ajb Posted October 7, 2013 Posted October 7, 2013 It’s quite different. Using light –years , doesn’t change the fact if we measure it in let say cm. Right, you are free to measure any distence you want in any units of length you want. Nothing will fundamentally depend on the units you use. However, some units are more suitied to some measurements than others. Sensiable units for measuring the distence between two cities include miles or kilometres. The numerical answer would look horrible in cm or inches, but you are free to do so. The same for measuring it in light-years, astronomical units, parsecs and so on. You could quote the distence between cities in those units, but it would seem rather unnatural. The same is true of electric charge. You have coulombs, ampere-hours, often just units of the fundamental charge are used, so that is e, or you could use units of the Planck charge. None of these change the fundamental physics, they just give sensiable units of charge for different situations.
Kramer Posted October 7, 2013 Author Posted October 7, 2013 Swansont "e^2 / (4*pi*ε*R) unequal h*C / (2*pi*R)" because e and qp are not the same. It's "illegal" to substitute terms when they aren't equal.----- What am I saying? Why Planck is allowed to substitute “e” with “Qpl”?! You are able to convert erg to joules. Why are you having difficulty in converting values of charge? What about converting inches to centimeters? It's conceptually the same problem. 1 inch is 2.54 cm. If the distance is 1 inch, you wouldn't try and use 1 cm in a formula.------- Swansont. I am sincerely puzzled : Is it really that you don’t understand discrepancy between common physic with alleged Planck area, (that creates Planck charge) ? It is about correlation of E.M. energy and photons energy? It is about constant of fine structure. Or you have any hidden argument to amaze me with? ajb Right, you are free to measure any distence you want in any units of length you want. Nothing will fundamentally depend on the units you use. However, some units are more suitied to some measurements than others. Sensiable units for measuring the distence between two cities include miles or kilometres. The numerical answer would look horrible in cm or inches, but you are free to do so. The same for measuring it in light-years, astronomical units, parsecs and so on. You could quote the distence between cities in those units, but it would seem rather unnatural.The same is true of electric charge. You have coulombs, ampere-hours, often just units of the fundamental charge are used, so that is e, or you could use units of the Planck charge. None of these change the fundamental physics, they just give sensiable units of charge for different situations. -----The debate is not about convenience. You can use whatever unity you want, but see what is the result. When you go to buy something you may use one dollar bill or a hundred dollar bill. This is not important, even though may be inconvenient. The important is the price.
swansont Posted October 7, 2013 Posted October 7, 2013 Swansont "e^2 / (4*pi*ε*R) unequal h*C / (2*pi*R)" because e and qp are not the same[/size]. It's "illegal" to substitute terms when they aren't equal. ----- What am I saying? Why Planck is allowed to substitute “e” with “Qpl”?! Because what is going on in coming up with Planck units is not simply replacing e with qp. You are able to convert erg to joules. Why are you having difficulty in converting values of charge? What about converting inches to centimeters? It's conceptually the same problem. 1 inch is 2.54 cm. If the distance is 1 inch, you wouldn't try and use 1 cm in a formula. ------- Swansont. I am sincerely puzzled : Is it really that you don’t understand discrepancy between common physic with alleged Planck area, (that creates Planck charge) ? It is about correlation of E.M. energy and photons energy? It is about constant of fine structure. Or you have any hidden argument to amaze me with? Any "discrepancy" is due to your misunderstanding. You can make up any value of charge you want and use it for convenience, as long as it has the right units. If, for some reason, it was useful to have a unit of charge that was 179.2e, you could make that up and use it in calculations. You are trying to set two things equal that are not equal, and nobody is claiming they are equal. You are manufacturing a controversy that doesn't actually exist in real physics. 1
ajb Posted October 8, 2013 Posted October 8, 2013 The debate is not about convenience. You can use whatever unity you want, but see what is the result. When you go to buy something you may use one dollar bill or a hundred dollar bill. This is not important, even though may be inconvenient. The important is the price. Now think about paying in another currency, say GBP or EUR.
