Martin 416 Posted January 10, 2009 (edited) Planck units like the mass and length come up all over the place in modern physics---they are not the only system, there are other convenient special purpose units, but they are worth getting familiar with. Using the google calculator we can take a kind of almost experimental approach to planck units, explore, build up our concepts, try them out. It is easy---a few posts will show what I mean. So here's a brief tutorial. 1. get to know hbar (Planck's constant) Think of hbar as being the product of amounts of energy and time. Check that by typing things like this into google hbar/minute hbar/year It will give you answers in energy terms----small amounts of energy. If you like BTU or calories or foot pounds you can always ask for the answer in those terms hbar/hour in foot pounds Personally I am most comfortable with ordinary metric units like second meter joule newton, but I want to emphasize that the google calculator works with a wide range of units and will give you answers in pretty much any units you happen to like. It knows how to convert and it does what you tell it. 2. Think of hbar as the product of (typically small) amounts of force x distance x time. Try pulling hbar apart using the calculator. Take hbar, pull out some length, then pull out some time (hbar/inch)/day It will always turn out to be a force. Let's have the answer come out in pounds of force: (hbar/inch)/day in pounds (hbar/foot)/hour in pounds Try putting the blue thing into google and see. More later Merged post follows: Consecutive posts merged This is the second post of this tutorial. The title of the post is Your first planck unit--the Force :-D 1. Try putting this into google c^4/G Again, to emphasize that there's no metric bias, put this in c^4/G in pounds c^4/G is a natural amount of force. It's the Planck force. Remember it. 2. Besides the Planck force we don't have very much we have to remember. You already pulled apart hbar and you know it is made up of force x length x time Well what other natural constant do you know that involves length and time? c c is length/time So what happens if you multiply force x length x time together with length/time? You get force x area. So let's try that. Multiply hbar*c and check to see if it is the product of some force with some area. The way we check is we put this in google: hbar*c/square inch That is, we multiply hbar and c together and then divide out any amount of area we please, square foot, square meter, square inch, square mile whatever. If hbar*c is the hidden product of some force with some area, then if we divide out some area we should get a force. Just to show we're not biased let's get it in pounds hbar*c/square inch in pounds Merged post follows: Consecutive posts mergedThis is the third post in this short tutorial. The title of the post is Your second planck unit---the Area OK we picked apart hbar a little and we know it acts like it's made of a bit of force multiplied by other stuff. And we teased apart hbar*c some and we know it acts like it is a force x area quantity... made of a small amount of force multiplied by some area. And we know c^4/G is a force. It is a natural amount of force that doesn't involve any man-made units like inch or hour or foot etc. So let's divide hbar*c by that natural force and see what we get. Put this into google hbar*c/(c^4/G) or even better put this in hbar*c/(c^4/G) in square inches That is a natural amoung of area that doesn't depend on man-made stuff like joules kilograms etc. It's expressed in square inches, OK, but that doesn't matter, you could choose any unit of area to express it in---square yards, square millimeters... The quantity of area itself is natural. And if you took highschool algebra you probably noticed that something could be simplified by canceling a c. hbar/(c^3/G) in square foot And we can make it look nicer by flipping the G up hbar*G/c^3 in square miles Your third Planck unit---the Length. Try this in google sqrt (hbar*G/c^3 ) in miles Edited January 10, 2009 by Martin Consecutive post/s merged. 0 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 12, 2009 I hear you and thank you, Martin. I pull up this stuff in my quantum GR studies. 0 Share this post Link to post Share on other sites

