A Geometric Rather than Algebraic Approach to Geometric Algebra and Calculus with Applications to Electrodynamics and Relativity David R. Rowland The University of Queensland, Brisbane QLD 4072, Australia Email:
[email protected]
Why study geometric algebra? In classical mechanics and classical electromagnetic theory, many general theoretical derivations are most conveniently done using vector algebra and vector calculus. Extending this approach to the four dimensions of spacetime is problematic however, since the vector cross product is only definable in 3D. Consequently, the usual approach for spacetime calculations is to either use the techniques of tensor analysis, which considers the components of vectors and tensors and which results in formulas with hard to discern geometrical content, or to use the techniques of exterior calculus and differential forms (or both). However, differential forms are rather abstract concepts and picturing them geometrically is very difficult except for the simplest case of a one-form [MTW: Secs. 2.5, 4.2, 4.3]. Geometric algebra (GA) addresses all the above-mentioned difficulties by providing a generalization of the vector cross product which is definable in any number of dimensions while also producing formulas with beautifully clear geometric interpretations. Geometric algebra is Last updated 24 September, 2016
also beautifully suited to spacetime calculations for the following reason. Because the four velocity of a particle has a constant magnitude, particle dynamics involves the rigid rotation of a particle’s four velocity, and GA provides a conceptually simple way of encoding hyperbolic as well as regular rotations in arbitrary planes. There is, unfortunately, no unique way to develop a geometric algebra for spacetime calculations. The two leading approaches are the “spacetime algebra” (STA) promoted by Hestenes (e.g. [H1-H3]), and the “algebra of physical space” (APS) promoted by Baylis [B, BS]. Since the goal of these notes is to develop and present a geometric algebra which is as simple a generalisation of 3D vector analysis as possible (to facilitate learning), APS will not be treated in these notes since it has more of a complex analysis feel than a vector algebra feel. These notes differ from other expositions in that standard expositions of GA/STA take the geometric product (to be defined in Ch. 3) to be fundamental and build up all results using the algebraic properties of the geometric product. While this is the most rigorous approach mathematically, and ultimately the approach which leads to the most general results, this approach appears to present a considerable barrier to those more familiar and comfortable with “standard” Gibbsian vector algebra and calculus. Consequently, while it has its limitations, the approach taken in these notes is to take the projection (parallel component) and rejection (orthogonal component) properties of the inner and outer products respectively as the guides to making calculations, with the geometric product “merely” being a useful tool for making some intermediate manipulations. This approach means that geometric visualization can guide calculations, and in particular allows one to readily use what one has learned from a study of standard vector algebra. As per 3D vector algebra and vector calculus, the result of the above approach is a table of identities which should be kept at one’s side when doing general calculations to facilitate those calculations. I should note though, that when it comes to deriving general results in an arbitrary number of dimensions and for arbitrary grade multivectors, eventually the standard approaches would be needed, but these 1
standard approaches are not generally needed for classical physics in 3D and 4D. After developing the mathematical theory of GA and geometric calculus (GC) in a geometrical way in Chapters 2-5, the value of the approach for studying physics in spacetime is illustrated in Chapters 6-8, where some standard results in special relativity, electrodynamics and general relativity respectively are derived using GA and GC. Although no new theoretical results are presented, a beautiful result of this approach is a simple geometric picture showing how a hyperbolic rotation of one’s reference frame axes causes a transformation of a pure electric filed into a mixture of electric and magnetic fields. Lest the above introduction lead the reader to believe these notes are more than they are, I should point out here that they are not meant to be a text book on the subject. They are in reality a set of reference / self-teaching notes on the spacetime algebra (STA) version of geometric algebra (GA) and calculus (GC) that I have written for myself as I’ve learned and explored the uses of GA and GC for spacetime physics analyses. I am posting them in case others find the notes a helpful resource of useful results. In places they are simply tables of results with derivations, while in others they have the flavour of lecture notes or a Schaum Outline Series type text. Nevertheless, some conceptual details and many sub-steps in calculations not found in other sources are provided, hopefully reducing the learning curve for beginners. I also hope that some might find the somewhat unconventional pedagogical approach taken in places facilitates their appreciation and learning of the mathematical formalism.
studied electromagnetic theory at a level where Maxwell’s equations, the Lienard-Wiechert potentials, and the electromagnetic energy-momentum tensor have been covered. The Special Theory of Relativity (SR) applications assume the reader has studied SR to a level where the Lorentz transformation of the electric and magnetic fields and Thomas precession have been covered and the inner product is defined with a mixed signature metric tensor. The General Theory of Relativity (GR) applications assume the reader has a basic understanding of the Einstein equations.
