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 1st March 2012, 07:23 AM #121 Molinaro Illuminator     Join Date: Dec 2005 Posts: 3,981 Originally Posted by Farsight Again you've evaded the undefined result at R=2M. I will reiterate, to understand why it's incorrect to do this you have to look at the empirical evidence of moving light, and what Einstein said, and what clocks do. That's funny, really, really funny. There is nothing incorrect about using a different coordinate system. It amazes me that you don't understand such a basic concept and yet think you are the one teaching people. __________________ 100% Cannuck!
 1st March 2012, 07:30 AM #122 Farsight Banned   Join Date: Mar 2008 Posts: 2,640 Originally Posted by sol invictus Clearly, you haven't understood anything about the waterfall analogy. The speed of sound can vary for all sorts of reasons, but here we're discussing a fluid with uniform pressure, salinity, temperature, etc. All that varies is the speed of the current. No we aren't, because the first line of the Visser abstract says this: It is a deceptively simple question to ask how acoustic disturbances propagate in a non-homogeneous flowing fluid. A little further down we see this: The acoustic metric governing the propagation of sound depends algebraically on the density, flow velocity, and local speed of sound. And we have our evidence from sonar in a non-flowing ocean, plus Shapiro and optical clocks demonstrating this: |-------------| |-------------| Originally Posted by sol invictus Now, are you ever going to get around the telling us (1) What's wrong with the math, or (2) What physical experiment gives results that differ between the waterfall and the black hole? OK, I'll look at the body of the paper. But you know full well I can't give you a physical experiment regarding black holes. We do not have any black holes available, and laboratory analogies such as this are not the real thing. I presume from your short response that you concede the point concerning photon energy.
 1st March 2012, 07:42 AM #123 Farsight Banned   Join Date: Mar 2008 Posts: 2,640 Originally Posted by Molinaro That's funny, really, really funny. There is nothing incorrect about using a different coordinate system. It amazes me that you don't understand such a basic concept and yet think you are the one teaching people. I understand these basic concepts. That's why I'm challenging your presumptions. The situation at the event horizon can be likened to a hypothetical observer travelling at c. Time dilation is infinite. He sees nothing. Read up on frozen stars, the name originally applied to black holes by Oppenheimer. Putting a frozen man in front of a frozen clock doesn't mean he sees it ticking.
 1st March 2012, 07:55 AM #124 sol invictus Philosopher     Join Date: Oct 2007 Posts: 8,613 Originally Posted by Farsight No we aren't, because the first line of the Visser abstract says this: It is a deceptively simple question to ask how acoustic disturbances propagate in a non-homogeneous flowing fluid. Yes - my previous post was written hastily and wasn't correct. What I meant to say is that in the fluid we are discussing pressure and density do vary, but they vary only when the current varies, not because of changes in pressure due to depth, or variations in salinity, temperature, etc. as you find in the ocean. Quote: OK, I'll look at the body of the paper. But you know full well I can't give you a physical experiment regarding black holes. We do not have any black holes available, and laboratory analogies such as this are not the real thing. I'm not asking you for a physical experiment, I'm asking you for a thought experiment. If the waterfall analogy is "absolutely wrong" and "a travesty of relativity", why can't you find even one thought experiment where it differs from relativity? Quote: I presume from your short response that you concede the point concerning photon energy. Of course not. That particular mistake of yours is wrong on such a basic level (it's wrong in every branch of physics, including Newtonian dynamics) that it will take us far off the topic of this thread, which is black holes, so it's not appropriate to discuss further in this thread.
 1st March 2012, 08:18 AM #125 Farsight Banned   Join Date: Mar 2008 Posts: 2,640 Originally Posted by sol invictus Yes - my previous post was written hastily and wasn't correct. What I meant to say is that in the fluid we are discussing pressure and density do vary, but they vary only when the current varies, not because of changes in pressure due to depth, or variations in salinity, temperature, etc. as you find in the ocean. Fair enough. I think more highly of you for saying that. Originally Posted by sol invictus I'm not asking you for a physical experiment, I'm asking you for a thought experiment. If the waterfall analogy is "absolutely wrong" and "a travesty of relativity", why can't you find even one thought experiment where it differs from relativity? I'll give it some thought. Maybe I can come up with something when I look at Vasser's paper. But for now I have to get on with something else. Originally Posted by sol invictus Of course not. That particular mistake of yours is wrong on such a basic level (it's wrong in every branch of physics, including Newtonian dynamics) that it will take us far off the topic of this thread, which is black holes, so it's not appropriate to discuss further in this thread. I'm not wrong, sol. If I was, the infinite time dilation would mean that the infalling photon acquires an infinite frequency and energy. It doesn't. You measure the photon frequency to be higher at a lower location because you and your clocks are going slower at that location than at a higher location, that's all. Not because the photon has gained energy in some magical mysterious fashion. Forget what your textbook tells you, think it through for yourself. Think principle of equivalence and conservation of energy. Energy is the one thing that can be neither created nor destroyed. But OK let's come back to it another time. Gotta go.
 1st March 2012, 08:45 AM #126 Maartenn100 Illuminator     Join Date: Nov 2010 Posts: 3,025 In the book of Stephen Hawking 'a brief history of time' a black hole is a massive object, like a star. Because of the huge mass of that object, light can't escape. The escapevelocity of light is too slow. So the light of that star bend back and it never reaches our eyes. So, a very massive object is enough to call it 'a black hole', if light cannot escape. But that implies: if we are in the vecinity of a black hole: it is not black anymore. We will be able to see the light, but we would not be able to escape. With this definition you can also say: A very massive cluster of stars can be so massive, light will bend back on this cluster. It will never reach our eyes. And that could explain the dark matter f.e. Last edited by Maartenn100; 1st March 2012 at 09:01 AM.
 1st March 2012, 10:44 AM #127 ben m Philosopher   Join Date: Jul 2006 Posts: 6,387 Originally Posted by Farsight Forget what your textbook tells you, think it through for yourself. Think principle of equivalence and conservation of energy. "Forget the textbook"? If you tell me to "forget what Pythagoras told me", and "think for myself" about the length of the hypotenuse of a right triangle, I'm going to re-create the conclusion that c^2 = a^2 + b^2. If I "think for myself" about conservation-of-energy issues for a photon falling into a black hole, I'm probably pick a coordinate system, to use GR to figure out the 4-momentum p_mu(t) of the photon in that coordinate system, and start looking at the p_0 term. Anyway, on page 657 of MTW this has been done, exactly the way Einstein wanted you to do it, and there is no GR-overturning energy-conservation problem. Seriously, Farsight, you want me to "think about it myself"? I did.
 1st March 2012, 12:07 PM #128 Reality Check Penultimate Amazing   Join Date: Mar 2008 Location: New Zealand Posts: 20,140 Originally Posted by Maartenn100 ...standard definition of black holes... And that could explain the dark matter f.e. Actually supermassive black holes cannot explain dark matter. For a start in general they are not dark! Supermassive black holes are some of the most energetic (bright) objects in the universe due to the heating of matter in orbit around into them. That is how they are detected - by the x-rays they cause. __________________ NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) Electric comets still do not exist!
 1st March 2012, 12:36 PM #129 Complexity Philosopher   Join Date: Nov 2005 Posts: 9,242 Originally Posted by Farsight Yes I'm an iconoclast destroying dogma, but I'm doing it by pointing out the issues with the mathematics and giving scientific evidence to back that up. Wow, you must be pretty great! (nonsense)
 1st March 2012, 03:36 PM #130 Maartenn100 Illuminator     Join Date: Nov 2010 Posts: 3,025 Originally Posted by Reality Check Actually supermassive black holes cannot explain dark matter. For a start in general they are not dark! Supermassive black holes are some of the most energetic (bright) objects in the universe due to the heating of matter in orbit around into them. That is how they are detected - by the x-rays they cause. ok, So Hawking was wrong. He was talking about the escapevelocity of light because of mass. He desinformed me obviously. So: thank you for the correct information. Last edited by Maartenn100; 1st March 2012 at 03:54 PM.
 1st March 2012, 04:13 PM #131 Roboramma Philosopher     Join Date: Feb 2005 Location: Shanghai Posts: 9,675 Originally Posted by Maartenn100 ok, So Hawking was wrong. He was talking about the escapevelocity of light because of mass. He desinformed me obviously. So: thank you for the correct information. Um, the brightness of black holes that RC is talking about is radiation originating from outside the event horizon of the black holes. There's nothing wrong with what Hawking said, nor is RC contradicting it. It's just that, counter-intuitively perhaps, the region near black holes is extremely bright, due to the acceleration of in-falling matter. __________________ "... when people thought the Earth was flat, they were wrong. When people thought the Earth was spherical they were wrong. But if you think that thinking the Earth is spherical is just as wrong as thinking the Earth is flat, then your view is wronger than both of them put together." Isaac Asimov
 1st March 2012, 04:15 PM #132 Reality Check Penultimate Amazing   Join Date: Mar 2008 Location: New Zealand Posts: 20,140 Originally Posted by Maartenn100 ok, So Hawking was wrong. He was talking about the escapevelocity of light because of mass. He desinformed me obviously. So: thank you for the correct information. Hawking was wrong about what? Where was he talk about "escapevelocity of light because of mass"? He has never said anything about dark matter being supermassive black holes. That was your idea. Light never escapes from within black holes. That is why they are black. What can be emitted from the outside of black holes, is the radiation from matter compressed by them which my post was about. Supermassive black holes compress the matter in their accretion disc. This heats the matter up so much that it emits x-rays. There is also Hawking radiation which is not measurable for the black holes that we have found. __________________ NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) Electric comets still do not exist!
 1st March 2012, 04:20 PM #133 Maartenn100 Illuminator     Join Date: Nov 2010 Posts: 3,025 Ok, thank you Roboramma, I think I get it.
 1st March 2012, 04:31 PM #134 Maartenn100 Illuminator     Join Date: Nov 2010 Posts: 3,025 Originally Posted by Reality-Check Hawking was wrong about what? Where was he talk about "escapevelocity of light because of mass"? First read his book "a brief history of time" and there you will find the page where he wrote about escapevelocity of light as the main cause of the 'blackness' of the blackhole. I deduced from that that clusters can be so heavy (or other big objects in space) that we can not see the light either. To call this 'dark matter' is not illogic at all, I think. It's vary logic. You didn't read the book of Hawking. So you jump into conclusions. I can't tell you wich page, because I have the dutch version 'een korte geschiedenis van de tijd". Last edited by Maartenn100; 1st March 2012 at 04:35 PM.
 1st March 2012, 04:39 PM #135 Reality Check Penultimate Amazing   Join Date: Mar 2008 Location: New Zealand Posts: 20,140 Originally Posted by Maartenn100 First read his book "a brief history of time" ... Do not need to read his book - that is what a black hole of any size is ! I have read actual GR textbooks. So once again: What was Hawking wrong about? My guess is that you still think that I said that light escapes from a black hole. That is wrong. All of the light being detected from the thousands of supermassive black holes that have been detected is emitted from outside of them. Originally Posted by Maartenn100 Ok, thank you Roboramma, I think I get it. And it took you a few minutes to lost it again __________________ NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) Electric comets still do not exist! Last edited by Reality Check; 1st March 2012 at 04:40 PM.
 1st March 2012, 04:45 PM #136 Maartenn100 Illuminator     Join Date: Nov 2010 Posts: 3,025 Reality check: check your reality: read the book, then you will see what I meant. First do your research.
 1st March 2012, 04:48 PM #137 ben m Philosopher   Join Date: Jul 2006 Posts: 6,387 Originally Posted by Maartenn100 I deduced from that that clusters can be so heavy (or other big objects in space) that we can not see the light either. To call this 'dark matter' is not illogic at all, I think. It's vary logic. Maartenn100, if you tried to take a supercluster and make it "so heavy" that we can't see light, that would simply be a black hole. It would not be "a normal cluster, but one that doesn't emit light". (Why not? Because this sort of ultrastrong gravity, if it is able to prevent light from escaping, will also crush all of the matter in the cluster into a single point.) We have evidence that dark matter is not in the form of supercluster-sized black holes.
 1st March 2012, 04:54 PM #138 Dinwar Penultimate Amazing   Join Date: Jul 2010 Posts: 16,668 Just to jump in here: I HAVE read Hawking's books (Brief History of Time and Universe in a Nutshell, anyway). He never discusses large clusters of...well, anything reaching densities that would cause them to appear as black holes. That's where one of your errors lies, by the way: it's density, not just mass, that's an issue. The Andromeda Galaxy weighs more than many black holes, but no one would presume that it therefore bends light to the point where light cannot escape it. We see it, therefore light MUST be able to escape. Gravitational lenses also argue against your idea. A gravitational lense is gravity so strong that it bends light around it, typically splitting images up into multiple smaller images. We know of a number of these; therefore, we can estimate a lower end of the density it'd require to do what you're talking about. Galactic clusters can't do it, therefore it's unlikely (to my mind anyway) that anything in the universe can.
 1st March 2012, 05:05 PM #139 dafydd Banned   Join Date: Feb 2008 Posts: 35,398 Originally Posted by ben m Maartenn100, if you tried to take a supercluster and make it "so heavy" that we can't see light, that would simply be a black hole. It would not be "a normal cluster, but one that doesn't emit light". (Why not? Because this sort of ultrastrong gravity, if it is able to prevent light from escaping, will also crush all of the matter in the cluster into a single point.) We have evidence that dark matter is not in the form of supercluster-sized black holes. Maarten doesn't do evidence.
 1st March 2012, 05:49 PM #140 Reality Check Penultimate Amazing   Join Date: Mar 2008 Location: New Zealand Posts: 20,140 Originally Posted by Maartenn100 Reality check: check your reality: read the book, then you will see what I meant. First do your research. Maartenn100: check your reality. You read the book and tell us what Hawking was wrong about. First do your own research and not just say Hawking was wrong about some unspecified thing. The facts are:Hawking was right about light not escaping from black holes. I never said that light escaped from black holes. The light that allows us to detect supermassive black holes is emitted from outside of black holes. So what did Hawking say that you think was wrong? __________________ NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) Electric comets still do not exist!
 1st March 2012, 05:57 PM #141 dafydd Banned   Join Date: Feb 2008 Posts: 35,398 Originally Posted by Reality Check So what did Hawking say that you think was wrong? This, whatever it means. What Hawking was wrong about is anybody's guess. ''ok, So Hawking was wrong. He was talking about the escapevelocity of light because of mass'' Last edited by dafydd; 1st March 2012 at 05:58 PM.
 1st March 2012, 06:30 PM #142 Reality Check Penultimate Amazing   Join Date: Mar 2008 Location: New Zealand Posts: 20,140 Originally Posted by dafydd This, whatever it means. What Hawking was wrong about is anybody's guess. ''ok, So Hawking was wrong. He was talking about the escapevelocity of light because of mass'' Maartenn100 thought that my post about detection of supermassive black holes through the radiation emitted from their accretion disks was somehow involved with what Hawking said about "escapevelocity of light because of mass''. That seems to be that light emitted from inside the event horizon cannot escape from black holes because of their mass, i.e. what everyone knows about black holes. I have pointed out several times that the accretion disks are outside the black holes. And he seemd to "get it": Originally Posted by Maartenn100 Ok, thank you Roboramma, I think I get it. But he keeps demanding that I read Hawking's book so it looks like Maartenn100 thinks that Hawking got something else wrong. My guess: Maartenn100 bad grasp of English means that he did not "get it", i.e. Hawking is not talking about the light emitted from outside of the black hole. __________________ NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) Electric comets still do not exist!
 1st March 2012, 06:32 PM #143 dafydd Banned   Join Date: Feb 2008 Posts: 35,398 Originally Posted by Reality Check Maartenn100 thought that my post about detection of supermassive black holes through the radiation emitted from their accretion disks was somehow involved with what Hawking said about "escapevelocity of light because of mass''. That seems to be that light emitted from inside the event horizon cannot escape from black holes because of their mass, i.e. what everyone knows about black holes. I have pointed out several times that the accretion disks are outside the black holes. And he seemd to "get it": But he keeps demanding that I read Hawking's book so it looks like Maartenn100 thinks that Hawking got something else wrong. My guess: Maartenn100 bad grasp of English means that he did not "get it", i.e. Hawking is not talking about the light emitted from outside of the black hole. Perhaps he should read the book in a Dutch translation. He might the point then.
