Kepler part 2

In a previous post on Kepler I came across an odd link between a calculation I had been performing on photon - electron/positron production and Keplers 3rd Law,

$ \frac {4 \pi^2} {T^2} = \frac {G M} {R^3} $    ...(1)

G- gravitational constant
M - mass of the larger body
R - distance between the center of mass of the two bodies
T - period

So I thought I would take a look at this in more detail and lo and behold a nice solution drops out, Take a look at this, set

T = $1 / \nu$
$\nu$ - frequency of the photon
M - $2m_e$ - mass of 2 electrons
$R^3 = V = l_p^2 d$
$l_p$ - Planck length
$d =  \lambda / 2\pi$
$\lambda$ - wavelength of photon

So we now have

$ 4\pi^2 \nu^2 =$ $ \frac {G 2 m_e} {l_p^2 d} $   ...(2)

sub in the value for d and rearrange so

$ \frac {4\pi^2 \nu^2 \lambda} {2\pi} $  $ l_p^2 = G 2 m_e   $   ...(3)

using

$c = \nu \lambda$    ...(4)

c- speed of light

$ l_p^2 = $  $\frac {\hbar G} {c^3} $   ...(5)

$\hbar = h / 2\pi $ - reduced Planck constant

subbing these into (3) give

$ \frac { 2\pi \nu c \hbar G} {c^3} =$  $ G 2 m_e $    ...(6)

dividing through by G and multiplying each side by $c^2$ gives

$2\pi \nu \hbar = 2 m_e c^2 $    ...(7)

or as it is more usually written

$ E = h \nu = 2 m_e c^2 $     ...(8)

I think that is lovely. It is the use of

$R^3 = V = l_p^2 d$    ...(9)

that generates the solution.

So, how do we arrive at (9)? Consider this, Newton's equation for gravitational force is given by

$F_{grav} = G M_1 m_2 / R^2$   ...(10)

Then we have, energy is force x distance, so say we try this

$E = mc^2 = F_{net} d$    ...(11)

Then lets say that

$F_{grav} = F_{net}$    ...(12)

(does this imply $F_{net}$ is a centripetal force?) this becomes

$\frac {GM_1 m_2} {R^2} = \frac {mc^2} {d}$   ...(13)

lets say

$M_1 = 2m_e$
$R = l_p$
$m_2 = m$
$c = 2\pi d \nu$

so

$F_{grav} =$  $ \frac {G2m_e m} {l_p^2}$     ...(14)

cancelling the m on both sides (13) becomes

$\frac {G2m_e} {l_p^2} = \frac {4 \pi^2 \nu^2 d^2} {d}$     ....(15)

cancelling the d on the right and rearranging gives

$\frac {G 2 m_e} {l_p^2 d} $ $=4\pi^2 \nu^2 $     ....(16)

which is just equation (2) above. So what does that tell us? Let's take a look at (14)

$M_1 = 2m_e$

This value is because we have an electron positron pair, both with a mass of $m_e$. The distance, $l_p$, is the Planck length, given in (5) above.

What is this saying? at the moment of creation of the e/p pair there is a mixed state photon in orbit around the electron/positron pair. There is a force associated with this photon that is equivalent to a gravitational force given by (14) ?

There are a few things that concern me here. If we sub (5) into (14) then the G divides out. Is it still a gravitational force even though G is no longer in the equation? ( I think there may be an argument that can be used here to say that it is still a Gravitational force, but I will cover it in another post.) for now...

$F_{grav} =$  $ \frac {2m_e m c^3} {\hbar}$     ...(17)

which becomes

$F_{grav} =$  $ \frac {E_{ep} m c} {\hbar}$      ...(18)

$E_{ep} = 2 m_e c^2$   - creation energy of an electron/positron pair
also, Compton gave us

$  m \frac {\lambda} {2\pi} = \frac {\hbar}{c} $   ...(19)

subbing this into (17) and using the value of d from earlier

$d = 2 \pi \lambda$

gives

$F_{grav} =$  $ \frac {E_{ep}} {d} $ ...(20)

which is the same form as  (11).

Another question is whether the inverse square law survives at these distances. There is some evidence that it may not be true for sub-millimeter distances and yet here it is at the Planck length! I'm sure Erik Verlinde will be shaking his head at this point. On the plus side, even though it is an incredibly short distance, the masses involved are considerably smaller than the Planck mass and so should obey Newtonian gravity rather than having to worry about general relativity effects. Wouldn't it be magnificent if Newtonian gravity extended all the way down to the Planck length!

What is the meaning of m in (14)? It cancels out so makes life easy, but what is it?

In the Kepler version of (2), $l_p = R$, $d = R$, giving $R^3$. In the version above

$d = \lambda / 2\pi$ - the wavelength of the photon, the equations and the numbers work with this value rather than $l_p$. While this sort of makes sense I am still a little worried.

So while we may have got to the bottom of why Kepler's third law seems to hold for sub atomic particles, there are still some outstanding questions to be resolved.

There is no time dependency discussed here . Once the e/p pair are created they go off in there separate directions and the photon is no more. The analysis described here can only be valid for a certain period of time. What is this? Is it related to the idea of the photon creation/destruction time mentioned in previous posts? This happens to be

$ \delta t = 1 / 2 \pi \nu  = 6.44044   10^{-23} s$
$\nu$ - frequency of the initial photon

Lastly,  I am arguing that Newton's gravitational law holds all the way down to Planck's length for an electron/positron pair. These are charged particles!

What about the Coulomb force?

If Coulomb's law holds, at the distances we are talking about it should be colossal and because electrons and positrons are oppositely charged it should be attractive. Definitely need to explain this away. Could it be that in the initial phase of electron/positron creation the charge has not separated enough to be +/-, or does the weak interaction take care of the Coulomb force, who knows? not me!

Wrote this while listening to this.