Calculating the Quantum Kepler Length of a Particle

[arivero]

Originally posted by Orion1
Solution #1:
Compton Wavelength:
r_1 = \frac{\hbar}{mc} = \overline{ \lambda}

Planck Mass:
m = m_p = \sqrt{ \frac{\hbar c}{G}}

Solution #2:
Planck Radius:
r_1 = \sqrt{ \frac{ \hbar G}{c^3}} = \overline{ \lambda}

K_k = 2.422*10^{-27} m^2s^{-1}
L = \hbar !

The "standard", or at least majoritary, definition for Planck length is "the compton length of a mass of plank". I supposse that you are pointing out that this particular case has a total gravitational action of exactly h. Is it?

In any case I think it is more important to stress that the Compton Length answer happens for any mass.

Also, it is usually told that m_p is the case where Swartzchild radius and Compton radius coincide. Not sure about the meaning of this, here.

 

[Orion1]

Binary Systems...



Compton Radius/Wavelength:
r_1 = \frac{\hbar}{mc} = \overline{ \lambda}

The Compton Radius is only a solution for a Keplerian-Planck System, for which m<<M.

As m approaches M in magnitude, the system becomes binaric, and all keplerian laws fail and no longer apply.

Kepler's Laws function only as derivatives to the Binary Theorem, therefore are not actually 'laws'.

Solutions to binary systems requires a modification to Newtons Gravitational Law using a center of mass integration.

dt_1 = dt_2

\frac{dA}{dt} = K_k = 2.422*10^{-27} m^2s^{-1}

Orion1 Binary System Theorem:
\frac{dA}{dt} = \frac{Gm_2^3t_1}{4 \pi (m_1 + m_2)^2 r_1} = K_k

r_1 = \frac{Gm_2^3t_1}{4 \pi K_k (m_1 + m_2)^2}

Limit t₁ = t_p (period)
r_1 = \frac{Gm_2^3}{2 K_k (m_1 + m_2)^2} \sqrt{ \frac{\hbar G}{c^5}}

Limit m₁ = m₂ = m_p
r₁ = 2.538*10^-35 m
r_p = 1.616*10^-35 m



Also, it is usually told that m_p is the case where Swartzchild radius and Compton radius coincide.
[/color]

Compton Radius:
r_c = \frac{\hbar}{mc} = \overline{ \lambda}

Schwarzschild Radius:
r_s = \frac{2Gm}{c^2}

r_c = r_s

\frac{\hbar}{mc} = \frac{2Gm}{c^2}

Compton-Schwarzschild Mass:
m_{cs} = \sqrt{ \frac{ \hbar c}{2G}

Close, but negative.
[/color]

 

[marcus]

Hi Orion
you said:"Also, it is usually told that [planck mass] is the case where Swartzchild radius and Compton radius coincide."

I have never heard that, what I have read has been that Planck mass is the case where half the Schwarzschild radius and the Compton wavelength coincide.

if you have a link to something online with the alternative definition you could post it, then anyone who is curious about the different defintion could check it out

but it is just a factor of two that we are worrying about

half the Schw. is GM/c^2

and that has to equal the Compton, which in reduced or angular format is hbar c/Mc^2

So one sets the two equal and solves

GM/c^2 = hbar c/Mc^2

I will try to put this in LaTex, to match your clear writing.

\frac{GM}{c^2} = \frac{\hbar c}{Mc^2}

M^2 = \frac{\hbar c}{G}

 

[arivero]

factor2

Sorry marcus it was me who raised the sloopy definition; there is a factor 2 (even 4) around all the thread, but I am not worried at this level; we are playing sort of dimensional analysis.

Orion, note that the initial setup does not ask for a whole revolution. Area law forks for a sector of the circle too. Answers are the same, although to look for a complete revolution is an interesting particular case, hmm.

