Martin Posted September 13, 2004 Posted September 13, 2004 Planck units are the standard quantities you get by setting the numerical values of the main physical constants equal to one. |G| = |c| = |hbar| = |k| = |e| = 1 setting the electron charge equal to unity is optional. When Planck described these units in 1899 he didnt bother with that. But it is one way of including electrical units (current, voltage) in the system. Relativists often use a variant where they set the numerical value of 8pi G =1 this gives you different Planck units but it makes many of the basic formulas even cleaner. The Einstein equation, the main equation of gravity, just becomes [math]G_{ab} = T_{ab}[/math] the LHS represents curvature and the RHS represents the energy density of matter, the "stress-energy tensor" Actually the fundamental constants c and 8pi G are still in the equation, which is really this: [math]G_{ab} = (\frac{8\pi G}{c^4}) T_{ab}[/math] but in these fashionable modified Planck units the constant in parens here has numerical value unity, so people often do not write it. Maybe the feeling is that this equation, relating the curvature in a region to the concentration of energy there, is so clean that it is a hint to us that we really should think of 8pi G as the fundamental constant, not G. At any rate, if you do that you get a bunch of natural units which differ from the conventional Planck ones by factors of about 5 or 25 that is by factors of the square root of 8pi, or of 8pi itself. Let's get acquainted with these units. Maybe some day they will be more widely used. I first encountered them on sci.physics.research a couple of years back when I saw them advocated by no less than John Baez, one of the moderators and a great explainer. he has some stuff about Gen Rel and about the meaning of Einstein's Equation at his website. People go to some lengths to make the equations clean because it really does make them easier to understand, when they arent so cluttered.
Martin Posted September 13, 2004 Author Posted September 13, 2004 if this is your first look at this system of units it is going to look cluttered no matter how I present it one thing is that many of the units are extreme, the speed unit is c the speed of light the temperature unit is essentially the temperature of the big bang (very hot) the force unit is about the strongest force anyone can imagine (the attraction between two black holes as they are merging---sort of irresistably huge) so applying these units you see a lot of powers of ten. just something to get used to, like the fact that we normally go about our business at speeds which are less than a millionth of the speed of light. Let's begin with the mass, length, duration units just as if we were going to the NIST fundamental constants website to look up the conventional Planck units. If you want to compare here is the NIST site http://physics.nist.gov/cuu/Constants/ http://physics.nist.gov/cgi-bin/cuu/Category?view=html&Universal.x=87&Universal.y=12 http://physics.nist.gov/cgi-bin/cuu/Value?plkl|search_for=universal_in! In calculating, I wont show error bounds or round off. I'll leave the numbers raw and round them later. M = mass unit = 4.34139 micrograms L = length unit = 8.10263 x 10-35 meters D = duration unit = 2.70275 x 10-43 seconds Temp = temperature unit = 0.282609 x 1032 kelvin If I remember right, it was in the same SPR posts where John Baez was advocating these different Planck units where he also speculated that there was a sense in which area was more basic than length. Sounds crazy enough to keep track of. So I am going to square the length unit here to get an area unit and call it one "John" of area, notated J. area unit = J = 65.65 x 10-70 square meters having that extra symbol will facilitate writing some of the others the way I want to---showing the dependence on hbar and c as well as the area unit. Now I will round off and get some equivalences [math]\text{energy unit} = \frac{\hbar c}{\sqrt J} = 0.390 \text{ Gigajoules}[/math] [math]\text{force unit} = \frac{\hbar c}{J} = 4.82\times 10^{42} \text{ newtons*}[/math] [math]\text{energy density unit} = \frac{\hbar c}{J^2}[/math] [math]\text{pressure unit} = \frac{\hbar c}{J^2} = 7.33\times 10^{110} \text{ pascals}[/math] [math]\text{mass unit} = \frac{\hbar }{c\sqrt J} = 4.34 \text{ micrograms*}[/math] [math]\text{mass density unit} = \frac{\hbar}{cJ^2}= 8.16 \times 10^{90}\text{ g/cc} [/math] *since a conventional pound of force is about 4.45 newtons, this large force unit amounts to roughly 1042 pounds force. and since a conventional pound mass is about 454 grams, the small mass unit here amounts to roughly 10-8 pounds mass. It should not be surprising, then, that the unit of acceleration here is around 1050 normal earth gee. If you dont think John Baez is a great enough mathematician to name a unit of area after, then you can imagine it is John von Neumann.
fuhrerkeebs Posted September 13, 2004 Posted September 13, 2004 I like how the plank length over the plank time is the speed of light. It would explain why light is the universal constant in the universe.