Sensei Posted October 8, 2013 Posted October 8, 2013 (edited) Right, you are free to measure any distence you want in any units of length you want. Nothing will fundamentally depend on the units you use. However, some units are more suitied to some measurements than others. Sensiable units for measuring the distence between two cities include miles or kilometres. The numerical answer would look horrible in cm or inches, but you are free to do so. The same for measuring it in light-years, astronomical units, parsecs and so on. You could quote the distence between cities in those units, but it would seem rather unnatural. The same is true of electric charge. You have coulombs, ampere-hours, often just units of the fundamental charge are used, so that is e, or you could use units of the Planck charge. None of these change the fundamental physics, they just give sensiable units of charge for different situations. That's true only if unit is not quantized at some level, and might be dividable infinitely. If you have quantized electric charge unit, you can have charges: 0, 1*e, 2*e, 3*e, 103834773*e, etc.etc, but never fraction of e. If you have fraction, you violated quantization.. Which obviously does Planck charge, which is 11.706 * e... Computer analogy: you cannot have half of bit (which is 0 or 1, mutually exclusive, values).. Edited October 8, 2013 by Sensei
swansont Posted October 8, 2013 Posted October 8, 2013 That's true only if unit is not quantized at some level, and might be dividable infinitely. If you have quantized electric charge unit, you can have charges: 0, 1*e, 2*e, 3*e, 103834773*e, etc.etc, but never fraction of e. If you have fraction, you violated quantization.. Which obviously does Planck charge, which is 11.706 * e... Computer analogy: you cannot have half of bit (which is 0 or 1, mutually exclusive, values).. From a calculation standpoint, nothing is changed if I decide on a unit that is a fraction of a fundamental charge. If it helped, one could throw a sqrt(2) in front of e, if it ended up making calculations easier, much like we have h and h-bar. Quantization is a separate concept.
Kramer Posted October 8, 2013 Author Posted October 8, 2013 Swansont Because what is going on in coming up with Planck units is not simply replacing e with qp. “What is going on in coming up with Planck units” was just what the lay-man didn’t understand. I asked, in the beginning of this thread, if this was a human flaw or a law of nature, in Planck area.With my shallow reasoning the only reason for this truck (by the Planck) was to put energy expressed with “ h*f “ in one foot with gravity energy, and with other kinds of energies:h*f = (h * C / (2*pi*R)) = G * Mpl ^ 2 / R and to derive Mpl. = ( h bar * C / G ) ^ 0.5 For Coulomb energy, Planck put the electric “charge” in the same foot with mass “m”. For me, this was wrong. Mass is a changeable amount, instead “e” is a constant of nature.So, putting instead of “e” this quantity of charge “Qpl” ( an arbitrary constant of nature) is “illegal”.Let see slowly:h*f = h * ( C / ( 2 * pi * R ) ) equation Planck’s energy, where f = C / ( 2 * pi * R ) e ^ 2 / ( 4 * pi * ε * R ) equation Coulomb’s energy, where “e” is a constant of nature.Sure, between them is a huge discrepancy. Even though Coulomb energy was steadily affirmed by experiment, Planck replaced “e” with it’s Planck charge “Qpl”. Why?Because it’s: E = h * f was too confirmed by experiment. So he can’t change “h”. He can’t change R the variable in all kind of energies. He can’t change “C” too.So the Planck charge was a desperate initiative. ---------------------------------Now let see a speculative variant, I noted in this thread above, about kind of frequencies ,without attire any comment by opponents.What is electro - magnetic frequency?(This is not at all a different thread , please don’t split it.)The Planck formula for frequency is f = C / (2 * pi * R ). In this formula I see “something” moving with velocity “C” in the same circle, with the same radius R. “The number” of circles realized in “unity of time” is frequency.This frequency happens in plane, as the circle is a plane. Dirak see differently: Something moving in a segment “R” with “C” velocity in a segment, that is with repetitive linear movement. The frequency is f = C / R.But what to do with “2*pi”. They go to “h” and voila: “h bar”.This frequency happens in linear movement. But we have the idea that particles (elementary) are not “plane structure”, not “linear structure”. They have a structure in volume, i.e. in sphere.In this case the frequency will be: f = C / ( ( 2 * pi / α ) * R )Without need to use Planck Charge.In this case would have ‘double bar h” i.e. h / (2*pi /α ) ! And will have “ Spherical structure of particles” Any "discrepancy" is due to your misunderstanding. You can make up any value of charge you want and use it for convenience, as long as it has the right units. If, for some reason, it was useful to have a unit of charge that was 179.2e, you could make that up and use it in calculations. ------- Give me please that “some reason” useful for “calculations’. You are trying to set two things equal that are not equal, and nobody is claiming they are equal. You are manufacturing a controversy that doesn't actually exist in real physics.-----I am not manufacturing controversy. There are a lot unsolved. Now think about paying in another currency, say GBP or EUR. ----- You may, if they are in not default status.
