Slinkey

Yes, I understand how interference is created. If you recall my thought experiment called for the entangled particles to be sent to separate double slit experiments one of which has a detector on one of the slits to see which slit the particle went through, and the other does not. As I understand it in the DS experiment, if we let a particle pass through the apparatus undetected we reveal its wave behaviour and an interference pattern will build up on the screen. If we detect which slit the particle went through then we don't get an interference pattern. This implies that we have destroyed the wave property of the particle. Agreed? So, if we have a pair of entangled particles, which as I understand it is described by a single wave function, then if we collapse the wave function by detecting which slit the particle passed through in the DS experiment with the detector, will that make the other entangled particle behave like a particle or will it behave like a wave when it passes through its DS experiment without a detector (experiment 2)?

I don't recall saying that it did. I am under the impression that when we measure some property of a particle we collapse its wavefunction. This explains why we don't get interference in the double slit experiment when we look to see which slit it went through. We've asked it to be a particle so we can see where it went, and the wave function collapses. Is this not correct? So if the wave function is describing an entangled state, when it collapses doesn't its collapse result in two particles?

I was under the impression that entangled particles are described by the same wave equation. I always took this to mean that the wave equation describes both particles and that if one particle collapses the wave equation then the collapse happens for the other particle as well. Thus in experiment 2 where we have the detected particle collapsing the wave equation first it will destroy the wave nature of the other particle. But as you have asked if these states are entangled it's made me think I have misunderstood something basic about entanglement.

Apologies if this has been asked before but as this thought experiment doesn't have a title that I'd know of it's hard to search for it to see if it has been asked. We have a pair of entangled particles. They fly apart toward a different experiment each. Particle A is flying towards the classic double slit experiment which reveals the wave nature of light and the classic interference pattern. Particle B is flying towards another double slit experiment but this one has a detector on one of the slits so we can see which slit the particle went through. There are three ways we can perform this experiment. 1. Particle A reaches its experiment first and we get interference. What happens with particle B? Do we get interference or no interference? Would particle A affect how particle B behaves in its experiment? 2. Particle B reaches its experiment first and we get no interference. What happens with particle A? Do we get interference or no interference? Would particle B affect how particle A behaves in its experiment? 3. They reach their respective experiment simultaneously. What happens? Thoughts?

Thanks for the link. Interesting to read the remarks.

Well, looking back on this thread I'm sad to see it fell into bickering. Apologies for my part in that bickering. I got a little frustrated at not getting my point across, and felt I was not being understood. I'd like to reignite this thread in a more congenial manner. We have fundamental disagreement on the existence of black holes, and looking above I could've used a different way of explaining myself to make my point clearer. First though I would like to hear thoughts on this article which made me come back to this thread. http://www.natureworldnews.com/articles/9187/20140925/are-black-holes-mathematically-impossible.htm

Tell it to get lost.

Err, no. It's precisely because of the different reference frames that this argument comes about and even the part you quoted states it. In the reference frame outside the EH an infinite amount of time must pass before we see anyone cross the EH. The other reference frame - falling in - crosses the EH and is obliterated. As the BH evaporates in less than infinite time then the person outside will never see anyone cross the EH. You can argue all you like about the person claimed to be falling in but it can never be observed from outside the EH as Susskind clearly states. Now we have two different histories. 1) we never see anyone cross the EH and the BH evaporates before they can for any observer outside the EH and they survive. 2) someone falls in and is obliterated. Your assumed flippancy of my reply is simply irrelevant. You just don't want to address what is staring you in the face. There's two histories and they cannot be mixed in any way. You can't both fall in and be crushed to death and someone never observe you fall in before the BH evaporates and survive. It seems that you want to ignore reference frames rather than me not understand them. It's the very fact that there are differing reference frames that this paradox is there. There is no known.... is our knowledge complete? No, it isn't. Thus this is an assumption. It's not selective quoting because you, being of a scientific mind, should know that the case is not proven. No, really? /sarcasm. The point I'm making, which you have skillfully evaded, is that a BH of that mass is far smaller than that radius. ie. it could conceivably still be an object that is not a BH because it can conceivably be larger than the required Schwarzchild radius. A BH is not a BH until it reaches that radius. As we have no direct evidence that the object is a Schwarzchild radius then this is an assumption. So what? Concensus does not make fact. Ok, I agree that my description of this is poor. It should be more correctly described as a particle-antiparticle pair being created at the EH with one escaping to infinity and the other not escaping. The energy that was used to create the pair would need to come from the BH in order for it to evaporate in this way. "If you ever did". Nice ad hominem. I've read it. Deal with it. I have no idea. Does that mean I am wrong? No.