Martin Posted September 13, 2004 Author Posted September 13, 2004 I like how the plank length over the plank time is the speed of light... I too,
Martin Posted September 13, 2004 Author Posted September 13, 2004 here's a rough equivalence to get a notion of the force unit [math]10^{-42}\text{ force unit} = 10^{-42}\frac{\hbar c}{J} = 4.82 \text{ newtons}[/math] = roughly one "poundforce" [math]10^{-50}\text{ acceleration unit} = 10^{-50}\frac{c^2}{\sqrt J} = 11.09 m/s^2[/math] = somewhat over one "gee" a normal earth gee is about 9.8 meters per second per second so this is roughly ten percent over.
Martin Posted September 14, 2004 Author Posted September 14, 2004 some equivalences for comparison [math]10^{33}\times \text{ length unit} = 10^{33}\times \sqrt{J} = 8.10 \text{ centimeters} [/math] = about a handwidth or 3 "inches" [math]10^{-106}\times\text{pressure unit} = 10^{-106}\times\frac{\hbar c}{J^2} = 7.33\times 10^{4} \text{ pascals} [/math] = 0.7 "atmospheres" [math]10^{8}\text{mass unit} = 10^{8}\frac{\hbar }{c\sqrt J} = 434 \text{ grams}[/math] = about a "pound" mass [math]10^{-91}\text{mass density unit} = 10^{-91}\frac{\hbar}{cJ^2}= 0.816 \text{ g/cc} [/math] = 0.8 "normal density" ---more about "baez units" added in edit---- It was a suggestion by Baez (at sci.physics.research) that instead of G = hbar = c = k = 1 one should be setting 8piG = hbar = c = k = 1 A corollary of that would be that in defining Planck units one should be doing everything just the same except with 8piG instead of G. One reason you might think that 8piG (or "Gbar") is more fundamental than G is that the 1915 Einstein equation, our premier equation about gravity, is often written: Gab = (8piG)Tab and you also see it where 8piG has been set equal to unity by adjusting units. When you see the main gravity equation written in this really clean form: Gab = Tab then you know that someone has adjusted the units so as to set not only c but also 8pi G = 1.
Severian Posted September 14, 2004 Posted September 14, 2004 As you point out the nice thing about natural units is that you simplify equations. But presumably one cannot simplify all equations at once, so it is best to simplify the ones which are most fundamental. So, for example, you still have c=1, which is good. But G is not really very fundamental. It is the gravitational coupling constant. It may seem fundamental when using GR, but eventually we will have a quantum theory of gravity where it becomes a renormalised coupling constant. We see this already in your beloved Quantum Loop Gravity. Renormalisation introduces an extra energy (or distance) scale into the equations (something called dimensional transmutation) and then coupling constants change with energy. The electromagnetic coupling [math]\alpha[/math] changes from 1/137 at eV scales to 1/128 at a GeV. So I expect that G changes too. So even if you pick one value of 'G' to set to one (or 8 pi or whatever) you will still generate numbers in your equations. I don't see the point in this. The gravitational constant is [math]G = 6.707 \times 10^{-39} \hbar c ({\rm GeV}/c^2)^{-2} = (1.221 \times 10^{19} {\rm GeV})^{-2}[/math] where I have set [math]\hbar c=1[/math]. This defines the Planck length and Planck energy (so G is the planck energy to the power of -2). But why should gravity be the force that we use for units? Why not use the Fermi constant: [math]G_F = 1.16639 \times 10^{-5} (\hbar c/{\rm GeV})^2 = (292.8 {\rm GeV})^{-2}[/math], which describes the strength of electroweak interactions (at a particlular scale)? In fact, our current energy unit (the eV) is nice because it is fundamental to two of the forces, electromagnetism (obviously) but also QCD, where conincidentally 1GeV is the hadronisation scale. Once we have a theory of everything, we may then have one force with one characteristic distance scale, but until them, isn't any choice we make rather arbitrary?