 

[Orion1]

Orion1 Binary Theorem...


the initial setup does not ask for a whole revolution. Area law forks for a sector of the circle too.
[/color]


dt_1 = dt_2

\frac{dA}{dt} = K_k = 2.422*10^{-27} m^2s^{-1}

Orion1 Binary System Theorem:
r_1 = \frac{4(m_1 + m_2)^2}{Gm_2^3} \left( \frac{dA}{dt} \right)^2

r_1 = \frac{}{Gm_2^3} \left( 2(m_1 + m_2) \left( \frac{dA}{dt} \right) \right)^2

r_1 = \frac{ (2K_k(m_1 + m_2))^2}{Gm_2^3}

Limit m₁ = m₂ = m_p
r₁ = 6.464*10^-35 m
r_p = 1.616*10^-35 m
r₁ = 4r_p


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"
[/color]

\frac{dA}{dt} = \frac{}{2} \sqrt{ \frac{\hbar G}{c}} = K_k = 2.422*10^{-27} m^2s^{-1}

Planck Mass Fusion:
Limit m₁ = m₂ = m_p
Limit r₁ = r_p
\frac{dA}{dt} = \frac{ \sqrt{Gm_2^3r_p}}{2(m_1 + m_2)} = \frac{}{4} \sqrt{ \frac{\hbar G}{c}} = K_1 = 1.211*10^{-27} m^2s^{-1}

Planck-Compton Radius: r₁ = r_p
r_p = \frac{\hbar}{m_pc} = \overline{ \lambda_p}

Gravitation 'quantum shutdown' below the Planck Radius: r₁ <= r_p

r₁ <= r_p
\frac{ \sqrt{m_2^3r_1}}{(m_1 + m_2)} = \frac{}{2} \sqrt{ \frac{\hbar }{c}}

r₁ <= r_p
\frac{m_2^3}{(m_1 + m_2)^2} = \frac{ \hbar}{4cr_1}

r₁ <= r_p
r_1 = \frac{ \hbar (m_1 + m_2)^2}{4cm_2^3}

[/color]

 

[arivero]

setting the record

Just to set the record straight, it is true that we must always keep the scent of two masses m_1,m_2. In fact in the first calculations from Orion the m in the angular moment is not the same m that under the square roots, but that is only a abuse of notation and the result is the right one.

Still, one does not need to go deep to Plank mass and radius too soon. For the area swept by the particle 1 around the center of mass, it suffices to substitute
m_2 \to {m_2^3 \over (m_1+m_2)^2}
and reciprocally for the particle 2. The analysis in terms of compton length is still possible.

I am unsure of which area to quantize in this case, if one or another or the sum, nor to speak of which reference frame to use. So lacking of a fundamental theory, it seems we aren't going to get some additional benefit from this additional freedom.

Perhaps a minor difference with the Keplerian case is that we have two masses to add for the total angular momentum. So the additional condition L=\hbar does not imply directly m=m_P now.

 

[Orion1]

Physical Break...



\frac{dA}{dt} = \frac{}{2} \sqrt{ \frac{\hbar G}{c}} = K_k = 2.422*10^{-27} m^2s^{-1}

Kepler's Second Law:
\frac{dA}{dt} = \frac{L}{2m} = K_k

m_2 \to {4m_2^3 \over (m_1+m_2)^2}

\frac{dA}{dt} = \frac{L(m_1 + m_2)^2}{8 m_2^3} = \frac{}{2} \sqrt{ \frac{\hbar G}{c}} = K_k

L = \frac{4m_2^3}{(m_1 + m_2)^2} \sqrt{ \frac{ \hbar G}{c}}

L = \hbar

\frac{4m_2^3}{(m_1 + m_2)^2} = \sqrt{ \frac{ \hbar c}{G}} = m_p

Kepler's Second Law fails at:
\frac{4m_2^3}{(m_1 + m_2)^2} &lt; m_p

L &lt; \hbar - impossible!

 

[marcus]

Originally posted by arivero
I have found this link

http://nedwww.ipac.caltech.edu/level5/Glossary/Essay_plancklt.html

which gives a interesting operational definition of Planck time.

There mass and time of Planck are defined as the quantitues simultanesly compatible with uncertainty principle and with gravitational collapse equation. IE a mass of plank spreaded in a plank volume takes a time of plank to collapse gravitationally into a point.

He says it is adapted from P. Coles' 1999 dictionary.

this is Barry Madore's "Level 5 Knowledgebase" for cosmology, a useful site, here are some more links to it
http://nedwww.ipac.caltech.edu/level5/cos_par.html
http://nedwww.ipac.caltech.edu/level5/toc.html
http://nedwww.ipac.caltech.edu/level5/Glossary/frames.html

 

[marcus]

Originally posted by arivero
I have found this link

http://nedwww.ipac.caltech.edu/level5/Glossary/Essay_plancklt.html

which gives a interesting operational definition of Planck time.

...