Martin 416 Posted January 12, 2009 Thanks for the response, Norman! Now I'll boil the first post down to essentials and move on. What I want is that we all are sufficiently familiar with how the Planck units arise in nature that we can construct them from scratch, without looking up the definitions. In this approach the only thing you really need to memorize is c^4/G. We all know c already and we know c is the natural unit of speed. So what we need to learn in addition is c^4/G is the natural unit of force. ======================= Before I was urging you try calculating things like hbar*c and c^4/G in terms of all sorts of units---inch, pound, foot, mile, etc etc. The natural unit quantities are built into nature and they are the same whatever random human units you choose to express them in. Now I've made that point, so let's forget it. We can do everything or almost everything just using the usual metric units that the google calculator uses in default mode, when you don't specify. ========================= So to proceed, you should know that hbar is a product of energy and time---often given in joule-seconds. And therefore hbar*c is a product of energy and distance----and thus equivalently a product of force and area. Anybody who has difficulty seeing that hbar*c is a force x area quantity please ask for discussion! I can give more explanation In fact, just as c is the natural unit of speed, hbar*c is the product of the natural unit of force multiplied by the natural unit of area. That's important to realize. What that does for us is it lets us divide out the natural unit of force and get the natural unit of area. You should do this hands-on, yourself, if you haven't already. Use Google to calculate hbar*c/(c^4/G) It should come out in square meters unless you tell the calculator otherwise. Now it is just basic algebra to simplify nature's unit of area to hbar*G/c^3 NOW WE CAN CALCULATE THE NATURAL UNIT OF ENERGY. The simplest way of imagining doing a given amount of work, applying a given amount of energy, is to push with a certain force for a certain distance. Our natural unit of force is c^4/G. Our natural unit of distance is sqrt(hbar*G/c^3) So this is a no-brainer. Just put this into the Google calculator. (c^4/G)sqrt(hbar*G/c^3) That will give you nature's unit of energy a.k.a. Planck energy. If you want Planck mass as well, then just use E = m c^2 to get it. Notice that the blue formula for the Planck energy unit can be simplified considerably. You can consider that homework. We will get to it next time. DOES ANYBODY HAVE ANY QUESTIONS SO FAR? 0 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 13, 2009 (edited) In the angular momentum Kerr metric field, a dimensionless ratio comes out: m/a where the two are as defined in General Relativity construction, as geometric mass and geometric angular momentum (density...) It seems to come out around E-44 and can be written as: [math] \frac {GM^2}{ Jc} [/math] for the electron. Note that it contains spin and not charge. The Reissner-Nordstrom metric becomes significant down around E-23 meters, IIRC. OK Martin the thing I need to ask you here is what of electron Schwarzschild radii of E-57? Is this beyond the pale of what may be described??? Normally I wouldn't care to be honest! Now, though, this ring singularity at E-13 meters is supposedly that thin and I suspend judgements... Merged post follows: Consecutive posts mergedAny other mass-energy states are not so far from the half-MEV of the electron (5-6 orders ?) that they, too, have a ring singularity of smaller radius, as [math]a_{GR} =J/Mc[/math]. The reason I say normally I would not Kerr <> about the Schwarzschild radius is that I figure things really do fall apart into quantum foam in the Planck regime. I trust that the essence that I portray in my inhomogeneous electron study is indeed fulfilled at that level at least. Merged post follows: Consecutive posts mergedI don't mean to muck up your presentation with needless ratios, but there are several length dimensions produced in GR and these are what concern me. Edited January 12, 2009 by Norman Albers Consecutive post/s merged. 0 Share this post Link to post Share on other sites

Martin 416 Posted January 13, 2009 Norman your stuff is more specialized/advanced than fits in to a basic tutorial like this. It could be that there is no interest---nobody who doesn't know the simplest Planck units, and wants to learn them, is reading. If there is no market I will just let it drop. If I don't let it drop I want to take things kinda slow and methodically. So far all we have done is the Planck force c^4/G and noted that hbar*c is a natural quantity of area multiplied by force. So that dividing hbar*c by c^4/G we get the Planck unit of area. And that gave us the Planck length, just taking square root of the area. Then multiplying force by length, we get the natural unit of energy. Planck energy is as far as we've come so far. What is another really basic Planck unit that is simple to write? What about the natural unit of pressure? Pressure is force divided by area, so to get Planck pressure all we need to do is divide two things we already have, force and area. 1 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 13, 2009 (edited) Cool, Martin, I shall enjoy this discussion in another locale. Carry on, please. You are giving me important tools of vision. Others will certainly appreciate this. Edited January 13, 2009 by Norman Albers 0 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 15, 2009 It could be that there is no interest---nobody who doesn't know the simplest Planck units, and wants to learn them, is reading. If there is no market I will just let it drop. If I don't let it drop I want to take things kinda slow and methodically. So far all we have done is the Planck force c^4/G and noted that hbar*c is a natural quantity of area multiplied by force. So that dividing hbar*c by c^4/G we get the Planck unit of area. If you rule your actions by immediate response you will not act. I get very little response from most of my posts. I do not offer them because my ego demands magnification. I do so because after long months of work I see things I did not previously see. Now Google on 'Kerr electron' and witness I am at top of the page along with A. Burinskii. He has worked here more than forty years. I hope to catch up with him. 0 Share this post Link to post Share on other sites

