Thursday, 3 October 2013

Moment of inertia of a photon

In a previous post I raised the question of whether energy can be accelerated. I also mentioned that a number of other interesting results come out of this analysis and idea. In this post I cover another of these. This one relates to inertia and starts out using Planck units. Consider the following

$ I_p = m_p l_p^2  $      .... (1)

Where mp is Planck mass and lp is Planck length. In this discussion Ip is Planck moment of inertia. It can be shown that this gives

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

tp - Planck time.

Equation (2) can also be represented by

$ I_p \omega_p = \hbar  $    ... (3)

Planck energy is defined as

$ E_p = m_p c^2 =$  $ \frac {\hbar} {t_p}   $     ...(4)

using (2) to eliminate Planck time and rearranging this gives

$ E_p I_p = \hbar^2    $     ....(5)

Also, by dividing (4) by (2)

$ \frac {E_p} {I_p} = \frac {1} {t_p^2} = \omega_p^2   $     ...(6)

Planck Momentum is defined by

$ p_p =$  $ \frac {\hbar} {l_p} = \frac {I_p \omega_p} {l_p} $ ....(7)

Angular Momentum is defined as

$ L = r  x  p  $     ....(8)

p - momentum, r -position vector?. Replacing r with Planck Length and p with Planck momentum gives

$ L_p = l_p   p_p  $ ....(9)

giving

$ L_p = l_p I_p $ $\frac {\omega_p} {l_p} =$ $ I_p \omega_p $  ...(10)

but from (3) this gives

$ L_p = \hbar   $     ...(11)

Ok. Let's stop there for a minute. For starters, I am a little uncomfortable with (9), we have replaced a position vector with a scalar length. Not really sure we can do that. Yet the result (11) is just the lowest level of the angular momentum of electrons around the atom given by Bohr Theory.

$ L = mvr = n\hbar  $ ... (12)

This is just a first-order approximation of the hydrogen atom using quantum mechanics. If that is the case then combining (10) and (6) gives

$ E_p = L_p \omega_p    $   ... (13)

can we argue that there is a more general term given by

$ E = L \omega  $  ... (14)

using a similar argument for (3) and (5) becomes

$ I \omega = \hbar $    ...(15)

and

$ E I = \hbar^2   $    ...(16)

respectively.

In the post on accelerating energy it was shown that

$ E c = \hbar g$    ...(17)

substituting this into (16) gives

$ I =$   $ \frac {\hbar c} {g} $    ...(18)

$g$ - acceleration

This can also be represented by

$ I = $  $\frac {\hbar^2} {m c^2}$    ...(19)

What does this mean? Earlier I mentioned that (11) is the angular momentum of an electron in the Bohr model. Could it be more than that? could it also be true for photons? Photons have angular momentum. Fine. Does (15) imply that a photon as a moment of inertia, I? Does (16) imply a relationship between moment of inertia of a photon and energy?

There are a number of assumptions made here. The first when we derive (11) and another when we go from Planck specific terms to general terms in (14), (15) & (16). I am not really comfortable with this and feel as though I have pulled a little bit of "smoke and mirrors".

That said, the result is rather interesting, photons have a moment of inertia, and a rather nice relationship between this and a photons energy. There is also the relationship between inertia and acceleration that is explored more in this post on acceleration.

Will ponder this one some more and see if I can justify my assumptions.

In the next post I'll show some more of the results that come out of this type of analysis.

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