"operational definition of Planck" quantities is what this thread is really about

you have given one or more and pointed to several
I recall that Baez gives a meaningful way to look at Planck length
in a short essay at his site
(it discusses what it means for a BH Schw. radius to be comparable to its Compton wavelength, if I remember right)

people sometimes ask "what IS Planck length?" wishing for some natural object to look at or some story to be told which has in it this length, or this mass.

(these quantities are also a system of units that makes as many as possible of the most basic constants equal to one and so it is the system of units that makes nature's favorite equations appear as simple as possible---a system of units that nature likes or that is intrinsic to nature---so one can say that, and forget about answering the question. why should telling a story be necessary when they just simply are the units inherent in the universe. and yet people still ask this naive and natural question...)

"what IS Planck length, or mass, actually?"


you have offered a novel answer involving the rate that a keplerian orbit sweeps out area

maybe in this thread we can come up with some more operational definitions or
"conceptual" or "visual" definitions of Planck units

 

[marcus]

Alejandro, it occurs to me that if one is to give a
conceptual idea of "what really are the Planck unities?" then
it should work in every dimension

asking that it work in other space dimensions besides 3 is a way of discovering what is a satisfactory approach

one might say that just as in spacedimension 3 one has
GM^2 = unit force x unit area = unit force x area = hbar c

by analogy in spacedimension 2 one should have
GM^2 = unit force x unit distance = unit energy = M c^2

(it is the analogy to the inverse-square law)

and so one gets (2D space) analogs of the Planck units

GM^2 = M c^2 which solves to

M = \frac{c^2}{G}

unit energy is
E = \frac{c^4}{G}

unit freq is
\omega = \frac{c^4}{G\hbar}

That makes unit time
T = \frac{G\hbar}{c^4}

and unit distance
L = \frac{G\hbar}{c^3}

--------------
do you happen to know if these are considered the "right" Planck units to use in 2D space?
(all my friends live in 3D space and I could not find anybody to ask or any websites by people living in 2D space)

 

[arivero]

Originally posted by marcus do you happen to know if these are considered the "right" Planck units to use in 2D space?
(all my friends live in 3D space and I could not find anybody to ask or any websites by people living in 2D space)

Well, I know people living a very flat life, but then they are not really interested on Nature...

You are right that the procedure is to look for a law equivalent to the inverse square distande of 3D, then extracting from it the dimension of G... and all the other coupling constants, consider flat "electromagnetism" and all that!

A more straighforward method is to look directly in the Lagrangian density, or in the action, instead of in the force law. This is the usual method in quantum field theory, when they need to know also the dimension of each field.

I'd add that sometime such analysis inspires ways to justify the dimensionality of space-time; but perhaps the rigorous justification will really need of a Long March through Galois Theory.

 

[marcus]

Originally posted by arivero
a Long March through Galois Theory.

maybe you just received a postcard from the future

"the mountains of Algebraic Geometry are very beautiful at
this season
while there is still snow on the heights of Mount Galois
the three dimensions of space are beginning to
appear on the lower slopes
wish you were here"

 

[arivero]

Old Grothendieck (born 28/May/1928) still keeps hidden, I supposse, at some mountain range, albeit I heard he had crossed the French border a few years ago. A pity he was never interested on physics.

 

[marcus]

Originally posted by arivero
Old Grothendieck (born 28/May/1928) still keeps hidden, I supposse, at some mountain range, albeit I heard he had crossed the French border a few years ago. A pity he was never interested on physics.

Grothendieck is the bald bartender in the last scene in the movie
"Zazie dans le Metro" which I believe is very good.
it is appropriate for great mathematicians to be secretive

 

[arivero]

Anyway, let's go back to the point you have raised, marcus. Yes, it is very interesting that if we model gravity with a force G {m m \over r^q}, then only for q=2 the coupling constant will cancel when imposing our "Kepler rule".

In general, asking A(t_P) to be a multiple n of A_P we have
2n=G^{\frac12-\frac1q} m^\frac12 r^{\frac32-\frac{q}2} <br /> c^{\frac3q-1} \hbar^{-\frac1q}

For q=3, R and c dissappear from the equation. For q=0, R and m could be canceled out. But only for q=2 we get to cancel Newton constant.

Thus, if you link q to the dimensionality of space, you can say that you have a hint of why should ST have a dimensionality 3+1, or why it should to compactify to 3+1 dimensions.

