# Page:AbrahamMinkowski1.djvu/13

Though, Lorentz's vectors $\mathfrak{E}$ and $\mathfrak{H}$ have an illustrative meaning. Namely, the excitations $\mathfrak{D}$ and $\mathfrak{B}$ can be split into two parts according to eq. (29), (30):

 (31) $\begin{cases} \mathfrak{D=E+P}, & \mathfrak{P}=(\epsilon-1)\mathfrak{E}';\\ \mathfrak{B=H+M}, & \mathfrak{M}=(\mu-1)\mathfrak{H}'.\end{cases}$

The first contribution of the electric and magnetic excitation, which is represented by $\mathfrak{E}$ and $\mathfrak{H}$, is interpreted by Lorentz as electric and magnetic excitation of the aether, and the second contribution, which is represented by the vectors $\mathfrak{P}$ and $\mathfrak{M}$ (electric and magnetic polarization), is interpreted as the electric and magnetic excitation of matter; the latter is set proportional to the electric and magnetic force $\mathfrak{E}'$ and $\mathfrak{H}'$, which acts upon the unit charges which is co-moving with matter.

We want to consider $\epsilon$ and $\mu$ in this paragraph as being (for a certain material point) dependent on velocity and time, although we reserve us the right to remove these confinements later.

To find the momentum density on the basis of relation (18), we calculate the quantities

 (31a) $\begin{cases} \mathfrak{E'\dot{D}-D\dot{E}'=E'\dot{E}-E\dot{E}'+E'\dot{P}-P\dot{E}'},\\ \mathfrak{H'\dot{B}-B\dot{H}'=H'\dot{H}-H\dot{H}'+H'\dot{M}-M\dot{H}'}.\end{cases}$

From (30) it follows

 $\begin{array}{l} \mathfrak{E'\dot{E}-E\dot{E}'=-q[\dot{E}H]+q[E\dot{H}]+\dot{q}[EH]},\\ \mathfrak{H'\dot{H}-H\dot{H}'=-q[E\dot{H}]+q[\dot{E}H]+\dot{q}[EH]},\end{array}$

Since the two other terms in (31) are vanishing according to (31), then relation (18) gives

 (32) $c\mathfrak{g=[EH]}$

as the value of the electromagnetic momentum density.

Now the question arises, whether this value at the same time satisfies the condition (18a)

$[\mathfrak{q}c\mathfrak{g}]=\mathfrak{[DE']+[BH']}$

According to (29), it is

$\mathfrak{[DE']+[BH']=[E'[qH]]-[H'[qE]]}$

From (30) it also follows

$\mathfrak{[DE']+[BH']=[E[qH]]-[H[qE]]}$

Due to the known identity

$\mathfrak{[q[EH]]=[E[qH]]-[H[qE]]}$

it will be seen, that expression (32) for the momentum density really satisfies condition (18a).

Now from (19), the value of the energy density follows

 (33) $\psi=\frac{1}{2}\mathfrak{E'D}+\frac{1}{2}\mathfrak{H'B}+\mathfrak{q[EH]}$

which one can also write

 (33a) $\psi=\frac{1}{2}\mathfrak{E}^{2}+\frac{1}{2}\mathfrak{H}^{2}+\frac{1}{2}\mathfrak{E'P}+\frac{1}{2}\mathfrak{H'M}$