index $ \def \MET {\quad \mbox{with} \quad} \def \EN {\quad \mbox{and} \quad} \def \SP {\quad \mbox{;} \quad} \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} \def \half {\frac{1}{2}} $

The **Compton wavelength**
of a particle is equal to the wavelength of a photon whose energy $E$ is the same
as the rest mass energy of that particle. The standard Compton wavelength, $\lambda$, of a particle here and now
is given by ($f$ = frequency):
$$
E_0 = hf_0 = h\frac{c}{\lambda_0} = m_0c^2 \hieruit \lambda_0 = \frac{h}{m_0\,c}
$$
with meaning of the constants as explained before. The reduced Compton wavelength is often preferred, though:
$$
\lambda_0 = \frac{\hbar}{m_0\,c} \MET \hbar = \frac{h}{2\pi}
$$
Suppose again that the elementary particle
(rest) mass shall be different / smaller in its own past / greater in its own future, say with a value $m$.
Then we have for the Compton wavelength $\lambda$ at that time:
$$
\lambda = \frac{\hbar}{m\,c} = \frac{m_0}{m} \frac{\hbar}{m_0\,c} \\ \Longrightarrow
\quad \lambda = \frac{m_0}{m} \lambda_0
$$
Conclusion: *the Compton wavelength ("size") of an elementary particle is inversely proportional with its
(varying rest) mass.*

Closely related to this is the De Broglie wavelength, which is also inversely proportional to elementary particle (rest) mass:
$$
\lambda = \frac{h}{mv}
$$
And the photon wavelength, where the mass $\,m\,$ is not a rest mass:
$$
mc^2 = h\nu = \frac{hc}{\lambda} \hieruit \lambda = \frac{h}{mc}
$$
The **Bohr radius** is $1/\alpha \approx 137$ times
the Compton wavelength (of an electron). According to the previous section it is
a measure for the size of a hydrogen atom. We have seen an alike result for the helium atom as well.
Let $\,a_0\,$,$\,m_0\,$ be the values of the Bohr radius and (electron rest) mass here and now. And let $\,a\,$,$\,m\,$
be their values at some (proper) time in the past or future. At least two arguments have come along now to ensure that:
$$
\boxed{\large a = \frac{m_0}{m} a_0}
$$
Conclusion: *the (Bohr) radius of a (hydrogen/helium) atom is inversely proportional with varying elementary
particle rest mass (of the electron).*

The **Electromagnetic Mass** of an electron
may be calculated as follows. The energy density in an electric field is given in general by
$\, w = \frac{1}{2} \epsilon_0 E^2 \,$ where the field strength $\, E \,$ at a distance $\, r\, $ is:
$$
E = \frac{q}{4 \pi \epsilon_0 r^2}
$$
with $\,q\,$ the electron charge and $\,\epsilon_0\,$ the dielectric constant of the vacuum. The total energy in the field is thus given by:
$$
U = \int_0^\infty\frac{1}{2}\epsilon_0 \left(\frac{q}{4\pi\epsilon_0 r^2} \right)^2 4 \pi r^2 dr \ = \\
\frac{q^2}{8 \pi \epsilon_0} \int_0^\infty \frac{dr}{r^2} = \infty
$$
Thus, in principle, there is an infinite outcome for the rest energy of the electron. A simple minded remedy is the following.
Replace the lower boundary $0$ by a finite radius $\,\sigma\,$. Also identify the electron's rest energy with the energy of
its rest mass. This results in:
$$
U = m c^2 = \frac{q^2}{8 \pi \epsilon_0} \int_\sigma^\infty \frac{dr}{r^2} = \frac{q^2}{8 \pi \epsilon_0} \frac{1}{\sigma} \hieruit
\sigma = \half \frac{1}{4 \pi \epsilon_0} \frac{q^2}{m c^2} = \half a
$$
Where $\,a\,$ is the so-called classical electron radius.
A similar calculation is in the Wikipedia page,
with $\,\sigma = 3/5\,a\,$ as a final outcome. Other calculations are feasible, such as in an article at the Mathematics forum named
Cauchy distribution instead of
Coulomb law? The following would have helped in understanding:
$$
\int_0^\infty \frac{r^2 dr}{(r^2+\sigma^2)^2} = - \frac{1}{2} \int_0^\infty r\;d\left(\frac{1}{r^2+\sigma^2}\right) =
\left[-\frac{r}{2(r^2+\sigma^2)}\right]_{r=0}^\infty + \frac{1}{2\sigma}\int_0^\infty \frac{d(r/\sigma)}{1+(r/\sigma)^2} =
\frac{1}{2\sigma}\left[\,\arctan(r/\sigma)\,\right]_{r=0}^\infty = \frac{\pi}{4\sigma}
$$
Because now it's easy to see that:
$$
U = \frac{q^2}{8 \pi \epsilon_0} \int_0^\infty \frac{r^2 dr}{(r^2+\sigma^2)^2} = \frac{q^2}{4 \pi \epsilon_0}\frac{\pi}{8\sigma} = mc^2
\hieruit \sigma = \pi/8 \, a
$$
Last but not least, we have calculated the electron self-energy with help of a convolution, as has been demonstrated in yet another article
at my website, called Electromagnetic Mass. And the outcome is:
$\,\sigma = 3\,\pi^2/20 \, a\,$. In all cases, since the classical *electron radius is inversely proportional to the electron rest mass*,
the Length Contraction law holds, by whatever means the "true" radius $\,\sigma\,$ shall be calculated:
$$
a = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{m c^2} \SP \sigma \sim a \hieruit \boxed{ \large \sigma = \frac{m_0}{m} \sigma_0}
$$
Now it's not such a great step to generalize the above results accordingly:

