Units based on c=G=hbar=k=e=1 turn out to be quick and convenient to use, and one can get used to the scales (a point damgo made earlier in the dimensionless units thread). Heres a kind of minimal toolkit for working with them including some handy comparisons with familiar units. E38 ----a mile E44-----a million miles E60-----1.7 billion years E6 ------22 grams the proton compton wavelength----2.103E-14 meter---13E18 the CMB temperature-----2.725 kelvin----------1.93E-32 the Hubble time----4.35E17 seconds------------8.06E60 average sunlight photon----2E-19 joules----------E-28 the distance to the sun----150 million km---------93E44 earth orbit speed------30 kilometer/second--------E-4 solar constant----1380 watts/sq.meter------------E-119 2.701k----3.73E-23 joule/kelvin-------------------2.701 sigma-----5.67E-8 watt/sq.meter kelvin^{4}--------pi^{2}/60 a (the aT^{4} law)---7.565E-16 joule/cub. meter kelvin^{4}---pi^{2}/15 I'll show some examples of the kind of things I want to do with this toolkit.
some CMB examples For example the energy (per volume) in the CMB is something we can find by raising T to the fourth and multiplying by a = pi^{2}/15 The energy of an average CMB photon is 2.701kT and since the temperature is 1.93E-32 and k=1, the average energy is just 5.2E-32. the energy density aT^{4} is just 9.13E-128 dividing by the average quantum energy gives the number of CMB photons per unit volume in space 1.7E-96 If we want to see this for a large volume, a cubic mile for instance is E114. So the number of CMB quanta in a cubic mile has to be 1.7E18 Just to mull this over a bit: there are 1.7 billion billion CMB photons in a cubic mile and how much energy have they lost since their emission (called "recombination" or "last scattering", believed to be 300,000 years since bang.) They were emitted at z = 1100, so they have lost all but 1/1100 of their original energy. The current CMB energy density was just calculated to be 9.13E-128. So the density of lost energy is 1100 times that: 1.004E-124, essentially E-124. The energy lost from a cubic mile of CMB is E114 times that or E-10-----two tenths of a joule. As yet no general global energy conservation theorem has been proved in General Relativity. Have to think about this two tenths of a joule---did it go anywhere or was it simply lost?
some sun examples as another example of using natural units, the sun's mass is M = RV^{2} where R is the 93 million mile distance to the sun and V is the average speed E-4 in the earth's roughly circular orbit-----it just happens to be a ten thousandth of the speed of light which is where the E-4 comes from 1. So the mass is 93E44 x E-8 = 93E36 this lets us do things like estimate the lifetime of the sun at its present rate of energy expenditure and calculate by how much the sun bends a ray of light if a ray passes by the sun, closest approach being a million miles (R = E44) then the bending angle is 4M/R 2. 4 x 93E36/E44 = 3.7 microradians. **** 3. Another sun example----what's the power falling on a square mile at this distance from the sun (but outside the atmosphere) The solar constant is E-119 and a square mile is E76 Multiplying the two gives E-43 4. How much energy is there is a cubic mile of sunlight at this distance? Again, the solar constant is E-119 and a cubic mile is E114 Multiplying give E-5---the energy in a cubic mile of sunlight expressed in planck units. 5. How many photons of sunlight are there in a cubic mile at this distance? Their energy is E-5 (about 5 food Calories actually) and the average quantum energy is E-28, so a cubic mile contains E23 photons.
some Hubble examples The Hubble time (reciprocal of Hubble parameter) is 8.06E60 E60 is what corresponds roughly to a billion years, more precisely to 1.7 billion years. The Hubble time 8.06E60 is 13.8 billion years. In the time 8E60 light travels 8E60----the speed of light is one. 1. What distance corresponds to a speed of recession equal to E-1, one tenth the speed of light? One tenth the Hubble distance, 0.8E60 or 1.38 billion lightyears. 2. What is the critical density? (3/8pi) divided by (8.06E60)^{2} this works out to 1.8E-123 Other energy densities of interest are often stated as fractions of the critical density.