Assumed knowledge The mathematical exposition assumes the reader has a knowledge of vector algebra and vector calculus. Some previous exposure to rank-2 and higher tensors and the Einstein summation convention would also be useful. The sections on electrodynamics, special relativity and general relativity are not meant to be pedagogical introductions to those topics, but rather an exposition of the uses of the GA formalism to facilitate calculations in those topics. Thus the electrodynamics applications assume the reader has 2
Contents 1. OVERVIEW AND REFERENCES ............................................................. 5 1.1 Structure of notes ................................................................................... 5 1.2 References .............................................................................................. 5 2. GEOMETRIC CONSTITUENTS OF GEOMETRIC ALGEBRA .............. 6 Table 1. Constituents of Geometric Algebra ................................................ 6 3. INNER, OUTER AND GEOMETRIC PRODUCTS ................................. 10 Table 2. Products for vectors in geometric algebra. ................................... 10 Table 3. Products involving higher grade elements in geometric algebra.. 13 Table 4. Applications ................................................................................. 20
6.4 Alternative form of a pure boost ........................................................... 41 6.5 Fermi Transport .................................................................................... 41 6.6 Fermi derivative .................................................................................... 42 6.7 Thomas precession................................................................................ 42 6.8 Rotor equation of motion ...................................................................... 44 6.9 Rindler Frame ....................................................................................... 44 6.10 Turntable Frame.................................................................................. 46 6.11 Length and orientation of a moving rod ............................................. 48 6.12 Relativistic Doppler effect .................................................................. 49 6.13 Aberration of light .............................................................................. 49 6.14 Boost in an arbitrary direction ............................................................ 49 6.15 Stress-energy tensor for a perfect fluid ............................................... 50
5. GEOMETRIC CALCULUS ....................................................................... 28 5.1 Gradient operator ................................................................................. 28 5.2 Coordinate and reciprocal bases and the metric tensor ........................ 28 5.3 “Div” and “Curl” .................................................................................. 29 5.4 Useful Differential Identities ................................................................ 29 5.5 Integral Theorems ................................................................................ 32 5.6 Multivector Calculus ............................................................................ 33
7. CLASSICAL ELECTRODYNAMICS ....................................................... 51 7.1 Units and Conventions.......................................................................... 51 7.2 Covariant Form of Maxwell’s Equations.............................................. 51 7.3 Covariant form of Lorentz Force Law .................................................. 53 7.4 Lienard-Wiechert Potentials ................................................................. 54 7.5 Fields of an arbitrarily moving charge .................................................. 55 7.6 Stress-Energy-Momentum Tensor for a Field in Vacuo ....................... 56 7.7 Separate conservation of the radiation and velocity field energies away from a charge .............................................................................................. 57 7.8 Back reaction on an accelerating charge............................................... 58 7.8.1. The Lorentz-Abraham-Dirac equation of motion......................... 58 7.8.2. Modified Lorentz-Abraham-Dirac equation ................................ 59 7.8.3. Another form ................................................................................ 60 7.9 Axiomatic “derivation” of Maxwell’s equations .................................. 61
6. SPECIAL RELATIVITY USING GEOMETRIC ALGEBRA .................. 35 6.1 Vectors in Minkowski spacetime ......................................................... 35 6.2 Bivectors in spacetime ......................................................................... 37 6.3 Lorentz Transformations ...................................................................... 37
8. GENERAL RELATIVITY ......................................................................... 62 8.1 Why expect gravity to be a theory of curved spacetime? ..................... 62 8.2 Standard Curved Spacetime General Relativity as a Gauge Theory .... 62 8.3 A bit of Riemannian geometry ............................................................. 63
4. LINEAR (TENSOR) ALGEBRA WITH GEOMETRIC ALGEBRA ....... 24 4.1 Basic Definitions .................................................................................. 24 4.2 Adjoint.................................................................................................. 25 4.3 Determinant .......................................................................................... 25 4.4 Inverse .................................................................................................. 26 4.5 Trace..................................................................................................... 26
