Kepler, Acceleration and some numerology

NOTE: since posting this I have found the explanation for the results shown here. You can find it in this post.

This is a very short post just to highlight a little oddity I noticed in my blog on acceleration. The title of the post was the conservation of acceleration, click on the link for details. This is just a bit of fun, but it is also very odd indeed.

In the post I derive a number of values including the time it takes to create/destroy a photon, $\delta t$. In the example given the photon has just enough energy to create an electron/positron pair.

$\delta t = 1 / \omega = 6.4404433   10^{-22} s  $
$ \omega = 2 \pi \nu $
$ \nu$ - frequency of photon

a further calculation is performed where a volume is derived and found to be

$ V = l_p^2 d =  5.0434295   10^{-83} $    ...(1)
$ d= \lambda / 2 \pi $
$ \lambda$ - wavelength of the photon
$l_p$ - Planck length

the last was someting relating to moment of inertia of a photon given by

$I = \frac {\hbar c} {g} $

I - moment of inertia - $6.7919   10^{-56}   kg  m^{-2}$
$\hbar$ - reduced Plancks Constant
c - speed of light in a vacuum
$g$ - photon creation acceleration - $4.654845  10^{29} ms^{-2}$

Now the mass of electron and gravitational constant are given by

$m_e - 9.1093829  10^{-31}  kg$
$G - 6.67384   10^{-11}     m^3 kg^{-1} s^{-2} $

Kepler's 3rd law of planetary motion is

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

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

rearrange this to give

$ R^3 4\pi^2 = G M T^2 $    ...(3)

now set
T = $2\pi \delta t$   - time period for frequency $\nu$
M = $2m_e$    - mass of e/p pair

$ R^3 = 2 G m_e \delta t^2$
$ R^3  = 5.043433   10^{-83}  $

but this is just the value of V from earlier, in other words

$R^3  = V $  ... (4)

$R = 3.694668    10^{-28}m$

taking this value of R and putting it into

$I_m = \mu R^2$  ...(4)

where $\mu$ is some reduced mass, we get a moment of inertia of

$I_m = 1.36506   10^{-55} \mu  $

dividing by the value of $I$ mentioned earlier gives

$I_m/I = 2.00983 \mu$    ...(5)

if we set (5) = 1 then

$2.00983 \mu = 1$

$\mu = 0.4975546 = \frac {m_0 m_1} {m_0+m_1} $ ...(6)

$m_1 \approx \frac {m_0} {2.00983m_0 -1}  $ ....(7)

So let's get this straight, we carried out a calculation involving the creation of an electron positron pair. As part of this calculation we determined a volume related to the Planck constant and the wavelength of the photon, equation (1).

We then found those values actually work in Kepler's 3rd Law! How cool is that?

In addition, an earlier idea about moment of inertia of a photon has comparable values to that obtained using a traditional moment of inertia calculation, though I have to say that this is a very weak idea, need to take more of a look at it.

Seeing as this is just a bit of fun and an odd coincidence I am going to abandon my usual caution and make some guesses as to what I think is going on.

NOTE: These ideas are just guesses, this is NOT real physics, this is a bit of fun! After all, we have not ven considered the Coulomb force!

What is the meaning of R in (4) ? In Kepler's 3rd law it is the distance between the center of mass of the two bodies. Is this the distance between the e/p pair? I don't think so.

I think that e/p pair represent the larger body, M in equation (3). So R could be the distance between the e/p pair and the center of mass of the photon that creates the e/p pair. Will ponder this more and then update this post.