Martin Posted September 14, 2004 Author Posted September 14, 2004 do go on. thanks bloodhound, I will after i think a bit about what Severian just said----he has a particle physics rather than gen rel viewpoint so it is worth thinking about because a really different perspective from the people who use Planck units. people who use planck units----or something analogous to that whether G=hbar=c=k=1 or 8piG =hbar=c=k=1 or something along those lines are mostly Relativists (not HEP people) and for them "electron" and "volt" and "electronvolt" is not so meaningful. BTW I am not sure right now how to continue. If anyone sees a question to ask or a direction to go in, suggest it!
Martin Posted September 14, 2004 Author Posted September 14, 2004 As you point out the nice thing about natural units is that you simplify equations. But presumably one cannot simplify all equations at once' date=' so it is best to simplify the ones which are most fundamental. So, for example, you still have c=1, which is good. But G is not really very fundamental. It is the gravitational coupling constant. It may seem fundamental when using GR, but eventually we will have a quantum theory of gravity where it becomes a renormalised coupling constant. [b']We see this already in your beloved Quantum Loop Gravity. Renormalisation introduces an extra energy (or distance) scale into the equations (something called dimensional transmutation)[/b] Severian, what extra energy or distance scale are you talking about? I am very eager to know! Please give me an arxiv link about this.
Martin Posted September 14, 2004 Author Posted September 14, 2004 As you point out the nice thing about natural units is that you simplify equations. But presumably one cannot simplify all equations at once' date=' so it is best to simplify the ones which are most fundamental. [b'] So, for example, you still have c=1, which is good.[/b].... Severian, are you sure that c does not change with energy? In gammaray astronomy one currently is looking for dispersion effects at very high energies over large distance and a more intensive search is planned with the Gammaray large array space telescope (GLAST) starting around 2007. Even if the speed of light changes slightly at very high energies one still has the constant c which is the limit at low energy. Likewise the fine structure constant, we still have the low energy limit of the coupling which is the usual 1/137.036...and works in all the myriad mundane applications-----even tho the coupling strengthens to around 1/128 at very short range. I wouldnt want to base UNITS on the fine structure constant because it is nice to consider it as dimensionless and the same number in any system. But the analogy may still be sound. We may have a useful (dimensionful) quantity in G, or 8piG if you please, which is the low energy limit of the coupling and works in almost all the situations we need it. (like the speed of light works, even tho there may be dispersion at very high energies which we currently dont see) any thoughts? any specific suggestion of another universal constant besides G to use instead?
Martin Posted September 14, 2004 Author Posted September 14, 2004 We see this already in your beloved Quantum Loop Gravity. Renormalisation introduces an extra energy (or distance) scale into the equations (something called dimensional transmutation) and then coupling constants change with energy.... are you possibly talking about something that happens in String? I dont think "dimensional transmutation" is mentioned in any of the standard treatements of LQG
Severian Posted September 14, 2004 Posted September 14, 2004 Severian' date=' what extra energy or distance scale are you talking about?I am very eager to know! Please give me an arxiv link about this.[/quote'] I am not sure I can give you an zrxiv link, since t is really text book stuff. This is quite a nice description of renormalization in general terms. To be honest, I don't know a lot about LQG so it is conceivable that you don't need renormalization. Are your integrals in LQG (ie. the loop integrals themselves) manifestly finite? If they are then maybe you don't need renormalization - if they are not, then you definitely do need it. And if you have renormalization, you have dimensional transmutation bringing in a new scale. I would be surprised to learn that G does not change with energy. (It is possible though, since gravity is so different from the other forces.) In contrast, I would be very surprised to find that 'c' changes with energy.
