How can the Planck length be claimed to be the smallest length?

[stevebd1]

Hello DonJStevens

The applicable time dilation factor [(3/2)^1/4 (Planck time/2pi seconds)^1/2] reduces the observable mass from (hc/12pi G)^1/2 kg to (h/4pi c) (c/3pi hG)^1/4 kg. We can now see that the observable "fundamental" mass value is the electron mass (9.10938x10^-31 kg).

I'd appreciate it if you could provide the source for the applicable time dilation factor demonstrated above.

regards
Steve

 

[DonJStevens]

Hi Steve,

The applicable time dilation factor is developed from the paper "Is the electron a photon with toroidal topology?" by J. G. Williamson and M. B. van der Mark, published in :Annales de la Foundation Louis de Broglie (1997).

In this paper, the electron is a self-confined, single-wavelength photon. It is confined in a double loop so that the loop radius is a photon wavelength divided by (4pi). Diffractive limit space curvature is required for this model to work.

The (required) limit space curvature is predicted at the electron mass photon sphere radius (3Gm/c^2). The length 4pi (3Gm/c^2) will be equal to a (gravitationally collapsed) photon size.

The length ratio 4pi (3Gm/c^2) divided by the electron Compton wavelength is equal to the ratio (3/2)^1/2 (Planck time/ 2pi seconds). The square root of this length ratio and the square root of this time ratio will be equal to the applicable time dilation (blue-shift) factor at the radius (3Gm/c^2).

The toroidal topology paper does not explain how the necessary space curvature is achieved. However, the pattern of equal ratios, found in post #18, is easily verified to be either precisely correct or a very close approximation of known factual relationships.

Evidence presented in post #26 indicates the pattern of equal ratios is precisely correct.

Don Stevens

 

[stevebd1]

Hi Don

Thanks for the reply. It appears the photon sphere is of significance, for a static black hole, this is at 3Gm/c^2, but for a rotating black hole, the photon sphere is at-

<br /> R_{ph}\ =\ 2M\left[1\ +\ cos\left(\frac{2}{3}cos^{-1}\mp \frac{a}{M}\right)\right]

The upper sign characterizes the prograde orbit (corotating with the black hole) and the lower sign holds for the retrograde orbit (counterrotating against the black hole).

For maximum spin (a/M=1) the prograde photon sphere orbit would be at 1M while the retrograde orbit would be at 4M. Would this have any implication?

Also, I often see that the black hole electron has angular momentum of ħ/2, where exactly does this figure come from?

regards
Steve

 

[DonJStevens]

Hi Steve,

When we define the electron as a gravitationally confined, single wavelength photon (or microgeon) we are into a part of physics that Brian Greene describes as "--the forefront of cutting-edge science". Many important questions are not yet answered.

The paper "Kerr Geometry as Space-Time Structure of the Dirac Electron" by Alexander Burinskii (2007) defines the electron as a microgeon.

If you have acces to the book "The Black Hole War" by Leonard Susskind, see page 382. Susskind writes "--there is a very special kind of charged black hole that is in perfect balance between gravitational attraction and electrical repulsion. Such a black hole is called extremal."

The microgeon extremal black hole electron is not quite like any black hole that is currently well defined. When its spin angular momentum energy is extracted, it has no residual (irreducable) mass. Its angular momentum accounts for its total mass energy. It does not have an event horizon.

In the black hole electron model, the spin (h bar/2) results from light velocity inertial frame dragging of its ring singularity at its photon sphere radius.

You may want to look at "Black hole electron" in Wikipedia, and associated Talk. The electron mass is clearly linked to the Planck mass.

Don Stevens

 

[DonJStevens]

Hi Steve,

You asked "-- where exactly does thie figure (h bar/2) come from?"

From many experimental tests (since 1925) the electron is known to have the spin angular momentum value (h bar/2). It "seems" to spin like a top, though this description is not considered acceptable.

Any useful electron model must have (h bar/2) angular momentum. Photon linear momentum is (E/c) or (h c/ wavelength) (1/c) or (h/wavelength). A photon in a one wavelength circumference loop will have angular momentum (h/wavelength) (wavelength/2pi) or (h/2pi) or (h bar). A photon in a two turn loop, with a 1/2 wavelength circumference will have a 720 degree spin cycle and a radius value (wavelength/4pi). With this radius, angular momentum is (h bar/2) as required for an electron model.

With gravitational collapse, angular momentum is a conserved property.

Don Stevens

 

[stevebd1]

Thanks again for your reply Don. I had a look at what the maximum angular momentum of an object with Planck mass would be based on the following equation which is often applied to Kerr black holes-

J_{max}=\frac{Gm^2}{c}

When incorporating Planck mass, the answer comes out at ħ, due to the fact that the square/square root, G & c elements cancel out. It could be said that this applies to an object of spin 1 and an object with maximum angular momentum (maximal) and spin 2 could be expressed as J_max/2, in the case of something with Planck mass, this would mean ħ/2.

 

[stevebd1]

The last sentence in post #36 should read-

'It could be said that this applies to an object of spin 1 and an object with maximum angular momentum (maximal) and spin 1/2 could be expressed as J_max/2, in the case of something with Planck mass, this would mean ħ/2.'

 

[DonJStevens]

Hi Steve,

We are defining the electron as a "perfect balance" or "extremal" black hole. The maximum angular momentum will be found when spin acceleration (photon path bending)
force is equal to gravitational force. This is required to be the condition when J is equal to (h bar/2). The electron cannot have more or any less angular momentum than the value (h bar/2).