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8.3.1 Coordinate basis vectors, covectors and the metric ...................... 63 8.3.2 Covariant (Intrinsic) Derivative .................................................... 64 8.3.3 Geodesic Equation ........................................................................ 65 8.3.4 Divergence and Curl ..................................................................... 67 8.3.5 Symmetry property of the connection coefficients ....................... 69 8.3.6 Calculation of connection coefficients.......................................... 69 8.3.7 Riemann Curvature Tensor ........................................................... 69 8.3.8 Ricci Curvature Tensor ................................................................. 73 8.3.9 Vector wave equation for the electromagnetic potential .............. 74 8.3.10 Einstein’s Equation ..................................................................... 74 8.3.11 Understanding Einstein’s equation geometrically ...................... 76 8.3.12 Lie Derivatives, Killing Vectors and Spactime Symmetries ...... 78 8.4 Hydrodynamics in a Curved Spacetime ............................................... 81 8.5 Gauss’ and Stokes’ Theorems in curved 4D spacetime ....................... 82 8.6 Komar mass .......................................................................................... 84 Appendix A. Proof of the usefulness of making the geometric product associative....................................................................................................... 87 Appendix B. Tensor – GA equivalencies ....................................................... 87
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1. OVERVIEW AND REFERENCES 1.1 Structure of notes Chapters 2-5 develop the mathematical theory of GA in a geometric way, while Chapters 6-8 provide applications to physics. Chapter 2 explains how GA extends the scalar and vector constituents of vector algebra to include bivectors, trivectors and so on. Chapter 3 explains the geometric meanings of the inner, outer and geometric products of GA. In particular, it shows how the vector cross product can be replaced by an outer product which can be defined in any number of dimensions, not just three. It also shows how to determine the inner and outer products of multivectors. Chapter 4 outlines geometric calculus, showing how vector calculus can be extended to any number of dimensions. Chapter 5 develops linear (tensor) algebra from the GA perspective. The special and general theories of relativity are described as being “geometric” theories of spacetime. In my view, GA and GC make the geometric nature of these theories much more transparent than does the traditional approach using tensor algebra, differential forms and coordinate transformations, and this is illustrated in Chapters 6 and 8. GA and GC are also used in Chapter 7 to bring out the geometric content of the electrodynamics of point charges in a vacuum, which is an inherently special relativistic theory.
1.2 References These notes are based on the references below from which I learnt GA and GC, however, in a number of places the pedagogical approach differs. In particular, it has been my aim to develop GA and GC in such a way that the leap from conventional vector algebra and vector calculus is as small as possible in the hope that it makes the value of GA and GC more easily
discernable, and that the learning hurdle for those already familiar with vector algebra and calculus is as small as possible. I do not know, however, whether ultimately this approach helps or hinders one hoping to do advanced research work using the formalism.
[B] W. E. Baylis, Electrodynamics: A Modern Geometric Approach (Birkhäuser, Boston, 1999). [BS] W. E. Baylis and G. Sobczyk, “Relativity in Clifford’s geometric algebras of space and spacetime,” Int. J. Theor. Phys. 43(10), 2061-2079 (2000). [DL] C. Doran and A. Lasenby, Geometric Algebra for Physicists (Cambridge University Press, Cambridge, 2003). [FK] M. R. Francis and A. Kosowsky, “Geometric algebra techniques for general relativity,” (2003), retrieved from http://arxiv.org/abs/gr-qc/0311007 [H1] D. Hestenes, “Proper particle mechanics,” J. Math. Phys. 15, 1768-1777 (1974). [H2] D. Hestenes, “Oersted Medal Lecture 2002: Reforming the mathematical language of physics,” Am. J. Phys. 71, 104-121 (2003). [H3] D. Hestenes, “Spacetime physics with geometric algebra,” Am. J. Phys. 71, 691-714 (2003). [H4] D. Hestenes, Spacetime Calculus for Gravitation Theory, retrieved from: http://geocalc.clas.asu.edu/pdf/NEW_GRAVITY.pdf [H5] D. Hestenes, Spacetime Geometry with Geometric Calculus, retrieved from: http://geocalc.clas.asu.edu/pdf/SpacetimeGeometry.w.GC.proc.pdf [H6] D. Hestenes, Gauge Theory Gravity with Geometric Calculus, retrieved from: http://geocalc.clas.asu.edu/pdf/GTG.w.GC.FP.pdf [J] B. Jancewicz, Multivectors and Clifford Algebra in Electrodynamics (World Scientific, Singapore, 1988). [MTW] C.W. Misner, K.S. Thorne and J.A. Wheeler, Gravitation (W.H. Freeman and Co. San Francisco, 1973). [P] E. Poisson, A Relativist's Toolkit: The Mathematics of Black-Hole Mechanics (Cambridge University Press, 2007). [R] D. R. Rowland, “On the value of geometric algebra for spacetime analyses using an investigation of the form of the self-force of an accelerating charge as a case study,” Am. J. Phys. 78(2), 187-194 (2010).