Martin 416 Posted January 15, 2009 (edited) Congratulations to you Norman. My project here is more casual/inconsequential. Edited January 15, 2009 by Martin 0 Share this post Link to post Share on other sites

Martin 416 Posted January 15, 2009 (edited) We have now derived formulas for several Planck units and in particular we have force (c^4/G) and the area (hbar*G/c^3) So let's divide and get the Planck unit pressure (c^4/G)/(hbar*G/c^3) which simplifies to c^7/(hbar*G^2) Let's put that into google to check that it is a pressure. Yes it comes out to 4.6 x 10^{113} newton per square meter (try it!) And newton per square meter is the same as joule per cubic meter---we have found the Planck unit of energy density, as well. Namely the Planck energy density is c^7/(hbar*G^2). Example 1 In contemporary quantum cosmology models where there is a bounce, the bounce commonly happens at about 40 percent of the Planck energy density. That would be 40 percent of 4.6 x 10^{113} joules per cubic meter. Just as a simple illustration, we have used the google calculator to find the energy density at the onset of expansion, in a common sort of QC model. Example 2 How much have volumes expanded since the bounce? For simplicity I'll ignore inflation. The present critical energy density is 0.85 nanojoule per cubic meter and excluding dark energy this is 0.23 nanojoule per cubic meter. What is the ratio of energy densities? I put this into google 0.4*4.6*10^113/(0.23*10^-9) and it came out with 8 x 10^{122} ==================== Norman, thanks for your interest, but I think this attempt at a google-calculator based tutorial on Planck units is not serving a useful purpose---except for yours, no comments, no questions. I am going to let it sit for a while and see. If no more interest then I will unsticky and let it drift off. Edited January 15, 2009 by Martin 0 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 15, 2009 (edited) I do not need your calculator. You have shown me relationships I needed to get to. Edited January 15, 2009 by Norman Albers 0 Share this post Link to post Share on other sites

Martin 416 Posted January 15, 2009 I do not need your calculator. You have shown me relationships I needed to get to. Wow. That's both gratifying and unexpected. 0 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 15, 2009 The bounce of the universe is "inconsequential"? 0 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 17, 2009 I suspect this is an inconsequential result but I was surprised by how close are the values of [math]\hbar[/math] and [math] c^-4[/math]. 0 Share this post Link to post Share on other sites

Primordial 10 Posted January 22, 2009 Norman: Who was the person that first derived the hbar constant? 0 Share this post Link to post Share on other sites

Norman Albers 27 Posted January 22, 2009 (edited) Good question, Primordial. Planck worked on the blackbody spectrum and published in 1901, but before then ??? What do you know? Martin I do not want to be offensive; I am expressing my frustration over trying to get to physics of quantum foam that do not produce the untenable results of our "first efforts" wherein we assume the vacuum has all possible modes in equal distribution. My brother scalbers sent me the calculator figure for Planck pressure. It is some very large number and I ask, so what??? What is? Our theory must make sense both in the small and in the large. Edited January 22, 2009 by Norman Albers 0 Share this post Link to post Share on other sites