 

[arivero]

related papers

Past yesterday I found a old 1949 paper from an outcast, MFM Osborne:
Phys. Rev. 75, 1579-1584 (1949) Quantum-Theory Restrictions on the General Theory of Relativity
http://link.aps.org/abstract/PR/v75/p1579 . It has passed mainly unnoticed in the literature, except for a footnote in Misner 1957 review.

Along this thread we (or at least, myself) have followed the "emerging way", expecting to get quantum mechanics out of quantum geometry. This is also the way of Asthekar (gr-qc/0207106, gr-qc/0211012) and Smolin(gr-qc/0311059, hep-th/0201031). On the contrary, Osborne takes an operative way of quantum measurement indeterminacy, he applyes it to general relativity and he concludes the existence of a Planck length and a Planck mass. And yes, also the Compton radius effect as in the beginning of the thread.

To Osborne, a Planck mass is the minimum mass whose curvature you can measure, given the traditional quantum uncertainty effects.

 

[arivero]

I am a bit slow to follow Orion1' calculations, so I have redone them, basically confirming the results. Now, it is interesting to look also to the intermediate steps, so let me play it again.

We have two bodies 1,2 circling around the center of mass, thus with a common angular velocity \omega such that \omega^2 R_i=G m_j/R^2. Here R is the sum of both radius. The equation is consistent with the center of mass condition R_1m_1=R_2m_2.

The sum of cases 1 and 2 let us to solve for \omega,
\omega=\sqrt{G{M\over R^3}}

Now we impose that the area A_i(t_P)must be a multiple n_i of Plank Area. This translates to
n_i=\frac12 \sqrt{cM\over\hbar} {R_i^2\over R^{3/2}}

Or, using the C.M. condition to substitute R,
R_i={4\hbar\over c} {M^2\over m_j^3} n_i^2

Note now that using again this condition over the already solved radiouses, we get a condition on the multiples of area, namely (m_1/m_2)^2=n_2/n_1. Or, say, m_1^2n_1=m_2^2n_2

Now let's go for the total angular momentum L=m_1\omega R_1^2+m_2\omega R_2^2. Substituting and after a little algebra we get
L={2 \hbar \over m_P} {M^{3/2}\over (m_1m_2)^{3/2} }<br /> {m_1^7 n_1^4+m_2^7n_2^4 \over(m_1^3n_1^2+m_2^3n_2^2)^{3/2}}

Which, using the relationship between n and m, simplifyes to
L={2\hbar\over m_P}{m_1+m_2\over m_1m_2} m_1^2 n_1

Orion' case L=\hbar, m_1=m_2\equiv m gives us, accordingly, m=m_P/4n

Also, if we take m1 a lot greater than m2, we recover the initial Compton formula for R2 and also we get a total angular momentum
L_{m_1&gt;&gt;m_2}\approx 2n_1\hbar {m_1^2\over m_P m_2}= 2n_2\hbar {m_2\over m_P}
which shows that plank mass keeps its role as a bound.

Last, a interesting mistake happens if we try to impose simultaneusly low quantum numbers (n1 and n2 small) and big mass differences (m1 a lot greater than m2). Then we are driven to write
L^{WRONG}_{m_1&gt;&gt;m_2}\approx 2 n_1 \hbar { m_1\over m_P} ({m_1\over m_2})^{\frac32}
that is not completely out of physical ranges, if for instance we put m1= 175 GeV then m2 is around some tenths or hundreths of eV, the maximum and minimum values of mass known in the standard model after the neutrino experiences.

The first post in the thread shows that Planck area and time can be used to refer to "low energy" scales. This calculation, althought wrong, shows that Planck mass could also be employed to refer to values in usual energy ranges.

(Personally I feel strange, because on my usual rules of thumbing I did not believe on space quantisation nor in neutrino masses, but the mathematics has invited us into these matters)

 

[marcus]

Originally posted by marcus
"operational definition of Planck" quantities is what this thread is really about

you have given one or more and pointed to several
I recall that Baez gives a meaningful way to look at Planck length
in a short essay...

... why should telling a story be necessary when they just simply are the units inherent in the universe. and yet people still ask this naive and natural question:

"what IS Planck length, or mass, actually?"


you have offered a novel answer involving the rate that a keplerian orbit sweeps out area

maybe in this thread we can come up with some more operational definitions...