*Wavelengths of electromagnetic radiation, sizes of elementary particles, atoms and molecules are inversely proportional
with varying elementary particle (rest) mass.*

This result is still more general. A rod or a rope is built from molecules and molecules are built
from atoms. These atoms have sort of a Bohr radius. If the Bohr radius changes, then it may be expected
that the length of the rod or the rope is changing as well, inversely proportional with the (varying rest)
mass of the particles it is made from. Thus all lengths $L$ of things built from molecules or atoms close
enough together - solid or fluid material - obey the law of **length contraction**:
$$
\large \boxed{L = \frac{m_0}{m} L_0}
$$
It is clear that if length contraction applies to lengths measured with a measuring rod
and if the measuring rod itself has become longer as well, then it's impossible to tell
the difference between two measurements. Much like in Special Relativity, there is no a
priory reason why *this* sort of length contraction would be inconsistent with observation.

Last but not least:
*it's reasonable to assume that distances in empty space do NOT show any sort of length contraction*.

This has a remarkable consequence. Any length that is bound to matter, whether that is a (measuring) rod or
a light ray cast back and forth to the moon, obeys the following law. The length of a length-measuring-device
is *inversely* proportional to varying rest mass: $\,L/L_0 = m_0/m\,$. However,
there do not exist other measuring devices than those bound to matter in such a way. But then we must necessarily
conclude that any length that is *measured by* a rod, or by a light ray cast back and forth is * observed
as being directly proportional* to varying rest mass, provided that the length that is measured

At several places in the book, it is claimed tat the **observed clumpiness in the universe**
can be described mathematically by a two-point AutoCorrelation function. But I just can't follow what Peebles
has been thinking about. To me, it looks like a sign of clum__s__iness rather than a description of clum__p__iness.
An impression that might be overruled by reading
*Clusters and Superclusters of Galaxies*
and doing some serious homework.
I think that a techique like the Delaunay tessellation field estimator (DTFE) is more effective. So *What is the
proper way to do DTFE*? Once a wrong turn has been made, there is no easy rollback of a large intellectual investment.

In Léon Brillouin's book
Relativity Reexamined we read on page 67 about **A New Approach to Special Relativity**.
That so-called new approach seems to be promising at first sight, but it uses the Doppler effect, which has been derived earlier
with relativistic mechanics, suggesting that there is a circular reasoning involved in the whole. Whatever. The "new approach"
starts with the mass-energy relation, and variable mass is in there from the very beginning:
$$
E = mc^2 \EN m = \frac{m_0}{\sqrt{1-v^2/c^2}}
$$
So we have, according to the findings above:
$$
\frac{L}{L_0} = \frac{m_0}{m} \hieruit L = L_0 \sqrt{1-v^2/c^2}
$$
Which is indeed Length Contraction according to Special Relativity.