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

[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.