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2. GEOMETRIC CONSTITUENTS OF GEOMETRIC ALGEBRA Table 1. Constituents of Geometric Algebra Object
Grade
Representations
Description
Additional Comments
Scalar
0
2.3, -0.1, m, q, k
- Oriented number with magnitude. - Oriented in the sense of being either positive or negative.
E.g. mass, charge, pure numbers.
Vector
1
a or a
- Oriented line segment. - Magnitude = “length”. - Orientation given by a direction along the line segment.
- Only “polar” or true vectors like displacement, velocity, acceleration, momentum, and force are vectors in GA. - Over arrows are used in these notes for the spatial part of vectors in 4D Minkowski spacetime.
Bivector
2
B or B , a b
- A “simple” bivector is an oriented plane segment which is also referred to as a “blade”. In 4 or more dimensions, it is possible to have non-simple bivectors. - Magnitude = “area”. In a Euclidean vector space, magnitude is given by the area of the parallelogram spanned by a and b. In a mixed signature space, magnitude is given by | a || b || a || b | , where a is the part of a orthogonal to b etc. As will be shown below, while a bivector has a magnitude (area), it has no particular shape. - Orientation given by a sense of circulation around the area.
- The plane spanned by a and b defines the plane in which a b lies. - Putting the tail of b to the tip of a gives the sense of circulation for a b . - “Axial vectors” like angular momentum ( L r p ), and torque are represented as bivectors in GA. - Antisymmetric rank 2 tensors like the electromagnetic field tensor are also treated as bivectors in GA (see Ch. 7). - Like vectors, bivectors can be split into components. E.g. a b a1 b a 2 b where a1 a 2 a .
Trivector
3
T, a b c
- A “simple” trivector is an oriented volume segment. - Magnitude = “volume” of the parallelepiped spanned by a, b and c. - Orientation given by a sense of circulation around the volume as defined by the order of the vectors in abc .
Chapter 2. Geometric Constituents of Geometric Algebra
- The vector sub-space spanned by a, b and c defines the sub-space in which a b c lies.
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Etc. Multivector
M a B T ...
- A formal sum of elements of different grades.
- If the sum of a scalar and a vector seems a nonsensical notion, think of it as being like the sum of the real and imaginary parts of a complex number. Or, although one can’t add “apples and oranges”, one can think of M as being a box which contain 5 apples, 6 oranges and one pear, and one can think of having 6 of those boxes or adding one box to another box which has different numbers of apples, oranges and pears. And as will be revealed below, since the product of blades of different grades can be defined, the product of two arbitrary multivectors can be defined, and this is why the formal sum is useful.
Reverse
~ M, M †
Reverses the order of the vectors in each blade of M. I.e. if M a b ... c , then ~ M c ... b a … (2.1)
If M a B T ... , then ~ M a B T ...
Pseudoscalar
Highest I unit pseudo-scalar grade multivector in vector space
I e1 e 2 ... e n … (2.2)
I 2 1 in general (The meaning of I2 will be defined below.)
Tensor
T(a) T(a b) T(a) T(b) ... (2.4)
if {e k } are an orthonormal basis for the ndimensional vector space.
[H4]: A tensor of degree k is a multilinear, multivector-valued function of k vectors. A tensor of degree k has a definite rank r if it is (r – k)-vector valued. A tensor of degree k and rank r is said to be of type r – k. Every multivector is a tensor of degree zero (e.g. a bivector is a type 2 – 0 tensor). To obtain expressions consistent with usual tensor algebra, we want T(a) T(a j e j ) Tij a j . (Note that Einstein’s summation convention for repeated raised and lowered indices is being
Chapter 2. Geometric Constituents of Geometric Algebra
I 2 1 in 2, 3 and 4D … (2.3) and hence is equivalent to a “geometric unit imaginary”, which is a very useful property. Examples of Tensors: (a) The energy-momentum tensor of the electromagnetic field in a vacuum takes as its argument the four-velocity of an observer and returns the negative of the four-momentum density of the field as a function of spacetime position relative to that observer: T( v; x) 12 F Fv ( v F) F where F(x) E(x) IB(x) is the Faraday bivector (see Section 7.2 for a full explanation). Thus T(v) is a tensor of type 2 – 1. (b) The electromagnetic field tensor can be considered to be a type 2 – 0 tensor given by the bivector F(x) = E(x) + IB(x) or as a type 2 – 1 tensor defined as follows: 7
used.)