Here is a partial quote from Burinskii paper, "Leading Role of Gravity in the Structure of Spinning Particle" (2005). Burinskii writes (regarding electrons) "Taking the parameters of the Kerr-Newman source, charge e, mass m and spin J equal to parameters of elementary particles, one obtains that the Kerr parameter a = J/m, which characterizes the radius of the Kerr singular ring, satisfies condition a >> m, when the Kerr's event horizon disappears, and the source represents a naked singular ring ---".

We can (with specific precise equations) relate the mass energy of one electron plus one positron to the photon wavelength that has energy equal to (2/3)^1/2 times the Planck mass energy. Three factors applied to the Planck mass will provide the electron mass value. The first is (1/2), the second is (2/3)^1/2, while the third is the time dilation factor, applicable at the electron mass radius (3Gm/c^2).

Don Stevens

 

[stevebd1]

Hi Don

Has a similar approach been applied to the quark? (though I imagine this would be more tricky due to the fact that the specifics of the quark are more difficult to define due to being confined within hadrons)

regards
Steve

 

[DonJStevens]

Hi Steve

The quarks are definitely more tricky. Leonard Susskind recently said; "There are many types of quarks with different electric charges and masses. What gives rise to these distinctions is a mystery; the internal machinery that underlies the differences is much too small to detect. So we call them elementary, at least for the moment, and like botanists give them different names."

An improved understanding of electrons (and muons later) will most probably lead to a better understanding of quarks. There has been some speculation that quarks could be gravitationally confined but I am not aware of equations or numbers that would give substance to this. The strong clue (that gravitational force is involved) is the close relationship found between the gravitational constant, G and the Planck constant, h.

Don Stevens

 

[DonJStevens]

Hi all,

Those who are not yet convinced that the Planck constant (h) can be derived from the gravitational constant (G) will want to examine the following equations. When the G value used is 6.671745574x10^-11 and the electron mass (m) value is 9.10938215x10^-31 , then excellent agreement is obtained (with current NIST values) when solving for electron Compton wavelength (Le) and for the Planck constant (h) as shown.

Le = 2(3Gm)^1/3 (2pi)^5/3 = 2.426310218x10^-12

h/2mc = (3Gm)^1/3 (2pi)^5/3

h = 2mc (3Gm)^1/3 (2pi)^5/3 = 6.626068964x10^-34

From the electron mass and light velocity, a relationship between the (quantum) Planck constant and the (classical) gravitational constant is specified. These equations are consistant with the pattern of equal ratios noted in post # 18. The electron is defined as a "qantum gravitational" mass particle.

Don Stevens

 

[DonJStevens]

Hi all,

Readers who have interest in the extremal black hole electron model will find the article "Naked Singularities" in the February 2009 issue of Scientific American is interesting.

A number of physicists have concluded that a naked singularity is allowed by the laws of nature, so the gravitationally confined electron concept is now more acceptable to some.

Don Stevens

 

[DonJStevens]

Hi all,

The CODATA gravitational constant value recommended in 1986 was 6.67259x10^-11 with standard uncertainty 0.00085x10^-11. When the standard uncertainty is subtracted from 6.67259x10^-11, the value found is 6.67174x10^-11. The electron properties imply that the correct value should be very close to 6.67174557x10^-11.

We may find that the 1986 value is more nearly correct than the current (2006 CODATA) value. The current recommended value is 6.67428x10^-11.

Don Stevens

 

[Bob_for_short]

...A number of physicists have concluded that a naked singularity is allowed by the laws of nature, so the gravitationally confined electron concept is now more acceptable to some.

Do you believe in a point-like electron?

 

[DonJStevens]

Hello Bob,

The electron is so small that it is very difficult to imagine how any entity can be so small and dense. It is a tiny ring with a ring radius equal to 3Gm/c^2, where the (m) value is the electron mass. If the electron has a radius greater than zero meters, it is not truly a point-like particle. Its ring radius is (1.5) times its Schwarzschild radius. Gravitational collapse is halted at this radius because photon path bending force and gravitational force are in perfect balance (at this radius).

Don Stevens

 

[Vanadium 50]

Le = 2(3Gm)^1/3 (2pi)^5/3 = 2.426310218x10^-12

This can't be right. G has units m³/kg/s², so GM has units m³/s². Take the cube root of that and you have ms^-2/3 on the left and m on the right. The units don't match.

The electron is so small that it is very difficult to imagine how any entity can be so small and dense. It is a tiny ring with a ring radius equal to 3Gm/c^2, where the (m) value is the electron mass.

This sounds very speculative. Is this published anywhere?

 

[DonJStevens]

Hi Vanadium 50,

The equation that looks so strange it seems that, "This can't be right" is developed from the pattern of equal length ratios shown below. Length definitions are listed following the pattern.

L1/L2=L2/L3=L4/L1=2(L1)/Le=Le/2(L3)=(L4/L2)^1/2= (L4/L3)^1/3
L1 = 2pi (Planck length)(3/2)^1/2
L2 = 1/2 (electron Compton wavelength) = 1/2 (Le)
L3 = (2pi seconds)(c)(2pi) = (2pi)^2 (c)
L4 = 2pi (3Gm/c^2), where the (m) value is electron mass

You can see the dimensionless length ratios shown are all equal when G equals 6.67174557x10^-11. You see also that (L4/L3)^1/3 equals (L2/L3). Then:

(L2/L3)^3 = L4/L3
(2pi)^5 (3Gm) = (L2)^3
L2 = (2pi)^5/3 (3Gm)^1/3
2(L2) = Le = 2 (2pi)^5/3 (3Gm)^1/3 = electron Compton wavelength

These equations have been shared with a number of theorists. The new equations are expected to be considered speculative until more evaluations are completed. I have found no reason to doubt that they are correct.

Don Stevens

 

[Vanadium 50]

They are not dimensionless ratios. L4 is not a length - it doesn't have dimensions of lengths.

Speculative theories need belong in the IR forum.