occam 11 Posted July 8, 2009 Norman: Who was the person that first derived the hbar constant? I believe it was Paul Dirac. “hbar” is usually known as the reduced Planck constant h/2pi associated with "spin" (which is another topic). Planck values arise from the interaction of the fundamental constants G, c, and h. However has anyone noticed the anomaly created by the use of hbar? Planck length and Planck time have the same relationship as wavelength to frequency, but after that things go downhill! This discrepancy can be shown by setting a wavelength as equal to the Planck length and comparing the values obtained by the standard equivalents, with those obtained using the Planck equations as Table 1.1: Table 1.1 wavelength (lambda) 1.6162524577E-35 Planck length (lp) √(ћG/c3) 1.6162524577E-35 frequency l/c 5.3912378866E-44 Planck time (tp)√(ћG/c5) 5.3912378866E-44 Energy hf 1.2290440710E+10 Planck energy (Ep) √ћc5/G 1.9560843918E+09 Mass E/c2 1.3674959545E-07 Planck Mass (Mp) √ћc/G 2.1764374082E-08 Temperature E/k 8.9019209427E+32 Planck temperature (Tp) √(ћc5/(G*k2) 1.4167847210E+32 If however the equations are reworked using the full value of h, the resulting equivalents are shown in Table 1.2: Table 1.2 wavelength (l) 4.0513441094E-35 Planck length (lp) √(hG/c3) 4.0513441094E-35 frequency l/c 1.3513829322E-43 Planck time (tp)√(hG/c5) 1.3513829322E-43 Energy hf 4.9031764441E+09 Planck energy (Ep) √hc5/G 4.9031764441E+09 Mass E/c2 5.4555195454E-08 Planck Mass (Mp) √hc/G 5.4555195454E-08 Temperature E/k 3.5513526408E+32 Planck temperature (Tp) √(hc5/(G*k2) 3.5513526408E+32 using hbar The product of “Planck Mass” andd “Planck Time” is 1.173369-51.Kg.seconds However using just “h” the value becomes 7.372496-51 kg.s, If mass is proportional to energy, as E = mc2 then “mass” should also have a value of “Planck’s constant”, which dimensionally would be Kilogram-seconds, which is necessarily h/c2. thus the two values are: “hE” relating to “Energy” has the value 6.62606896 x 10-34 J.s “hM” relating to “mass” has the value 7.37249599975914 x10-51.kg.s. Isn’t that interesting? Occam 0 Share this post Link to post Share on other sites