I always read the entire reply and choose what I see as the relevant statements. No, that's incorrect. According to Hawking the mass loss of the BH is due to negative energy crossing the EH. If positive energy crosses then it gains weight. In fact I can think of a way round this. The EH does not have a precise location. If it did it would defy Heisenberg Uncertainty. If a virtual particle-antiparticle pair is created within that region then one could appear beneath the horizon and the other above it. If the positive particle is above the horizon then the BH loses mass due to the negative particle forming below it. Thus it can still evaporate. Point of fact is physicists say that you will never see anything cross the EH due to the infinite time dilation, yet they also accept HR. So, their scenario is the one that says the BH will evaporate before anything crosses the EH. So in a way you have pointed out the flaw in their logic as well. In fact the more I think about it, it doesn't matter if photons (or massless particles) do cross the EH. Anything falling towards it cannot cross the EH as it takes an infinite amount of time to do so for any observer above the EH, and the BH exists for a finite time. All this would do is make me change my argument slightly and the main point still remains. Negative energy/mass.

If you are asking do I have an alternative theory, then no I don't, but I do have a question about the article. It states "In fact, recent observations[15] indicate that the radius is no more than 6.25 light-hours, about the diameter of Uranus' orbit." I worked out the Schwarzchild radius for 4.1 million solar masses and it came out 12,109,072,691.52m, or 40.39 light seconds (feel free to check that), so if the mass is not confined within that radius it isn't a BH. Agreed? 6.25 light hours is substantially larger than 40.39 light seconds. So, you tell me. Are you sure there's a BH there? Anywhere outside the EH. I thought that would be obvious. Edit: Didn't answering the "how" bit. I'm looking through a McDonalds straw because it's fun. Feel free to answer my question from post 129.

You accept Hawking radiation, yes? You have already stated that you will never see Alice cross the EH due to the infinite time dilation. BHs evaporate due to HR. As you will never see Alice cross the EH, when the BH has evaporated, where is Alice? I'll have to re-listen to the audio book and find the part.

Hawking radiation. He says it quite early on. I have the audio book so don't have a page number.

Chris, I'm inclined to think they don't exist at all.

The event of falling toward the EH is outside the EH and is in principle observable by anyone outside the EH. And you also accept that BHs evaporate, yes? So if you never see Alice cross the EH and the BH evaporates then on which side of the EH was/is Alice? It's not a fact. It's a conjecture that has never been evidenced except by math that is known to be incomplete. I suggest you read Leonard Susskind's "The Black Hole War" because he disagrees with you. In fact the only way he can reconcile it is by invoking "black hole complementarity" in the same vein as particle-wave complementarity. Actually this is still hotly debated. So what you're arguing is that the wavelength of a photon can be longer than the photon itself? So we have a wavelength of 1 meter and you are saying the photon would not actually be 1 meter long? Please explain that one to me. An infinite wavelength would have a non-zero frequency. Period.

