When deriving the Maxwell-Garnett equation for composite systems (host material with dielectric particles disposed in it), the steps generally taken are
Step 1) Equate the Clausius-Mossotti equation and equal it to the polarisability of a dielectric sphere:
$$ \frac{\epsilon_{eff} -1}{\epsilon_{eff} +2} = p_i \frac{\epsilon_{i} -1}{\epsilon_{i} +2} \tag{E.1}$$
where $\epsilon_{eff}$ is the effective permittivity of the medium/mixture, $\epsilon_i$ is the permittivity of the particles, and $p_i$ is the volumetric percentage of these particles in the mixture.
Step 2) Multiply it all by the vacuum permittivity
$$ \frac{\epsilon_0\epsilon_{eff} - \epsilon_0}{\epsilon_0\epsilon_{eff} +2\epsilon_0} = p_i \frac{\epsilon_{i}\epsilon_0 -\epsilon_0}{\epsilon_0\epsilon_{i} +2\epsilon_0} \tag{E.2}$$
Soon after, the sources I found state something like "As the model of Maxwell Garnett is a composition of a matrix medium with inclusions we enhance the equation" (this quote is taken from wikipedia, but similar things can be found in D. Schurig (2017) and in T. Don et al. (2021). To do this, a background permittivity seems to be multiplied to the last element in each numerator and denominator of E.2, and the vacuum permittivity is subsequently removed, resulting in the final equation.
Final equation)
$$ \frac{\epsilon_{eff} -\epsilon_{m}}{\epsilon_{eff} +2\epsilon_m} = p_i \frac{\epsilon_{i} -\epsilon_m}{\epsilon_{i} +2\epsilon_m} \tag{E.3}$$
What is the maths behind this last step? Nowhere seems to explain it.
Edit:
I know that a way to get the LHD of equation (E.3) is by equaling the displacement field, $D$ as
$$ D = \epsilon_0 E + P \tag{E.4}$$
where $P$ is the moment, and it can be written as
$$D = \epsilon_{eff} E \tag{E.5}$$
and that when in a composite material, E.4 becomes
$$D = \epsilon_m E + P \tag{E.6}$$
by substituting the vacuum permittivity with the permittivity of the host material. By playing with this equation we can define $P$, as
$$P = (\epsilon_{eff} - \epsilon_m) E\tag{E.7}$$
and based on the definition of a localised electric field,
$$E_{loc} = E + \frac{P}{3\epsilon_0} \tag{E.8}$$
and $$P = \alpha \frac{E_{loc}}{V} \tag{E.9}$$
where $V$ is the volume of the field and $\alpha$ is the polarisability of the media.
From this, I can get the following equation for the dielectric sphere,
$$ \frac{\epsilon_{eff} -\epsilon_m}{\epsilon_{eff} +2\epsilon_m} = \frac{\alpha}{3 \epsilon_0 V}\tag{E.10}$$
the LHS of equation $E.3$ is the same as the LHS of equation $E.10$, but how do I get the RHS of equation $E.3$ ?
For me to get the RHS of $E.3$, I believe that I would need equation $E.11$ to be true.
$$P = (\epsilon_i - \epsilon_m)E$$
How can this be so?
Edit 2 (sorry for the excess): I found another way to interpret this, but this other p.o.v. still does not clear everything.
Normally, the polarisation for a macroscopic field in the vacuum would be
$$\tag{E.11} P = \epsilon_0 (\epsilon_{eff} - 1)E$$
but in the case of particles embedded in a host medium, $\epsilon_0$ is substituted by the $\epsilon_m$ of the material in question. Now, using equations $E.8$ and $E.9$ (and substituting $\epsilon_0$ for $\epsilon_h$ in $E.8$) we can manipulate this to result in
$$\tag{E.12} \frac{\epsilon_{eff} -1}{\epsilon_{eff} +2} = \frac{N\alpha}{3\epsilon_m V}$$
This can also be done if the local electric field equation also substitutes the vacuum permittivity, $\epsilon_0$ for the $\epsilon_m$ permittivity of the host material in question. The LHS of this equation is what I was looking for, but then two questions arise:
How can I describe the total permittivity of the system relative to the permittivity of each of its elements? Before it was that $\epsilon = \epsilon_0 \chi_e + \epsilon_0\epsilon_m$ (link).
I still do not know what leads to the RHS of $E.3$.