 1st March 2012, 10:37 PM #144 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 3,236 some homework problems for Farsight, part 4 It's time to offer more homework problems. As before (except for exercise 8), these exercises assume nothing beyond basic skill in differential calculus and high school algebra. This sequence of exercises began in a different thread, so you might want to reviewthe basic definitions of charts and manifolds, the first eight exercises, which apply those definitions to a 2-sphere such as the earth's surface (but note this important correction to one of the inequalities), the second set of eight exercises, which introduce Schwarzschild charts outside the event horizon of a black hole, and exercises 17 through 23, which use Lemaître charts to prove the Schwarzschild singularity at the event horizon is a mere coordinate singularity. References. In epilogue 2, I cited and linked to an English translation of Lemaître's 1933 paper. Today's exercises are derived fromMatt Visser. Acoustic black holes: horizons, ergospheres, and Hawking radiation. Classical and Quantum Gravity, volume 15, number 6, 1998. http://dx.doi.org/10.1088/0264-9381/15/6/024 http://iopscience.iop.org/0264-9381/15/6/024Andrew J S Hamilton and Jason P Lisle. The river model of black holes. American Journal of Physics, volume 76, issue 6, June 2008, page 519ff. http://ajp.aapt.org/resource/1/ajpia...sAuthorized=no http://arxiv.org/abs/gr-qc/0411060 Why more exercises are needed. Farsight made the case for more exercises: Originally Posted by Farsight Painlevé-Gullstrand coordinates suffer from the same problem to Kruskal-Szekeres coordinates which we've discussed previously. Farsight's problem is that neither Painlevé-Gullstrand nor Kruskal-Szekeres coordinates exhibit the coordinate singularity that Farsight believes to be a real singularity of spacetime. He rejects those coordinate systems because they prove him wrong. Rejecting mathematics because you're afraid it will prove you wrong is not a good way to understand physics. It's better to learn what the math has to teach you. That is not Farsight's way: Originally Posted by Farsight There you go hiding behind mathematics again. It cuts no ice, sol. Originally Posted by Farsight Change the record, Clinger. Taking refuge behind you don't understand the mathematics just isn't good enough. Originally Posted by Farsight The two situations are not the same. You're misinterpreting the mathematics used to describe a black hole because you're ignoring the undefined result at R=2M. Then you're claiming that the mathematics supports your description when it doesn't. Note well that Farsight has made mathematical claims about the event horizon (where r=2M). Note also that Farsight has made an implicit claim that he understands the math well enough to know others are misinterpreting the math. Because Farsight has made both explicit and implicit claims about the math, I don't see any reason to take him seriously when he dismisses math's relevance. Before we get to the exercises, however, let's take care of these red herrings: Originally Posted by Farsight I've provided actual evidence. Optical clocks do go slower when they're lower. The Shapiro delay is a delay, the radar signal goes slower when it goes past the sun. We have hard scientific evidence that the speed of light varies... Et cetera. Farsight has mentioned plenty of empirical facts, but they are not evidence for his idiosyncratic views because the mainstream interpretation of general relativity predicts those observations as well or better than his. In particular, I have already proved (in exercises 20 and 21) that Lemaître charts predict every physical observation that can be predicted using Farsight's beloved Schwarzschild charts. Originally Posted by Farsight Changing to Lemaître coordinates removes the coordinate singularity at the Schwarzschild radius, which is present in Schwarzschild coordinates. Again you've evaded the undefined result at R=2M. I will reiterate, to understand why it's incorrect to do this you have to look at the empirical evidence of moving light, and what Einstein said, and what clocks do. No. Lemaître coordinates explain everything that Schwarzschild coordinates explain, and more. That's a mathematical fact. Had Farsight done his homework (exercises 20 and 21), he'd know that. He hasn't, doesn't, and probably won't. The "undefined result at R=2M" occurs only in Schwarzschild coordinates. That's a purely mathematical defect of Schwarzschild coordinates, with no more physical significance than the coordinate singularity at the north pole of spherical coordinates. People are often willing to tolerate that defect because Schwarzschild coordinates are convenient when you want to ignore a star or black hole's negligible far-field gravity. That purely mathematical, non-physical defect can be avoided by using Lemaître, Painlevé-Gullstrand, Eddington-Finkelstein, or Kruskal-Szekeres coordinates. All of those coordinate systems are in perfect agreement with Schwarzschild coordinates on the submanifold that Schwarzschild coordinates can describe, but (unlike Schwarzschild) are well-behaved and smooth all the way down to the black hole's central singularity. Don't take my word for it. Prove it to yourself. Exercises 20 and 21 established that Lemaître charts agree with Schwarzschild charts on the submanifold covered by Schwarzschild charts. The following exercises prove the analogous result for Painlevé-Gullstrand coordinates. Painlevé-Gullstrand coordinates. With the notational conventions of the previous 23 exercises, the Painlevé-Gullstrand metric around a black hole of mass M is $ $ds^2 = - dt_f^2 + (dr + \beta dt_f)^2 + r^2 d\Omega^2$$where the spatial coordinates are the same as in Schwarzschild coordinates, tf is the proper time of a free-falling object, and $ $\beta = \sqrt{\frac{2M}{r}}$$Those are equations (1) and (2) of the Hamilton and Lisle paper; Visser's equation (49) is basically the same as Hamilton&Lisle's equation (1). Their equation (5) relates the free-fall time tf to Schwarzschild time t and radius r: $ $t_f = t - \int_r^\infty \frac{\beta}{1 - \beta^2} dr$$ Exercises. Exercise 24 is Visser's equation (52) for an infalling object: Exercise 24. Prove: $ $dt_f = dt + \frac{\beta}{1 - \beta^2} dr$$Hint: Differentiate both sides of equation (5) with respect to r.Exercise 25. Prove: Given any Schwarzschild chart, there exists a Painlevé-Gullstrand chart whose restriction to r > 2M is isometric to the Schwarzschild chart. Hint: trivial algebra, as in exercise 20.Exercise 26. Prove: The manifold covered by Schwarzschild charts is (isometric to) a proper submanifold of the manifold covered by Painlevé-Gullstrand charts. Hint: see exercise 21.Exercise 27. Find a minor and inconsequential miscalculation in the arXiv version of Visser's paper. Hint: see section 7.
 2nd March 2012, 01:32 AM #145 Farsight Banned   Join Date: Mar 2008 Posts: 2,640 All: as ever Clinger has no argument, does not respond to the issue I raised, and tries to bury the discussion with mathematics that most readers don't follow. Don't buy it. It's smoke and mirrors, it's Emperor's New Clothes. Insist on scientific evidence and factual English. If somebody offers a mathematical expression to support this, fair enough, but demand a list of terms. If somebody gives you a complex expression without a list of terms, be sceptical. They may be trying to pull the wool over your eyes. RC: Hawking is a celebrity physicist, a "high priest" whose medical condition insulates him from media criticism, and whose actual contribution to physics is scant. Don't think his word is gospel. We should start a thread on Hawking radiation some time. Originally Posted by ben m "Forget the textbook"? If you tell me to "forget what Pythagoras told me", and "think for myself" about the length of the hypotenuse of a right triangle, I'm going to re-create the conclusion that c^2 = a^2 + b^2. Don't use specious examples, ben. I've previously told you how the Lorentz factor is derived from Pythagoras' theorem, it is not an issue. Originally Posted by ben m "If I "think for myself" about conservation-of-energy issues for a photon falling into a black hole, I'm probably pick a coordinate system, to use GR to figure out the 4-momentum p_mu(t) of the photon in that coordinate system, and start looking at the p_0 term. Anyway, on page 657 of MTW this has been done, exactly the way Einstein wanted you to do it, and there is no GR-overturning energy-conservation problem. Seriously, Farsight, you want me to "think about it myself"? I did. Don't pick a coordinate system! It's an artefact of measurement, and we all know that we measure an infalling photon to be blueshifted. Focus instead on the photon itself, it's in space, there are no other particles present, and it's moving through a gravitational field, directly towards our black hole. Does the photon actually gain energy, and if so where does that come from? What loses energy to the photon?
 2nd March 2012, 01:39 AM #146 Roboramma Philosopher     Join Date: Feb 2005 Location: Shanghai Posts: 9,675 Originally Posted by Farsight RC: Hawking is a celebrity physicist, a "high priest" whose medical condition insulates him from media criticism, and whose actual contribution to physics is scant. Don't think his word is gospel. We should start a thread on Hawking radiation some time. I don't see what that has to do with anything that RC said. I don't think he thinks "his word is gospel". He simply doesn't think Marten knows what he's talking about and is looking for him to actually make clear what he thinks Hawking was wrong about. That light can't escape from (beyond the event horizon of) black holes? That's true, and I'm pretty sure you agree with it, so where do you think RC is taking Hawking's word as gospel? __________________ "... when people thought the Earth was flat, they were wrong. When people thought the Earth was spherical they were wrong. But if you think that thinking the Earth is spherical is just as wrong as thinking the Earth is flat, then your view is wronger than both of them put together." Isaac Asimov
 2nd March 2012, 09:08 AM #147 Complexity Philosopher   Join Date: Nov 2005 Posts: 9,242 Originally Posted by Farsight I presume from your short response that you concede the point concerning photon energy. Nonsense. Never make the mistake of assuming that someone agrees with you merely because they didn't respond to something you wrote. What a nasty little trick.
 2nd March 2012, 01:30 PM #149 ben m Philosopher   Join Date: Jul 2006 Posts: 6,387 Originally Posted by Farsight All: as ever Clinger has no argument, does not respond to the issue I raised Yes he does. He posted a correct derivation, using precisely correct GR, that explicitly contradicts your "physics stops at the horizon" assertion. Quote: and tries to bury the discussion with mathematics that most readers don't follow. I follow it just fine. What's your problem?
 2nd March 2012, 04:05 PM #150 ben m Philosopher   Join Date: Jul 2006 Posts: 6,387 I have a question for Farsight. You're certainly trying to project the feeling of being so obviously and transparently correct, that your handful of opponents here must be unusually and especially moronic. You're so certain that their math is wrong that you can't even bother parsing it out to find the error. What if that's true? That would require two failures: (a) that we're too stupid to derive it ourselves AND (b) you're incompetent to explain it in English, and untrained to explain it mathematically. But, Farsight, you've got years and years of internet trolling. Surely you've met at least one person who "understood you" in terms of (a), and who knows enough GR to overcome (b). You would remember such a person, right? Someone who said, "Aha! Farsight, you're right. Your falling-light-clocks argument invalidates equation X which Misner used in Y to derive Z, because the Reggeon doesn't have a Hopf fibration with an odd Hausdorff (i.e., math stuff you didn't understand but that you agreed with.)" Here's a suggestion. Whether or not Complexity, WC, RC, and myself are morons or not, you are definitely incompetent to explain your theory to us. All you need to do is go find the GR-math-conversant expert who agrees with you, and ask that person to explain your idea by proxy. (Nothing wrong with this. Oliver Heaviside was a much better explainer of Maxwell's Equations than Maxwell himself was.) (Oh, wait, what's that? You don't have a proxy who can express your argument in standard GR terminology? Of your crowd of non-moronic supporters-who-lurk-in-the-wings, NOT ONE ever went to college, took 40 credit-hours of GR, and used that knowledge to defend you? Why not?)
 2nd March 2012, 04:25 PM #151 Dinwar Penultimate Amazing   Join Date: Jul 2010 Posts: 16,668 I've got a question for the physicists here: Let's say I'm in a ship and I fall into a black hole, normal to the event horizon. Let's say I also have two laser pointers--one pointed in the direction of motion, and one pointed in the direction I came (ie, outside the black hole). Outside the event horizon, the light would hit both walls. What would happen once I crossed the horizon? My intuitive understanding is that it would still hit both walls. Things CAN go in the opposite direction of gravity at lower than escape velocity, after all. I just want to be sure that my intuitive understanding matches what the physics says. And I figure a minor derail like this isn't too big of an issue.
 2nd March 2012, 04:36 PM #152 ben m Philosopher   Join Date: Jul 2006 Posts: 6,387 Originally Posted by Dinwar I've got a question for the physicists here: Let's say I'm in a ship and I fall into a black hole, normal to the event horizon. Let's say I also have two laser pointers--one pointed in the direction of motion, and one pointed in the direction I came (ie, outside the black hole). Outside the event horizon, the light would hit both walls. What would happen once I crossed the horizon? My intuitive understanding is that it would still hit both walls. Things CAN go in the opposite direction of gravity at lower than escape velocity, after all. I just want to be sure that my intuitive understanding matches what the physics says. And I figure a minor derail like this isn't too big of an issue. Yes, you're in an ordinary inertial frame, and light from one pointer hits the one wall and the other pointer hits the other wall. This is true before, during, and after the (uneventful) horizon crossing.
 2nd March 2012, 04:50 PM #153 Dinwar Penultimate Amazing   Join Date: Jul 2010 Posts: 16,668 And from the outside observer's perspective, it appears that the light pointed out of the black hole is going more slowly than the wall of the ship, and therefore being hit by the ship, right? Or is the concept of an outside observer inapplicable to falling into a black hole?
 2nd March 2012, 04:52 PM #154 Ziggurat Penultimate Amazing     Join Date: Jun 2003 Posts: 38,311 Originally Posted by Dinwar My intuitive understanding is that it would still hit both walls. Things CAN go in the opposite direction of gravity at lower than escape velocity, after all. I just want to be sure that my intuitive understanding matches what the physics says. And I figure a minor derail like this isn't too big of an issue. Your intuitive understanding is half-right. The laser will indeed hit both walls, but not because the light is moving out at less than escape velocity. Inside a black hole, nothing can move outwards at all. Things can only fall inwards, and that includes light. What happens is that you can fall inwards faster or slower. And the "inward-facing" light will fall in faster than your ship, while the "outward-facing" light will fall in more slowly. But it is always still falling inwards. I used quotes because inside, everything points towards the singularity, because the singularity is the inevitable future no matter where you are or which direction you move. __________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law
 2nd March 2012, 04:57 PM #155 Brian-M Daydreamer     Join Date: Jul 2008 Posts: 8,044 Originally Posted by Dinwar My intuitive understanding is that it would still hit both walls. Things CAN go in the opposite direction of gravity at lower than escape velocity, after all. I just want to be sure that my intuitive understanding matches what the physics says. And I figure a minor derail like this isn't too big of an issue. Assuming you mean that the event horizon is between you and the wall, that's pretty much what I was saying at the end of post 10, only with radio waves striking a probe instead of light striking a wall. That idea got slapped down pretty quick. However... Originally Posted by ben m Yes, you're in an ordinary inertial frame, and light from one pointer hits the one wall and the other pointer hits the other wall. This is true before, during, and after the (uneventful) horizon crossing. So what's going on here? Would the ship have to be falling in fast enough, and time dilation great enough, that the other side of the wall would be inside the event horizon before it gets to the wall, or is there something else going on that I don't understand? ETA: Originally Posted by Ziggurat Inside a black hole, nothing can move outwards at all. Things can only fall inwards, and that includes light. What happens is that you can fall inwards faster or slower. And the "inward-facing" light will fall in faster than your ship, while the "outward-facing" light will fall in more slowly. But it is always still falling inwards. I used quotes because inside, everything points towards the singularity, because the singularity is the inevitable future no matter where you are or which direction you move. Ah, okay. That makes sense. From your frame of reference the photons would be moving away from the singularity, while from the singularity's frame of reference, the light would be traveling toward it. But wouldn't this mean that the singularity would appear to be moving towards you at faster than the speed of light? (Not that you'd be able to actually observe it.) __________________ "That is just what you feel, that isn't reality." - hamelekim Last edited by Brian-M; 2nd March 2012 at 05:11 PM.
 2nd March 2012, 06:40 PM #156 Dinwar Penultimate Amazing   Join Date: Jul 2010 Posts: 16,668 Originally Posted by Ziggurat What happens is that you can fall inwards faster or slower. Right--the "outside" wall (closest to the event horizon) is falling faster than the light, so it hits the light. This would be much easier with pictures.... ASCII to the rescue! | is the wall of the ship, - is laser light, and [] is the laser/observer. Distance from the colon is distance from the event horizon. Light hits at T=5 t=0:..|_____-[] T=1:......|____--[] T=2:..........|___---[] T=3:..............|__----[] T=4:..................|_-----[] T=5:......................|------[] It's not really right--the laser/observer would move faster than the wall, being closer to the black hole (it's not noticeable on Earth, but when g>c it becomes a major factor), but it shows what I'm thinking, I think. Originally Posted by Brian-M Assuming you mean that the event horizon is between you and the wall, Flawed assumption. I'm talking about what happens when both are on the same side of the event horizon. I'd like to make sure my logic is correct within each frame of reference before I start mucking around with what happens between them. Last edited by Dinwar; 2nd March 2012 at 06:42 PM. Reason: Fixing my sad, sad attempt at a pictoral representation of my logic
 2nd March 2012, 08:54 PM #157 Ziggurat Penultimate Amazing     Join Date: Jun 2003 Posts: 38,311 Originally Posted by Brian-M Ah, okay. That makes sense. From your frame of reference the photons would be moving away from the singularity, while from the singularity's frame of reference, the light would be traveling toward it. It's weirder than that. Once you're inside, there is no direction to the singularity. The singularity is everywhere. But it's everywhere in the future. You don't need to move to get to the singularity, you just need to wait. Motion only changes how long the wait is. Quote: But wouldn't this mean that the singularity would appear to be moving towards you at faster than the speed of light? No, because the singularity is no longer moving towards you through space, it's now moving towards you through time. __________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law
 2nd March 2012, 09:16 PM #158 Zeuzzz Banned   Join Date: Dec 2007 Posts: 5,211 Black holes dont exist, there a good reason why the color "black" was chosen in their name to explain them, considering the universe is black; you cant see it. If they were called purple holes would make things a lot easier, and I would agree with their existance.
 2nd March 2012, 09:34 PM #159 Dinwar Penultimate Amazing   Join Date: Jul 2010 Posts: 16,668 Originally Posted by Zeuzzz Black holes dont exist, there a good reason why the color "black" was chosen in their name to explain them, considering the universe is black; you cant see it. If they were called purple holes would make things a lot easier, and I would agree with their existance. That's like saying Galathea enuiensis doesn't exist, because obviously no crab is boring.
 2nd March 2012, 10:34 PM #160 MrSkunkwork100 Scholar     Join Date: Sep 2011 Posts: 67 Taken from NASA's website: "Black holes are the evolutionary endpoints of stars at least 10 to 15 times as massive as the Sun. If a star that massive or larger undergoes a supernova explosion, it may leave behind a fairly massive burned-out stellar remnant. With no outward forces to oppose gravitational forces, the remnant will collapse in on itself. The star eventually collapses to the point of zero volume and infinite density, creating what is known as a "singularity." Around the singularity is a region where the force of gravity is so strong that not even light can escape. Thus, no information can reach us from this region. It is therefore called a black hole, and its surface is called the "event horizon." But contrary to popular myth, a black hole is not a cosmic vacuum cleaner. If our Sun was suddenly replaced with a black hole of the same mass, Earth's orbit around the Sun would be unchanged. Of course, Earth's temperature would change, and there would be no solar wind or solar magnetic storms affecting us. To be "sucked" into a black hole, one has to cross inside the Schwarzschild radius. At this radius, the escape speed is equal to the speed of light, and once light passes through, even it cannot escape." http://imagine.gsfc.nasa.gov/docs/sc...ack_holes.html Basically, a black hole is just a situation where a once active star has collapsed in on itself and the gravitational forces that were associated with that star are still working while there is no longer any volume. Black holes have a threshhold across which matter cannot escape the extreme, massive, gravitational pull--once that thresshold has been crossed. They are not holes so much as they are just altered celestial bodies from which we can't see what is going on beyond the threshhold. __________________ Just another blue-collar guy