the equilibrium temperature of the earth ignoring albedo and greenhouse effects, which tend to cancel, the equilibrium temperature of the earth can be estimated as that of a dark sphere at this distance from the sun Since the solar constant is E-119 and the sphere's area is 4 times its cross section the sphere must reach a temperature where its surface radiation is one quarter of the solar constant. sigma T^{4} = 2.5E-120 sigma is pi^{2}/60 which is about 1/6 T^{4} = 15E-120 By taking fourth root of 15 we see the temperature is 1.97E-30 This is not terribly far from the measured value of the global average surface temperature----it's about 6 celsius. these are just back-of-envelope guesstimates designed to test-drive the natural units---try them out in various unexpected contexts (where one expects to find them is in quantum gravity and related areas) Maybe I should give some examples of more expected uses of these units---black hole temperature, Unruh acceleration temperature, etc.
some black hole examples All black holes considered here are non-rotating and uncharged. In natural units, the black hole radius is 2M Remember that the sun's mass is 0.93E38 If a black hole has mass E38 (slightly more than sun's) then its radius is 2E38 which is 2 miles so the units let one see immediately the size of a solarmass black hole. The surface gravity g of a black hole is 1/4M the hawking temperature is g/2pi Therefore the temperature is 1/8piM The radiative power per unit area is sigma T^{4} as we saw earlier, and sigma is pi^{2}/60 In fact the total power is 1/960A, where A is the surface area (dependence on area is algebraically a bit cleaner than on mass in this case). Have to go, so must proofread later. Cheers, anybody
A Casimir force example Natural units make a lot of things easier to calculate and the Casimir force formla simpliefies to F = (pi^{2}/240) A/L^{4} the constant in front is about 1/24, since pi-sqare is about ten. With all these formulas if you want to work them in metric you have to put the hbars and cees and maybe some Gees back in first. In other words make them more messy before you can use them. Oh, and sometimes kays too. However the cost of having radically clean formulas with c=G-hbar=k=1 is that the scale is unfamiliar. For example E38 is a mile and E35 is a pace---a thousandth of a mile, around 1.6 meters---and E30 is 16 microns. E-50 is about a micronewton of force---more exactly 1.2 micronewton. So you could say "It is hopeless, I will never be able to learn this new scale----E44 is a million miles and E-50 is a micronewton---it's impossible" But it's actually OK and one does get used to the scale and the clean simplicity of the formulas is a big plus. Anyway: think of two plates 1.6 meters on a side so E35 on a side, area is E70. And imagine the separation is 16 microns which is E30. the force is easy to calculate, just the area (E70) divided by the fourth power of the sep(E120) which gives E-50 that micronewton thing, and you have to include the numerical constant which is about 1/24. Maybe I should include that tiny force of 1.2 micronewtons in the dictionary: E60-----1.7 billion years E44-----a million miles E38 ----a mile E6 ------22 grams E-50----1.2 micronewtons Here's the list of constants etc, for review: the proton compton wavelength----2.103E-14 meter---13E18 the CMB temperature-----2.725 kelvin----------1.93E-32 the Hubble time----4.35E17 seconds------------8.06E60 average sunlight photon----2E-19 joules----------E-28 the distance to the sun----150 million km---------93E44 earth orbit speed------30 kilometer/second--------E-4 solar constant----1380 watts/sq.meter------------E-119 2.701k----3.73E-23 joule/kelvin-------------------2.701 sigma-----5.67E-8 watt/sq.meter kelvin^{4}--------pi^{2}/60 a (the aT^{4} law)---7.565E-16 joule/cub. meter kelvin^{4}---pi^{2}/15 I see I havent yet used the proton compton. Maybe that should be next.