Martin Posted September 14, 2004 Author Posted September 14, 2004 ...To be honest' date=' I don't know a lot about LQG so it is conceivable that you don't need renormalization...[/quote'] that is right. LQG does not use renormalization so the "dimensional transmutation" new scale does not come into it. (I knew it sounded out of place in LQG context! ) LQG is a nonperturbative theory so it does not have a perturbation series. The loop integrals (holonomies of the connection variable) are manifestly finite. I would be surprised to learn that G does not change with energy. In LQG there is a dimensionless number called the Immirzi parameter which appears in the spectrum of the area operator (the observable associated with the area of a given surface) and this number may relate the macroscopic G to a "bare" G. MAY (I dont think it is known) but this does not introduce a new length scale or new energy scale. The dimensionless number takes care of that. There are currently attempts to determine the value of this number using BH entropy and radiation as guides. Perhaps you could interpret this as G changing but that is not how it is usually thought of. In contrast' date=' I would be very surprised to find that 'c' changes with energy [/quote'] I agree! It would be extremely surprising! Some versions of string theory suggest this happens and also some versions of LQG seem actually to predict it. I can get some arxiv links if you wish. It is even surprising that they are even going to look for this effect! It is something expected to be revealed only in UHECR and in high energy gammaray. the effect is suppressed by planck scale so it is so small that it can only be revealed (if it is there) in a brief gammaray burst that has traveled for, say, a billion years. In such a long time, it is conjectured, the much higher energy photons will have had time to get out ahead of the rest just enough to measure. So the Satellite Telescope called GLAST is supposed to look for this microscopic dispersion. It will really be a shocker if they find it! Some search has already been made with other instruments that observe Gammaray Burst. The dispersion was not found at the accuracy available with those instruments. If you are interested I should get the arxiv links.
Severian Posted September 15, 2004 Posted September 15, 2004 that is right. LQG does not use renormalization I find that surprising. I had thought one of the major problems of quantum gravity was that it had too many non-renormalizable divergences. There are very few theories which need no renormalization (N=4 supersymmetry is one). Maybe I should go read up on LQG... (strangely enough it is a subject which I hear very little about). In LQG there is a dimensionless number called the Immirzi parameter which appears in the spectrum of the area operator (the observable associated with the area of a given surface) and this number may relate the macroscopic G to a "bare" G. That sounds very like renormalization under a different name....
Martin Posted September 15, 2004 Author Posted September 15, 2004 Maybe I should go read up on LQG... (strangely enough it is a subject which I hear very little about). ... the best short read-up about LQG is currently a paper by Lee Smolin that came out last month called An Invitation to Loop Quantum Gravity http://arxiv.org/hep-th/0408048 It is written for physicists in other specialties' date=' as a survey with a FAQ, a list of main results, a list of open problems. That sounds very like renormalization under a different name.... It would be amusing if you found that loop quantum gravitists were actually doing renormalization all this time, but just not talking about it in those terms.
Martin Posted September 16, 2004 Author Posted September 16, 2004 I will go back to post #6 in order to pick up where we left off with the "different Planck units" topic. I am still tinkering with this system of units to see what can be gotten from it. A system of everyday-size units which are simply the Planck ones scaled by appropriate powers of ten can serve to give us a handle on this version of Planck units. I started constructing such a human-scale system in another thread. Am now referring to 8pi G as "Gbar" by analogy with hbar. Gbar units gradually taking shape. I added electric units (charge current and voltage) and gave the energy unit a name. The names are still placeholders, in case better name ideas show up. the system is essentially a version of Planck units using Gbar = 8pi G instead of the newtonian G constant as Planck originally did.e is the elementary charge. Our charge unit is exactly 1018 e, the charge on a billion billion electrons. the unit of energy (called a jot) is about 1/100 of a calorie. The units are defined by assigning these values to the constants Gbar = 10-7 hand3count-2pound-1 c = 109 hand count-1 hbar = 10-32 jot count k = 10-22 jot degree-1 e = 10-18 charge unit [think of the word "jot" as a placeholder, it is this word for the energy unit that I'm asking you to think of a replacement for] In conventional metric, c = 2.99792458 x 108 meter second-1 and the other constants are even messier, so there's some attraction to Gbar units getting simple powers of ten for the constants. with the above powers of ten stipulated then the time unit (count) comes out 222 to the minute. the temperature degree turns out to be about half a Fahrenheit the mass unit pound comes out to 434 grams, roughly one pound. the length unit hand is 8.09 centimeters (around 3 and 1/4 inches) the force unit dram is 0.4816 newton, a couple of ounces of force. the energy unit jot works out to around 0.04 joule or 1/100 of a calorie. the unit current is about 2/3 of a conventional ampere and the unit voltage is about 1/4 of a conventional volt. the power unit is approx. 1/6 watt. Now it might be good to try some applications of these units to see how quickly one can get used to them or how handy they are in practical situations.