f qF( v) qF v
Since T(a) is a linear function of its vector argument (a tensor field is also a function of position), we can write:
where f is the four-force on a particle with charge q. Often no distinction is made between the linear function F and the bivector F. (c) The Riemann curvature tensor takes a bivector as its argument and returns a bivector and so is a type 4 – 2 tensor. (d) The metric tensor g is a scalar-valued bilinear function
T(a) T(a j e j ) a j T(e j ) … (2.5)
Now the result of T on the vector ej, can be written as a linear sum of basis vectors, with the fact that the linear sum depends on j being indicated as: T(e j ) Tij e i , for some set of scalars, Tij ,
of vectors (see Sec. 8.3). In frame independent form: g(a, b) a b . … (2.10)
and where the {e i } are the reciprocal basis vectors which satisfy
Relative to a coordinate frame,
e i e j ij … (2.6),
where ij is the Kronecker delta symbol and
where the {g } are the reciprocal basis vectors to the coordinate basis vectors {g } . Equation (2.11) has the
the dot product will be defined in Table 2. Consequently, as required,
value in separating the metric tensor from its inputs. In other words, in MTW tensor notation adapted to GA form,
T(a) T(a j e j ) a j T(e j ) Tij a j e i …
(2.7) (Note that the index of the basis vector being acted on goes in the second position in the tensor components.) Note also that ei T(e j ) ei Tkj e k Tkj ik Tij … (2.8) and that T(a) can be written explicitly as a linear function of a as
g(a, b) g a b g g a g b , … (2.11)
the metric tensor g g g g g x x . … (2.12) The metric tensor can also be considered to be a vectorvalued linear function of single vectors. I.e.
g(..., a) g g g a g a g a g a so in a sense, the metric tensor is an identity map. But, it is a map from a coordinate basis expansion to a reciprocal coordinate basis expansion.
T(a) Tijei e j a … (2.9) Expansions like (2.9) are useful when considering a derivative of the tensor which is to act on the tensor only but not its Chapter 2. Geometric Constituents of Geometric Algebra
8
argument a. (See Appendix B for a greater explication of the link between usual tensor expressions and those that arise in GA. Note that this appendix depends on material from later sections.)
Chapter 2. Geometric Constituents of Geometric Algebra
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3. INNER, OUTER AND GEOMETRIC PRODUCTS Table 2. Products for vectors in geometric algebra. Product
Representation
Geometrical meaning in a 3D Euclidean vector space
Geometrical meaning in a vector space of arbitrary signature
Comments
Inner
a b (“a dot b”)
| a || b | cos … (3.1) where is the angle between a and b.
| a|| || b | | a || b|| | … (3.2)
(3.2) might seem a bit complicated, but once a metric is defined for a given coordinate basis, it is all quite straightforward. See for example, Ch. 6. In a Euclidean vector space, a|| a bˆ bˆ , where bˆ is a
where a || is the component of a parallel to b and vice versa, and 1 depending on whether a || is in the same or opposite direction to b and the signature of b b (e.g. in Minkowski spacetime, whether b is a spacelike or timelike vector).
unit vector in the direction of b. In a mixed signature space, this formula might give the wrong direction though, so in general it should be replaced by a|| a bb / b 2 … (3.3) where b 2 b b (Note that if b2 is negative, as can be the case in a Lorentzian manifold, then a|| a bˆ bˆ ). For vectors, the inner product is symmetric: a b b a … (3.4) To generalize to higher grade multivectors, note that the inner product involves taking projections and is grade lowering (i.e. the inner product of two vectors is a scalar).
Outer
ab (“a wedge b”)
ˆ … | a || b | sin B
ˆ | a || b | B ˆ … (3.5) | a || b | B
(3.4) where is the angle between a and b and Bˆ is a plane segment with unit area in the plane spanned by a and b and with the same orientation as a b .
where a is the component of a orthogonal to b and vice versa.
Chapter 3. Inner, outer, commutator and geometric products
Replaces the vector cross product. Advantage is that it is definable in an arbitrary number of dimensions. Consistent with the vector cross product, the outer product is antisymmetric: a b b a … (3.6) and so a a 0 … (3.7). To get a sign change on changing the order of a and b in an outer product, relate sign to a sense of rotation defined by the order of a and b (see figure).
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a a a || a a bb / b 2 b ba / b 2 a bb / b 2
(a b) b / b 2 by (3.21) … (3.8) To generalize to higher grade multivectors, note that the outer product involves taking orthogonal components (rejections) and is grade raising (i.e. the outer product of two vectors is a bivector). To relate to the familiar cross product in 3D, note that a i e i b j e j (a1b 2 a 2 b1 )e1 e 2 (a 2 b 3 a 3 b 2 )e 2 e 3 (a 3b1 a1b 3 )e 3 e1 Cf. a b (a 2 b 3 a 3 b 2 )e1 (a 3 b1 a 1b 3 )e 2
(a1b 2 a 2 b1 )e 3 where denotes the usual 3D vector cross product written with an overarrow to distinguish it from the commutator product of GA (see e.g. (3.38)). Thus, in terms of components, e1 e 2 e 3 , e 2 e 3 e1 and e 3 e1 e 2 (note the cyclic rotation of subscripts. See also (3.50)).