Alejandro has found online the historically first description of Planck units in a reproduction of a Spring 1899 document of the
"Royal Prussian Academy of Sciences"

http://www.bbaw.de/bibliothek/digital/struktur/10-sitz/1899-1/jpg-0600/00000494.htm

I have seen this in hardcopy some years ago but did not know that it is online. Thanks Alejandro. It is wonderful to have this online!

There is something here I did not realize. The 1899 units differ by a factor of something like
\sqrt{2\pi}
(about 2.5) from the units we use now

the original 1899 Planck length, I see in this "fax"
is 4.13 x 10^-35 meters
so one must divide by about 2.5
to get the
1.6 x 10^-35 meters
we use nowadays

anyway that is what it looks like so I will check the others

 

[marcus]

so I see I have been mistaken about a factor of
sqrt 2 pi.
for some years I have thought that the Planck units
we use now are the same as those presented to the
Prussian Academy in 1899.
but this is not right.

the time unit in the original paper is also (just as
you would expect) too big by about a factor of 2.5

Planck gave its value as 1.38 x 10^-43 second
and we now say 0.539 x x 10^-43 second

his "Einheit der Temperatur" is bigger (by about factor 2.5)
than what we call Planck temperature

his "Einheit der Masse" is correspondingly bigger than
what we call Planck mass.

 

[arivero]

Originally posted by marcus
so I see I have been mistaken about a factor of
sqrt 2 pi.

Marcus, do not put attention to it.

The issue of h versus \hbar plagues all the scientific literature, and for sure this thread. Before Planck, it was already infamous with Heaviside units. It ultimatelly comes from Fourier transform, which you can normalise in some different ways, including or not a \sqrt{2\pi} factor.

 

[arivero]

I am still a bit worried about
L^{WRONG}_{m_1&gt;&gt;m_2}\approx 2 n_1 \hbar { m_1\over m_P} ({m_1\over m_2})^{\frac32}
There is not a way to put another wrong to do a right?

Perhaps some argument exchanging particles or so. Hmm.

 

[Orion1]

Harmonic Gravitons...



Conditions:
\frac{dA}{dt} = \frac{}{2} \sqrt{ \frac{\hbar G}{c}} = K_k = 2.422*10^{-27} m^2s^{-1}

dt_1 = dt_2
d \omega_1 = d \omega_2 = \omega_t
r_1m_1 = r_2m_2

M_t = \frac{R_t^3 \omega_t^2}{G} = (m_1 + m_2)
R_t = \sqrt[3]{ \frac{GM_t}{ \omega_t^2}} = (r_1 + r_2)
\omega_t = \sqrt{ \frac{GM_t}{R_t^3}}

L_t = (m_1 \omega_1 r_1^2 + m_2 \omega_2 r_2^2)
d \omega_1 = d \omega_2 = \omega_t
L_t = \omega_t (m_1r_1^2 + m_2r_2^2)
r_1m_1 = r_2m_2
L_t = \omega_t m_{1,2} r_{1,2} (r_1 + r_2)
L_t = \omega_t m_{1,2} r_{1,2} R_t
L_t = \omega_t m_{1,2} r_{1,2} \sqrt[3]{ \frac{GM_t}{ \omega_t^2}}
L_t = m_{1,2} r_{1,2} \sqrt[3]{GM_t \omega_t}

Impose that at Planck scales, gravitational exchanges occur via a 'vibrating space-time string' called a 'harmonic graviton':

harmonic graviton wave velocity:
c = \sqrt{ \frac{F_g}{\mu}}
F_g - Gravitational Force
\mu - Energy per distance

E_g = \frac{Gm_1m_2}{R_t}
\mu = \frac{E_g}{R_tc^2} = \frac{Gm_1m_2}{(R_tc)^2}

harmonic graviton normal mode frequency:
f_n = \frac{c}{ \lambda_n} = n \frac{c}{2R_t}
n - harmonic graviton vibration mode

normal modes:
n = \frac{2f_nR_t}{c} = \frac{2R_t}{\lambda_n}

R_t = \sqrt[3]{ \frac{GM_t}{ \omega_t^2}}

f_n = n \frac{c}{2} \sqrt[3]{ \frac{ \omega_t^2}{GM_t}}

Limit M_t = 2m_p
Limit r₁ = r_p

harmonic graviton Planck Frequency/Wavelength:
f_p = n \frac{c}{2} \sqrt[3]{ \frac{c^2}{2Gm_pr_p^2}}
f_p = 7.361*10^{42} Hz
\lambda_p = \frac{c}{f_p} = 4.072*10^{-35} m

harmonic graviton Planck quantum number:
E_g = n_g \hbar f_p
n_g = \frac{E_g}{ \hbar f_p}
n_g = 1.26