Martin Posted September 18, 2004 Author Posted September 18, 2004 one application is calculating how brightly a hot object glows this uses the Stefan-Boltzmann constant sigma [math]\sigma = \frac{\pi^2}{60}\frac{k^4}{\hbar^3 c^2}[/math] let's calculate what that constant is in these units have to go, back later
Martin Posted September 19, 2004 Author Posted September 19, 2004 [math]\sigma = \frac{\pi^2}{60}\frac{k^4}{\hbar^3 c^2}[/math] this is trivial to calculate, just plug in 10-22 for k to get 10-88 and plug in 10-32 and 109 for hbar and c to get 10-76 and divide, to get 10-10 So then [math]\sigma = \frac{\pi^2}{60}\times 10^{-10} jot count^{-1} hand^{-2} degree^{-4}[/math] it tells the radiant power (jot per count) per unit area (hand2) per degree4
jordan Posted September 19, 2004 Posted September 19, 2004 What might be some useful classes to take when I go to college next year to learn about this stuff Martin? How do you know so much about it? As of now, I can follow the math, but the words are still meaningless as far as a real-life application. I'm interested in learning more about it though.
Martin Posted September 19, 2004 Author Posted September 19, 2004 I still have to get back to Severian's post on this thread where he suggests using the Fermi (electroweak coupling) constant instead of Gbar. what I'm going on here is that I see a lot of research papers where Planck units are used (so convenient to have simple values, like one, of important conversion factors) and in recent papers I see a shift towards "Gbar" units. Gbar is my name for the 8piG constant that appears in the main GR equation, I have heard other people suggest it but can't recall a source. What I mean by "Gbar units" is the usual Planck units but with Gbar replacing G. It looks to me as if Relativists and especially Quantum Gravitists (a small but growing research sector) are going to start using Gbar units, are in fact already using them piecemeal via various notations and conventions. I am just looking for this to solidify and for notation to get more uniform. So I want to appreciate the Gbar system of units and assimilate and size them up. One way to do that is to take some power-of-ten scale multiples of the units and apply them to everyday-size familiar stuff. Like sunlight. that is about as everyday as it gets. I am curious, now, about the size of sunlight in Gbar units. So let's calculate a little about that in the next post. BTW I disagree with Severian's idea about using the Fermi constant instead of the gravity coupling as a basis for natural units. that is because the gravitational field is space----gravity is not just another force in QFT but has a special status as the geometry in which or upon which other fields are defined. but this is very much a matter of opinion. Particle theorists like Severian will differ from relativists and from quantum gravitists (especially) about this. the QFT and GR perspectives are so different! So I will have to postpone replying, Severian, until I feel more certain how to.
Martin Posted September 19, 2004 Author Posted September 19, 2004 What might be some useful classes to take when I go to college next year to learn about this stuff Martin? How do you know so much about it? As of now, I can follow the math, but the words are still meaningless as far as a real-life application. I'm interested in learning more about it though. Jordan! you are incredible! As an entering college freshman you should not be worrying about quantum gravity! for god's sake learn all the basic first year college physics that you can. It is so beautiful---the most beautiful human invention (besides great music) that exists. Indeed a lot of the first 3 semesters of college physics are simply devoted to learning how to use a halfdozen basic constants like G, and c, and hbar, and Boltzmann k. For 2 years or so, you learn how to use each of those 4 constants separately as indeed you should. Savor the power of each proportion in nature. Only later would you put those 4 constants together, in cosmology say-----hmmmmm, astronomy. yes. astronomy puts all 4 together earlier than other subjects. in understanding how a star works, and figuring out how far away it is by looking at its light----that uses G, hbar, c, and k all at once. But the main thing is, don't rush it! Take two years to learn what each separate constant of nature will do for you. Put them together when you understand each one. k is for anything to do with heat and temperature and the thermal glow things give off---planck radiation law, kinetics, etc. G is for newtonian gravity c is for special rel and also electricity/magnetism hbar is initially just for the simple quantum mechanics that relates the undulation frequency of light to its energy (maybe you are a nerd highschool student who knows college physics already, but you see the kind of timetable I have in mind, put it together when you have a solid grasp of each piece. this is not meant to be a put-down. It is great to find you interested. glad you are here!) yes, the short answer is astronomy. in a beginning astronomy course you meet everything in several years of physics, but simplified and incidentally it will probably be in astronomy that the first tests of quantum gravity will come, as gammaray bursts are measured more accurately and better understood