Geometric
ab
a b a b … (3.9)
Since a b is a symmetric product and a b is an antisymmetric product, this suggests that they might be considered to be the symmetric and antisymmetric parts of a more general product called the “geometric product”. As will be seen below, the value in introducing this new product is that it can be taken to be associative (i.e. a(bc) = (ab)c ; see App. A), which is key to defining more complex products such as a (b c) . The geometric product also has the advantage of being an invertible product. (If a 1 a / a 2 … (3.10), then aa 1 aa / a 2 1 as required.)
Chapter 3. Inner, outer, commutator and geometric products
11
Since a b is a scalar and a b is a bivector, their sum is treated as a formal sum like the real and imaginary parts of a complex number. Inner product in terms of geometric products
Outer product in terms of geometric products
a b
12 (ab ba ) … (3.11)
in general. = a||b = ab|| … (3.12) (This provides a geometric way of generalizing products to higher grade multivectors.) = ab = ba when a is parallel to b ... (3.13).
ab
By definition, a b 0 , so a b (a|| a ) b a|| b a||b by (3.9) and using
a || b 0 .
12 (ab ba ) … (3.14)
in general (note that this shows algebraically that a b b a )
a b can be taken to be the antisymmetric part of the geometric product. By definition, a || b 0 , so
a b ab … (3.15)
a b (a|| a ) b a b a b
(This provides a geometric way of generalizing products to higher grade multivectors.) = ab = –ba when a is orthogonal to b … (3.16).
by (3.9) and using a b 0 .
The key to generalising the inner and outer products to higher grade objects is to realize that the inner product is a product involving parallel components which is grade lowering, while the outer product involves orthogonal components and is grade raising. Thus for example, if B1 a b and B 2 c d , then splitting a and b into components parallel and orthogonal to B2, one finds that B1 a|| b|| a b (a|| b a b|| ) and so there are three possible products in general (more details are given in Table 3): (a) the inner product (a scalar): B1 B 2 (a|| b|| )B 2 | (a|| b|| ) || B 2 | , where the sign is determined by the signature of the vectors involved; Chapter 3. Inner, outer, commutator and geometric products
Since a b is a symmetric product and a b is an antisymmetric product, a b can be taken to be the symmetric part of the geometric product.
(b) the outer product (a quadvector and so only possible in 4D or higher): ˆ is a unit fourˆ where Q B1 B 2 (a b )B 2 (a b ) || B 2 | Q volume which lies in the vector sub-space spanned by a, b, c and d; and (c) the commutator product (a sum of bivectors): ˆ | (a b ) || B | B ˆ , where Bˆ and Bˆ B1 B2 | (a|| b ) || B2 | B 3 4 3 || 2 4 are unit bivectors obtained from a|| b and a b|| respectively, by rotating them through 90o in the B2 plane (the sense of orientation will be explained in Table 3). These ideas will be made precise in Table 3. 12
Table 3. Products involving higher grade elements in geometric algebra. Product
Representation
Geometrical meaning in a 3D Euclidean vector space
Geometrical meaning in a vector space of arbitrary signature
Comments
Inner product between a vector and a simple bivector or blade
aB a (b c)
| a || B | cos uˆ … (3.17) where is the angle between a and the component of a in the plane of B and uˆ is a unit vector in the plane of B which is orthogonal to a and such that a|| uˆ has
a|| B | a|| || B | uˆ … (3.19)
(b c) a b a c c a b a (b c) … (3.21) So a B is an antisymmetric product which can thus also be written as: a B 12 (aB Ba ) … (3.22).
the same orientation as B. In a 3D Euclidean space, a (b c ) a (b c ) … (3.18) where signifies the Gibbsian vector cross product
where again uˆ is a unit vector in the plane of B which is orthogonal to a. a || uˆ has the same orientation as given by B if a||2 > 0, otherwise it is in the opposite orientation. (Proof: Can always write B a || b a ||b where b is orthogonal to a||. Thus a||B = a||2b, and so is either parallel or antiparallel to b depending on the sign of a||2.)
a (b c) a b c a c b … (3.20) which is a an extremely useful result.
