Magnetization v/s Polarization: Why are they weirdly dissimilar?

Question

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1. What the student is asking (in plain language)

The student sees two pairs of macroscopic fields

Electric: ( \mathbf E,\;\mathbf D)
Magnetic: ( \mathbf B,\;\mathbf H)

and the corresponding material responses

Polarization: (\mathbf P) and Magnetization: (\mathbf M).

The textbook formulas read

[ \mathbf P =\varepsilon_0\chi_e\mathbf E ,\qquad \mathbf M = \chi_m\mathbf H , ]

[ \mathbf D =\varepsilon_0\mathbf E+\mathbf P ,\qquad \mathbf H =\frac{\mathbf B}{\mu_0}-\mathbf M . ]

The student wonders why

( \mathbf P) is written as proportional to the applied electric field (\mathbf E) while (\mathbf M) is written as proportional to the auxiliary field (\mathbf H);
we do not write (\mathbf M =\chi_m\mathbf B) (or (\mathbf P = \chi_e \mathbf D));
the “field produced by the magnetization’’ appears with a factor (1/\mu_0) instead of (1/(\mu_0\chi_m));
the total magnetic induction is written as (\mathbf B =\mu_0\mathbf H) (or (\mu\mathbf H)) rather than (\mathbf B = \mu \mathbf B_{\text{applied}}).

In short: Why do the electric and magnetic constitutive relations look asymmetrical?

The answer lies in how we define the four macroscopic fields and in the separation of free versus bound sources in Maxwell’s equations.

2. Microscopic vs. macroscopic fields

2.1 Microscopic Maxwell equations

At the microscopic level the fields (\mathbf e(\mathbf r,t)) and (\mathbf b(\mathbf r,t)) obey

[ \begin{aligned} \nabla!\cdot!\mathbf e &= \frac{\rho_{\text{tot}}}{\varepsilon_0},\[2pt] \nabla!\times!\mathbf b &= \mu_0\mathbf j_{\text{tot}}+\mu_0\varepsilon_0\frac{\partial\mathbf e}{\partial t}, \end{aligned} ]

where (\rho_{\text{tot}}) and (\mathbf j_{\text{tot}}) contain both the free charges/currents we control and the bound charges/currents bound inside atoms and molecules.

2.2 Averaging → macroscopic fields

We smooth (average) the microscopic fields over a volume large compared with atomic dimensions but small compared with the wavelength of interest. The result are the macroscopic fields (\mathbf E,\,\mathbf B). They obey the same Maxwell equations if we also introduce effective source densities

[ \rho_{\text{free}},\quad \mathbf j_{\text{free}} ]

and define the bound sources through the material response:

Bound electric charge density
[ \rho_{\text{bound}} = -\nabla!\cdot!\mathbf P . ]
Bound current density (including the “magnetization current’’)
[ \mathbf j_{\text{bound}} = \nabla\times\mathbf M +\frac{\partial\mathbf P}{\partial t}. ]

With these definitions the macroscopic Maxwell equations become

[ \boxed{ \begin{aligned} \nabla!\cdot!\mathbf D &= \rho_{\text{free}}, & \mathbf D &\equiv \varepsilon_0\mathbf E+\mathbf P,\[4pt] \nabla!\cdot!\mathbf B &= 0,\[4pt] \nabla\times\mathbf E &= -\frac{\partial\mathbf B}{\partial t},\[4pt] \nabla\times\mathbf H &= \mathbf j_{\text{free}}+\frac{\partial\mathbf D}{\partial t}, & \mathbf H &\equiv \frac{\mathbf B}{\mu_0}-\mathbf M . \end{aligned}} ]

Key point:

(\mathbf D) and (\mathbf H) are auxiliary fields that contain only the free sources.
(\mathbf E) and (\mathbf B) contain both free and bound contributions.

Because of this definition the equations look “asymmetric’’ when we write the material constitutive relations.

3. Linear, isotropic, homogeneous media

For many common materials the response is linear and direction‑independent:

[ \mathbf P = \varepsilon_0 \chi_e \mathbf E ,\qquad \mathbf M = \chi_m \mathbf H . ]

These are definitions of the electric susceptibility (\chi_e) and magnetic susceptibility (\chi_m).
Why does (\mathbf M) involve (\mathbf H) and not (\mathbf B)?

By definition (\chi_m) is the proportionality constant between the magnetization and the field that would exist if the bound currents were absent, i.e. the free‑current field (\mathbf H).
If we tried to write (\mathbf M = \tilde\chi_m\mathbf B), the constant (\tilde\chi_m) would have different dimensions (because (\mathbf B) already contains (\mu_0)). The conventional choice keeps (\mathbf M) and (\mathbf H) having the same units (A m⁻¹), which makes (\chi_m) dimensionless, just like (\chi_e).