Martin Posted September 19, 2004 Author Posted September 19, 2004 What might be some useful classes to take when I go to college next year to learn about this stuff Martin? How do you know so much about it? As of now, I can follow the math, but the words are still meaningless as far as a real-life application. I'm interested in learning more about it though. BTW jordan there is another possible answer. Yes in one sense what im talking about is hard because before you see why you put Gbar, hbar, c, and k together to make a set of units you want to thoroughly understand what those 4 constants are. but in another sense, you dont need college courses and all that! If you simply take these constants on faith then it is not all that complicated to put them together to make a set of units. If all you want to do is see how they fit together into a system of units, then I can probably explain that adequately. You just have to keep asking and saying what you dont understand until I figure out some workable explanations. I was wondering about sunlight, and going to size it up in these units. see if it is understandable and if not ask
Martin Posted September 19, 2004 Author Posted September 19, 2004 so we have Gbar units and also practical-size versions of them and the length unit is a handwidth (8.09 cm) the Gbar-Planck length is 8.09 x 10-33 centimeters but I dont want to carry the extra powers of ten so I park them and just talk handwidths. the conventional physicsbook says visible light is 400-700 nm. to convert meters to hands you divide by 0.0809 4.9 and 8.6 microhand that is the cycle-length---the wavelength corresponding to a full cycle (it's also good to know the length corresponding to one radian of phase, which is 1/2pi of a full cycle, but postpone this) now I want to learn the photon energies of visible colors of light the constant I need is hbar c, or actually 2pi times that. Our hbar is E-32 and our c is E9, so hbarc= E-23 jot hand. multiplying by 2pi we get 2piE-23 jot hand. to get the photon energy for red light, wavelength 8.6E-6 hand, I just divide 2piE-23 jot hand by 8.6E-6 hand and get 7E-18 jot Likewise for blue light, dividing by 4.9E-6 we get 13E-18 jot. photon energies: 7E-18 jot for reddest red 13E-18 jot for bluest blue now we need a unit of voltage---which means energy per unit charge the conventional metric volt is one joule per coulomb (a coulomb is some hideous number of electrons like 6.2415E18) and our charge unit is a flat E18, a quintillion electrons. So our unit of voltage has to be one JOT per quintillion electrons----call it a QUARTERVOLT. Our voltage unit is one jot of energy distributed over one charge unit, of E18 electrons. It is right around a quarter of a conventional volt. so an eQ, pronounced "eeQUEUE", or "electronquartervolt" is simply E-18 of our energy unit. that is, each electron's share. it is a quintillionth of a jot. so we also know the energy of a red photon, or a blue photon, in electronjolt terms: 7 eQ for reddest red 13 eQ for bluest blue I guess all that could have been said a lot simpler but too much trouble to rewrite it now. so that was 3 times more words than necessary but next time maybe I will know better
jordan Posted September 19, 2004 Posted September 19, 2004 Indeed a lot of the first 3 semesters of college physics are simply devoted to learning how to use a halfdozen basic constants likeG' date=' and c, and hbar, and Boltzmann k.[/quote'] Yes indeed. I didn't expect to learn all of this freshman year (though that would be nice). I am currently in AP physics. If you don't know what AP is, they're classes designed by the same people who do the SAT's and they're intended to be like an intro level college course. So, though this class has just begun, it and previous physics work have taught me basic kinematics. Some charge stuff with Coulombs and Volts. Then there was power and watts and joules. We recently started in on deriving graphs of velocity and displacement from not-constant accelerations. maybe you are a nerd highschool student who knows college physics already Nope. Just what I said above. If all you want to do is see how they fit together into a system of units, then I can probably explain that adequately. You just have to keep asking and saying what you dont understand until I figure out some workable explanations. Alright. If you don't mind then: hbar is a constant it seems. What exactly is the number, and why was that number chosen? Was it modeled after a physical observation or was it provided through mathematical equations? How does it relate to this: |G| = |c| = |hbar| = |k| = |e| = 1 How can they all equal 1?
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