explaining the accelerated expansion of the universe the first Friedmann equation is actually about the acceleration in the expansion of the universe. That is what its left-hand side is. Since in physics "explanation" and "mechanism" often refer to a differential equation, one should probably start with the 1922 Friedmann equations, as opposed to a verbal substitute. In natural units they are simple to state. I shall state them in the case where the spatial curvature k = 0, the now generally accepted case: a_{t,t}/a = - (4pi/3) (rho + 3p) (a_{t}/a)^{2} = (8pi/3) rho *a* is the universe's distance scale, used in defining the metric It is normalized so that it is equal to one at the present time. the LHS of the second equation is the square of the Hubble parameter by definition. a_{t}/a indeed is how cosmologists define the Hubble parameter----an expansion rate the subscript zero in H_{0} refers to t=0 which is the present moment, meaning that the derivative of *a* has been evaluated at the present moment. Note that the second equation does not depend on pressure. The LHS of the first equation is the acceleration----except for normalization it is simply the second derivative of the scale parameter *a*. The minus sign is significant. It means that there cannot be any positive acceleration unless one can introduce negative pressure. The pressure must be so negative as to overcome the positive contribution of rho. On a large scale the pressure due to matter is negligible. Both the density and pressure of ordinary radiation----light---are small components and can be neglected to a first approximation. The special feature of dark energy is that its pressure is minus its density. Putting all these things together we have: rho = rho_{matter} + rho_{dark energy} + some small terms p = - rho_{dark energy} + some small terms a_{t,t}/a = - (4pi/3) (rho + 3p) = - (4pi/3) (rho_{matter} - 2rho_{dark energy} ) I cannot underemphasize how important the 3 is to this, and the fact that the pressure from dark energy is the negative of its density (not mass energy but energy density---they are different types of quantity--energy density being formally the same as pressure). These things allow cancelation and something that is overall negative to appear. Then, with the minus sign in front of the 4pi/3, one can have a positive acceleration. The discussion would look the same as this if one did it in metric units except that there would be a lot of Gees and cees in the equations---obscuring otherwise simple relationships and being bothersome for me to type and generally being obnoxious. this is why, on the web, when one sees the Einstein equation or the two Friedmann equations derived from it they are often written in natural units, without the Gees and cees. I still need to do some examples using the proton Compton 13E18, or its reciprocal the proton mass 1/(13E18).
a speed of sound example At the altitude of many commercial flights---above convection and the cloud layer---above the tropopause----temperature is remarkably uniform and according to the US atmospheric model in handbooks temp is around 225 kelvin. This is typical of 6 miles or 30 thousand feet or 10 kilometers or whatever---everybody has flown a bunch at this altitude. What is the speed of sound where you do most of your flying? If you are limited to using metric units in your approach to physics then you should be able to guesstimate the speed of sound in air at 225 kelvin in meters per second. In natural units, we calculate it as a fraction of the speed of light. The temperature is 1.6E-30-----or 1.60E-30 if it makes you feel good to see more decimal places. The average molecular weight of air is 29. This is just what it happens to be and you could guess it from 20% oxygen (32) and 80% nitrogen (28). The formula is real simple. The square of the speed of sound = (7/5) x 13E18 x T/29 I will explain the formula in a moment but first, T = 1.6E-30 so you can just plug it in and get the speed of sound. It comes out to E-6, one millionth of the speed of light. Or if you are a stickler for accuracy, 1.006E-6, but this is back of envelope physics. The 7/5 is because air is bi-atomic, so that is the ratio of its two heat capacities (const pressure vs const vol) The 29 is atomic mass units and 13E18 is converting atomic mass units to planck mass units. kT is an energy-----the kay is omitted because it's one----and energy divided by mass is a velocity squared! (as in Einstein's proverbial c^{2} = E/m). So this formula says to divide a certain energy (the temperature) by a certain mass (the molecule) and multiply by 7/5. Some algebraic equivalent of this usually comes in the Freshman's Second Semester, but it will be carried out with metric paraphernalia and take a bit longer to compute. What other uses of the proton mass/compton number come to mind? I want to explore the things about the world that can be quickly calculated in natural units from a very limited batch of initial information---roughly sketched out in this quote from earlier post:
