Vector Product

(A.36) |

Is a proper vector? Suppose that , . In this case, . However, if we rotate the coordinate axes through about then , , and . Thus, does not transform like a vector, because its magnitude depends on the choice of axes. So, previous definition is a bad one.

Consider, now, the *cross product* or *vector product*:

(A.37) |

Does this rather unlikely combination transform like a vector? Let us try rotating the coordinate axes through an angle about using Equations (A.20)-(A.22). In the new coordinate system,

(A.38) |

Thus, the -component of transforms correctly. It can easily be shown that the other components transform correctly as well, and that all components also transform correctly under rotation about and . Thus, is a proper vector. Incidentally, is the only simple combination of the components of two vectors that transforms like a vector (which is non-coplanar with and ). The cross product is

(A.39) |

distributive,

(A.40) |

but is not associative,

(A.41) |

The cross product transforms like a vector, which means that it must have a well-defined direction and magnitude. We can show that is perpendicular to both and . Consider . If this is zero then the cross product must be perpendicular to . Now,

(A.42) |

Therefore, is perpendicular to . Likewise, it can be demonstrated that is perpendicular to . The vectors , , and form a right-handed set, like the unit vectors , , and . In fact, . This defines a unique direction for , which is obtained from a right-hand rule. (See Figure A.8.)

Let us now evaluate the magnitude of . We have

(A.43) |

Thus,

(A.44) |

where is the angle subtended between and . Clearly, for any vector, because is always zero in this case. Also, if then either , , or is parallel (or antiparallel) to .

Consider the parallelogram defined by the vectors and . (See Figure A.9.) The scalar area of the parallelogram is . By convention, the vector area has the magnitude of the scalar area, and is normal to the plane of the parallelogram, in the sense obtained from a right-hand circulation rule by rotating on to (through an acute angle). In other words, if the fingers of the right-hand circulate in the direction of rotation then the thumb of the right-hand indicates the direction of the vector area. So, the vector area is coming out of the page in Figure A.9. It follows that

(A.45) |

Suppose that a force is applied at position , as illustrated in Figure A.10. The torque about the origin is the product of the magnitude of the force and the length of the lever arm . Thus, the magnitude of the torque is . The direction of the torque is conventionally defined as the direction of the axis through about which the force tries to rotate objects, in the sense determined by a right-hand circulation rule. Hence, the torque is out of the page in Figure A.10. It follows that the vector torque is given by

The angular momentum, , of a particle of linear momentum and position vector is simply defined as the moment of its momentum about the origin: that is,

(A.47) |