reason I am also questioning the significance of crossing the EH ... You say here there is no Hawking Radiation and no Unruh ... And you agree that in an elevator you would not be able to detect crossing the event horizon ... So I think we are in agreement. Not sure about what we are in disagreement about.We are in disagreement because you are questioning the significance of crossing the event horizon, based on a spurious argument. You've been told why the event horizon is still present and detectable for infallers, and you've been told why crossing it is significant. Making six scattershot posts in which you simply repeat yourself doesn't change matters.

Grant Hutchison

[WayneFrancis]

Light can not hover. Light can only got at light speed. So one photon falls in the other photon has a light cone where it escapes the the black hole.

Can someone draw the light cones for him so he understands?

Tommac is just repeating what I said, but missing out the coordinate choice. In Schwarzschild coordinates, light emitted outwards at the event horizon does "hover" (its radial coordinate remains fixed), while light emitted inwards moves rapidly towards the singularity. (Hence the parable about "space" moving inwards at lightspeed at the event horizon, with outward directed photons swimming against this current.) The lightcone at the event horizon is tilted right over so that its outward-facing slope is parallel to the time axis.
A freefaller moves inwards with this apparent spacial flow, and so can see a sheaf of photons emitted in the vicinity of the event horizon as his head passes through them. Thus, he has a view of his feet all the way through the event horizon.

Grant Hutchison

Fair enough but in all practical terms nothing can hover right above the event horizon, the energy required is just to great.

A photon, to "hover" at the event horizon has to
1) be emitted right at the event horizon
2) be emitted with a vector exactly opposite in direction of the singularity
3) the black hole can not increase in size because that would then have the black hole's event horizon bubble out past the point where the photon was racing along.

A photon not on a vector exactly opposite the centre of the singularity will very quickly find its orbit plunging very deeply into the black hole.

A photon forever stuck at the EH is for all intensive purposes just as lost to the rest of the universe as one racing towards the singularity.

But again you've put it in better terms then I. Tommac ignores that fact that while he's doing his toe touches that even when he straightens up his over all vector for his head remains in a positive direction towards the singularity and that is we he can still see his toes.