Gravitational Constant
Key reference is the monograph by Barry
Setterfield: Cosmology
and the Zero Point Energy
(Natural Philosophy Alliance Monograph series, No.1, 2013, ISBN 978-1-304-19508-1).
On page 147 in Setterfield's book we read about The Newtonian Gravitational Constant,
G. Let's repeat some of the derivation given there. Formula (45) comes from a paper by
Haisch, Rueda and Puthoff (HRP for short), with a factor $(2/3)$ replaced by $(4/3)$:
$$
m = (4/3)(\alpha)(m_P^2/m^*) \qquad (44) \quad \Longrightarrow \quad mm^* = 4/3\,\alpha\,m_P^2
$$
Here $m$ is "our atomic mass" (HRP's atomic particle's inertial mass $m_i$) , $m^*$ is "our
intrinsic mass" (HRP's bare mass $m_0$) and $\alpha$ is the fine structure constant given by
$$
\alpha = (e^2/\epsilon)(1/2hc) \qquad (46)
$$
The quantity $m_P$ is the Planck mass given by
$$
m_P = h\omega_P/(2\pi c^2) \qquad (47) \quad \Longrightarrow \quad \omega_P = 2\pi c^2 m_P/h
$$
Where $\omega_P$ is the Planck frequency. A dimensionless conversion factor $k$ is introduced,
which is the quotient of the Planck frequency and the Compton frequency. Let's check it out:
$$
\frac{\omega_P}{\omega_e} = \frac{2\pi c^2 m_P/h}{2\pi c^2 m_e/h} = \frac{m_P}{m_e} = k
$$
Alas, with the Wikipedia values for the Planck mass $\,m_P \approx 2.17651\times 10^{-8} kg\,$ and the
electron mass
$\,m_e \approx 9.11\times 10^{-31} kg\,$ , we have not been able to reproduce the
(first) value as written in the book:
$$
k = 3.334 \times 10^{21} \qquad (48) \qquad \ne 2.389\times 10^{22}
$$
With equation (49), it is proposed to replace $\omega_P$ by $\omega_P/k$ , "in order to get
the correct results". With (44) and $k=m_P/m_e$ we have:
$$
mm^* = 4/3\,\alpha\,(m_P/k)^2=4/3\,\alpha\,m_e^2
$$
The gravitational constant is then calculated in the book with the following formula:
$$
G = (2/3\pi)(e^2/\epsilon)(1/k^2)\left[1/(mm^*)\right]
$$
Instead of plugging in numbers, as Setterfield does, we shall leave the algebra intact.
Note that $k=m_P/m_e$ and use the last formula for $mm^*$:
$$
G = (2/3\pi)\left[(e^2/\epsilon)(1/2hc)\right](2hc)\left[1/(m_P/m_e)^2\right]
\left[1/(4/3\,\alpha\,m_e^2)\right]
$$
According to Wikipedia,
the Planck mass is defined by:
$$
m_P = \sqrt{\frac{\hbar c}{G}} \quad \Longrightarrow \quad m_P^2 = \frac{h c}{2\pi G}
$$
So there we go; also use (46) for the fine structure constant.
$$
G = \frac{2\;\alpha\;2hc\;m_e^2\;3}{3\pi\;m_P^2\;4\;\alpha\,m_e^2} = \frac{hc}{\pi\,m_P^2}
= \frac{hc\;2\pi\,G}{\pi\,h c} = 2 G
$$
Upon dividing by $G$ we get:
$$
1 = 2
$$
How is this possible? Did I make a mistake somewhere?
Let's give Setterfield the advantage of the doubt. Suppose that Barry's hand-held calculator
has been different from mine. Or maybe there is just another "insignificant" error in his math. Or I have been goofing myself indeed. Whatever. To his defense,
suppose that finally there is the result we're all hoping for:
$$
G = G
$$
Upon subtracting $G$ on both sides we get a famous beginners "result" in mathematics:
$$
0 = 0
$$
This means that:
$$
G = 6.670 \times 10^{-11} \qquad (60) \quad \mbox{on page 148}
$$
Proof: Add $\,G\,$ to the left side and $\,6.670 \times 10^{-11}\,$ to the right
side of $\,0 = 0$ .
Then Setterfield writes on top of page 149:
This value is in accord with the current experimental evidence.
Yes, if course it is: you have put it in, hence you get it out : OUTPUT = INPUT .
It should be noticed that Haisch, Rueda & Puthoff,
where Setterfield has this theory from (?), wisely do NOT calculate
the gravitational constant.
Therefore, according to my not so humble opinion, the pages 147 and 148
simply can be safely cut out from the book, without missing anything.
But, wait a moment! One further clue is found in Chapter 7: The ZPE and Relativity.
There on page 273 we find the following formulas:
$$
m_P = \frac{h\omega_P}{2\pi c^2} \qquad (35) \\
\omega_P = \left[\frac{4\pi c^5}{h G}\right]^{1/2} \qquad (38)
$$
Consequently:
$$
\omega_P = \frac{2\pi c^2}{h} \sqrt{\frac{h c}{\pi G}} = \frac{2\pi c^2}{h} m_P
\\ \mbox{with} \quad m_P = \sqrt{\frac{h c}{\pi G}} \quad \mbox{instead of} \quad
m_P = \sqrt{\frac{h c}{2\pi G}}
$$
Great relief! This at least resolves the issue with $1=2$ instead of $1=1$.
So Barry is right in his claim that $G=G$.
In retrospect, Barry Setterfield is not the person to be blamed for this nonsense. In Gravity as a zero-point fluctuation force by Harold Puthoff we have, among the simpler
formulas in this article:
$$
G = \frac{\pi}{2}\frac{c^5}{\hbar\int_0^{\omega_c}\omega\,d\omega}
\; ,\quad \omega_c = \left[\frac{\pi c^5}{\hbar G}\right]^{1/2}\qquad (2)
$$
When working out the integral we see, indeed, that $G=G$:
$$
G = \frac{\pi}{2}\frac{c^5}{\hbar\left[\pi c^5/(\hbar G)\right]/2} = G
$$
Brilliant! But not really. Instead of this vicious circle result, I'll give an example of a formula
that does say something. It's the fourth power thermal radiation law and the expression for
the radiation density
constant $\,a\,$ was derived by Max Planck in his book
$$
u = a T^4 = \frac{8\pi^5}{15}\frac{(kT)^4}{(hc)^3} \quad \Longrightarrow \quad
a = \frac{8\pi^5}{15}\frac{k^4}{(hc)^3}
$$
With $k=$ Boltzmann constant,
$c=$ speed of light, $h=$ Planck constant. Somebody has noticed the difference with $G=G$ ?
I sincerely hope so.