# Kepler length.

by arivero
Tags: kepler, length
 PF Gold P: 2,934 One knows that in classical gravity a orbiting point sweeps equal areas at equal times. It can be seen that for macroscopic distances the area swept in a plank time is a lot greater than the minimum quantum of area, which is about (plank length)^2. Now, I ask, given a particle of mass m, for which radius will a circular gravitational orbit around the particle to have the property of sweeping one plank area in exactly one plank time unit? Below that radius, it should be posible to use "plank time beats" to divide area into regions smaller that previous. So a fundamental break of physics will happen at "quantum kepler length of the particle m" ;-) Of course you know which other name this length receives, do you? [SPOILERS FOLLOW IN NEXT ANSWER. So padding inserted here] . . . . . . . . . . . . . . . . . . . .
Astronomy
PF Gold
P: 23,271
 Originally posted by arivero ...\ ;-) Of course you know which other name this length receives, do you?
compton wavelength
nice
Astronomy
PF Gold
P: 23,271
 Originally posted by marcus compton wavelength nice
maybe a little discussion would be appropriate

this is a really nice pensee

(physics haiku?)

in natural units the (reduced) compton a particle with mass M is 1/M

the circular orbit velocity around such a particle is
$$\sqrt{\frac{M}{R}}$$

so the rate area is swept out is

$$R\sqrt{\frac{M}{R}}= \sqrt{MR}$$

setting that equal to one (as Alejandro requires) tells us
that R = 1/M

this is the (reduced) compton

(when dealing with G = hbar = c = 1
units one prefers the reduced wavelength which is the cyclic
one divided by 2pi, as well as the angular format for frequency)

 P: 174 Kepler length. How does the compton value relate to the Red/Orange line 655nm with no mass? LPF
 Astronomy Sci Advisor PF Gold P: 23,271 Alejandro something is catchy or piquant about the implied argument. I like it. Alejandro says, in effect, that a circular orbit cannot sweep out area at a rate that is slower than the planck rate. He says it must be true for any circular orbit that: $$\sqrt{MR}> 1$$ because say the sweeping of area is slower than unity, say that it takes 3 time units to sweep out 1 area unit then one can divide the time interval up and this leads to being able to describe an area smaller than the area quantum. (one can use planck "beats" to divide the indivisible) so he looks for a paradox he says fix some mass M and make the orbit radius R so small that $$\sqrt{MR}< 1$$ now the rate of sweeping of area is less than unity what prevents this? and Alejandro discovers that what prevents this situation is that the orbit radius is less than the Compton for that mass. (the central gravitating body is being localized more restrictively than quantum mechnanics permits it to be localized---every particle is spread out by at least its compton)
Astronomy
PF Gold
P: 23,271
 Originally posted by arivero One knows that in classical gravity a orbiting point sweeps equal areas at equal times. It can be seen that for macroscopic distances the area swept in a plank time is a lot greater than the minimum quantum of area, which is about (plank length)^2. Now, I ask, given a particle of mass m, for which radius will a circular gravitational orbit around the particle to have the property of sweeping one plank area in exactly one plank time unit? Below that radius, it should be posible to use "plank time beats" to divide area into regions smaller that previous. So a fundamental break of physics will happen at "quantum kepler length of the particle m" ;-) Of course you know which other name this length receives, do you?
If it is correct, Loop Quantum Gravity implies that the planck quantities (the "natural units" as Planck called them) are real features of the environment
that they are built into spacetime just as the speed of light (which is the planck unit of speed) is built in.

the theory pinpoints the place at which new physical effects can be expected, namely at Planck energy (2 billion joules)
and it derives quantized spectrums for area and volume where
the minimum measurable amounts of area and volume are order-one multiples of planck area and planck volume.

among the new physical effects expected at planck scale are
modifications of Lorentz symmetry

It now seems that these and other planck scale effects may be possible to probe in the next few years leading to tests of LQG.
If it checks out this will among other things infuse a greater sense of solidity into planck units. There will be reason to think that they really are nature's units.

Alejandro's puzzle (why cant area be swept out at slower than planck unit rate?) is an example of a "Planck Parable", as I would call it.
Or a "Planck units Haiku".

I will try to think of another such story or puzzle, to respond in kind.

If Loop Gravity develops successfully and tests out we will probably see more Planck haiku growing up as a kind of physics folklore.
 Astronomy Sci Advisor PF Gold P: 23,271 to understand this you need to go to the NIST website and look up Planck temperature or take my word for it that it is 1.4168e32 kelvin that is the temp T such that kT is planck energy and if planck energy is built into spacetime as an intrinsic threshold of some new effects then the temp is also (in the same way) a built in intrinsic feature of nature On that scale room temperature is about 2e-30 and a nice oven temperature is about 3e-30 (in fact if you work it out 3e-30 of the natural unit temperature is 425 kelvin, which is 305 Fahrenheit qualifying as a "slow to moderate" oven. Well in this short short story there is a Wizard who wants his stove to always be at 3e-30. Hope notation is clear, I mean 3x10-30 have to go for now, continue later
 Astronomy Sci Advisor PF Gold P: 23,271 Once there was a Wizard who lived in a cave that he'd hollowed out in the core of a comet and the wizard was always cold He wanted a potbelly stove that always stayed at a steady temperature of 3E-30 natural. This would take the chill out of the cave and be convenient for baking. What he got to stoke the stove with is interesting. He got a supply of the usual kind of black holes each of which was always at the steady temperature of 3E-30 natural. (that is 425 kelvin or 152 celsius or fahrenheit 305 depending on which conventional human scale you like) the question is, what was the mass of each of the black holes. and how big were they?
PF Gold
P: 2,934
Yes, it is the typical Haiku one finds in elementary quantum mechanics book. Perhaps sometime we will see it in elementary LQG books :-) By the way, I had never seen it, but I do skip a lot of literature on this.

