Page:AbrahamMinkowski1.djvu/15

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From the standpoint of the system used by us, the task also arises to derive the momentum density from relation (18). It follows from (36)

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

Thus the right-hand side of (18) becomes

(38) \left\{ \begin{array}{c}
\mathfrak{E'\mathfrak{\dot{D}}-D\dot{E}'+H'\dot{B}-B\dot{H}'=\dot{q}\left\{ [E'H]+[EH']\right\} }\\
+\mathfrak{q\left\{ [E'\dot{H}]+[\dot{E}H']-[\dot{E}'H]-[E\dot{H'}]\right\} }\end{array}\right.

We express, on the basis of (37), \mathfrak{EH} and \mathfrak{\dot{E}\dot{H}} by the vectors arising in the main equations, and we find

(38a) \mathfrak{[E'H]+[EH']=}2\mathfrak{[E'H']+q(E'D)-D(qE')+q(H'B)-B(qH')}
(38b) \begin{cases}
\mathfrak{\qquad[E'\dot{H}]+[\dot{E}H']-[\dot{E}'H]-[E\dot{H'}]}\\
=\mathfrak{\dot{q}(E'D)-D(\dot{q}E')+\dot{q}(H'B)-B(\dot{q}H')}\\
+\mathfrak{q\{E'\dot{D}-D\dot{E}'+H'\dot{B}-B\dot{H}'}\}\\
-\left\{ \mathfrak{\dot{D}(qE')-D(q\dot{E}')+\dot{B}(qH')-B(q\dot{H}')}\right\} .\end{cases}

By inserting (38a,b) into (38), we obtain

(38c) \begin{cases}
\qquad\mathfrak{E'\dot{D}-D\dot{E}'+H'\dot{B}-B\dot{H}'}\\
=2\mathfrak{\dot{q}}\left\{ \mathfrak{[E'H']+q(E'D)+q(H'B)-D(qE')-B(qH')}\right\} \\
+\mathfrak{(\dot{q}D)(qE')-(qD)(\dot{q}E')-(q\dot{D})(qE')+(qD)(q\dot{E}')}\\
+\mathfrak{(\dot{q}B)(qH')-(qB)(\dot{q}H')-(q\dot{B})(qH')+(qB)(q\dot{H}')}\\
+\mathfrak{q}^{2}\{\mathfrak{E'\dot{D}-D\dot{E}'+H'\dot{B}-B\dot{H}'}\}.\end{cases}

However, now it follows from (36)

\begin{array}{l}
\mathfrak{-(q\dot{D})(qE')-(qD)(q\dot{E}')=(\dot{q}D)(qE')+(qD)(\dot{q}E')}\\
\mathfrak{-(q\dot{B})(qH')-(qB)(q\dot{H}')=(\dot{q}B)(qH')+(qB)(\dot{q}H')}\end{array}

thus the second and third row of the right-hand side of (28c) assume the values

\begin{array}{l}
2\left\{ \mathfrak{(\dot{q}D)(qE')-(qD)(\dot{q}E')}\right\} =2\left(\mathfrak{[\dot{q}q][DE']}\right),\\
2\left\{ \mathfrak{(\dot{q}B)(qH')-(qB)(\dot{q}H')}\right\} =2\left(\mathfrak{[\dot{q}q][BH']}\right).\end{array}

If it indeed holds, as required by (18a)

(39) [\mathfrak{q}c\mathfrak{g}]=\mathfrak{[DE']+[BH']},

then the second and third row are providing together:

2\left(\mathfrak{[\dot{q}q][\mathfrak{q}c\mathfrak{g}]}\right)=2\mathfrak{[\dot{q}q][\mathfrak{q}c\mathfrak{g}]}-\mathfrak{q}^{2}(\mathfrak{q}2c\mathfrak{g})

Therefore it eventually follows from (18)

(39a) c\mathfrak{g=[E'H']+q(E'D)+q(H'B)-D(qE')-B(qH')+q(q}c\mathfrak{g)}

The comparison with (20) gives the important relation

(40) \mathfrak{g=}\frac{\mathfrak{S}}{c^{2}}