Tuesday, 15 October 2013

Stoney, Planck and Schrödinger

Recently I have been looking at the idea of universal or natural units. The two best known seem to be examples given by Stoney and Planck, but there are others, a good post can once again be found on the fabulous Wikipedia.

While doing this I also came across Schrödinger units. I will list them here along side Stoney and Planck


Quantity Stoney Planck Schrödinger
Length(L) $l_s=\sqrt \frac{Ge^2}{c^4 4\pi \epsilon_0}$ $l_p = \sqrt \frac{\hbar G} {c^3} $ $l_\psi = \sqrt \frac {\hbar^4 G (4\pi \epsilon_0 )^3} {e^6} $
Mass(M) $m_s=\sqrt \frac {e^2} {G 4 \pi \epsilon_0}$ $m_p=\sqrt \frac {\hbar c} {G} $ $m_\psi=\sqrt \frac {e^2} {G 4 \pi \epsilon_0}$
Time(T) $t_s = \sqrt \frac {Ge^2} {c^6 4 \pi \epsilon_0}$ $t_p = \sqrt \frac {\hbar G} {c^5}$ $t_\psi = \sqrt \frac {\hbar^6 G (4 \pi \epsilon_0)^5}{e^{10}}$
Electric Charge (Q) $q_s = e$ $q_p=\sqrt {4 \pi \epsilon_0 \hbar c}$ $q_\psi = e$
Temperature ($\theta$) $T_s = \sqrt \frac {c^4 e^2} {G 4 \pi \epsilon_0 k_B^2}$ $T_p = \sqrt \frac {\hbar c^5}{G k_b^2}$ $T_\psi = \sqrt \frac {e^{10}} {\hbar^4 (4 \pi \epsilon_0)^5 G k_b^2}$

First things first, I have not been able to confirm the Schrödinger units. The various sites out there look like they are quoting a single source. The danger of course is that if that source is wrong, then so is everyone else!

Let's proceed as if all is well and the values given above are correct. We can derive a number of results.

Length 

It can be shown that

$l_p^4 = l_s^3 l_\psi$     ...(1)

$\frac {l_s} {l_\psi}$ $ = \alpha^2$    ...(2)

$\frac {l_s^2} {l_p^2}$ $ =\alpha$    ...(3)

Mass

It can be shown that

$\frac {m_s^2} {m_p^2}$ $ =\alpha$  ...(4)

and

$m_s = m_\psi$    ...(5)

I think (5) introduces are a rather interesting problem, consider the following Planck result

$\frac {l_p m_p} {q_p^2}$  $= 10 ^{-7}  $   ...(6)

the Stoney version is

$\frac {l_s m_s} {q_s^2}$ $  = 10 ^{-7}  $   ...(7)

for an electron it is 

$\frac {r_e m_e} {e^2}$ $  = 10 ^{-7}  $   ...(8)

$r_e$ - classic electron radius, $m_e$ - mass of electron , $e$ - charge on an electron

where as for Schrödinger it is not the case 

$\frac {l_\psi m_\psi} {q_\psi^2}$ $  \not= 10 ^{-7}  $   ...(9)

The obvious place to look is at the Schrödinger mass term, use an alternative version where (9) now works

$\frac {l_\psi m_\psi^{'}} {q_\psi^2}$  $= 10 ^{-7}  $     ...(10)

Dividing (10) by (7) and using $q_s = q_\psi = e$

$ \frac {l_\psi m_\psi^{'}} {l_s m_s}$  $= 1$    ...(11)

using (2) and rearranging this gives

$ m_\psi{'} = m_s \alpha^2 = m_\psi \alpha^2 $   ...(12)

This would give the Schrödinger mass a value of 
$m_\psi{'} = 9.9   10^{-14}  kg$ 

which I think is a more appropriate value. I call this the Alternative Schrödinger mass.

Going forward and using (4) gives

$m_s^5 = m_p^4 m_\psi^{'}$    ...(13)

this can be substituted back into (10) along with the result obtained in  (1) gives

$\frac {l_\psi m_\psi^{'}} {e^2} = \frac {(\frac {m_s^5 l_p^4}{m_p^4 l_s^3})}{e^2} = \frac {l_s m_s} {e^2}$   ...(14)

taking the second half of the equation this just becomes

$\frac { m_s^5 l_p^4} {e^2 m_p^4 l_s^3} = \frac {m_s l_s}{e^2}$   ...(15)

which simplifies to

$m_s l_p = m_p l_s $    ...(16)

Using the Schrödinger mass derived in (12) we have 

$m_\psi{'} = $ $\sqrt \frac {e^{10}} {G (4 \pi \epsilon_0)^5 \hbar^4 c^4} $ $= m_p \alpha^{5/2}$   ...(17) 

This is probably the best place to wrap up this post, but before I do, I did notice something a little odd though when I was doing this, but I will cover it in another post.

Wrote this while listening to this.

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