steevey 45 Posted January 21, 2011 Planck units like the mass and length come up all over the place in modern physics---they are not the only system, there are other convenient special purpose units, but they are worth getting familiar with. Using the google calculator we can take a kind of almost experimental approach to planck units, explore, build up our concepts, try them out. It is easy---a few posts will show what I mean. So here's a brief tutorial. 1. get to know hbar (Planck's constant) Think of hbar as being the product of amounts of energy and time. Check that by typing things like this into google hbar/minute hbar/year It will give you answers in energy terms----small amounts of energy. If you like BTU or calories or foot pounds you can always ask for the answer in those terms hbar/hour in foot pounds Personally I am most comfortable with ordinary metric units like second meter joule newton, but I want to emphasize that the google calculator works with a wide range of units and will give you answers in pretty much any units you happen to like. It knows how to convert and it does what you tell it. 2. Think of hbar as the product of (typically small) amounts of force x distance x time. Try pulling hbar apart using the calculator. Take hbar, pull out some length, then pull out some time (hbar/inch)/day It will always turn out to be a force. Let's have the answer come out in pounds of force: (hbar/inch)/day in pounds (hbar/foot)/hour in pounds Try putting the blue thing into google and see. More later Merged post follows: Consecutive posts merged This is the second post of this tutorial. The title of the post is Your first planck unit--the Force :-D 1. Try putting this into google c^4/G Again, to emphasize that there's no metric bias, put this in c^4/G in pounds c^4/G is a natural amount of force. It's the Planck force. Remember it. 2. Besides the Planck force we don't have very much we have to remember. You already pulled apart hbar and you know it is made up of force x length x time Well what other natural constant do you know that involves length and time? c c is length/time So what happens if you multiply force x length x time together with length/time? You get force x area. So let's try that. Multiply hbar*c and check to see if it is the product of some force with some area. The way we check is we put this in google: hbar*c/square inch That is, we multiply hbar and c together and then divide out any amount of area we please, square foot, square meter, square inch, square mile whatever. If hbar*c is the hidden product of some force with some area, then if we divide out some area we should get a force. Just to show we're not biased let's get it in pounds hbar*c/square inch in pounds Merged post follows: Consecutive posts mergedThis is the third post in this short tutorial. The title of the post is Your second planck unit---the Area OK we picked apart hbar a little and we know it acts like it's made of a bit of force multiplied by other stuff. And we teased apart hbar*c some and we know it acts like it is a force x area quantity... made of a small amount of force multiplied by some area. And we know c^4/G is a force. It is a natural amount of force that doesn't involve any man-made units like inch or hour or foot etc. So let's divide hbar*c by that natural force and see what we get. Put this into google hbar*c/(c^4/G) or even better put this in hbar*c/(c^4/G) in square inches That is a natural amoung of area that doesn't depend on man-made stuff like joules kilograms etc. It's expressed in square inches, OK, but that doesn't matter, you could choose any unit of area to express it in---square yards, square millimeters... The quantity of area itself is natural. And if you took highschool algebra you probably noticed that something could be simplified by canceling a c. hbar/(c^3/G) in square foot And we can make it look nicer by flipping the G up hbar*G/c^3 in square miles Your third Planck unit---the Length. Try this in google sqrt (hbar*G/c^3 ) in miles I know that some scientists say the universe is moving along increments of Planck time, but Planck time is just the shortest amount of time that matters in our current physics, why would scientists think its the shortest amount of time possible? Couldn't something have happened in the creation of the universe that took less than a unit of Planck time? 0 Share this post Link to post Share on other sites

DonJStevens 10 Posted February 4, 2011 Hello Martin, When you have an opportunity, would you look at "Ring singularity" at: http://www.absoluteastronomy.com/discussionpost/Electron_as_a_ring_singularity_56595 In this concept, the fundamental length is (3/2) exponent 1/2 times the Planck length because this length is based on a photon orbit radius rather than a Schwarzschild radius. An equation can then be developed that defines an electron mass value that has a specific relationship to the Planck mass. The quantized electron mass value found is (h/4pi c) (c/3pi hG) exponent 1/4 kilogram. Don Stevens 0 Share this post Link to post Share on other sites

Daedalus 329 Posted July 4, 2011 Hello Martin, When you have an opportunity, would you look at "Ring singularity" at: http://www.absolutea...ngularity_56595 In this concept, the fundamental length is (3/2) exponent 1/2 times the Planck length because this length is based on a photon orbit radius rather than a Schwarzschild radius. An equation can then be developed that defines an electron mass value that has a specific relationship to the Planck mass. The quantized electron mass value found is (h/4pi c) (c/3pi hG) exponent 1/4 kilogram. Don Stevens That's awesome. Helped me realize what I was imagining in my head, except I had no way of describing it properly. I got a lot to think about and even more to learn. Thank You! 0 Share this post Link to post Share on other sites

