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Grassmann Algebra

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Exploring<strong>Grassmann</strong><strong>Algebra</strong>.nb 30<br />

S � <strong>Grassmann</strong>Solve�Q, �Α, Β, x��<br />

��Α �C2 � y ��31 � 36 C1<br />

2 � 11 C2� �������������������������������� ��������������������� ,<br />

12 C2<br />

18 �<br />

����������������������<br />

�11 � C2 � y ����<br />

2 �<br />

2 � C2<br />

108 C1<br />

� ����������������������������<br />

��11 � C2 2<br />

2� �<br />

12<br />

��������������������<br />

�11 � C2 2 �<br />

6C2 �<br />

�������������������� ���� �11 � C2 2 �<br />

,<br />

x � yC1 � 1 2<br />

���� �5 � C2��, �Α �yC3, Β�<br />

6 18 203 y 5 31 y<br />

������� � �������������� ,x� ���� � �����������<br />

11 121 6 36 ��<br />

Again we get two solutions, but this time involving some arbitrary scalar constants.<br />

��Q �. S���Simplify<br />

��True, True�, �True, True��<br />

9.7 Exterior Division<br />

� Defining exterior quotients<br />

An exterior quotient of two <strong>Grassmann</strong> numbers X and Y is a solution Z to the equation:<br />

X � Y � Z<br />

or to the equation:<br />

X � Z � Y<br />

To distinguish these two possibilities, we call the solution Z to the first equation X�Y�Z the left<br />

exterior quotient of X by Y because the denominator Y multiplies the quotient Z from the left.<br />

Correspondingly, we call the solution Z to the second equation X�Z�Y the right exterior<br />

quotient of X by Y because the denominator Y multiplies the quotient Z from the right.<br />

To solve for Z in either case, we can use the <strong>Grassmann</strong>Solve function. For example if X<br />

and Y are general elements in a 2-space, the left exterior quotient is obtained as:<br />

�2; X� Create<strong>Grassmann</strong>Number�Ξ�;<br />

Y � Create<strong>Grassmann</strong>Number�Ψ�;<br />

<strong>Grassmann</strong>Solve�X � Y � Z, �Z��<br />

��Z � Ξ0<br />

�������<br />

Ψ0<br />

� e1 ��Ξ1 Ψ0 �Ξ0 Ψ1�<br />

�������������������������������� �������������<br />

Ψ2 �<br />

0<br />

e2 ��Ξ2 Ψ0 �Ξ0 Ψ2�<br />

�������������������������������� ������������� �<br />

Ψ2 0 ��Ξ3 Ψ0 �Ξ2 Ψ1 �Ξ1 Ψ2 �Ξ0 Ψ3� e1 � e2<br />

�������������������������������� ��������������������������������<br />

Ψ2 0<br />

Whereas the right exterior quotient is obtained as:<br />

2001 4 5<br />

������������������������ ��

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