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# zbMATH — the first resource for mathematics

On class-preserving Coleman automorphisms of finite separable groups. (English) Zbl 1302.20028
One says that the normaliser problem has a positive answer for a finite group $$G$$ if the only invertible elements of the integral group ring $$\mathbb ZG$$ that normalise the group $$G$$ are the obvious ones, i.e. the elements that belong to $$GZ(\mathcal U(\mathbb ZG))$$, where $$Z(\mathcal U(\mathbb ZG))$$ denotes the centre of the unit group $$\mathcal U(\mathbb ZG)$$ of $$\mathbb ZG$$. Equivalently, the group $$\text{Out}_{\mathbb Z}(G)=\operatorname{Aut}_{\mathbb Z}(G)/\text{Inn}(G)$$ is trivial, where $$\operatorname{Aut}_{\mathbb Z}(G)$$ is the subgroup of the automorphism group $$\operatorname{Aut}(G)$$ consisting of those automorphisms of $$G$$ that are induced by an inner automorphism of $$\mathcal U(\mathbb ZG)$$. A result due to Krempa says that $$\text{Out}_{\mathbb Z}(G)$$ is an elementary Abelian $$2$$-group. As shown by M. Hertweck [Ann. Math. (2) 154, No. 1, 115-138 (2001; Zbl 0990.20002)], there exist finite groups for which the normaliser property has a negative answer.
In this context, the following two groups play a crucial role: $$\text{Out}_c(G)=\operatorname{Aut}_c(G)/\text{Inn}(G)$$ and $$\text{Out}_{\text{Col}}(G)=\operatorname{Aut}_{\text{Col}}(G)/\text{Inn}(G)$$, where $$\operatorname{Aut}_c(G)$$ consists of the (conjugacy) class-preserving automorphisms of the group $$G$$ and $$\operatorname{Aut}_{\text{Col}}(G)$$ consists of those automorphisms of $$G$$ whose restriction to each Sylow subgroup of $$G$$ equals the restriction to some inner automorphism of $$G$$ (the so called Coleman automorphisms of $$G$$). It is well known that $$\text{Out}_{\mathbb Z}(G)\subseteq\text{Out}_c(G)\cap\text{Out}_{\text{Col}}(G)$$. So, the normaliser problem has a positive answer for $$G$$ in case the latter intersection is an odd group.
In this paper the following two results are proved for a finite group $$G$$.
Theorem A: Assume $$G=NA$$, where $$N$$ is a nilpotent normal subgroup of $$G$$ and $$A$$ is an Abelian subgroup of $$G$$. If the Sylow $$2$$-subgroup of $$N$$ is cyclic, then $$\text{Out}_c(G)\cap\text{Out}_{\text{Col}}(G)$$ is of odd order.
Theorem B: Assume $$G=A\rtimes P$$, a semidirect product of an Abelian group $$A$$ and a dihedral or generalised quaternion $$2$$-group $$P$$. If the Sylow $$2$$-subgroup of $$A$$ is cyclic, then $$\text{Out}_c(G)\cap\text{Out}_{\text{Col}}(G)$$ is of odd order.
An example of Z. S. Marciniak and K. W. Roggenkamp, [in: Algebra – representation theory. Proceedings of the NATO Advanced Study Institute, Constanta, Romania, August 2–12, 2000. NATO Sci. Ser. II, Math. Phys. Chem. 28, 159-188 (2001; Zbl 0989.20002)] shows that if $$G=NA$$, with $$N$$ a nilpotent normal subgroup of $$G$$ and $$A$$ an Abelian subgroup of $$G$$, then in general it is not the case that $$\text{Out}_c(G)\cap\text{Out}_{\text{Col}}(G)$$ is of odd order.

