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Figure from:Physics from geometry

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Figure 2.1: The vector a can be decomposed into the sum of a component perpendicular to the plane bivector B and a component in that plane. The bivector B can be written as the product a Ab, with b normal to ay. three vectors, producing a trivector; this has the geometric interpretation of an oriented volume (see Fig. 1.2). Just as the bivector was the unique area element of two-dimensional space, the trivector is the unique volume element of three-dimensional space. This is the highest grade element and it is unique up to scale (or volume) and handedness (or sign). According to our established conventions, this is called the pseudoscalar of the algebra. In three dimensions the algebra is spanned by — Figure 2 1: The vector a can be decomposed into the sum of a component perpendicular to the plane bivector B and a component in that plane. The bivector B can be written as the product a Ab, with b normal to ay. three vectors, producing a trivector; this has the geometric interpretation of an oriented volume (see Fig. 1.2). Just as the bivector was the unique area element of two-dimensional space, the trivector is the unique volume element of three-dimensional space. This is the highest grade element and it is unique up to scale (or volume) and handedness (or sign). According to our established conventions, this is called the pseudoscalar of the algebra. In three dimensions the algebra is spanned by

Related Figures (15)

‘igure 1.1: The bivector: the outer product of a and 0 is a directed area element of mag- nitude |a||b| sin @. The orientation of the parallelogram is defined by whether the area is generated by dragging b over a (direct) or a over b (reverse).

gure 1.2: The trivector: the outer product of aAb/Ac is a directed volume and is associative; the sign remains unchanged by circular permutations of the 3 vectors.

Figure 2.4: The vector a is rotated onto b by first reflecting in the plane perpendicular to h and then in the plane perpendicular to b. All the vectors have unit length. We may need to find the rotor responsible for rotating a unit vector a into another unit vector b, leaving unchanged the vectors perpendicular to the plane aAb. In order to do this we have to reflect a in a plane perpendicular to a vector h that bisects the angle between a and 0b; this produces a vector opposite to 6, which must then be further reflected in a plane perpendicular to 6b (see Fig. 2.4). The vector h can be found by

Figure 3.1: The outer product a/Ab is independent of the shape of the parallelogram formed by the two vectors; the two vectors b and b! generate the same outer product. So we can set \ = a-b/a?; then we have a-b! = 0 and it is

to an r— 1 blade and expand the term inside the sum as follows: This derivation makes use of Eq. (3.32). Substituting in Eq. (3.36) provides the proof of Eq. (3.35) for a grade-r blade, assuming it is true for grade r — 1; since it had been oroven for grade-1 in Eq. (3.34) it is then proven for all blades and for all multivectors. The outer product definition is extended by And the geometric product between a vector and a grade-r multivector is

The inner and outer products are reserved for the lower and higher grade parts, respec- tively

These formulae explain the rationale of the indexing scheme: for basis vectors we use lower indices for the main frame and upper indices for the reciprocal frame; conversely vectors’ components get upper indices when referred to the main frame and lower indices when referred to the reciprocal frame. TN — i. TH ae: ogg ol 1/nHnHaAt1ner?d>0} my

Recalling the discussion in Sec. 3.5 we note that if we take any unitary Hermitian multivector h from the algebra, two idempotents can be formed by

These idempotents are mutually orthogonal because the product of any two of them returns zero. The four idempotents can be added among themselves, resulting in new idempotents, so that we have, with zero, a total of 16 idempotents, arranged in five categories:

The projection of the Hermitian basis vectors 01, 72, 73 produces the following repre- sentations The remaining Hermitian basis elements either project to zero, the projector itself, or reproduce the previous ones. The top left 2 x 2 corner of the matrices reproduces Pauli matrices, which are the generators of SU(2) group, the group of unitary matrices with dimension 2 and unit determinant. The generators of SU(2) are actually given by reproduce the previous ones. The top left 2 x 2 corner of the matrices reproduces Pauli which shows that this is indeed a representation of a 3-dimensional space. The pro-

We look at the transformation for the » vector components; the p vector is Only the momentum along c® and total energy are transformed, however, all components of 3D velocity get transformed in the process. In order to get the velocity components we

awe A surface integral is the limit of a sum over a triangulated surface (see Fig. 7.1). A set of points is chosen over the surface and triangles or simplices are defined by sets of three adjacent points. The number of points is then increased until, in the limit, one obtains the original surface. Each simplex has an attached orientation such that adjacent simplices have the common edge traversed in reverse directions. In this way the orientation for the entire surface is built and the boundary is all traversed in the same direction. For some surfaces it is impossible to define a consistent orientation; one example is the Mébius strip. For such surfaces it is impossible to define a directed integral, so we restrict our presentation to orientable surfaces.

Figure 7.2: A spacetime contour, closed at spatial infinity. boundary data if we try to construct an interior solution with arbitrary boundary data. We must start by setting up the Green’s function. This can be done via its Fourier transform. With x = opx° + o,2! we find

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