A zero frequency does not equal an infinite wavelength. Zero frequency equals zero energy and zero wavelength. So nothing passes the EH. As a photon moves toward an EH it's wavelength tends towards infinity. No, at infinite wavelength you have an undefined but non-zero frequency. At zero frequency you simply have nothing. Tunneling occurs when barriers get so small that their own wave properties come into play and aren't barriers in the classical sense. At quantum dimensions "barriers" is not a meaningful term. This is all moot anyway because you don't see photons unless they hit your retina or equipment and this whole discussion is based upon an entirely ambiguous point which I really should've known better than to discuss. So you want to argue that the size of a photon has nothing to do with its wavelength? You want to think of a photon as a point particle that has a wavelength that is longer than a point? As we are talking about an unmeasured photon it is in a super-position. ie. it is completely uncertain and its size could be anything from planck scales up to infinity. We can't say anything about a photon until we measure it and we measure it by its energy content from which we can infer its wavelength and frequency. ie. when it hits a detector and deposits its energy we can work out its frequency and wavelength from how much energy it registers (or with the eye by what colour you "see"). You may now wonder how a photon can deposit its energy all in one go when it has a "length" which would seem to imply that the energy would take time - proportional to its wavelength - to deposit all its energy, and the longer the wavelength the longer it takes to deposit its energy. Well, a "photon" exists at every point from where it was emitted to where it lands, but "once it lands" the wave-function collapses to that point in no time at all. Thus you could argue that a photon is as big as the distance between where it was emitted to where it is absorbed. Its wavelength could then be described as the distance between these two points divided by a whole number. No. That wasn't my point. No, it's not. It's actually where you are failing because you are still failing to understand that you will never see anyone reach the EH if you are outside the EH. It will take an infinite amount of time for them to reach the EH from any observer outside the EH. ie. never. No, that's an assumption that you have no evidence for and the point under discussion. Consider this: I measure the mass of a BH and find it is 5 solar masses. I take Hawking's equations and work out that the BH will exist for 8.268976e+85 seconds. I then dive towards the BH. As I fall toward the BH I look back to where I came from and see that as I get closer to the BH the Universe is speeding up relative to me. At some point before I reach the EH - where time dilation is infinite - the dilation will be so great that 8.268976e+85 seconds will pass in a tiny fraction of a second as measured by me and the BH will have had enough time to evaporate. So when did I cross the EH? I'm sure you agree that 8.268976e+85 seconds is a lot less than infinite, no? I'm well aware that clocks in the same reference frame tick at the same rate. I think rather it is you that is not understanding what I am saying.

Err, no, it's not. The event is outside the EH thus it is in principle observable from any reference frame outside the EH. Secondly, to all outside observers an entity will never be seen to cross the EH the event of crossing the EH does not happen for any observer outside the EH. To argue that in the reference frame of the entity falling towards the EH that he experiences the event of crossing the EH is to argue there are two distinct histories for the entity falling towards the EH. History 1, for all observers outside the EH, and history 2, for the observer falling towards the EH are at odds with each other. Remember the premise is that an object falling toward an EH will continue to fall towards the EH at an ever decreasing rate for all outside observers and will never be seen to cross the EH. As BHs evaporate then the object will not cross the EH before the BH evaporates. So, you have to reconcile two distinct and contradictory histories. And non-zero. Basic physics also tells us that a photon cannot have an infinite wavelength, hence my point that it cannot cross the EH horizon as at that point its wavelength would be longer than the width of the Universe. Which means it ceases to exist. Thus preserving Heisenberg Uncertainty. Sure, I understand that it's a convention to do with the math. That's precisely the point and hence my statement "I'll leave you to figure out what negative energy is". The one property I am concerned with here is an energy <0. Cool.

I don't quite see how that follows as anything but an assumption on your part. A photon cannot have zero energy as that would defy Heisenberg Uncertainty. An infinitely long wavelength would still have an energy value. It would be infinitely small but not zero. The maximum wavelength anything can obtain cannot be larger than what it is contained within. Thus a wavelength longer than the "width" of the Universe would be less than a whole wavelength and thus cannot exist. Wavelengths don't come in fractions. The wavelength of the photon cannot be longer than the "width" of the Universe. Your assumption is that the photon crosses the EH (and thus adds its energy to the BH) but you yourself say it ceases to exist at the EH. It can't do both. Claiming that it "must go to the black hole" doesn't really cut it from where I'm standing. Which isn't correct either as you will never see anything reach the EH (even if you could see it). As I said above, an object falling toward a black hole would appear to stop moving but it won't actually stop moving. It will continue approaching the EH but never reach it. The closer it gets the slower it gets and because time is infinitely warped at the EH it will take an infinite amount of time for an observer to see anything cross the EH. ie. never. In the case of a photon as it approaches the EH it gets more and more redshifted but it cannot reach infinite wavelength as it is bounded by the width of the Universe. Thus it's energy must always lie outside the EH for to cross it would need to have a wavelength longer than the "width" of the Universe. Nor can the EH horizon "absorb" a photon as the EH is not a surface. Well, my reading of HR seems to differ slightly to yours. I just dug out my copy of "A Brief History of Time" and will paraphrase: When virtual antiparticle-particle pairs are created near the EH one has positive energy and the other "negative" energy. The one with positive energy becomes a real particle and flies away from the BH whilst the one with negative energy becomes a real particle when it cross the EH (as, apparently, negative virtual particles can become real particles when they cross the EH). Thus a BH loses mass by the ingestion of negative energy ie. energy that is less than zero. The BH doesn't repay any debt to the Universe. The virtual particles becoming real particles - one negative energy, the other positive energy - balances the energy scales for the Universe. I'll let you figure out what the hell negative energy is! This brings up the "trans-planckian" problem. If we measure a photon that has emanated from near the EH and it has a definite value then it has a wavelength smaller than the Planck length when it was created. I made two conjectures initially and that was one of them. My other conjecture was that they cannot form at all, and as I learn more about the subject I am increasingly favouring the latter and that they cannot form at all. This is nothing but argument from authority and a known fallacy of argument. As to GR helping you. It is GR that brings up the absurdity of watching something fall towards an EH forever which is what actually prompted my argument in the first place. As Hawking added to the mix that BHs radiate HR then it seems quite plausible that a BH will evaporate before we can see anything cross the EH because we can only ever watch something fall towards the EH but never cross it. No, it's nothing to do with reference frames. Events happen regardless of reference frame. We might disagree about their duration, position, or order, but events still happen.