Expected/Predicted:
n_g = 1/1.26

P = 1.26^-1 = 79.365% accuracy
T_l = +,- 20% tolerance

harmonic graviton Classical Planck Frequency/Wavelength:
f_p = \frac{Gm_p^2}{2 \hbar r_p}
f_p = 9.274*10^{42} Hz
\lambda_p = \frac{c}{f_p} = 3.232*10^{-35} m
n_g = 1

Normalization:
f_n = n n_g \frac{c}{2} \sqrt[3]{ \frac{ \omega_t^2}{GM_t}}
n_g = 1.26

 

[marcus]

Originally posted by marcus
Alejandro has found online the historically first description of Planck units in a reproduction of a Spring 1899 document of the
"Royal Prussian Academy of Sciences"

http://www.bbaw.de/bibliothek/digital/struktur/10-sitz/1899-1/jpg-0600/00000494.htm

I think we should try to understand how Planck could have thought of the Planck units in 1899
when the Radiation Law (which has the constant h or hbar in it)
only occurred to him in late 1900. The Radiation Law paper was
published only in December of 1900.

So in March 1899, when he presented this paper to the Academy, he did not have "Planck's constant" as we understand it.
But he could still smell something universal in some radiative thermodynamics quantities he was finding and he got rather excited
as one can hear in the unusual tone of voice on page 600 of the
document which Alejandro has given us.

I have been trying to imagine what the old Prussians of the
Koenigliche Preussiche Akademie would have been thinking
when the 40-year not-so-famous-yet Planck started telling them
about some universal units intrinsic to nature.

(even after the Radiation Law discovery and acceptance of the h or hbar constant, the scientific community continued to disregard the Planck units. they were largely forgotten by everybody but Planck himself for 50 years or more)

 

[marcus]

...the historically first description of Planck units in a reproduction of a Spring 1899 document of the
"Royal Prussian Academy of Sciences"
(Königlich Preußischen Akademie der Wissenschaften)

http://www.bbaw.de/bibliothek/digital/struktur/10-sitz/1899-1/jpg-0600/00000494.htm
...

Here are a couple of exerpts from Planck's March 1899 paper, from around page 480 of the document Alejandro gave the link to:

----------------------

...Diese Grössen behalten ihre natürliche Bedeutung, so lange bei, als die Gesetze der Gravitation, der Lichtfortpflanzung I am Vacuum und die beiden Hauptsätze der Wärmtheorie in Gültigkeit bleiben,sie müssen also, von den verschiedensten Intelligenzen nach den verschiedensten Methoden gemessen, sich immer wieder als die nämlichen ergeben...

...ihre Bedeutung für alle Zeiten und für alle, auch ausserirdische und ausser menschliche Culturen nothwendig behalten und welche daher als "natürliche Maasseinheiten" bezeichnet werden können...

-----a rough translation into English---

...These quantities retain their natural meaning as long as the Law of Gravitation, the Speed of Light in Vacuum and the Laws of Thermodynamics remain valid, they must therefore appear as such, again and again, measured by the most various intelligences and by the most various methods...

...they therefore necessarily retain their meaning for all times and for all, even extraterrestrial and non-human, civilizations---and hence may be referred to as "natural units"...


------------------

"natürliche Maasseinheiten"
I believe Maass means "measure" in this context
and "Einheiten" are "unities" or "units".
So Maasseinheiten means "measurement-units" or "units of measure"

 

[Orion1]

Modern Translation...



Modern Equasion Translation:

I translated the equations into modern SI units, however for nostalgia, I normalized and kept the old standard units in brackets.