##### MSC:
 20D45 Automorphisms of abstract finite groups 20C05 Group rings of finite groups and their modules (group-theoretic aspects) 16S34 Group rings 16U60 Units, groups of units (associative rings and algebras) 20D20 Sylow subgroups, Sylow properties, $$\pi$$-groups, $$\pi$$-structure
Full Text:
##### References:
 [1] Coleman D. B., Proc. Amer. Math. Soc. 5 pp 511– (1964) [2] DOI: 10.1007/s11425-011-4298-2 · Zbl 1234.20029 · doi:10.1007/s11425-011-4298-2 [3] DOI: 10.1006/jabr.2001.8760 · Zbl 0993.20017 · doi:10.1006/jabr.2001.8760 [4] DOI: 10.1016/S0022-4049(00)00167-5 · Zbl 0987.16015 · doi:10.1016/S0022-4049(00)00167-5 [5] DOI: 10.1007/s006050200029 · Zbl 1004.20011 · doi:10.1007/s006050200029 [6] Hertweck M., J. Group Theory 12 pp 157– (2009) [7] DOI: 10.1007/s002090100318 · Zbl 1047.20020 · doi:10.1007/s002090100318 [8] DOI: 10.1081/AGB-120029897 · Zbl 1072.20030 · doi:10.1081/AGB-120029897 [9] DOI: 10.1016/S0021-8693(02)00102-3 · Zbl 1017.16023 · doi:10.1016/S0021-8693(02)00102-3 [10] DOI: 10.1007/978-94-010-0814-3_8 · doi:10.1007/978-94-010-0814-3_8 [11] Rose J. S., A Course on Group Theory (1978) [12] Sehgal S. K., Units in Integral Group Rings (1993) · Zbl 0803.16022
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

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# Definition of isonormal Gaussian process

The definition of an isonormal Gaussian process (from Nualart's book: The Malliavin Calculus and related topics) is as follows:

My question is: why we want the space $$H$$ to be a real separable Hilbert space?

If you read the rest of the first chapter, there is a paragraph entitled "Wiener's Chaos Decomposition". This Decomposition is only valid if the surrounding space is separable . I don't have the book with me.

I think the real restriction is not that essential. It's just simpler. One does not have complex conjugates all over the place in the definitions. It is also related to the desire to model real-valued processes like standard Brownian motion. The separable hypothesis is more important. That makes $$H$$ a Polish space which is nice for probability. Note however that here the space that really needs to Polish is typically a Banach space that is bigger than $$H$$. See the notion of abstract Wiener space.

