Schwere, Elektricität und Magnetismus:382

Bernhard Riemann: Schwere, Elektricität und Magnetismus
Seite 10
<< Zurück	Vorwärts >>
fertig
Fertig! Dieser Text wurde zweimal anhand der Quelle Korrektur gelesen. Die Schreibweise folgt dem Originaltext.

 

VECTOR ANALYSIS.

$a={\frac {\rho \cdot \beta \times \gamma }{\alpha \cdot \beta \times \gamma }},\quad b={\frac {\rho \cdot \gamma \times \alpha }{\alpha \cdot \beta \times \gamma }},\quad c={\frac {\rho \cdot \alpha \times \beta }{\alpha \cdot \beta \times \gamma }}.\,$

(3)

By substitution of these values, we obtain the identical equation,

$(\alpha \cdot \beta \times \gamma )\rho =(\rho \cdot \beta \times \gamma )\alpha +(\rho \cdot \gamma \times \alpha )\beta +(\rho \cdot \alpha \times \beta )\gamma$

(4)

(Compare No. 31.) If we wish the four vectors to appear symmetrically in the equation we may write

$(\alpha .\beta \times \gamma )\rho -(\beta \cdot \gamma \times \rho )\alpha +(\gamma \cdot \rho \times \alpha )\beta -(\rho \cdot \alpha \times \beta )\gamma =0.$

(5)

 If we wish to express $\rho$ as a sum of vectors having directions perpendicular to the planes of $\alpha \,$ and $\beta \,$, of $\beta \,$ and $\gamma \,$, and of $\gamma \,$ and $\alpha \,$, we may write

$\rho =e\beta \times \gamma +f\gamma \times \alpha +g\alpha \times \beta .\,$

(6)

To obtain the valnes of $e,\,f,\,g$, we multiply directly by $\alpha \,$, by $\beta \,$, and by $\gamma \,$. This gives

$e={\frac {\rho \cdot \alpha }{\beta \cdot \gamma \times \alpha }},\quad f={\frac {\rho \cdot \beta }{\gamma \cdot \alpha \times \beta }},\quad g={\frac {\rho \cdot \gamma }{\alpha \cdot \beta \times \gamma }}.$

(7)

Substituting these values we obtain the identical equation

$(\alpha .\beta \times \gamma )\rho =(\rho .\alpha )\beta \times \gamma +(\rho .\beta )\gamma \times \alpha +(\rho .\gamma )\alpha \times \beta .\,$

(8)

(Compare No. 32.)

 38. Reciprocal systems of vectors.—The results of the preceding section may be more compactly expressed if we use the abbreviations

$\alpha '={\frac {\beta \times \gamma }{\alpha \cdot \beta \times \gamma }},\quad \beta '={\frac {\gamma \times \alpha }{\beta \cdot \gamma \times \alpha }},\quad \gamma '={\frac {\alpha \times \beta }{\gamma \cdot \alpha \times \beta }}.\,$

(1)

The identical equations (4) and (8) of the preceding number thus become

$\rho =(\rho \cdot \alpha ')\alpha +(\rho \cdot \beta ')\beta +(\rho \cdot \gamma ')\gamma .\,$

(2)

$\rho =(\rho \cdot \alpha )\alpha '+(\rho \cdot \beta )\beta '+(\rho \cdot \gamma )\gamma '.\,$

(3)

We may infer from the similarity of these equations that the relations of $\alpha ,\,\beta ,\,\gamma$, and $\alpha ',\,\beta ',\,\gamma '$ are reciprocal; a proposition which is easily proved directly. For the equations

$\alpha ={\frac {\beta '\times \gamma '}{\alpha '\cdot \beta '\times \gamma '}},\quad \beta ={\frac {\gamma '\times \alpha '}{\beta '\cdot \gamma '\times \alpha '}},\quad \gamma ={\frac {\alpha '\times \beta '}{\gamma '\cdot \alpha '\times \beta '}}$

(4)

are satisfied identically by the substitution of the values of $\alpha ',\,\beta ',$ and $\gamma '\,$ given in equations (1). (See Nos. 31 and 34.)

 Def.—It will be convenient to use the term reciprocal to designate these relations, i. e., we shall say that three vectors are reciprocals of three others, when they satisfy relations similar to those expressed in equations (1) or (4).