swansont Posted October 8, 2013 Posted October 8, 2013 “What is going on in coming up with Planck units” was just what the lay-man didn’t understand. I asked, in the beginning of this thread, if this was a human flaw or a law of nature, in Planck area. You have given only two choices. This is a false dichotomy. It's not a flaw, and it's not a law of nature. It's a unit of convenience. With my shallow reasoning the only reason for this truck (by the Planck) was to put energy expressed with “ h*f “ in one foot with gravity energy, and with other kinds of energies: h*f = (h * C / (2*pi*R)) = G * Mpl ^ 2 / R and to derive Mpl. = ( h bar * C / G ) ^ 0.5 For Coulomb energy, Planck put the electric “charge” in the same foot with mass “m”. For me, this was wrong. Mass is a changeable amount, instead “e” is a constant of nature. e is a constant of nature. But we are not talking about e. We are talking about Q, i.e. an amount of charge. Q is a variable: how much charge do you have. You can charge up a piece of metal, and it will not have a charge of e on it, it will have a charge of many times that. The amount of charge in a system is a variable amount. The planck mass is also a constant, not a variable. It has a fixed value. Same for all of the planck units. So, putting instead of “e” this quantity of charge “Qpl” ( an arbitrary constant of nature) is “illegal”. Let see slowly: h*f = h * ( C / ( 2 * pi * R ) ) equation Planck’s energy, where f = C / ( 2 * pi * R ) e ^ 2 / ( 4 * pi * ε * R ) equation Coulomb’s energy, where “e” is a constant of nature. No! The general equation does not use e, a constant. It uses Q, a variable. If you put e in there you are specifically talking about something that has one fundamental unit of charge. But the general equation has a variable for each of the charges. ANY value can be put into the equation. In this case, we are using the planck charge as the amount of charge we have. Sure, between them is a huge discrepancy. Even though Coulomb energy was steadily affirmed by experiment, Planck replaced “e” with it’s Planck charge “Qpl”. Why? Yes, they are different. So? Nobody said they were the same. You have yet to explain why this matters. Why? To make a self-consistent set of units. Because it’s: E = h * f was too confirmed by experiment. So he can’t change “h”. He can’t change R the variable in all kind of energies. He can’t change “C” too. So the Planck charge was a desperate initiative. No, you can't change c. But c is a speed, and speed is a variable. I can pick any value of speed I want. I can define vuEs as the speed of an unladen European swallow and solve problems in terms of that if I want. Any "discrepancy" is due to your misunderstanding. You can make up any value of charge you want and use it for convenience, as long as it has the right units. If, for some reason, it was useful to have a unit of charge that was 179.2e, you could make that up and use it in calculations. ------- Give me please that “some reason” useful for “calculations’. I didn't say I had a use for it. I said it was allowed, as an example. There's nothing that prevents it.
Kramer Posted October 9, 2013 Author Posted October 9, 2013 Swansont You have given only two choices. This is a false dichotomy. It's not a flaw, and it's not a law of nature. It's a unit of convenience.-----Sorry but I don’t see it as a “convenience”. It was an absolute necessity, because Planck energy did not fit with Coulomb energy in Planck area, where was supposed that all kind of energies have the same value. e is a constant of nature. But we are not talking about e. We are talking about Q, i.e. an amount of charge. Q is a variable: how much charge do you have. You can charge up a piece of metal, and it will not have a charge of e on it, it will have a charge of many times that. The amount of charge in a system is a variable amount.------- That right. But our debate is about why it was needed? Another participant in this debate has expressed his rebut about discreteness; I think his rebut is right. Qplanck presume some fraction of “e” which is strange.The planck mass is also a constant, not a variable. It has a fixed value. Same for all of the planck units.------ Planck mass doesn’t merit to be a constant. As it depend by an arbitrary Qpl. Here is the gist of debate: Is “Qpl” arbitrary or has a physical meaning? No! The general equation does not use e, a constant. It uses Q, a variable. If you put e in there you are specifically talking about something that has one fundamental unit of charge. But the general equation has a variable for each of the charges. ANY value can be put into the equation. In this case, we are using the planck charge as the amount of charge we have. --------I don’t think so. Using Qpl as a variable, there becomes meaningless other Planck constants. Yes, they are different So? Nobody said they were the same. You have yet to explain why this matters.. -------- It maters a lot. It gives theory a distorted direction. Why? To make a self-consistent set of units.--------- Self - consistent? Using an arbitrary charge, instead of natural ones? Strange! No, you can't change c. But c is a speed, and speed is a variable. I can pick any value of speed I want. I can define vuEs as the speed of an unladen European swallow and solve problems in terms of that if I want.--------- “C” in Planck area is a steady unchanged unity. (I can define vuEs as the speed of an unladen European swallow and solve problems in terms of that if I want.)This sentence is very subtle for me. Sorry that I can’t answer. I didn't say I had a use for it. I said it was allowed, as an example. There's nothing that prevents it. ------ Anyway.I think that all this debate has to do with CONSTANT OF FINE STRUCTURE—alpha.In Common Physic ------ alpha = (e^2 / (4*pi*epsilon) ) / (h-bar * C) = 1 / 137.036.In Planck area Physic --- alpha = ( Qpl^2 / (4 * pi * epsilon ) / ( h-bar * C ) = 1Why in Planck area ALPHA, this important constant of nature, is eliminated? And I think artificially!