You seem so intolerably fascinated by natural units -- why? I think it's funny that you're the only one who's posted anything to this thread -- are you just talking to yourself? Personally, I detest the way you do not even specify what type of quantity you're talking about -- I'm fine with you calling "10^44 natural length units" a mile -- or even just "10^44 length" -- but I strongly dislike the way you just refer to a mile as just "E44." E44 WHAT? 10^44 natural length units, of course. These numbers your throw around are NOT dimensionless! - Warren
So marcus, I'm thinking about all this natural unit stuff, and it's kinda blowing my mind. Please let me know if I'm making any sense of it: 1) If you pick c = h-bar = 1, you still have one degree of "unit freedom." You can pick an energy, say, a MeV, and define units of length and time as MeV^{-1}. 2) If you pick c = h-bar = G = k = 1, you have no more degrees of freedom. You cannot pick any units. They are all "picked for you." 3) As a consequence of the fact that you have no degrees of freedom, it turns out that all units are the same. Length, mass, time, acceleration, force, energy -- they're all represented by the same unit! 4) Since they're all the same unit, you can just stop referring to units at all, and just call quantities by their number alone. So the statement 1 mile = 10^44... sounds extremely troubling to me. But 1 mile = 10^44 units of length = 10^44 units of time = 10^44 units of acceleration and so on... This really is one of the most remarkable things I've thought about in a really long time... I still don't completely understand it. - Warren
there are some interesting discussions on Usenet. Here is a quote from John Baez in sci.physics.research. I quote it partly because the wording is kind of funny. [[Message 11 in thread From: John Baez (baez@galaxy.ucr.edu) Subject: Re: Perturbing fundamental constants Newsgroups: sci.physics.research Date: 2003-03-12 22:12:47 PST In article <BA92DC5B.A835%rbj@surfglobal.net>, robert bristow-johnson <rbj@surfglobal.net> wrote: >If length, time, or mass are not dimensionless quantities, then neither are >c, hbar, nor G, no matter *what* set of units you define. Right. Ergo, Lodder must have been talking about units where length, time and mass *are* dimensionless quantities. There is no angel from on high that comes down and punishes you if you decide to use units where hbar, c and G are 1 and are dimensionless. Nor is there one that comes down and punishes you if you decide to use units where hbar, c and G are 1 but have dimensions of ML^2/T, L/T and L^3/MT^2. The angels only get fluttered if you flip-flop and change conventions within a single given calculation. Apart from that, it's up to you to choose your system of units, including how many independent "dimensions" you want, and what they are. It's common to work with units where length, time and mass are the 3 basic dimensions. It's common to take the units of these quantities to be the Planck length, Planck time and Planck mass. It's also common to work with units where everything is dimensionless. It's also common to work with units where there are more than 3 basic dimensions - for example, people often take temperature and charge as dimensions in addition to length, time and mass.]] Chroot, I'm back. That is the end of the Baez quote. Maybe it is not so deep but I liked the way he expressed some of the thoughts. Keep me posted on your thoughts about natural units. I have nothing particular to teach you (at your physics level) but we can share perceptions. Everything you said in the rest of your post seems right to me. Sometimes people use Planck UNITS (little dimensioned quantities) just like metric units only different sized. Then the unit speed is "one planck length per planck time" Sometimes they use dimensionless units and then the speed of light is simply the number 1. You are quite right that if you say c = hbar = G = k =1 then there are no more degrees of freedom---all your scales of measurement (except perhaps electrical) are determined. In research papers this way is sometimes chosen. There was a thread here at PF on dimensionless units. Damgo and Hurkyl had things to say. I dont remember the conclusions if any. When you use dimensionless scales of measurement it is up to you to keep track of the type of quantity things are so that you COULD put back the Gees and cees and kays and hbars into the equations if you needed to or wanted to. Meanwhile, there is less to write. This is just my take. Other people, like in the QFT thread where they use dimensionless or semi-dimensionless units and set things equal to one, those people would have a different take. Maybe they will post and give you some of their ideas
Sorry if it sounded ironical, I meant it in a complimentary way Anyway, if you get into natural units or related areas, do keep me posted as to how things look to you. Of course I will try to answer any questions as best I can.