[This is the standard way that a B is expressed. It has the advantage of being valid even if B is not a blade, but has the big disadvantage of having no obvious geometrical meaning. The usual derivation of the expansion of a (b c) “follows” from this expansion by next writing B b c 12 (bc cb ) and then being really clever in re-expresing, rearranging and grouping terms (e.g. [DL: pp. 31-32]) . In contrast, by using the approach shown below, geometry can be used to guide the derivation.] Note that if a lies in the plane of B, then we can write B ab for some vector b orthogonal to a. –1
Consequently, Ba aba aab aB and thus in this case, (3.22) can be simplified to a B aB … (3.22’). a (b c) generalises the vector triple cross product to any number of dimensions. a (b c) a b c a c b can be determined using geometric products as follows (the main idea is to expand vectors in parallel ( || ) and orthogonal ( ) components): (i) b c b c b c by (3.15) where b b b|| b b cc / c 2 .
(ii) a|| a b b / b 2 a cc / c 2 (iii) Chapter 3. Inner, outer, commutator and geometric products
13
a (b c) a || (b c) a b b b c / b 2 a cccb / c 2 a b c a cb a (b b cc / c 2 )c a c(b b cc / c 2 ) a b c a c b.
From above, a a a bb / b 2 and a a.1 a(b b / b 2 ) , so a a b b / b 2 a bb / b 2 (a b) b / b 2 … (3.23) by (3.21).
Outer product between a vector and a simple bivector
aB a (b c)
Inner product between two simple bivectors (or blades)
B1 B 2
| B1 || B 2 | cos
(a b) (c d)
where is the angle between the two bivectors and the product is negative if the two bivectors have the same relative orientation.
ˆ … (3.24) | a || B | T
where a is the component of a orthogonal to B and Tˆ is a trivector of unit magnitude in the vector subspace spanned by a and B.
| B1|| || B 2 | … (3.26a)
If a and b are split into components parallel and orthogonal to the (c d) plane, then where B1|| is the part of B1 which lies a b (a|| a ) (b|| b ) , which gives a component in B2 and 1 depending on the a|| b|| B1|| which lies wholly within the plane of B2 signatures of the vectors involved and the relative orientation of the and hence is the only contributor to the inner product; a two bivectors; see next formula. component a b which is wholly orthogonal to the (a|| b|| )(c d)
a db c a cb d … (3.26b) a (b (c d)) … (3.26c) using (3.20), so
(a b) B a (b B) (a B) b (B a ) b
… (3.27) From (3.27), it follows that if:
B B e j ek [Einstein 1 2
jk
summation convention for repeated Chapter 3. Inner, outer, commutator and geometric products
Two interchanges of the order of the outer products in a (b c) with a change in sign each time gives a B B a , so this product is symmetric. It can thus be expressed as: a B 12 (aB Ba ) … (3.25).
plane of B2 and hence is the only contributor to the outer product; and components a|| b a b|| which lie both within B2 and orthogonal to it and hence requires the definition of a new product to describe it (see commutator product below). (3.26): Now since c c c d / d 2 is the component of c orthogonal to d, then a|| a c c / c 2 a d d / d 2 and
b|| b c c / c 2 b d d / d 2 . Consequently, a|| b|| a c b dc d / c 2 d 2 a db c d c / c 2 d 2 14
indices assumed], then
.Thus, (a|| b|| )(c d) involves evaluating
B jk (e k e j ) B (e j e k ) B
c d c d d c c d d c c d c 2 d 2 and
(e j B) e k
… (3.28) and the reciprocal basis vectors {ek} are defined by: e k e j kj … (3.29),
where jk is the Kronecker delta symbol, i.e.
1, if j k … (3.30) 0, if j k In terms of geometric products: B1 B 2 B1 ||| B 2 … (3.31)
kj
Using (3.28) to expand two bivectors in terms of coordinate basis vectors {g j } with
g j g k g jk , it follows from (3.26b) that
B C 12 B jk e j e k 12 C mne m e n 12 B jk C jk ... (3.31' )
d c c d d c c d c 2 d 2 . Substituting these results and c gives (3.26b).