Sensei Posted October 9, 2013 Posted October 9, 2013 (edited) e constant was measured in oil drop experiment in 1909 and published to public in 1913. http://en.wikipedia.org/wiki/Oil_drop_experiment Planck units were developed in 1899. Another participant in this debate has expressed his rebut about discreteness; I think his rebut is right. Qplanck presume some fraction of “e” which is strange. Max Planck simply didn't know about e. Edited October 9, 2013 by Sensei 3
swansont Posted October 9, 2013 Posted October 9, 2013 --------I don’t think so. Using Qpl as a variable, there becomes meaningless other Planck constants. I didn't say use Qpl as a varaible. I said "charge", in general (given by q or Q) is a variable. Anyway, Sensei has provided a very salient detail, that the planck units predate the discovery of a fundamental charge and all of quantum mechanics.
John Cuthber Posted October 10, 2013 Posted October 10, 2013 Anyway, Sensei has provided a very salient detail, that the planck units predate the discovery of a fundamental charge and all of quantum mechanics. I wonder if Kramer is going to postulate time travel as a way round that. otherwise it seems pretty watertight.
Kramer Posted October 10, 2013 Author Posted October 10, 2013 Sensey e constant was measured in oil drop experiment in 1909 and published to public in 1913. Planck units were developed in 1899. Max Planck simply didn't know about e. -------- Thanks for your interesting historic data. That means once more that the Planck constants were and are out-dated. That means that using Planck quanta as a “ ghost protagonist “ of everything in physics, ignoring “electric charge” and it’s nemesis “ gravity constant”, ( I think--- those are real protagonists of reality) , this gave an unilateral direction in physic and a wrong ones. In my lay-man’s shallow reasoning, Planck quanta is a measuring means of action by really other sounding players in physic.So I think that Planck constants are a controversy with Einstein constants. Swansont. I didn't say use Qpl as a varaible. I said "charge", in general (given by q or Q) is a variable. Anyway, Sensei has provided a very salient detail, that the planck units predate the discovery of a fundamental charge and all of quantum mechanics.------- This doesn’t change a jota in the gist of this thread:Planck constants are wrong, in flagrant controversy with real data of physic.Instead, the Einstein constants are free by any controversy, they gave a reasonable interpretation where was the mistake in Planck constants. And, if I had prerogative of physicist, I would proposed “them” as universal constants of physics. Thinking now about historic data and for “salient detail” given by Sensey, I don’t blame Planck (who am i ), this giant of physic in the same stature as Einstein, but those “quantum physicist” that have used his reasonable mistakes .
swansont Posted October 10, 2013 Posted October 10, 2013 ------- This doesn’t change a jota in the gist of this thread: Planck constants are wrong, in flagrant controversy with real data of physic. Instead, the Einstein constants are free by any controversy, they gave a reasonable interpretation where was the mistake in Planck constants. And, if I had prerogative of physicist, I would proposed “them” as universal constants of physics. Thinking now about historic data and for “salient detail” given by Sensey, I don’t blame Planck (who am i ), this giant of physic in the same stature as Einstein, but those “quantum physicist” that have used his reasonable mistakes . Planck units can't be "wrong"; they don't refer to a physical object. Nobody is claiming that they represent a fundamental constant. There is no mistake involved, and no controversy. It's merely a different set of units. Miles instead of kilometers.
