# Re: Planck Length

```This is the basic argument. The Compton wavelength or equivalently the de
Broglie wavelength with v = c is equal to the Schwarzschild radius. That is
how to derive the Planck length. The argument that nothing smaller exists
just means the Heisenberg uncertainty principle can't isolate something
smaller than its wavelength, a Fourier transform version of the Nyquist
frequency, and for general relativity anything smaller than a black hole is
not observable. There is then some odd equivalency between black hole
physics or general relativity and quantum physics. This means one is not
able to isolate a quantum bit in some region smaller than a Planck area, or
volume. The event horizon of a black hole is then a system of Planck are
pixels or units of area. The Bekenstein formula is that the entropy of a
black hole is```
```
S = k A/4ℓ_p^2

for ℓ_p = sqrt{Għ/c^3} the Planck length. The area of the black hole
horizon is A = 4πr^2 and r = 2GM/c^2. This Schwarzschild horizon area is
then some integer multiple of the Planck areas,  A_p = πℓ_p^2, A =  4Nℓ_p^2
and we find S = Nk. It is an equipartition result.

LC

On Monday, January 7, 2019 at 3:25:16 PM UTC-6, John Clark wrote:
>
> On Mon, Jan 7, 2019 at 8:03 AM <agrays...@gmail.com <javascript:>> wrote:
>
> *> How does one calculate Planck length using the fundamental constants G,
>> h, and c, and having calculated it, how does one show that measuring a
>> length that small with photons of the same approximate wave length, would
>> result in a black hole? TIA, AG*
>
>
> In any wave the speed of the wave is wavelength times frequency and according
> to
> Planck E= h*frequency  so E= C*h/wavelength.  Thus the smaller the
> wavelength the greater the energy. According to Einstein energy is just
> another form of mass (E = MC^2) so at some point the wavelength is so
> small and the light photon is so energetic (aka massive) that the escape
> velocity is greater than the speed of light and the object becomes a Black
> Hole.
>
> Or you can look at it another way, we know from Heisenberg that to
> determine the position of a particle more precisely with light you have to
> use a smaller wavelength, and there is something called the  "Compton
> wavelength" (Lc) ; to pin down the position of a particle of mass m to
> within one Compton wavelength would require light of enough energy to
> create another particle of that mass. The formula for the Compton
> Wavelength is Lc= h/(2PI*M*c).
>
> Schwarzschild told us that the radius of a Black Hole (Rs), that is to
> say where the escape velocity is the speed of light  is:  Rs= GM/c^2. At
> some mass Lc will equal Rs and that mass is the Planck mass, and that Black
> Hole will have the radius of the Planck Length, 1.6*10^-35 meters.
>
> Then if you do a little algebra:
> GM/c^2 = h/(2PI*M*c)
> GM= hc/2PI*M
> GM^2 = hc/2*PI
> M^2 = hc/2*PI*G
> M = (hc/2*PI*G)^1/2    and that is the formula for the Planck Mass , it's
> .02 milligrams.
>
> And the Planck Length turns out to be (G*h/2*PI*c^3)^1/2 and the Planck time
> is the time it takes light to travel the Planck length.
>
> The Planck Temperature Tp is sort of the counterpoint to Absolute Zero, Tp
> is as hot as things can get because the black-body radiation given off by
> things when they are at temperature Tp have a wavelength equal to the
> Planck Length, the distance light can move in the Planck Time of 10^-44
> seconds. The formula for the Planck temperature is Tp = Mp*c^2/k where Mp
> is the Planck Mass and K is Boltzmann's constant and it works out to be
> 1.4*10^32 degrees Kelvin.  Beyond that point both Quantum Mechanics and
> General Relativity break down and nobody understands what if anything is
> going on.
>
> The surface temperature of the sun is at 5.7 *10^3  degrees Kelvin so if
> it were 2.46*10^28 times hotter it would be at the Planck Temperature, and
> because radiant energy is proportional to T^4 the sun would be 3.67*10^113
> times brighter. At that temperature to equal the sun's brightness the
> surface area would have to be reduced by a factor of 3.67*10^113, the
> surface area of a sphere is proportional to the radius squared, so you'd
> have to reduce the sun's radius by (3.67*10^113)^1/2, and that is
> 6.05*10^56.
> The sun's radius is 6.95*10^8   meters and  6.95*10^8/ 6.05*10^56  is
> 1.15^10^-48 meters.
>
> That means a sphere at the Planck Temperature with a radius 10 thousand
> billion times SMALLER than the Planck Length would be as bright as the sun,
> but as far as we know nothing can be that small. If the radius was 10^13
> times longer it would be as small as things can get and the object would be
> (10^13)^2 = 10^26 times as bright as the sun. I'm just speculating but
> perhaps that's the luminosity of the Big Bang; I say that because that's
> how bright things would be if the smallest thing we think can exist was as
> hot as we think things can get.
>
> John K Clark
>
>

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