Similarly for the electric case we could write (\mathbf P = \varepsilon_0\tilde\chi_e \mathbf D) but that would just introduce an extra factor of (\varepsilon_0) and hide the physical meaning of (\chi_e).

4. Deriving the familiar “(\mathbf B = \mu \mathbf H)” relation

Starting from the definition of (\mathbf H),

[ \mathbf H = \frac{\mathbf B}{\mu_0} - \mathbf M, ]

and substituting the linear relation (\mathbf M = \chi_m\mathbf H),

[ \mathbf H = \frac{\mathbf B}{\mu_0} - \chi_m\mathbf H \;\;\Longrightarrow\;\; \mathbf B = \mu_0(1+\chi_m)\mathbf H \equiv \mu\,\mathbf H . ]

Thus the total magnetic induction (\mathbf B) is the sum of two parts:

Free‑current part: (\mu_0\mathbf H) (the field that would exist without bound currents);
Bound‑current part: (\mu_0\mathbf M = \mu_0\chi_m\mathbf H).

So the field produced by the magnetization is (\mu_0\mathbf M), not (\mathbf M/\mu_0). The textbook formula

[ \boxed{\mathbf B = \mu_0(\mathbf H + \mathbf M)} ]

is simply the rearranged version of the definition of (\mathbf H).

5. Parallel electric formulas

The electric side works analogously:

[ \mathbf D = \varepsilon_0\mathbf E + \mathbf P, \qquad \mathbf P = \varepsilon_0\chi_e\mathbf E, ]

which gives

[ \mathbf D = \varepsilon_0(1+\chi_e)\mathbf E \equiv \varepsilon\mathbf E . ]

Here the applied field is (\mathbf E) (the field that would exist without bound charge), while (\mathbf D) is the auxiliary field that responds only to free charge. The symmetry is thus present; the only difference is the conventional naming of the “applied’’ field:

Quantity	Applied (free‑source) field	Auxiliary field containing only free sources
Electric	(\displaystyle \mathbf E)	(\displaystyle \mathbf D)
Magnetic	(\displaystyle \mathbf H)	(\displaystyle \mathbf B)

Because (\mathbf B) already includes (\mu_0), the magnetic auxiliary field is (\mathbf H); because (\mathbf E) already includes (\varepsilon_0), the electric auxiliary field is (\mathbf D). This explains the apparent asymmetry.

6. Summary of the answers to the explicit questions

Question	Answer
Why is (\mathbf M) proportional to (\mathbf H) and not (\mathbf B)?	(\mathbf H) is defined as the magnetic field produced only by free currents. The susceptibility (\chi_m) is defined with respect to that free‑current field, giving a dimensionless constant and keeping (\mathbf M) and (\mathbf H) in the same units. Writing (\mathbf M = \chi_m \mathbf B) would mix free and bound contributions and would require a different (non‑dimensionless) constant.
Why not write (\mathbf P = \chi_e \mathbf D)?	It would be equivalent but would hide the physical meaning of (\chi_e). The conventional definition (\mathbf P = \varepsilon_0\chi_e\mathbf E) makes (\chi_e) dimensionless and directly ties the polarization to the electric field that would exist without bound charge.
Why does the “field produced by magnetization’’ have the factor ( \mu_0) (or (1/\mu_0) in the definition of (\mathbf H))?	From (\mathbf H = \mathbf B/\mu_0 - \mathbf M) we get (\mathbf B = \mu_0(\mathbf H + \mathbf M)). The contribution of the magnetization to (\mathbf B) is therefore (\mu_0\mathbf M). The factor (\mu_0) appears because the magnetic field intensity (\mathbf H) was defined to subtract the bound‑current term (\mathbf M) from (\mathbf B/\mu_0).
Why is the total (\mathbf B) written as (\mu_0\mathbf H) (or (\mu\mathbf H)) and not (\mu\,\mathbf B_{\text{applied}})?	The “applied’’ magnetic field is (\mathbf H); the total induction is (\mathbf B = \mu \mathbf H). If you tried to call (\mathbf B_{\text{applied}}) the field without magnetization you would actually be referring to (\mu_0\mathbf H). The notation already incorporates the material’s permeability (\mu = \mu_0(1+\chi_m)).
Is there a hidden symmetry?	Yes. The electric and magnetic sets of equations are symmetric once you remember that (\mathbf D

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