Schwere, Elektricität und Magnetismus:377

Bernhard Riemann: Schwere, Elektricität und Magnetismus
Seite 5
<< Zurück	Vorwärts >>
fertig
Fertig! Dieser Text wurde zweimal anhand der Quelle Korrektur gelesen. Die Schreibweise folgt dem Originaltext.

 

VECTOR ANALYSIS.

Direct and Skew Products of Vectors.

 13. Def.—The direct product of $\alpha \,$ and $\beta \,$ (written $\alpha \cdot \beta \,$) is the scalar quantity obtained by multiplying the product of their magnitudes, by the cosine of the angle made by their directions.

 14. Def.—The skew product of $\alpha \,$ and $\beta \,$ (written $\alpha \times \beta \,$) is a vector function of $\alpha \,$ and $\beta \,$. Its magnitude is obtained by multiplying the product of the magnitudes of $\alpha \,$ and $\beta \,$ by the sine of the angle made by their directions. Its direction is at right angles to $\alpha \,$ and $\beta \,$, and on that side of the plane containing $\alpha \,$ and $\beta \,$ (supposed drawn from a common origin), on which a rotation from $\alpha \,$ to $\beta \,$ through an arc of less than 180° appears countcr-clock-wise.

 The direction of $\alpha \times \beta \,$ may also be defined as that in which an ordinary screw advances as it turns so as to carry $\alpha \,$ toward $\beta \,$.

 Again, if $\alpha \,$ be directed toward the east, and $\beta \,$ lie in the same horizontal plane and on the north side of $\alpha \,$, $\alpha \times \beta \,$ will be directed upward.

 15. It is evident from the preceding definitions that

$\alpha \cdot \beta =\beta \cdot \alpha ,\,$ and $\alpha \times \beta \ =-\beta \times \alpha .\,$

 16. Moreover,

$\lbrack n\alpha \rbrack \cdot \beta =\alpha \cdot \lbrack n\beta \rbrack =n\lbrack \alpha \cdot \beta \rbrack ,\,$

and	$\lbrack n\alpha \rbrack \times \beta =\alpha \times \lbrack n\beta \rbrack =n\lbrack \alpha \times \beta \rbrack .$

The brackets may therefore be omitted in such expressions.

 17. From the definitions of No. 11 it appears that

$i\cdot i=j\cdot j=k\cdot k=1,\,$ $i\cdot j=j\cdot i=i\cdot k=k\cdot i=j\cdot k=k\cdot j=0,\,$ $i\times i=0,\quad j\times j=0,\quad k\times k=0,\,$ $i\times j=k,\quad j\times k=i,\quad k\times i=j,\,$ $j\times i=-k,\quad k\times j=-i,\quad i\times k=-j.\,$

 18. If we resolve $\beta \,$ into two components $\beta '\,$ and $\beta ''\,$, of which the first is parallel and the second perpendicular to $\alpha \,$, we shall have

$\alpha \cdot \beta =\alpha \cdot \beta '\,$ and $\alpha \times \beta =\alpha \times \beta ''.\,$

 19. $\alpha \cdot \lbrack \beta +\gamma \rbrack =a\cdot \beta +\alpha \cdot \gamma \,$ and $\alpha \times \lbrack \beta +\gamma \rbrack =\alpha \times \beta +\alpha \times \gamma \,$.

 To prove this, let $\sigma =\beta +\gamma \,$, and resolve each of the vectors $\beta ,\,\gamma ,\,\sigma$ into two components, one parallel and the other perpendicular to $\alpha \,$. Let these be $\beta ',\,\beta '',\,\gamma ',\,\gamma '',\,\sigma ',\,\sigma ''$. Then the equations to be proved will reduce by the last section to

$\alpha \cdot \sigma '=\alpha \cdot \beta '+\alpha \cdot \gamma '\,$ and $\alpha \times \sigma ''=\alpha \times \beta ''+\alpha \times \gamma ''\,.$