Assumed black holes we "see" in the Universe have jets associated with them which are obviously firing a great deal of mass away from the black hole at near light speed. These jets clearly cannot emanate from beyond the EH thus a lot of matter is clearly not reaching the EH. This could be enough to halt mass reaching black hole density and extended the EH as you conjecture.

No problem. I'm at odds with it too but I'm finding it very hard to come to a definite conclusion because of the presumed nature of black holes, and the logic that is entailed from the presumed nature of them. Indeed. But here's the point - science is about observation. Thus if we do not see something enter a BH then nothing has entered the BH. When science leaves the realm of observation we enter a grey area of postulates that we cannot prove in any way shape or form except maybe in mathematics which (according to Godel) cannot be self-consistent. Yep, that's the assumption. There's a pretty good diagram in Kip Thorne's "Black Holes and Time Warps". Actually, no it's not and you explain why it's not below (which I hadn't considered because I was ignoring redshifting for the purposes of the thought experiment). Ok, I'll now pull the observation card on you here. You cannot see a photon enter a black hole you can only see photons that enter your eye or your equipment thus you are arguing an assumption. You are actually arguing the "trans-plancking problem" in reverse. ie. a photon emanating from a black hole (Hawking radiation) that we measure to have a finite wavelength must've had an infinitely small wavelength at the EH. What you could be doing here is questioning the nature of photons and what we think they are (no one knows what they are - we just measure an effect). You're contradicting yourself. If according to physics a photon reaches an infinitely long wavelength then can it be said to actually exist? What you're actually saying is that black holes exist and they logically create a scenario where they defy conservation of energy - they are redshifted to infinity ie. a wavelength longer than the "width" of the universe - but in order to sidestep it you demand that the photon must've entered the BH (crossed the EH) without actually showing how it can exist at the EH. Which is what Lawrence Krauss argues. If you check above in this thread you'll find the link to his article. With respect, that's what we're discussing and you are now assuming the conclusion as part of the premise. ie. begging the question. It might also violate the conservation of information law as outlined in Leonard Susskind's "The Black Hole War", which I highly recommend. His answer to this paradox, to my mind, is also highly unsatisfactory. He calls it "black hole complementarity" in a similar vein to particle-wave complementarity. ie. something falling toward a black hole both enters and does not enter the BH. Edit: Literally, a black hole creates two histories for anything that falls towards it.