Kramer Posted October 11, 2013 Author Posted October 11, 2013 Swansont Planck units can't be "wrong"; they don't refer to a physical object. Nobody is claiming that they represent a fundamental constant. There is no mistake involved, and no controversy. It's merely a different set of units. Miles instead of kilometers. -------- Thanks for your patience, Swansont. So let it be: You in yours “in box teaching”, I in mine lay man’s stubborn doubts.
imatfaal Posted October 11, 2013 Posted October 11, 2013 Swansont Planck units can't be "wrong"; they don't refer to a physical object. Nobody is claiming that they represent a fundamental constant. There is no mistake involved, and no controversy. It's merely a different set of units. Miles instead of kilometers. -------- Thanks for your patience, Swansont. So let it be: You in yours “in box teaching”, I in mine lay man’s stubborn doubts. That's a pretty ridiculous way to end a thread after the amount of information you have been given by a number of highly qualified physicists. Arguments from personal incredulity and lack of clear knowledge are rarely edifying but your position and refusal to take on board completely valid criticisms has taken the biscuit. And finishing with an accusation of dogmatism is just insulting. May I remind you that the denial of a valid argument due to misunderstanding of its basic tenets does not fall within the parameters of healthy scepticism. There is an excellent pinned thread on the planck units with their derivations by Martin in the main forum - I would suggest you read it carefully and then you may wish to come back to this thread to discuss any further points you do not understand
swansont Posted October 11, 2013 Posted October 11, 2013 Swansont Planck units can't be "wrong"; they don't refer to a physical object. Nobody is claiming that they represent a fundamental constant. There is no mistake involved, and no controversy. It's merely a different set of units. Miles instead of kilometers. -------- Thanks for your patience, Swansont. So let it be: You in yours “in box teaching”, I in mine lay man’s stubborn doubts. If "insiders" tell you it's no big deal, and you have no cause to doubt that, you should accept it as truth.
ajb Posted October 12, 2013 Posted October 12, 2013 If "insiders" tell you it's no big deal, and you have no cause to doubt that, you should accept it as truth. Absolutely there is no controversy here, just now and again they get misinterpreted.
Kramer Posted October 12, 2013 Author Posted October 12, 2013 Imatfaal --- That's a pretty ridiculous way to end a thread after the amount of information you have been given by a number of highly qualified physicist. Arguments from personal incredulity and lack of clear knowledge are rarely edifying but your position and refusal to take on board completely valid criticisms has taken the biscuit. And finishing with an accusation of dogmatism is just insulting. May I remind you that the denial of a valid argument due to misunderstanding of its basic tenets does not fall within the parameters of healthy scepticism. ------ Now the debate has taken a serious and in the same time ridiculous sense. Your high stylistic criticism, toward a lay - man, that you charge for self - boasting is very funny, and baseless. On the other hand I am not able to dispute with you, cause of many “my lacks”.I admit that I have personal incredulity, for many weird statements of modern physics.You my call it baseless, the same as Swanson says that I fabricate imaginary “controversy”. At least in this thread I began to doubt in the power of reasoning on both sides. The thread is very - very simple:In Planck case we have one equation that is in flagrant diversity with an inequation in real physic: Qpl^2 / ( 4*pi*ε R) = h–bar* C / R and (1) e^2 / (4*pi*ε* R) < h-bar * C / R (2)Simple question: Which is true? First or second?Is told by Physicists that both are true: Multiply “ e^2” with 137.036 and you will have equation (1), or Divide “Qpl^2” with 137.036 and you will have inequation (1)Let leave aside, even though is very reasonable, the rebut of Sensey that “Qpl” can‘t be a multiple of “e” because 137.036 is not a full number.What I considered as a controversy was question:In Planck area Which is true? Equation (1) or inequation (2)? Because is without doubt that Planck considered all kind of energies equivalents, (Coulomb energy, Newton energy, Einstein energy, Planck energy , Boltzmann energy etc…) toward constants of nature : G, C, h, Q , Boltzmann constant…. And from this consideration are derived Planck constant : Mpl, Lpl, Tpl , Planck Temp.. etc…Ye. I made speculations that instead Planck constant “h” as the main protagonist, I used “e” as main protagonist and derived a little different constants in Planck area , I called them Einstein constants. This is …a little out of what “you call it dogma”.The specialists disregarded the possibility of different constants in Planck area, and avoided debate about them, seems to me cause out of box.Here is nothing insulting. There is an excellent pinned thread on the planck units with their derivations by Martin in the main forum - I would suggest you read it carefully and then you may wish to come back to this thread to discuss any further points you do not understand----- I will try for curiosity to find recommended source. So you my close this thread until I be able to debate with specialists. Swanson If "insiders" tell you it's no big deal, and you have no cause to doubt that, you should accept it as truth.---- I think is big deal. It has to do with two view - point about reality. ajb Absolutely there is no controversy here, just now and again they get misinterpreted-----Absolutely there must be a controversy about which we have not a clear concept.