K_b = Boltzmann's Thermal Constant.

a = \frac{ \hbar}{K_b} = 7.638*10^{-12} K*s (Kelvin*Seconds)
b = \hbar (m^2*kg*s^-1)
c = c (m*s^-1)
f = G (m^3*kg^-1*s^-1)

r_p = \sqrt{ \frac{bf}{c^3}} = \sqrt{ \frac{ \hbar G}{c^3}}
m_p = \sqrt{ \frac{bc}{f}} = \sqrt{ \frac{\hbar c}{G}}
t_p = \sqrt{ \frac{bf}{c^5}} = \sqrt{ \frac{ \hbar G}{c^5}}
T_k = \frac{a}{t_p} = a \sqrt{ \frac{c^5}{bf}} = \frac{ \hbar}{K_b} \sqrt{ \frac{c^5}{ \hbar G}}

Interesting to note that Planck used Planck Time to calculate Planck Temperature, as a modern Physicist would use Planck Mass to calculate Planck Temperature, which would have been the next step:
T_k = \frac{ \hbar}{K_b} \sqrt{ \frac{c^5}{ \hbar G}} = \frac{}{K_b} \sqrt{ \frac{ \hbar c^5}{ G}}

Planck units:
\hbar = aK_b (j*s) (m^2*kg*s^-1)

However note that due to the time base in Planck units, explains this approach.

Reference:
http://www.bbaw.de/bibliothek/digital/struktur/10-sitz/1899-1/jpg-0600/00000494.htm

 

[marcus]

Orion this looks right
exept for the factor of 2pi.
What you say about Planck's a,b,c,f
in this paper is all clear and makes
good sense exept for his b turning out
not to be hbar but being h = 2pi hbar.

that had me confused earlier when I hadnt
noticed it

still don't understand how he can have come up
with b
when it wasnt even the year 1900 yet and he
had not figured out the Radiation Law in which
that constant (I thought) first made its appearance

thanks for the clear exposition of the formulas
in Planck's paper!

 

[arivero]

Originally posted by marcus
"the mountains of Algebraic Geometry are very beautiful at
this season. While there is still snow on the heights of Mount Galois
the three dimensions of space are beginning to appear on the lower slopes. Wish you were here"

By the way, the quotes are just stilistic, or were you actually quoting someone?

Lets inspect a bit more the lower slopes. As I put a QM argument in the other thread, and we have followed that area quantization plus classical gravity implies at least some bit of quantum mechanics, let's try now from this point of view.

We had that for a force G {m m&#039; \over r^q}, the Area/time rule implies
2n=G^{\frac12-\frac1q} m^\frac12 r^{\frac32-\frac{q}2} <br /> c^{\frac3q-1} \hbar^{-\frac1q}

G, h and c are constants, but mc a momentum, and the other variable object is the distance r. So we obtain
p \approx {1 \over r^{3-q}}
And directly from r we have a natural interval t=r/c. Joining both equations we have a natural force
f=Dp/Dt\approx {1\over r^{4-q}}
If this force is to be identifyed with our initial force \approx {1 /r^q}, we get
4-q=q
And the only consistent force (excluding q=0, where the equation at h becomes undefined) is q=2, inverse square law.

The argument works for any area quantization, ie the G in Planck length does not need to coincide with the G in Newton force. Still, it is weaker than the one directly from quantum mechanics (at new thread) because here the momentum is not defined for the mediating particle, but for the kepler centre. Thus a bit of flaw.

[Edited]If we aim for a force from m', we can do a distintion between inertial mass and gravitational mass. Then the two forces can not be of the same exponent anymore; if d=2 (gravity as input) the new force is a confining F=K mm', with K=c^3/h. The magnitude is not unphysical (order of QCD string tension) but the meaning is, er, lacking. Surely one should concentrate in mediating particles.

 

[Orion1]

Planck Genesis...


This formula is the first mention of the universal constant (b) via reverse tracing the constant (b) in this thesis.

Pg. 465 - Paragraph 2 - Translation:
[/color]
"The entropy S of a resonator with rotational oscillation frequency (v) and the energy (U) we definition the following measurement:

S = -\frac{U}{av}log_e \frac{U}{bv} (41)

"Whereby (a) and (b) designate two universal positive constants, whose total value in the absolute C.G.S. system in the following section (\delta 25)(pg. 478-479) on thermodynamic pathway is determined; (e), which is the source of the natural logarithms, only from out formulation purpose massively superthermal mass turbulent gases course joins."