Aguirre 0 Posted August 7, 2011 We have now derived formulas for several Planck units and in particular we have force (c^4/G) and the area (hbar*G/c^3) So let's divide and get the Planck unit pressure (c^4/G)/(hbar*G/c^3) which simplifies to c^7/(hbar*G^2) Let's put that into google to check that it is a pressure. Yes it comes out to 4.6 x 10^{113} newton per square meter (try it!) And newton per square meter is the same as joule per cubic meter---we have found the Planck unit of energy density, as well. Namely the Planck energy density is c^7/(hbar*G^2). Example 1 In contemporary quantum cosmology models where there is a bounce, the bounce commonly happens at about 40 percent of the Planck energy density. That would be 40 percent of 4.6 x 10^{113} joules per cubic meter. Just as a simple illustration, we have used the google calculator to find the energy density at the onset of expansion, in a common sort of QC model. Example 2 How much have volumes expanded since the bounce? For simplicity I'll ignore inflation. The present critical energy density is 0.85 nanojoule per cubic meter and excluding dark energy this is 0.23 nanojoule per cubic meter. What is the ratio of energy densities? I put this into google 0.4*4.6*10^113/(0.23*10^-9) and it came out with 8 x 10^{122} ==================== This is indeed very interesting what You present. I can use it for my one work. Thank You. 496=496 In der Ruhe liegt die Kraft Sig Norman, thanks for your interest, but I think this attempt at a google-calculator based tutorial on Planck units is not serving a useful purpose---except for yours, no comments, no questions. I am going to let it sit for a while and see. If no more interest then I will unsticky and let it drift off. 0 Share this post Link to post Share on other sites

derek w 6 Posted March 9, 2012 do not know if my understanding is correct?If Pl=plank length. Then pl^3 x c=smallest amount of energy possible? 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted March 9, 2012 From my basis knowledge the planck energy would be the energy equivalent of the planck mass (which is actually fairly big at 2.17x10-8 kg - an eyelash is about the same order of magnitude) and similarly from the energy time uncertainty principle it is hbar divided by the planck time. Your dimensions would not fit - energy does not equal distance cubed times velocity- the exact definition is here The planck scale is not entirely a boundary beyond which you cannot go - there are myriad examples of things smaller than the planck mass - it is scale of energies(very high), lengths(very small), times (very short) etc at which quantum field theory encounters problems due to the effects of gravity. The energies required to experiment at the planck scale are way beyond those we can create artificially. We can look at the fossil cosmology from the period after the big bang and use these remnants to try and understand what happened and more about sub-planck scale physics 0 Share this post Link to post Share on other sites

derek w 6 Posted March 10, 2012 Sorry I not being clear.I was thinking of Pl^3 as a volume of stuff,unknown stuff. so my Pl^3 volume of unknown stuff travelling at the speed of light = kinetic energy. If my volume of unknown stuff is spinning,and each spin produces h amount of energy. then I can say E=f x h. My question would be is the plank length anything to do with the energy of a photon? 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted March 12, 2012 Sorry I not being clear.I was thinking of Pl^3 as a volume of stuff,unknown stuff. so my Pl^3 volume of unknown stuff travelling at the speed of light = kinetic energy. "Stuff" does not go at the speed of light - only massless particles go that fast If my volume of unknown stuff is spinning,and each spin produces h amount of energy. then I can say E=f x h. Energy = Reduced Planck's Constant (h bar) times frequency: this is the energy of a photon - not to the best of my knowledge a spinning massive object. I would have assumed any massive object would have the mass as part of the equation - ie same/similar to the classical formula - Rotational Energy = 1/2 Moment of Inertia times the angular speed squared My question would be is the plank length anything to do with the energy of a photon? the planck length is just a term of measurement. I really would recommend re-reading the first few pages of this thread and then reading wikipedia on planck units. 0 Share this post Link to post Share on other sites

Ant Sinclair 70 Posted March 24, 2014 Is Plancks' Constant 694444.44recur There had to be a creator and his patterns of creation so evident; Take Plancks' Constant - c = h x f from Pythagoras/Angles of pentagram we get 432 (base frequency) 300000000/432 = 694444.44recur which just happens to be the radius of the sun. Again take 432 x 5 = 2160 which is sum of angles of a cube but also the six circle flower formation pattern, everything has been designed around the numbers 5 & 9. 2160 also happens to be the length of time of each Zodiacal precession 2160 years!!!! Planck is my favourite for showing me Gods' hand at work! Look at Jupiters radius - is it not the vacuum cleaner of the solar system with a radius of 69910, after all the asteroids/meteors and other space debris it has collected since creation of the solar system could it not had an originally had a radius of 69444.44 recur -1 Share this post Link to post Share on other sites