• As I understand we need this more or less in order to construct Gaussian measures on Banach spaces. Because of the connection between Hilbert spaces and abstract Wiener spaces. – homomathematicus Oct 15 '18 at 11:50


text:
for Journals by Title or ISSN for Articles by Keywords help
 Subjects -> MATHEMATICS (Total: 864 journals)     - APPLIED MATHEMATICS (68 journals)    - GEOMETRY AND TOPOLOGY (19 journals)    - MATHEMATICS (643 journals)    - MATHEMATICS (GENERAL) (40 journals)    - NUMERICAL ANALYSIS (19 journals)    - PROBABILITIES AND MATH STATISTICS (75 journals) MATHEMATICS (643 journals)                  1 2 3 4 | Last

1 2 3 4 | Last

 Facta Universitatis, Series : Mathematics and Informatics   [0 followers]  Follow       Open Access journal    ISSN (Print) 0352-9665 - ISSN (Online) 2406-047X    Published by U of Niš  [11 journals]
• A COMBINED SDLINAR(p) MODEL AND IDENTIFICATION AND PREDICTION OF ITS
LATENT COMPONENTS

• Authors: Miodrag Srboljub Djordjevic
Pages: 919 - 946
Abstract: The main subject of this paper is a combined integer-valued autoregressive time series with both positive and negative values, based on a new thinning operator. Some important properties are analyzed. Estimators of the unknown parameters are derived and their asymptotic behavior are analyzed. A simulation and an
application on real-data are, also, shown. At the end, a mechanism for identification and prediction of latent dimensions of the model are presented.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• Soft connected properties and irresolute soft functions based on b-open
soft sets

• Authors: Alaa Mohamed Abd El-latif
Pages: 947 - 967
Abstract: In this paper, we introduce another application on the b-open soft sets [10], by
studying the notion of connectedness based on it. We gave the basic denitions and
theorems about it. Further, we introduce the notion of b-irresolute soft functions as a
generalization to the b-continuous soft functions and study their properties in detail.
Finally, we show that, the surjective b-irresolute soft image of soft b-connected space
is also soft b-connected.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• FIXED POINT THEOREMS FOR TWO PAIRS OF MAPPINGS IN PARTIAL METRIC SPACES

• Authors: Valeriu POPA, Alina-Mihaela PATRICIU
Pages: 969 - 980
Abstract: In this paper a general fixed point theorem for two pairs of weakly compatible mappings satisfying a \phi - implicit relation different than the type from [14] is proved. As applications, we obtain the sufficient conditions for the existence of fixed
points for a sequence of mappings in partial metric spaces.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• JENSEN’S TYPE TRACE INEQUALITIES FOR CONVEX FUNCTIONS OF SELFADJOINT
OPERATORS IN HILBERT SPACES

• Authors: Sever S. Dragomir
Pages: 981 - 998
Abstract: Some Jensen.s type trace inequalities for convex functions of self-adjoint operators in Hilbert spaces are provided. Applications for some convex functions of interest are also given.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• On maximal function and V-conjugation

Pages: 999 - 1009
Abstract: We proved that on the 3-series field H1
cannot be defined by means of the V-conjugation. In other words, the norms kf k
H1 and k~f k L1 are not equivalent. This answers the question formulated in [12]. Also, we prove that the mean value of the function on the coset IN 1(x) is dominated by maximal operator, either by MN 1 or MN on some translated element.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• On the measurement of growth properties of entire and meromorphic
functions focusing their relative type and relative weak type

• Authors: Dr Sanjib Kumar Datta, Tanmay Biswas
Pages: 1011 - 1028
Abstract: In this paper we investigate some growth properties of entire and meromorphic functions on the basis of their relative type and relative weak type.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• h-exponential change of Finsler metric

• Authors: Manish Kumar Gupta, Anil K. Gupta
Pages: 1029 - 1039
Abstract: In this paper, we studied a Finsler space whose metric is given by an \textsl{h}-exponential change and obtain the Cartan connection coefficients for the change. We also find the necesssary and sufficient condition for an h-exponential change of Finsler metric to be projective.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• ON THE BASE SPACE OF AN ALMOST PARACONTACT SUBMERSION

• Authors: Tshikunguila Tshikuna-Matamba
Pages: 1041 - 1049
Abstract: The purpose of this note is to describe the base space of an almost paracontact submersion. Here the base space is an almost para-Hermitian manifold. So, the paper intertwines paracontact and para-Hermitian structures via the theory of submersions.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• ON GENERALIZED M-PROJECTIVE -RECURRENT TRANS-SASAKIAN MANIFOLDS

• Authors: Jai Prakash Jaiswal, Arjun Singh Yadav
Pages: 1051 - 1060
Abstract: The aim of the present paper is to study generalized M-projective phi-recurrent trans-Sasakian manifold and its various geometric properties. First, we nd the sucient condition for generalized M-projective phi-recurrent trans-Sasakian manifold to become Einstein. Then non-existence of generalized M-projective phi-recurrent
trans-Sasakian manifold has been shown under some certain conditions.
At the last, the sucient condition for super generalized Ricci-recurrent
was also established.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• On decomposable and warped product generalized quasi Einstein manifolds

• Authors: SAHANOUS MALLICK, Prajjwal Pal
Pages: 1061 - 1071
Abstract: The object of the present paper is to study decomposable and warped product generalized quasi Einstein manifolds.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• General decay of solution for coupled system of viscoelastic wave
equations of Kirchhoff type with density in $\mathbb{R}^{n}$

• Authors: Abbes Benaissa, Abderrahmane Beniani, Khaled Zennir
Pages: 1073 - 1090
Abstract: System of viscoelastic wave equations of Kirchhoff type isconsidered. For wider class of relaxation functions, we use a spaces weighted by density function to establish a very general decay rate of solution.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

• SUPRA HOMEOMORPHISM IN SUPRA TOPOLOGICAL ORDERED SPACES

• Authors: Mohamed Abo-elhamayel, Tareq M. Al-shami
Pages: 1091 - 1106
Abstract: The purpose of this paper is to introduce the concepts of x-supra continuous (open, closed,homeomorphism) maps in supra topological ordered spaces for x  in {I,D,B}. We study the relationship among these types with the help of examples and investigate the equivalent conditions for each concept. In particular, we present the sucient conditions for maps to preserve some of separation axioms.
PubDate: 2017-01-10
Issue No: Vol. 31, No. 5 (2017)

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# Prove (lim n -> infinity) (1 + 1/n)^n = e

1. Jan 13, 2005

### kishtik

(lim n -> infinity) (1 + 1/n)^n = e ?

2. Jan 13, 2005

### Galileo

For me, that is the definition of the number e.

So if you want to prove it, you have to tell what your definition of e is to show it is equivalent to that limit.

If for example:

$$\mbox{ e is the number such that:} \quad \lim_{h \to 0} \frac{e^h-1}{h}=1$$
is your definition. Then take the derivative of ln(x) at the point 1. (Expressed as a limit)

Last edited: Jan 13, 2005
3. Jan 13, 2005

### HallsofIvy

Staff Emeritus
Using what preliminaries? That is sometimes used as the definition of e!

Lacking that, the simplest way is to use L'Hopital's rule. Strictly speaking L'Hopital's rule only applies to limits of differentiable functions but the limit of (1+1/x)x, as x-> infinity, must be the same as the limit of the sequence.

At x= infinity this is of the form "1infinity" so let y= (1+ 1/x)x and take the logarithm: ln(y)= x ln(1+ 1/x) is now of the form "infinity*0". We can write that as $\frac{ln(1+1/x)}{x}$ and differentiate numerator and denominator separately. The derivative of ln(1+1/x) is $\frac{1}{1+1/x}(-\frac{1}{x^2})= -\frac{1}{x^2+x}$. The derivative of 1/x is -1/x2. The ratio of those two is $\frac{x^2}{x^2+x}= \frac{x}{x+1}= \frac{1}{1+\frac{1}{x}}$ and the limit of that, as x-> infinity, is 1.

Since the limit of ln y(x)= 1, and ln is continuous, the limit of y itself is e.

4. Jan 13, 2005

### kishtik

Thanks but I couldn't understand the latex code.

5. Jan 13, 2005

### vincentchan

Okay, I'll use the definition in calculas to PROVED that.....

$$\frac{d}{dx} e^x = e^x$$ -------->>>difinition of e
substitude $$e = \lim (1+1/n)^n$$ into $$e^x$$

we have $$e^x = \lim (1 + 1/n)^{nx}$$..

let $$u=nx$$,

we got $$e^x = \lim (1 + 1/n)^{nx}=\lim (1 + x/u)^u=\lim (1 + x/n)^n$$

$$\frac{d}{dx}\lim (1 + 1/n)^{nx} = \frac{d}{dx}\lim (1 + x/n)^n = \lim n(1+x/n)^{n-1}*(1/n) = \lim (1 + x/n)^n=\lim (1 + 1/n)^{nx}$$

so..$$e=\lim (1 + 1/n)^n$$

6. Jan 13, 2005

### HallsofIvy

Staff Emeritus
Oh blast! I forgot which forum I was in!

At x= infinity this is of the form "1infinity" so let y= (1+ 1/x)x and take the logarithm: ln(y)= x ln(1+ 1/x) is now of the form "infinity*0". We can write that as $\frac{ln(1+1/x)}{x}$ and differentiate numerator and denominator separately. The derivative of ln(1+1/x) is $\frac{1}{1+1/x}(-\frac{1}{x^2})= -\frac{1}{x^2+x}$. The derivative of 1/x is -1/x2. The ratio of those two is $\frac{x^2}{x^2+x}= \frac{x}{x+1}= \frac{1}{1+\frac{1}{x}}$ and the limit of that, as x-> infinity, is 1.

Since the limit of ln y(x)= 1, and ln is continuous, the limit is e.

Last edited: Jan 13, 2005


text:
# What is the Transport Equation?

The transport equation describes how a scalar quantity is transported in a space. Usually, it is applied to the transport of a scalar field (e.g. chemical concentration, material properties or temperature) inside an incompressible flow. From the mathematical point of view, the transport equation is also called the convection-diffusion equation, which is a first order PDE (partial differential equation). The convection-diffusion equation is the basis for the most common transportation models.

## Mathematical derivation

The transport equation (or convection-diffusion equation) can be seen as the generalization of the continuity equation1

${}^{1}$

. While the continuity equation (extensively described in the article about incompressible flow) usually describes the conservation of mass, the convection-diffusion equation describes the continuity/conservation of any scalar field in any space. Let’s consider an infinitesimal portion of space and its boundaries, as described in figure 1.

The continuity principle states that the rate of change for a scalar quantity in any differential control volume is given by flow and diffusion into and out of that part of the system, along with any generation or consumption inside the control volume. In practice, it means that the variation of concentration of a certain quantity in the volume is given by the balance of this quantity flow across the boundary and the amount of quantity produced or removed in the volume. From the mathematical point of view, this balance is expressed by the following equation:

ct+j=S(1)

$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial }c}{\mathrm{\partial }t}+\mathrm{\nabla }\cdot j=S\end{array}$

where c

$c$

is the scalar field to be analyzed, j

$j$

is the flux of c

$c$

through the boundary, and S

$S$

is the source/sink term inside Ω

$\mathrm{\Omega }$

. Equation 1 is nothing more than a balance of a scalar quantity inside the volume: the first term (ct

$\frac{\mathrm{\partial }c}{\mathrm{\partial }t}$

) represents the variation of quantity of the scalar field inside the control volume, the second term (j

$\mathrm{\nabla }\cdot j$

) represents the net balance of the quantity of field which enters and exits the control volume, and the third term (S

$S$

) represents the amount of the scalar quantity “created” or “destroyed” inside the volume.

Equation 1 can be further detailed by developing its terms:

• The flux j
$j$

can be divided into two terms: the convective and the diffusive terms. The convection term is the quantity of the transported field which moves across the boundaries because of the flow; thus it is proportional to the velocity and can be written as jconvection=cu

${j}_{convection}=cu$

, where c

$c$

is the transported scalar quantity and u

$u$

is the velocity of the means which transports this quantity. The diffusion term is the transportation of the scalar quantity according to its gradient, so jdiffusion=Dc

${j}_{diffusion}=D\mathrm{\nabla }c$

, where D

$D$

is the diffusivity.