The original premise was that black holes cannot be entered; specifically, that nothing can be seen to cross the event horizon (for the purposes of this thought experiment we were ignoring gravitational redshifting and tidal forces) by an outside observer due to the time dilation when an infalling object nears the event horizon. So, in order to do this thought experiment correctly you have to take that into account when talking about the CMBR. As the premise was that nothing can be seen to enter a black hole this would include the CMBR. ie. a black hole does not absorb the CMBR as it defies the postulate being put forward. Thus black holes cannot gain mass they can only "evaporate". As to the evaporation itself. I haven't read up on it for a long time but from what I remember the mass loss of a black hole is due to the creation of particle-antiparticle pairs at the event horizon allowing one of the pair to fly off to infinity and the other to fall back into the black hole(!) (whereas normally they would annihilate in short order) creating a mass deficit for the black hole and thus reducing its mass (correct me if I am wrong here. I always welcome correction as the purpose of this discussion was/is to learn more). However, this view of Hawking radiation would defy the postulate of the thought experiment also. Thus when there is particle-antiparticle creation at the event horizon, according to the thought experiment, they must both escape to infinity. Conclusion: the time it takes for a black hole to "evaporate" is less than that postulated by Bekenstein and Hawking. This also means that evaporation of black holes is not subject to the temperature of the CMBR (which is after all just photons at a specific wavelength) and would seemingly be in contradiction to the laws of thermodynamics. This leads to another interesting question: as photons do not fill every point in spacetime what is the actual temperature of spacetime when there is no CMBR photon present? The temperature we measure is not spacetime itself but photons that are whizzing about within it. Further, we call a photon a massless particle and one of the laws of physics is that massless particles must travel at the speed of light. You would therefore assume that if you "pointed" photons in the direction of a black hole they would cross the EH in pretty short order. However, as I reiterated above the premise was that nothing can cross the EH, and a photon is something. So, I postulate that something is happening to spacetime at the EH that allows a photon to continue moving at light speed but to never cross the EH. My idea is that from an outside observers point of view time would appear to slow down for an object as it approaches the EH but not only that but spacetime is being "compressed" as well. ie. as an object falls towards an EH it finds there is an infinite distance to be crossed before it can get there, and in that time the BH will have evaporated before it can actually cross the EH. Call me crazy if you like, but I find BHs to be so bizarre as to be impossible and what we call a BH is actually something that was approaching infinite density but evaporates before it can get there. It's the whole reason I have embarked on a physics degree because this question needs to be answered. I'm not saying that I will ever have a recognised contribution to this question, only that it has intrigued me so much that I intend to get educated enough to try and answer my own questions.

That's what the article says, yes.

As I understand it, Krauss didn't produce a new theory. He explained why the "old" one didn't explain black holes. As I said, science doesn't need a new theory to show that an old one is wrong. With respect you'd have to show me the link for me to comment. However, I will ask, how do we see x-rays coming from a BH? This is because matter is colliding at relativistic speeds. We see x-rays but these weren't x-rays when they were created they were far more energetic nearer to the event horizon. As I said, Krauss showed how the old theory doesn't lead to BHs. Thus the old theory doesn't explain what you have said either.

Science doesn't work like that. It is not necessary to have an alternative hypothesis in order to show that an existing hypothesis is incorrect. The observations we have indicate something very heavy there, of that there is little doubt. However, whether the gravitating body is a black hole is another question which, if Krauss is right, is not shown by any current theory. According to the BH hypothesis it is an object where at the Event Horizon the escape velocity is equal to the velocity of light. This means accretion disks, jets, x-ray emissions and the effects we association with quasars, are produced outside the EH because if they were produced at the EH (or inside) they could not escape into space. So, you could argue that these effects are not conclusive evidence of a BH and could be the result of an extremely heavy body which is not a BH. As to the observed movement of stars in the vicinity. What details do we have about the actual orbits? ie. what is the smallest orbital radius of an observed object there? Anyone know?

Royston, It appears that I am not alone in my reservations about BHs. Luminaries in the field of physics such as Lawrence Krauss also have reservations. "Physicist Lawrence Krauss and Case Western Reserve colleagues think they have found the answer to the paradox. In a paper accepted for publication in Physical Review D, they have constructed a lengthy mathematical formula that shows, in effect, black holes can't form at all. The key involves the relativistic effect of time, Krauss explains. As Einstein demonstrated in his Theory of General Relativity, a passenger inside a spaceship traveling toward a black hole would feel the ship accelerating, while an outside observer would see the ship slow down. When the ship reached the event horizon, it would appear to stop, staying there forever and never falling in toward oblivion. In effect, Krauss says, time effectively stops at that point, meaning time is infinite for black holes. If black holes radiate away their mass over time, as Hawking showed, then they should evaporate before they even form, Krauss says. It would be like pouring water into a glass that has no bottom. In essence, physicists have been arguing over a trick question for 40 years." http://news.sciencemag.org/sciencenow/2007/06/21-01.html?rss=1 I only found this article today (hence why I have come back to this discussion), so I don't know what has happened in the intervening time and whether Krauss has been refuted or not, but, obviously, I agree with him.

Pioneer, As we are inside the Universe by definition it will be from our reference frame. Unless you are outside the Universe there is no other reference frame for the Universe.