the Chandrasekhar limit example The Chandrasekhar mass limit is a limit on the mass of white dwarves and depends on the chemical composition of the central region of the star through the parameter Z/A atomic number/atomic weight essentially the number of electrons compared with that of nucleons this because the exclusion principle for electrons is what is holding up the matter and preventing its collapse, so having a high Z/A like carbon (where it is 1/2) is "better" for the star than having a lower Z/A like iron (where it is 26/56). since the chemical composition can be mixed these numbers represent averages. the mass limit depends on the square of Z/A and on the number 13E18 which we know as the proton compton or reciprocal mass. For a useful, two or three figure, approximation: M_{Chandra} = pi (Z/A)^{2} (13E18)^{2} It is somewhat tedious to do the same calculation in metric because one must insert a lot of Gees, hbars, cees and the arithmetic is considerably increased. This formula was derived from the version presented in Frank Shu's "Physical Universe" text----which depends on Z/A and the proton mass. Using this formula one quickly sees that the limit for a carbon dwarf is about 1.3E38 The sun, on the other hand, is 0.93E38, so the chandra limit of 1.3E38 is around 1.4 times the mass of the sun.
the sun's lifetime example a French person named ant284 and I recently calculated the lifetime of the sun essentially from general knowledge plus three numbers: the proton mass reciprocal 13E18, the solar constant E-119, and 5.5E-22, which is the energy released per H nucleus during fusion to helium. It was some thread in PF "Homework Help" The earthorbit radius is 93 million miles (93E44) and speed is E-4, so the sun mass (RV^{2}) is 93E36 Ant said 75% of its mass is H and it will stay on main sequence until 12% of that is used up---then starts having fits. 12% of 75% of 93E36 is 84E35 Multiplying that by 13E18 will tell how many hydrogens will be consumed! One sees that 1.1E56 hydrogens will be consumed in the course of the sun's normal life, and each one yielding 5.5E-22 means the total lifetime energy output will be 6E34. How fast is this energy being released? The solar constant E-119 applies to a sphere of radius 93E44. To get the grand total energy output of the sun one simply multiplies E-119 by the area (4piR^{2}) and gets a power of about E-26. With more attention to accuracy, it comes out 1.07E-26. But for back-of-envelope purposes the problem comes down simply to asking how long a power E-26 takes to deliver an amount of energy 6E34. This is clearly 6E60 on the time scale. But E60 is our "billion year" time scale. In fact E60 is 1.7 billion years. So the sun's normal lifetime of 6E60 translates to some number like 10 billion years. With slightly more accuracy one gets 9 point something. So what we needed for this, besides general knowledge like 93 million miles and E-4 orbit speed are the three numbers 13E18, and E-119 (essentially the brightness of sunlight at this distance, the solar constant), and 5.5E-22 the fusion yield from one H atom. Maybe you like it maybe you dont. this is test driving these units "off road" to see how they work. An "on road" test would be to use them where they are already in common use---quantum gravity, cosmology, inflation, some kinds of quantum field theory (Baez gives a general view of this in the passage quoted above). But an "on road" test wouldnt prove anything since we already know the units are preferable to metric in those contexts where experts use them. I have to take them onto unexpected terrain to see how they do. I think they do OK. It is liberating to not have to write units all the time because the choice is made for you by deciding that c = G =hbar = k = 1. And it is nice to have formulas without so many Gees and cees and so on. But there are pedagogical drawbacks, such as mentioned by Hurkyl in the other thread. Have to balance things.
the green light example Somehow in the Religion forum, god only knows, an issue came up of how much inertia is added to a box when you put 3E35 green photons into it. The typical energy of a green photon is 2E-28 so you just multiply by the number of photons and get 6E7 or 60E6 thats the total energy of all those photons and adding them to the box increases its inertia by 60E6 mass units. Our dictionary says E6 on the mass scale is 22 grams. The National Institute of Standards and Technology website would say 21.767 grams (their value for planck mass is 21.767 micrograms) but approximately 22. 60 x 22 is 1300. So the light gives the box 1300 grams of additional inertia Extra facts: the green frequency is 2E-28 the vacuum wavelength is (1/2)E28 angular format used throughout. The green frequency is easy to remember if you know, from the dictionary, that the average quantum energy in sunlight is E-28. The bulk of sunlight photons are infrared and the visible part is in the high end, numberwise. It just happens that green is about twice the average for a sunlight photon. I will repost the dictionary/constants list I'm using for reference. In metric, converting between wavelength and frequency and eevee-energy and joule-energy and kilogram-mass-equivalent, all this takes some arithmetic. Here it is automatic. In metric you would be using the speed of light and some other conversion factors. The energy conversion factors are tabulated at the NIST website but it is still a nuisance and doesnt add anything to one's understanding.