Formula: (41) - C.G.S. to SI conversion:
S = -\frac{U}{av}log_e \frac{U}{bv} = -\frac{UK_b}{ hf}ln \frac{U}{hf} = -\frac{UK_b}{ \hbar \omega}ln \frac{U}{\hbar \omega}

S = -\frac{UK_b}{ \hbar \omega}ln \frac{U}{\hbar \omega}

E_p = K_bT_k = \hbar \omega

a = \frac{T_k}{\omega} = \frac{\hbar}{K_b}

E_i = E_p

dS = - \frac{dU}{T_k} ln \frac{U_f}{E_i}

\hbar = \frac{E_p}{\omega}

S - resonator entropy
U - potential energy
E_p - Planck Thermodynamic Energy
f, \omega - rotational oscillation frequency
e - natural base e

Planck Mass Entropy:
T_k = \frac{}{K_b} \sqrt{ \frac{ \hbar c^5}{ G}}
E_i = E_p = \sqrt{ \frac{ \hbar c^5}{ G}}

dS_p = - \frac{dU}{T_k} ln \frac{U_f}{E_i} = -dUK_b \sqrt{ \frac{ G}{ \hbar c^5}} ln \left(U_f \sqrt{ \frac{G}{ \hbar c^5}} \right)

dS_p = -dUK_b \sqrt{ \frac{ G}{ \hbar c^5}} ln \left(U_f \sqrt{ \frac{G}{ \hbar c^5}} \right)

Universe Entropy:
dU = E_p
U_f = K_bT_u = K_b(2.725 K) = 3.762*10^{-23} J
S_u = K_b ln \left(K_bT_u \sqrt{ \frac{G}{ \hbar c^5}} \right)

S_u = 1.008*10^{-21} J*K^{-1}

Reference:
http://www.bbaw.de/bibliothek/digital/struktur/10-sitz/1899-1/jpg-0800/00000479.htm
[/color]

 

[marcus]

A.R.,
the quotes around the imagined text
of that postcard from the future
were stylistic, in fact, as you suggested

Orion, you seem to have found the
place in Planck's paper where he
first encounters his h constant, which
at that point in time he was calling "b".
It is very obscure to me where this "b"
constant (which Planck recognized as universal
in some sense) came from, or if it would
even make sense in the light of presentday
understanding. Maybe he stumbled on
Planck's Constant a year early by "mistake"?

 

[Orion1]

Universe Entropy...



'Expanding' the Planck Mass Entropy equation for volume expansion results in the following equasion:

Planck Mass Entropy equation for spherical universe containing an ideal gas:
S_u = K_b \left( ln \left(K_bT_u \sqrt{ \frac{G}{ \hbar c^5}} \right) + ln \left( \frac{r_f}{r_i} \right)^3 \right)

r_i = r_p = \sqrt{ \frac{ \hbar G}{c^3}}
r_f = \frac{c}{H_o}

H_o - Hubble Constant

\Delta S_{u1} = K_b \left( ln \left(K_bT_u \sqrt{ \frac{G}{ \hbar c^5}} \right) + ln \left( \frac{c}{H_o} \sqrt{ \frac{c^3}{\hbar G} \right)^3 \right)

\Delta S_{u1} = 4.810*10^{-21} J*K^{-1}

T_i = T_p - Planck Temperature
T_f = 3000 K CBR photo-transparency temperature
r_i = r_p = \sqrt{ \frac{ \hbar G}{c^3}}
r_f = 7*10^5 Ly CBR photo-transparency range

\Delta S_{u2} = K_b \left( ln \left(K_bT_f \sqrt{ \frac{G}{ \hbar c^5}} \right) + ln \left(r_f \sqrt{ \frac{c^3}{\hbar G} \right)^3 \right)

\Delta S_{u2} = 4.487*10^{-21} J*K^{-1}

T_i = 3000 K CBR photo-transparency temperature
T_f = 2.725 K CBR temperature
r_i = 7*10^5 Ly CBR photo-transparency range
r_f = \frac{c}{H_o} = 1.761*10^{10} Ly Universe range

Classical entropy equation for spherical universe containing an ideal gas:
\Delta S_{u3} = K_b \left( ln \frac{T_f}{T_i} + ln \left( \frac{r_f}{r_i} \right)^3 \right) = K_b \left( ln \frac{T_f}{T_i} + ln \left( \frac{c}{H_o r_i} \right)^3 \right)

\Delta S_{u3} = 3.230*10^{-22} J*K^{-1}

\Delta S_{u1} = \Delta S_{u2} + \Delta S_{u3}

[/color]

"We cannot order men to see the truth or prohibit them from indulging in error." - Max Planck, Philosophy of Physics, 1936