# Maxwell-equations E or D

1. Jun 6, 2015

### Backdraft

Hi!

I am learning electromagnetism, and I think I understand the differences between the various vector fields (E-D, B-H), but I can't figure out one thing:
3 of the Maxwell-equations are always written with D and H, but Faraday's law is always written with E and B.
So my question is why?
Is $$curl \ D=-\frac{\partial H}{\partial t}$$ simply not true?

2. Jun 6, 2015

### axmls

Because it's off by a factor. Why is $F = 3ma$ not true? Faraday's law is given as
$$\nabla \times E = -\frac{\partial B}{\partial t}$$
If you used $D$ instead, then you would have (in free space) introduced a factor of $\epsilon_0$ which doesn't belong.

Also, it's not always the case that the other three are always written in terms of the $D$ and $H$ fields. It depends on the context.

3. Jun 6, 2015

### stevendaryl

Staff Emeritus
According to Wikipedia,
• $\nabla \cdot \vec{D} = \rho_f$
• $\nabla \times \vec{H} - \dfrac{\partial D}{\partial t} = \vec{J}_f$
• $\nabla \cdot \vec{H} = 0$
• $\nabla \times \vec{E} + \dfrac{\partial B}{\partial t} = 0$
(According to SI units--cgs units throw in factors of $c$ so that $\vec{E}$ and $\vec{B}$ have the same units.)

When the medium is linear, we assume that:
$\vec{D} = \epsilon \vec{E}$
$\vec{H} = \frac{1}{\mu} \vec{B}$
which means that the last equation can be written as:

$\nabla \times \vec{D} +\epsilon \mu \dfrac{\partial H}{\partial t} = 0$

In vacuum, according to SI units, $\epsilon \mu = \frac{1}{c^2}$, so this can be written as

$\nabla \times \vec{D} +\frac{1}{c^2} \dfrac{\partial H}{\partial t} = 0$

On the other hand, in cgs units, $\epsilon = \mu = 1$ in vacuum, so you do have:

$\nabla \times \vec{D} + \dfrac{\partial H}{\partial t} = 0$

(but that's only true in vacuum)

4. Jun 6, 2015

### andresB

That is weird. It should be ∇⋅ B=0 instead of ∇⋅H=0.

5. Jun 6, 2015

### stevendaryl

Staff Emeritus
You are right. I goofed up.

You can define $\vec{E}$ and $\vec{B}$ in terms of vector and scalar potentials $\vec{A}$ and $\Phi$ as follows:

$\vec{B} = \nabla \times \vec{A}$
$\vec{E} = - \dfrac{\partial \vec{A}}{\partial t} - \nabla \Phi$

Then two of the equations are true automatically:

$\nabla \cdot \vec{B} = 0$
$\nabla \times \vec{E} + \dfrac{\partial \vec{B}}{\partial t} = 0$

6. Jun 7, 2015

### vanhees71

On a fundamental level there is only one and only one electromagnetic field, described by $\vec{E}$ and $\vec{B}$. Now in a wide class of problems with electromagnetics at the presence of bulk matter you can treat the mutual influence of the bulk matter (consisting of electrically charged particles) and the electromagnetic field in linear-response approximation. What comes out is that the electric and magnetic polarization of the medium are linear to the external electromagnetic field. Usually matter is electrically neutral, and thus you can lump the response of the matter to the electric field into electric and magnetic polarization and consider only the external charges, added to the matter in equilibrium.

Then you are at the level of the macroscopic Maxwell equations quoted in #3 with the very important correction in #4. The fields there are not the microscopic fields but averaged over macroscopically small but microscopically large regions of the medium. In this averaging the homoegeneous Maxwell equations look like the microscopic ones, while the inhomogeneous ones change, because you lump the part of the charge-current distribution which is due to the linear response of the equilibrium medium into the polarizations.

Now because of a historical misunderstanding, mixing up $\vec{B}$ and $\vec{H}$, you have the following relations (written in the ugly SI units, which mask the beautiful relationship between the field components due to the underlying relativistic structure of the theory):
$$\vec{D}=\epsilon_0 \vec{E}+\vec{P}, \quad \vec{H}=\frac{1}{\mu_0} \vec{B}-\vec{M}.$$
While the electric polarization $\vec{P}$ of the medium comes with the physically sensible sign, it's opposite for the magnetic polarization $\vec{M}$ due to the mentioned misunderstanding in the early days of Maxwell theory. The ugly conversion factors $\epsilon_0$ and $\mu_0$ are mere conversion factors without much physical significance enforced by the choice of SI units. SI units are good for electrical engineering but mess up the physics.