• The source term can be divided into a pure source term and a reaction term. The pure source term (SS
${S}_{S}$

) represents the creation/destruction rate of the field inside the volume. The reaction term (SR

${S}_{R}$

) describes the creation/destruction of the transported quantity as a reaction to this quantity itself; it is, therefore, proportional to the transported field and can be written as SR=f(c)

${S}_{R}=f\left(c\right)$

, where f(c)

$f\left(c\right)$

is a function of the transported scalar field. SR

${S}_{R}$

is quite uncommon for engineering applications, so it is often neglected.

Equation 1 can thus be re-written in its fully developed form as:

ct+(Dc)+(uc)=SS+SR(2)

$\begin{array}{}\text{(2)}& \frac{\mathrm{\partial }c}{\mathrm{\partial }t}+\mathrm{\nabla }\cdot \left(D\mathrm{\nabla }c\right)+\mathrm{\nabla }\cdot \left(uc\right)={S}_{S}+{S}_{R}\end{array}$

## Heat Transfer

During thermal simulations, the temperature field (which is scalar) is transported according to the convection diffusion equation. In this specific case, the following notation is commonly used by the science community:

(ρcpT)t+(kT)+(uT)=Q+ϵσA(T4hotT4cold)(3)

$\begin{array}{}\text{(3)}& \frac{\mathrm{\partial }\left(\rho {c}_{p}T\right)}{\mathrm{\partial }t}+\mathrm{\nabla }\cdot \left(-k\mathrm{\nabla }T\right)+\mathrm{\nabla }\cdot \left(uT\right)=Q+ϵ\sigma A\left({T}_{hot}^{4}-{T}_{cold}^{4}\right)\end{array}$

where:

• ρ
$\rho$

is the material density

• cp
${c}_{p}$

is the heat capacity

• T
$T$

is the temperature

• k
$k$

is the thermal conductivity

• Q
$Q$

is the volumetric heat flux

• ϵ
$ϵ$

is the emissivity

• σ
$\sigma$

is the Stefan-Boltzmann constant

• A
$A$

is the boundary surface in which heat is exchanged by radiation.

Equations 2 and 3 differ only for the notation and for the complexity of the reaction term, coming from the physical modelling of heat transfer phenomena3

${}^{3}$

.

Equation 3 is often simplified in common engineering applications thanks to the following hypotheses:

• Homogeneous material (i.e. constant and uniform material parameters):

ρcp(T)tkΔT+(uT)=Q+ϵσA(T4hotT4cold)(4)

$\begin{array}{}\text{(4)}& \rho {c}_{p}\frac{\mathrm{\partial }\left(T\right)}{\mathrm{\partial }t}-k\mathrm{\Delta }T+\mathrm{\nabla }\cdot \left(uT\right)=Q+ϵ\sigma A\left({T}_{hot}^{4}-{T}_{cold}^{4}\right)\end{array}$

• Incompressible flow (i.e. u=0
$\mathrm{\nabla }\cdot u=0$

):

ρcp(T)tkΔT+uT=Q+ϵσA(T4hotT4cold)(5)

$\begin{array}{}\text{(5)}& \rho {c}_{p}\frac{\mathrm{\partial }\left(T\right)}{\mathrm{\partial }t}-k\mathrm{\Delta }T+u\cdot \mathrm{\nabla }T=Q+ϵ\sigma A\left({T}_{hot}^{4}-{T}_{cold}^{4}\right)\end{array}$

• Neglection of heat exchange by radiation:

ρcp(T)tkΔT+uT=Q(6)

$\begin{array}{}\text{(6)}& \rho {c}_{p}\frac{\mathrm{\partial }\left(T\right)}{\mathrm{\partial }t}-k\mathrm{\Delta }T+u\cdot \mathrm{\nabla }T=Q\end{array}$

$\frac{\mathrm{\partial }T}{\mathrm{\partial }t}=0$

):

kΔT+uT=Q(7)

$\begin{array}{}\text{(7)}& -k\mathrm{\Delta }T+u\cdot \mathrm{\nabla }T=Q\end{array}$

## Chemical concentration

Transportation models are commonly used to analyze the dispersion of a certain chemical component in a fluid — some pollution particles, for instance. In this case, the transport is defined as “passive transport”, because the presence of the chemical concentration does not affect the fluid flow. Referring to equation 2, the convection term represents the transportation of the chemical component with the fluid, while the diffusive term represents a chemical reaction or molecular diffusion phenomena which could occur in the flow. When a simple recirculation of a certain amount of chemical components is modeled, no explicit boundary conditions are needed for the simulation and only the initial chemical concentration distribution is required (figure 2).

On the other side, explicit boundary conditions are needed when the injection of a chemical component in a region is simulated, e.g. a chimney stack (figure 3); in this case, a boundary condition on the inlet boundary is needed to be imposed (for instance c=1

$c=1$

, if pure smoke is emitted into the atmosphere from the chimney or c=0.5

$c=0.5$

if the chimney output is a mixture of 50% smoke and 50% air).

## Level-set method

The level set is a particular family of transportation models in which a distance function (named level-set function) is transported 2,3

${}^{2,3}$

. This distance function is computed with respect to an interface (surface for 3D problems, line for 2D problems, and point for 1D problems) and must have 2 characteristics:

• it must be an eulerian function, i.e. ||φ||=1
$||\mathrm{\nabla }\phi ||=1$

where φ

$\phi$

is the level-set function

• it is signed, so it is positive on one side of the interface and negative on the other side.

For the sake of clarity, examples of the 1D and 2D distance function to be transported are shown in figure 4 and figure 5 respectively.

Figure 4: Level-set function computed in an 1D domain with respect to the interface Γ=[x=1x=2]

$\mathrm{\Gamma }=\left[x=1\cup x=2\right]$

C=(0,0)

$C=\left(0,0\right)$

$r=0.5$

Level-set methods follow the principles of the classic transport equation, but they require a further computation step (called re-initialization) in order to maintain the Eulerian distance function condition. The main advantage of the use of a level-set function is the possibility to solve the transport equation just once, but to obtain many results in the post-processing. For instance, for multiphase flows (see the following section), it is possible to compute the transported viscosity and density fields, the surface tension, and to impose a mixing law across the interface with very little computational effort. In addition, the use of the level-set function allows dealing with fields which show high gradients (e.g. the density field for the simulation of a water droplet falling through the air). The transport of the density field would lead to numerical instabilities across the interface, where the density gradient is very high, while the transport of the level-set is more stable and enables computing the density as post-processing. The transportation of a distance function has several engineering applications; in the following sections, we will report the most common ones.

### Image segmentation

The purpose of level-set methods in image processing is to identify and export the contour of a certain profile in the image. This process (called image segmentation) is widely used in computer graphics and medical applications4

${}^{4}$

. It is commonly used to analyze ultrasounds and tomographies’ outputs in the localization of tumors; in figure 6 it is shown the segmentation process for an ultrasound of lungs. By the segmentation process, it is possible to identify objects inside a digital image (i.e. the lungs) through the identification of its boundaries and, eventually, use this information for the analysis of the pathology.

In the case of image segmentation, convection and source terms in equation 3 are not considered; the only means of transport is diffusion according to the gradient of pixels density of the image.

### Multiphase flow

The level-set method is also commonly used to simulate multiphase flows. In this case, the transport model is defined as “active transport” because convection/diffusion of the interface (through the distance function) affects the flow. The distance from the interface influences the flows because it is used to compute the material parameters in the computational domain; given a flow composed by two fluids characterized by different material properties η1

${\eta }_{1}$

and η2

${\eta }_{2}$

, the value in any given position (or element, in the case of numerical simulation) is given by:

η=η1H(φ)+η2(1H(φ))(8)

$\begin{array}{}\text{(8)}& \eta ={\eta }_{1}H\left(\phi \right)+{\eta }_{2}\left(1-H\left(\phi \right)\right)\end{array}$

where H(ϕ)

$H\left(\varphi \right)$

is the Heaviside function defined as

H(φ)=1φ0(9)

$\begin{array}{}\text{(9)}& H\left(\phi \right)=1\phantom{\rule{2em}{0ex}}\mathrm{\forall }\phi \ge 0\end{array}$

In this way, any complex multiphase flow with non-uniform material properties can be modeled through the simple transport of a scalar field (the level-set function) and the interface can be tracked implicitly as the zero Isocontour of this function. In figure 7, the results of the simulation of a rising bubble are shown; in this case, the interface represents the φ=0

$\phi =0$

isoline, while the region where φ>0

$\phi >0$

is colored in blue and the region where φ<0

$\phi <0$

is colored in red.

### Microstructure analysis

Similarly to the simulation of multiphase flows, the level-set method is used for the modelling of material crystallization. The transport of the level-set function is the starting point to analyze the phase change and crystals growth within a material. In figure 8, an example of a 2D representation of the crystal structure of a material through level-set method is presented: each microstructural grain is identified through its boundaries (the zero-level of the level-set function), while the color is based on the level-set function in order to distinguish the grain.

6

${}^{6}$


text:
# 7.14. Character classification¶

It is often useful to examine a character and test whether it is upper or lower case, or whether it is a character or a digit. C++ provides a library of functions that perform this kind of character classification. In order to use these functions, you have to include the header file ctype.h.

char letter = 'a';
if (isalpha(letter)) {
cout << "The character " << letter << " is a letter." << endl;
}

You might expect the return value from isalpha to be a bool, but for reasons I don’t even want to think about, it is actually an integer that is 0 if the argument is not a letter, and some non-zero value if it is.

This oddity is not as inconvenient as it seems, because it is legal to use this kind of integer in a conditional, as shown in the example. The value 0 is treated as false, and all non-zero values are treated as true.

Technically, this sort of thing should not be allowed—integers are not the same thing as boolean values. Nevertheless, the C++ habit of converting automatically between types can be useful.

Other character classification functions include isdigit, which identifies the digits 0 through 9, and isspace, which identifies all kinds of “white” space, including spaces, tabs, newlines, and a few others. There are also isupper and islower, which distinguish upper and lower case letters.

Finally, there are two functions that convert letters from one case to the other, called toupper and tolower. Both take a single character as a parameter and return a (possibly converted) character.

char letter = 'a';
letter = toupper (letter);
cout << letter << endl;

The output of this code is A.

As an exercise, use the character classification and conversion library to write functions named stringToUpper and stringToLower that take a single string as a parameter, and that modify the string by converting all the letters to upper or lower case. The return type should be void.