the CMB dipole example----hotspot in Leo The temperature of the CMB and its dipole have been measured with great accuracy, see for example Lineweaver, Tenorio, Smoot et al http://arxiv.org/PS_cache/astro-ph/pdf/9601/9601151.pdf "The Dipole Observed in the COBE DMR Four-Year Data" The CMB has a hotspot in Leo and a coldspot 180 degrees opposite Leo. See this picture: http://aether.lbl.gov/www/projects/u2/ The dipole is 3.358 millikelvin----that is how much hotter it is in Leo over the average. And how much colder is the coldspot on the opposite side of the sky, in Aquarius. The overall CMB temperature is 2.726 kelvin. If you are limited to metric units you probably should know the speed the solar system is moving relative to CMB in meters or kilometers per second. It is our motion relative to the cosmic rest frame so it is sort of a basic thing to know about the universe. Think about how you would calculate it, if you havent already. Its real easy even in metric. In natural units our speed, in the Leo direction, relative to the Hubble flow or the CMB (however you think about it) is 1.23E-3---------1.23 thousandths of the speed of light You calculate it merely by dividing 3.358 mK by 2.726 K. It is simply the ratio of the two temperatures. And it is clearly 12 times the earth's orbital speed (E-4) around the sun, so the position and size of the dipole are essentially uneffected by the earth's orbital motion (it has only a small percentagewise effect). It is an absolute velocity. Cosmology is not like Special Relativity where there are no absolute velocities. Cosmologists have a preferred rest frame given by the expansion of space. The COBE researchers (Lineweaver et al) gave the direction of the dipole in galactic coordinates (264 degrees, +48 degrees) and in celestial coordinates (11 h 12 m, -7.22 degrees) These were given in much higher precision, with confidence intervals but I have rounded off for easy writing. This confirms that the direction of the hot spot and the solar system's motion is in Leo. A figure for our speed relative to CMB that they give is (quoting verbatim) 1.231 +/- 0.008 x 10^{-3} This agrees with the 1.23E-3 which I calculated earlier.
the example of the light in the garden I just went out into the garden and there was a lot of light---and some things flying around pollinating the fruit trees or whatever they do. And I thought of the Income Tax Form called the "Ten-Ninetyone" Because ten-ninetyone is the occupancy of space by sunlight. There are 10^{-91} photons per unit volume of sunlight at this distance. It is clear because 10^{-119} is the solar constant and the average energy per photon is 10^{-28} and one subtracts 28 from 119 and gets 91. Trivial arithmetic. How many photons are in a cubic meter, if you like working with them? So if I want to know the photons per cubic mile I just multiply by the cube of E38, which is E114, and get (since 114 - 91 is 23) that there are E23 photons in a cubic mile of sunlight. Or if I'm looking at a square pace of ground---E35 by E35---and wonder how many photons would impinge on it in a minute---E45---in vertical sunlight....Then I just add those exponents---E115---and multiply by the occupancy "ten-ninetyone" number E-91. That tells me E24-----a trillion trillion---photons would land on that square patch of ground in a minute. Oh, E45 is 54 seconds, so only 9/10 of a minute. Had better add it to the dictionary. ********* Here's the current dictionary: E60-----1.7 billion years E45-----9/10 of a minute E44-----a million miles E38-----a mile E35-----a pace E6 ------22 grams E-50----1.2 micronewtons Here's the list of constants etc, for review: the proton compton wavelength----2.103E-14 meter---13E18 the CMB temperature-----2.725 kelvin----------1.93E-32 the Hubble time----4.35E17 seconds------------8.06E60 average sunlight photon----2E-19 joules----------E-28 the distance to the sun----150 million km---------93E44 earth orbit speed------30 kilometer/second--------E-4 solar constant----1380 watts/sq.meter------------E-119 2.701k----3.73E-23 joule/kelvin-------------------2.701 sigma-----5.67E-8 watt/sq.meter kelvin^{4}--------pi^{2}/60 a (the aT^{4} law)---7.565E-16 joule/cub. meter kelvin^{4}---pi^{2}/15