Anyway, then the inhomogeneous equations read
$$\vec{\nabla} \cdot \vec{D}=\rho_{\text{free}}, \quad \vec{\nabla} \times \vec{H}-\partial_t \vec{D}=\vec{j}_{\text{free}}.$$
One should NOT put the discplacement current on the right-hand side, because it's not a source term but part of the dynamical field equations. This is another misunderstanding, causing often confusion, discussen in this forum.

7. Jun 7, 2015

### stevendaryl

Staff Emeritus
I think that the reason for the difference in signs between the definitions of $H$ and $D$ is because typically, in a medium,

$|B| > |H|$ (because magnetic dipoles tend to point in the same direction as the magnetic field, increasing its strength)
$|E| < |D|$ (because electric dipoles tend to point in the opposite direction as the electric field, decreasing its strength)

With the sign choices, its typically the case in a medium that
$\epsilon > \epsilon_0$
$\mu > \mu_0$

So, choosing the opposite signs for the constitutive relationships leads to the same signs for the macroscopic parameters $\epsilon$ and $\mu$.

8. Jun 7, 2015

### vanhees71

Well, but if our ancestors had known the proper (relativistic) formulation of electromagnetism they would never have written the consitutive equations so incoherently but in some way similar, i.e., like (for a medium at rest and for one Fourier mode of the fields)
$$\vec{D}=\epsilon \vec{E}, \quad \vec{H}=\tilde{\mu} \vec{B},$$
which of course you still can do, but unfortunately nobody does this actually but instead uses the historical relation $\vec{B}=\mu \vec{H}$, using $\mu=1/\tilde{\mu}$. I guess we have to live with this, as we unfortunately have to live since the 3rd edition of Jackson's textbook with the bad habit to teach E+M in SI units; if I had to say something, I'd rather simply change from the old non-rational Gaussian units to the more modern rationalized Heaviside-Lorentz units, as it is done in the HEP/Nuclear Physics theory community. But that's another story.

9. Jun 9, 2015

### DrDu

In this context I want to mention that in solid state physics one often uses the convention $B=H$, so that all effects of the medium are lumped up in P or D.
In fact, P is nothing but a consequence of the conservation of the charges $\rho_\mathrm{int}$ which make up the medium (as opposed the external charges and currents which are produced by the experimentator). Combining the internal charges and currents into a four vector j, and the polarisation P and magnetization M into an antisymmetric tensor $\Pi$, we can write the conservation of charge as $d*j=0$ which can always be fulfilled if $*j=d\Pi$. The latter equation does not fix $\Pi$ completely. So we are free to choose e.g. M=0. This is only inconvenient at $\omega=0$, i.e. in electrostatics. Where we are back in the field of electrical engineering.

10. Jun 11, 2015

### Backdraft

Thank you very much for the answers! It really helped, I think I get it

11. Jun 12, 2015

### vanhees71

But that's for sure an approximation, because there are magnetization effects in linear response. It's only that for many materials they are quite negligible and thus you can work with $\mu=1$ (in Gaussian or Heaviside-Lorentzt units).

Note that in general the $\epsilon$ and $\mu$ are tensor fields and not simple constants!

12. Jun 12, 2015

### DrDu

No, it isn't an approximation. The magnetization effects are encoded in the difference between the longitudinal and transverse part of the dielectric tensor.
Namely $\epsilon(\omega)=\epsilon_\parallel (\omega,0)$ and
$1-\frac{1}{\mu(\omega)}=\lim_{k\to 0} \frac{\omega^2[\epsilon_\perp(\omega,k)-\epsilon_\parallel(\omega,k)]}{c^2k^2}$.
Hence magnetization is a special case of spatial dispersion, i.e. the dependence of $\epsilon$ on k.

This is quite nicely discussed in Landau & Lifshitz, Vol. 8. Landau gives also arguments why the definition of magnetization becomes ambiguous at higher frequencies.

Last edited: Jun 12, 2015
13. Jun 12, 2015

### vanhees71

I see, great! Is it really Vol. 5 or rather Vol. 8?

14. Jun 12, 2015

### DrDu

You are right, I corrected the volume.

15. Jun 12, 2015

### vanhees71

Good, I'll have a look at it this evening. Now I don't have this book available.