Try writing the stringToUpper and stringToLower functions in the commented sections of the active code below. Both functions take a single string as a parameter and have return type void. stringToUpper should convert the string to uppercase, and stringToLower should convert the string to lowercase. Some functions that you might find useful include isalpha, isupper, islower, toupper, and tolower. If you get stuck, you can reveal the extra problems at the end for help.


text:
location:  Publications → journals → CMB
Abstract view

# Equivariant Cohomology of $S^{1}$-Actions on $4$-Manifolds

Let $M$ be a symplectic $4$-dimensional manifold equipped with a Hamiltonian circle action with isolated fixed points. We describe a method for computing its integral equivariant cohomology in terms of fixed point data. We give some examples of these computations.


text:
12:37 PM
I am reading the proof of the Implicit Function Theorem in Geometry and Topology by Bredon; I am having trouble with a few parts.
Here's the statement of the theorem: Let $g : \Bbb{R}^n \times \Bbb{R}^m \to \Bbb{R}^m$ be $C^1$ and let $\xi \in \Bbb{R}^n$, $\eta \in \Bbb{R}^m$ be given with $g(\xi, \eta) = 0$.
Assume that the differential of the composition $y \mapsto (\xi, y) \mapsto g(\xi, y)$ is onto at $\eta$. [This is equivalent to the statement that the Jacobian determinant $J(g_i,y_j) \neq 0$ at $(\xi, \eta)$]. Then there are numbers $a > 0$ and $b > 0$ such that there exists a unique function $\phi : A \to B$ (with $A,B$ as in Lemma 1.4), with $\phi (\xi ) = \eta$, such that $g(x,\phi (x)) = 0$ for all $x \in A$.
First, how is the differential of $y \mapsto (\xi, y) \mapsto g(\xi, y)$ defined? I ask for the general definition because later in the proof that want to compute the differential of $y \mapsto f(\xi , y)$ at $\eta$, where $f(x,y) = y - L^{-1}(g(x,y))$ and $L : \Bbb{R}^m \to \Bbb{R}^m$ is the linear map represented by the Jacobian(?) And what does it mean to say that it is onto at $\eta$?

12:54 PM
I think I'll post this on the main...

1:08 PM
@MartinSleziak Is the Jacobian of a function the differential of that function?


text:
# Curl of the Polarization (Electrostatics)

1. Jul 24, 2013

### ManuJulian

I've been reading Griffith's "Introduction to Electrodynamics" and I've got to this part where it says:

"When you are asked to compute the electric displacement, first look for symmetry. If the problem exhibits spherical, cylindrical, or plane symmetry, then you can get $\vec{D}$directly from Gauss's equation (for the displacement) in integral form. (Evidently in such cases ∇x$\vec{P}$ is automatically zero, but since symmetry alone dictates the answer you're not really obliged to worry about the curl.)"

Now, why is it that the curl of the polarization is always zero in those cases where there is symmetry?

Is it just because in every case that exhibits symmetry the polarization is perpendicular to the boundary?

Last edited: Jul 24, 2013
2. Jul 24, 2013

### Andy Resnick

Are you considering electrostatics or electrodynamics?
In what sense do you mean 'symmetry': a symmetric homogenous body, or some sort of crystal symmetry?

3. Jul 24, 2013

### ManuJulian

Electrostatics. Symmetry meaning the first one, a symmetric homogeneous body.

4. Jul 24, 2013

### Andy Resnick

Since ∇x E = -∂B/∂t = 0 for statics and D = E + P, the result immediately follows.


text:
+0

help

0
109
1

If we divide a 2-digit positive integer by the sum of its digits, we get the quotient and remainder of 4 and 3, respectively.

If we divide the same 2-digit positive by the product of its digits, we get quotient and remainder of 3 and 5, respectively.

What is the 2-digit integer?

Jul 6, 2020

#1
+25597
+2

If we divide a 2-digit positive integer by the sum of its digits,
we get the quotient and remainder of 4 and 3, respectively.
If we divide the same 2-digit positive by the product of its digits,
we get quotient and remainder of 3 and 5, respectively.
What is the 2-digit integer?

The 2-digit integer $$ab$$ is $$10a + b$$

$$\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the sum of its digits} \\ \end{array} \\ \hline (1): & \mathbf{\dfrac{10a + b}{a+b}} &=& \mathbf{4 + \dfrac{3}{a+b}} \quad | \quad \times (a+b) \\\\ & 10a + b &=& 4(a+b) + 3 \\ & 10a + b &=& 4a+4b + 3 \\ & 10a-4a + b-4b &=& 3 \\ & 6a-3b &=& 3 \quad | \quad : 3 \\ & 2a-b &=& 1 \\ & \mathbf{b} &=& \mathbf{2a-1} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the product of its digits} \\ \end{array} \\ \hline (2): & \mathbf{\dfrac{10a + b}{ab}} &=& \mathbf{3 + \dfrac{5}{ab}} \quad | \quad \times (ab) \\\\ & 10a + b &=& 3ab+5 \quad | \quad \mathbf{b=2a-1} \\ & 10a + 2a-1 &=& 3a(2a-1) + 5 \\ & 12a-1 &=& 6a^2-3a + 5 \\ & 6a^2 -15a +6 &=& 0 \quad | \quad : 3 \\ & \mathbf{2a^2 -5a +2} &=&\mathbf{ 0 } \\\\ & a &=& \dfrac{5\pm \sqrt{5^2-4*2*2} }{2*2} \\ & a &=& \dfrac{5\pm \sqrt{9}}{4} \\ & a &=& \dfrac{5\pm 3}{4} \\\\ & a &=& \dfrac{5+ 3}{4} \\\\ & a &=& \dfrac{8}{4} \\\\ & \mathbf{a} &=& \mathbf{2} \\\\ \text{or} & a &=& \dfrac{5- 3}{4} \\\\ & a &=& \dfrac{2}{4} \qquad \text{no integer }! \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{b} &=& \mathbf{2a-1} \quad | \quad \mathbf{a=2} \\\\ b &=& 2*2-1 \\ \mathbf{b} &=& \mathbf{3} \\ \hline \end{array}$$

The 2-digit integer ab is $$\mathbf{23}$$

Jul 7, 2020


text:
# Pasch axiom and Pythagorean field condition?

I am looking for a reference for the claim that the Pasch axiom is equivalent to the Pythagorean field condition, and with respect to what base theory this should be true.

Since posting the question, somebody directed me to

Szmielew, Wanda. The role of the Pasch axiom in the foundations of Euclidean geometry. Proceedings of the Tarski Symposium (Proc. Sympos. Pure Math., Vol. XXV, Univ. California, Berkeley, Calif., 1971), Amer. Math. Soc., Providence, R.I., 1974, pp. 123–132.

as being the source, but I don't have a copy of this yet.

Note. The reason the rationals are not enough is because the hypothenuse of a right-angle triangle with sides $1$ and $a$ needs to be in the base field in order to be able to do Euclidean geometry over it.

• Is this clear? Plane geometry over $\mathbb Q$ also satisfies Pasch's axiom, but $\mathbb Q$ is not Pythagorean. Maybe instead of a Pythagorean field, you want an ordered field? – Sebastian Goette May 4 '16 at 11:35
• This depends on what other axioms you assume. You are not going to be able to do much work with circles over the rationals :-) – Mikhail Katz May 4 '16 at 11:48
• What is the base theory over which the equivalence is claimed? – Todd Trimble May 4 '16 at 11:58
• It doesn't really answer your question (in fact, it casts some doubt upon it), but you might like to read this account I wrote on Reddit(!) a few years ago. – Gro-Tsen May 4 '16 at 12:10
• I'm inclined to think that "Pythagorean" is indeed needed for working with circles, but the Pasch axiom looks entirely linear to me and so should not require a Pythagorean field. – Andreas Blass May 4 '16 at 21:52

WANDA SZMIELEW's article consider Tarski's axiom system for geometry, exposed into: What is elementary geometry ? (1957/58).

[Tarski] axiom system consists only of twelve short first order $\Pi \Sigma$ sentences and the continuity axiom, in case of the second order geometry, or the continuity axiom schema, in case of the first order (i.e., elementary) geometry.

The first version from 1930, published by Tarski only in 1948 as a note in the monograph A decision method for elementary algebra and geometry, still consists of eighteen elementary axioms, but six of them turned out step by step to be superfluous before Tarski's paper was published. After that two more axioms (identity and connectivity axioms for betweenness) were proved to depend on the remaining ones, and then the only ten left are: transitivity axiom for betweenness; Pasch's axiom:

$P: \ \ \exists a'[B(apa') \land B(ba'c)] \to \exists b'[B(bpb') \land B(ab'c)];$

identity, reflexivity, and transitivity axioms for equidistance; axiom of segment construction; five segments axiom; two dimension axioms (lower and upper); and Euclid's axiom. [...] After such a change let us denote the ten axioms by $A_1,\ldots, A10$.

The Pasch axiom is independent of all the remaining axioms together with the continuity axiom. [...] The independence model is constructed as follows: Let

$$\mathcal R=(R,+,\times,R^+)$$

be the ordered field of real numbers, $R^+$ being the positive cone of $\mathcal R$. Let the function $f$ be any one of the discontinuous one-to-one mappings of $R$ onto itself which satisfies the condition

$$f(x + y)=f(x)+f(y).$$

Then the Cartesian plane constructed over the so-called semi-ordered field

$$\mathcal R_f=(R,+,\times,f(R^+))$$

of reals is the independence model for the Pasch axiom.

Let us notice that the set $P = f(R^+)$, called the semipositive cone of $\mathcal R_f$ satisfies the conditions

(i) $x \in P \lor -x \in P$,

(ii) $x \in P \land -x \in P \to x = 0$,

(iii) $x \in P \land y \in P \to x + y \in P$,

but does not satisfy the condition

(iv) $x \in P \land y \in P \to x \times y \in P$.

In general, given a field $(F, +, \times)$ and a subset $P$ of $F$, we refer to the algebraic system

$$\mathcal F=(F,+,\times,P)$$

as a semi-ordered field iff the set $P$ (the semi-positive cone of $\mathcal F$) satisfies the conditions (i), (ii), (iii) (but not necessarily (iv)). In other words, a semi-ordered field is a field with the additive group ordered.

Since the Pasch axiom turns out to be independent, the next question arises: What is Euclidean geometry without the Pasch axiom? The answer is given by Szczerba and Szmielew in [On the Euclidean geometry without the Pasch axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.18 (1970), 659-666, MR43#2596.].

Let us denote by $\mathcal E^P$ and call weak Euclidean geometry the theory based upon the axioms $A_1,\ldots, A10$. Now, let us drop the Pasch axiom ($A10$), but instead let us add to the remaining nine axioms two linear statements:

$A10_1: \ \ B(abc) \to B(cba)$

$A10_2: \ \ \exists b \ [B(abc) \land D(abbc)]$,

in the proofs of which, in Tarski's system, the Pasch axiom is involved. $A10_1$ is the commutativity axiom for betweenness; $A10_2$ is the axiom of midpoint existence.

Let us denote by $\mathcal E$ and call weak Euclidean Pasch-free geometry the theory based upon the eleven axioms just listed.

It is well known that $\mathcal E^P$ is the theory of Cartesian planes over arbitrary ordered (thus formally real as well) Pythagorean fields; in other words, the models of $\mathcal E^P$ coincide up to isomorphism with the Cartesian planes over such fields. Now $\mathcal E$ turns out to be the theory of Cartesian planes over arbitrary semi-ordered formally real Pythagorean fields.

Moreover, in theory $\mathcal E$ the Pasch axiom occurs to be just equivalent to the condition (iv), stating that in the semi-ordered scalar field, constructed on an arbitrarily chosen line, the semi-positive cone is closed under multiplication.