 Originally posted by marcus (physics haiku?) (when dealing with G = hbar = c = 1 units one prefers the reduced wavelength which is the cyclic one divided by 2pi, as well as the angular format for frequency)
It becomes even more rythmical if you allow for G, as then the one coming from plank length simplifies against the one coming from newton universal gravitation law. At the end, Compton length contains, of course, m, h, and c, but no G around. As for author trickery, I confess that the paradox has been choosen a posteriori :-)

I am -paying time- at a cybercafe now, so I hope that by monday the wizard will have all his charcoal supply issues solved!!
Astronomy
PF Gold
P: 23,271
 Originally posted by arivero ... I confess that the paradox has been choosen a posteriori :-) I am -paying time- at a cybercafe now, so I hope that by monday the wizard will have all his charcoal supply issues solved!!
A significant confession!

the wizard reflected to himself that the temperature of the usual kind of black hole (that they sell for stoves) is

$$\frac{1}{8\pi M}$$

where M is the mass of the hole
 Astronomy Sci Advisor PF Gold P: 23,271 at this time a tribe of gypsies was passing through the solar system telling fortunes, selling black holes and love potions, and stealing as gypsies always do. So the wizard solved the following equation for the mass M, so that he could ask the gypsies for holes of that mass: $$\frac{1}{8\pi M} = 3*10^{-30}$$
 Astronomy Sci Advisor PF Gold P: 23,271 $$\frac{1}{8\pi M} = 3*10^{-30}$$ $$8\pi M = \frac{1}{3}*10^{30}$$ $$M = \frac{1}{24\pi}*10^{30}$$
 Astronomy Sci Advisor PF Gold P: 23,271 this means that each piece of "stove charcoal" would be the size of a mote of dust and would have a mass about a twenty-thousandth of that of the earth and if it would not be too tedious I will work that out from the previously derived value of M
 Astronomy Sci Advisor PF Gold P: 23,271 the mass calculated was M = 1.3E28 planck and the earth's mass is 2.75E32 planck that's why I said roughly a twenty-thousandth----1/20,000 earth mass. ------if you like everything metric------ the planck mass unit is 22 micrograms as can be seen at NIST site, so that E9 planck is 22 kilograms I get that in kilograms M = 29E19 kg = 2.9E20 kg. and that is, in fact, roughly a 20,000th of the mass of earth which is like about 6E24 kg. ---------------------------------------- the radius of an ordinary kind of black hole is 2M so the ones for the Wizard's stove have a radius of 2.6E28 this is the size of a tiny mote of dust or pollengrain because E28 is about one sixth of a micron so we are talking about half a micron radius or micron diameter
 Astronomy Sci Advisor PF Gold P: 23,271 black holes this size are always nice and warm in fact fahrenheit 305, OK for baking cornbread and homefry potatoes, macaroni-and-cheese, such as a wizard might want for supper.
 PF Gold P: 2,934 Fine parable. I like this kind of effort to grasp the scale of the units and objects we use; in fact some months ago I did the example of putting the elementary particle masses in amu (atomic mass units) and then to look for the nucleus having the same weight! As you have completely given the answer to your history, let me to complete mine too. We have got Compton Length. Now, can we get quantum mechanics from it, using similar arguments? Yes, we can: the Bohr-Sommerfeld quantum condition can be formulated, at least for circles, via a a Newton(*)-Kepler principle: any bound particle sweeps a multiple of Compton Area in a unit of Compton Time. This principle does not need gravity; it works for any central force. The usual way, instead of this, is to use the obtained wavelength to check for destructive interference, most arguments in this line invoque De Broglie instead of Compton(**), so perhaps our whole thread could need a "fine tuning". [EDITED:] In fact, note that the argument here above does not depend of $$c$$. Very much as the argumeng starting the thread does not depend of $$G$$. (*)http://members.tripod.com/~gravitee/booki2.htm (**) Indeed this was the point raised, subtly, by LPF earlier in the thread.
Astronomy
 Astronomy Sci Advisor PF Gold P: 23,271 every lady has some idea of her weight--she may wonder if it is too much or too little, or she may be content with it, but she has some idea the moon thinks of her weight as a force measured by scales at her surface* Now Alejandro you were just talking about circular orbit speed. The moon knows the circular orbit speed right down at her surface and she thinks of it in planck terms (that is as a fraction of the speed of light) because that is natural she thinks of every physical quantity that way, in those units. what is the relation between her weight and the fourth power of that speed? ---------------------- * it is hard to imagine the moon standing on bathroom scales placed on her own surface, but she thinks of the force which is her weight as a million times what one millionth of her would weigh on scales at her surface. or perhaps a billion times a billionth, I'm not sure which. -------------------- EDIT: hello A. using edit button as suggested to clarify problem. the key thing is "she thinks of every physical quantity in planck terms" so the question is, what is the relation between two numbers: the fourth power of the surface orbit speed and her weight (both expressed in the planck system of units). no big deal, kind of a silly problem actually, I should solve it probably just as a check that I am stating it right. what I like is the moon having an idea of her own weight. --------------------- MORE EDIT: well,you didnt take me up on this one, so I will say the answer. They are the same number. in planck units the circ orbit velocity at surface is $$\sqrt{\frac{GM}{Rc^2}}$$ you see I divided the dimensionful speed by c to get the speed as a fraction of c And so the fourth power of the speed is $$(\frac{GM}{Rc^2})^2$$ this is the same as her weight at her own surface $$(\frac{GM^2}{R^2})$$ divided by the unit of force belonging to planck units $$\frac{c^4}{G}$$