Endy0816 Posted October 12, 2013 Posted October 12, 2013 Qpl^2 / ( 4*pi*ε R) != h–bar* C / R (3) This is true. If two things were not equal in the first place they won't be equal in the end. You doing the equivalent of saying: 1 = 2. I don't know any other way to put this. Whatever you think, the two terms are referring to different values. They are not the same. You cannot set them equal.
John Cuthber Posted October 13, 2013 Posted October 13, 2013 ajb Absolutely there is no controversy here, just now and again they get misinterpreted -----Absolutely there must be a controversy about which we have not a clear concept. Actually, We have quite a clear idea about this. It's only you who is missing it.
Kramer Posted October 13, 2013 Author Posted October 13, 2013 Andy0816 Qpl^2 / ( 4*pi*ε R) != h–bar* C / R (3) This is true. If two things were not equal in the first place they won't be equal in the end. You doing the equivalent of saying: 1 = 2. I don't know any other way to put this. Whatever you think, the two terms are referring to different values. They are not the same. You cannot set them equal.------ I am not sure I understand your rebut. Why you think, I do equivalent 1=2 ? What are two terms that, you think, are referring to different values?I am posting about Planck constants, and his idea that in Planck area all kind of energies are equivalent. This equivalence, after Planck, is based in the value of Planck constant “h” , quanta of energy, and other fundamental constants : C , G , Boltszmann constant…and an “invented” fundamental Planck constant of electric charge “ Qpl ”, which is not at all a constant. AS we know the fundamental electric charge is ”e”.And this put all Planck constants in a controversy: Are they legitimated? Because changing one fundamental constant you must change “all the panorama” accordingly.Kutbeer Actually, We have quite a clear idea about this. It's only you who is missing it.------ With “we”, you are alluding: only your opinion, your and ajb’s, the forum staff’s, the world’s scientific forum’s?I would liked your scientific arguments, about the roles of “h” and “e” in physic, about their intertwined relation in Planck area, and, indeed, about my missing’s.
John Cuthber Posted October 13, 2013 Posted October 13, 2013 "I am not sure I understand your rebut." There may be a reason for that. The reason might be related to my earlier post.
swansont Posted October 14, 2013 Posted October 14, 2013 I am not sure I understand your rebut. Why you think, I do equivalent 1=2 ? What are two terms that, you think, are referring to different values? I am posting about Planck constants, and his idea that in Planck area all kind of energies are equivalent. This equivalence, after Planck, is based in the value of Planck constant “h” , quanta of energy, and other fundamental constants : C , G , Boltszmann constant…and an “invented” fundamental Planck constant of electric charge “ Qpl ”, which is not at all a constant. AS we know the fundamental electric charge is ”e”. And this put all Planck constants in a controversy: Are they legitimated? Because changing one fundamental constant you must change “all the panorama” accordingly. It's equivalent because you keep insisting that qp should be equal to e, and they are not equal, like 1 ≠ 2. qp is a constant, not a variable. Since it's defined in terms of fundamental constants, it must be a constant. What do you mean that "all kind of energies are equivalent"? That does not mean that all values of energy are equal. 1
ajb Posted October 14, 2013 Posted October 14, 2013 I would liked your scientific arguments, about the roles of “h” and “e” in physic, about their intertwined relation in Planck area, and, indeed, about my missing’s So you cobble together the fundamental constants you have into an expression that has the units of length squared. It is the interpretation of what this means and where it enters physics that you have questions about? Well, at one level it is nothing more than a possibily useful unit of area. The interepretation is that it gives us an area scale at which the effects of quantum gravity cannot be ignored. One place that the Planck area does explicitly enter physics is in the Bekenstein–Hawking formula, which related the entropy of a black hole to its surface area.
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