In turn, let us extend both the theories $\mathcal E^P$ and $\mathcal E$ by the continuity axiom $Co$. In this way we get full Euclidean geometry $(\mathcal E^P)^{Co}$ and Euclidean Pasch-free geometry $\mathcal E^{Co}$ both second order theories. As is shown by Szczerba in [The Pasch-free geometry with the continuity axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.19 (1971), 613-616], the continuity axiom loses a lot of its power when the Pasch axiom is lacking. In fact, $(\mathcal E^P)^{Co}$ is known to be the theory of the Cartesian plane over the field $\mathcal R$ of real numbers, but $\mathcal E^{Co}$ turns out to be the theory of Cartesian planes not only over continuously semi-ordered fields of real numbers, but over continuously semi-ordered arbitrary formally real and Pythagorean fields. Any such field can be obtained up to isomorphism from the field $\mathcal R$ by means of a suitable modification $\circ$ of multiplication, and thus has the power of continuum.

Conversely, every field of characteristic $0$ and power of the continuum may be continuously semi-ordered (characteristic $0$ is a necessary condition for a field to be semi-orderable). In conclusion, with regards to the scalar field (formally real and Pythagorean) treated as a reduct of the semi-ordered scalar field (constructed on a fixed line), the continuity axiom without the Pasch axiom determines only that its power is that of the continuum.

Finally, let us extend the theories $\mathcal E^P$ and $\mathcal E$ , instead by the continuity axiom, only by the circle axiom

$C: \ \ B(abc) \to \exists c'[B(pbc') \land D(acac')]$

This axiom says that a half-line with the origin $b$ inside or on a circle cuts this circle in a point $c'$.

$(\mathcal E^P)^C$, called geometry of elementary constructions, is known to be the theory of Cartesian planes over arbitrary ordered Euclidean fields. What is the theory $\mathcal E^C$ ? One might expect that with $\mathcal E^C$ the class of the semi-ordered formally real Euclidean fields is connected. No; it is not the case. The theories $(\mathcal E^P)^C$ and $\mathcal E^C$ turn out just to coincide with each other

$$(\mathcal E^P)^C=\mathcal E^C$$

since, as is shown by Szmielew in [The Pasch axiom as a consequence of the circle axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.18 (1970), 751-758], the Pasch axiom is a theorem of $\mathcal E^C$. In other words, the Pasch axiom is dependent (thus superfluous) in the axiom system of geometry of elementary constructions. At first glance the situation seems to be paradoxical, since it is common to say that the circle axiom is a particular
consequence of the continuity axiom, but, at the same time,

$$P \in \mathcal E^C \ \text {and} \ P \notin \mathcal E^{Co}.$$

In fact, in $(\mathcal E^P)^{Co}$, $C$ can not be proved without the use of $Co$, however in the proof of $C$ besides $Co$ also some other axioms are involved ; now it is clear that one of them must coincide with $P$.

In conclusion, the circle axiom, known to restrict the class of fields underlying the models from Pythagorean to Euclidean, in fact restricts also the semi-ordered fields to ordered ones.


text:
0
Research Papers

# A Convection/Radiation Temperature Control System for High Power Density Electronic Device Testing

[+] Author and Article Information
Matthew Sweetland

Department of Mechanical Engineering, Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139-4307sweetlan@alum.mit.edu

John H. Lienhard V

Department of Mechanical Engineering, Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139-4307lienhard@mit.edu

Alexander H. Slocum

Department of Mechanical Engineering, Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139-4307slocum@mit.edu

The value of the device minimum test temperature and tolerance vary among manufacturers and across device types.

Solid thermal interface materials generally have liquid silicon or some other liquid imbedded in them, which leaves a residue when removed.

With component sizes shrinking, devices that may have had open areas on the back side are becoming covered with external interconnects.

Calculation based on fully turbulent flow over the entire width of the device.

The number of nozzles for each TTV is different, but the spacing, nozzle diameter, and offsets are identical, so for the same Reynolds number (or manifold pressure) the convection coefficient is the same.

A fast IR bulb typically has a 1–3 s settling time to reach 90% of the steady state.

The actual high temperature point varied over the range of devices due to external losses. The range varied from $70°C$ to $75°C$ with an air forcing temperature of $85°C$.

Supply voltage was known to an accuracy of $±50 mV$(13).

Teradyne, Inc. has a patent pending on this system.

J. Electron. Packag 130(3), 031012 (Aug 12, 2008) (10 pages) doi:10.1115/1.2966437 History: Received October 02, 2007; Revised February 10, 2008; Published August 12, 2008

## Abstract

Active control of die-level temperature is required during production testing of high power microprocessors in order to ensure accurate performance classification, but control is becoming more difficult as the device power densities increase. With power densities approaching $100 W/cm2$, the current passive control systems are no longer able to maintain the required temperature tolerance for production testing. This paper describes the design and testing of a temperature control system that combines high performance impingement cooling with higher power laser heating with application to packaged integrated circuit devices under dynamic testing conditions. Also presented are system design concepts and experimental results for typical microprocessor thermal test vehicles.

<>

## Figures

Figure 1

Typical cross section of a high power microprocessor device. Not all devices will contain all shown components.

Figure 2

TTV images with superimposed die and measured RTD positions

Figure 3

Base line test power profile. Peak power is 46.6 W. Test sequence consists of 0.1 s square waves.

Figure 4

Single and stacked nozzle modules. Number of nozzles in the array could be varied by changing the number of modules and by blocking off specific nozzle mounting holes.

Figure 5

Manifold pressure versus h¯c for the prototype system. A 5×5 nozzle array was used with the HPLD1 TTV and a 3×3 nozzle array was used with the HPD2 TTV.

Figure 6

Volumetric flow rate versus the manifold pressure under standard conditions (101.36 kPa and 21.4°C) for the 5×5 nozzle array used with the HPLD1 TTV and the 3×3 nozzle array used with the HPD2 TTV

Figure 7

Side view and cross sectional view of assembled laser/convection system. Manifold system and base support structures are not shown.

Figure 8

Schematic diagram of the control and data acquisition systems for laser/convection prototype system

Figure 9

Uncontrolled die temperature for a HDP2 TTV subject to a 23.3 W peak die power test sequence with a 41.37 kPa manifold pressure for the nozzle cooling system. The die temperature data are from all ten RTD sensors.

Figure 10

Uncontrolled die temperature for a HDP2 TTV subject to various peak die power scaled test sequences with a constant manifold supply pressure of 41.37 kPa. Data are from a single die centered RTD channel for peak die powers of 14 W, 23 W, 28 W, 46 W, and 56 W.

Figure 11

Uncontrolled die temperature for a HDP2 TTV subject to the same 23.3 W peak die power test sequence with nozzle cooling system manifold pressures of 20.69 kPa, 41.37 kPa, 62.05 kPa, 82.74 kPa, and 103.4 kPa. Data are from a single die centered RTD channel for each pressure.

Figure 12

Uncontrolled and controlled die temperatures for a HDP2 TTV subject to a 46.7 W peak die power test sequence with a 41.37 kPa nozzle cooling system manifold pressure. Data represent a single die centered RTD channel.

Figure 13

Controlled die temperatures for a PDP2 TTV subject to a 23.3 W peak die power test sequence at cooling nozzle manifold pressures of 20.68 kPa, 41.37 kPa, and 62.06 kPa. Data represent a single die centered RTD channel.

Figure 14

Controlled die temperature for HDP2 TTV subject to a 23.3 W peak die power test sequence with a 41.37 kPa manifold pressure for the nozzle cooling system. The die temperature data are from all ten RTD sensors.

Figure 15

Uncontrolled and controlled die temperatures for a HPLD1 TTV to a 46.7 W peak die power test sequence with 41.37 kPa cooling nozzle manifold pressure. Data represent a single die centered RTD channel.

Figure 16

Detailed view of controlled die temperature for a HPLD1 TTV to a 46.7 W peak die power test sequence with 41.37 kPa cooling nozzle manifold pressure. Data represent a single die centered RTD channel.

Figure 17

Control power sequence for HPLD1 TTV controlled temperature response shown in Fig. 1

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text:
# Equivalent Gradients in Kernelized SVM

Let $\varphi: \mathcal{X} \to \mathcal{H}$ a mapping with corresponding kernel $K:\mathcal{X}\times\mathcal{X}\to \mathbb{R}$ (that is, $K\left(x,x'\right) = \left<\varphi\left(x\right), \varphi\left(x'\right)\right>$), and consider the soft-SVM problem over the dataset $\left\lbrace x_i, y_i\right\rbrace_{i=1}^m$

$$\min_{w\in \mathcal{H}} \left(\frac{\lambda}{2} \left\Vert w \right\Vert^2 + \frac{1}{m} \sum_{i=1}^{m} \max\left\lbrace 0, 1-y_i \left<w, \varphi\left(x_i\right)\right> \right\rbrace \right)$$

The representer theorem states that the solution admits the following form $$w^* =\sum _{j=1}^{m} \alpha_j^{*}\varphi\left(x_i\right)$$

Let G be the Gramm matrix $G_{ij} = \left<\varphi\left(x_i\right), \varphi\left(x_j\right)\right>$. Then, we can equivalently write the problem as

$$\min_{\alpha \in \mathbb{R}^m } \left(\alpha^T G \alpha + \frac{1}{m} \sum_{i=1}^{m} \max\left\lbrace 0, 1- y_i\sum_{j=1}^m\alpha_j G_{ji} \right\rbrace \right)$$ And we are guaranteed to have the same solution.

The problem is convex (both in $\alpha$ and $w$), so we can solve it using stochastic gradient descent. However, optimizing on $w$ and optimizing on $\alpha$ seem to result in two very different rules for $\alpha$. Let $x_i,y_i$ be a single random sample.

• The gradient of the loss function of first equation (on a single, random, sample) w.r.t to $w$ is $- y_i \varphi\left(x_i\right) 1_\left\lbrace y_i\left<w,\varphi\left(x_i\right)\right><1\right\rbrace$. This means that each iteration we update only one coordinate of $\alpha$. (This is the version that appears in the books)
• The gradient of the loss function is the second equation w.r.t alpha is, instead, $-y_i G e_i 1_\left\lbrace y_i\left<w,\varphi\left(x_i\right)\right><1\right\rbrace$ (where $e_i$ is the standard basis element). This means, that every step we update all the coordinates of $\alpha$.

The two gradients are related, but by multiplication by $G^{-1}$. Thus seems to me that it can't be that they are equivalent.

Is the second method valid, or is it a mistake to use these updates?


text:
Maths / Direct and Inverse Variation / Direct Variation

QUESTION

A travels 432 Km on 48 liters of petrol. How far would it travel on 20 litres of petrol?

EXPLANATION
Explain TypeExplanation Content
Text

$\dpi{120} \fn_jvn \large \frac{432}{48}=\frac{x}{20}$

$\dpi{120} \fn_jvn \large x=\frac{432\times 20}{48}$

$\dpi{120} \fn_jvn \large x=180$

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