Page:AbrahamMinkowski1.djvu/13

 Though, '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):

The first contribution of the electric and magnetic excitation, which is represented by $$\mathfrak{E}$$ and $$\mathfrak{H}$$, is interpreted by 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

From (30) it follows

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

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

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

From (30) it also follows

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

Due to the known identity

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

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

which one can also write