Note that the product is symmetric, so B1 B 2 12 B1B 2 B 2 B1 … (3.32) where means the scalar part. Note that if B1 = B2 = B and a is orthogonal to b, then B B B2 abab a 2b 2 (using ba ab when a and b are orthogonal), which in a Euclidean vector space is the negative of the square of the area of B. In Minkowksi spacetime, B2 could be positive if one of a or b is a timelike vector. ˆ be a unit spacelike* bivector blade so that Let B 2 ˆ 1 … (3.33). B
ˆ behaves like a geometric unit imaginary This means B ˆ ) in a Taylor series, one finds and so expanding exp(B ˆ ) cos B ˆ sin … (3.34) that exp(B
which is useful when considering rotations of vectors. Similarly, if Bˆ is a unit timelike* bivector so that ˆ 2 1 … (3.35), B ˆ ) in a Taylor series gives expanding exp(B ˆ ) cosh B ˆ sinh … (3.36) exp(B
which is useful when considering hyperbolic rotations (Lorentz boosts). * Spacelike bivector: there exists a particle worldline orthogonal to the bivector. Timelike bivector: a particle worldline can lie in the plane of the bivector. (See Ch. 6.) Chapter 3. Inner, outer, commutator and geometric products
15
Outer product between two simple bivectors
B1 B 2 (a b) (c d)
(a b )(c d) … (3.37)
Since a a 0 , this product can only be non-zero in a vector space with four or more dimensions. Magnitude is the volume of the parallelepiped spanned by a, b, c and d. By successively swapping the order of pairs of outer products and swapping sign each time, one finds B1 B 2 is a symmetric product so B1 B 2 12 B1B 2 B 2 B1 4 … (3.37) where . 4 means the grade 4 part.
Commutator product between two bivectors
B1 B 2 (a b) (c d)
(a|| b a b|| )(c d) …
(3.38)
b (a|| (c d)) a (b|| (c d)) a cb d a db c (3.39a) b ca d b da c
Is an antisymmetric product so B1 B 2 12 B1B 2 B 2 B1 2 … (3.40) though 2 is not really necessary. Note that the commutator product results in a bivector, and so is grade preserving. Consider (a b|| ) (c d) . WLOG, assume c d and b|| || c . Thus
[(a b) c] d c [(a b) d]
(a b|| ) (c d) a b||cd
… (3.39b)
| b|| || c | a d which is a b|| rotated by 900 in the plane of c d and dilated by | c d | .
Geometric product between two bivectors
B1B2
Inner product between a vector and a trivector
aT a (b c d)
B1 B 2 B1 B 2 B1 B 2 … (3.41)
Chapter 3. Inner, outer, commutator and geometric products
= sum of scalar, bivector and quadvector parts. (Quadvectors can’t exist in 3D, so the quadvector is only a possibility in 4D and higher.) Rearranging (3.43), (B a) b a bB a (B b) … (3.44)
a bc d a cb d a db c … (3.42) In Euclidean 3D, a must lie within the space spanned by Thus using results from above, T, and hence it must be possible to expand T as a (b B) a bB b (a B) (3.43) T a b c abc where a, b and c are all mutually 16
orthogonal. ˆ , where In this case, a T aT aabc | a |2 | b || c | B
ˆ is a unit bivector orthogonal to a which has the same B
orientation as b c . In 4D and higher, it is possible for a to have a component which lies outside the space spanned by T, so in this case, ˆ … (3.45). a T | a|| |2 | b || c | B Geometric product between a bivector and a trivector
TB
T B T B T B … (3.46)
By definition, T B can only be non-zero in 5D or higher. In 3D, B must lie wholly within the space spanned by T so it must be possible to write
TB (ab1b 2 )(b1b 2 ) with a, b1 and b2 all mutually orthogonal. Recalling that the order of orthogonal vectors in a geometric product can be swapped with a change of sign, it follows that –1
ab1b 2 b1b 2 ab1b 2 b 2 b1 | b 2 | 2 ab1b1 | b1 | 2 | b 2 | 2 a
and hence TB T B | B |2 a … (3.47) where a is orthogonal to B and T a B . In 4D, it is possible for one of the components of B to have a component orthogonal to the space spanned by T. I.e. TB (ab 1b 2 )(b1 (b 2 b )) ab 1b 2 b1b 2 ab1b 2 b1b T B | b1 | 2 ab 2 b
and thus T B | b1 |2 ab 2 b … (3.48) which is a trivector which is composed of a plane lying Chapter 3. Inner, outer, commutator and geometric products
17
within T and a vector orthogonal to T. Note that T B lowers the grade of T by two, T B raises the grade of T by two, while T B preserves the grade of T. Inner product between a bivector and a trivector
BT B1 (B 2 a)
= B||T ... (3.45) where B|| is formed by projecting components of B onto the components of T.
B1 B 2 a (B1 a) B 2 … (3.49) Geometric products with the unit pseudoscalar – duality transformations
a b I 3 a b or a b I 3a b
… (3.50)
Here, I 3 the unit pseudoscalar in a 3D Euclidean space.
II I 2 1 ... (3.53) in 2D, 3D and Minkowksi spacetime, and so I acts like a geometric unit imaginary in these vector spaces. From [DL: Sec. 4.1.4, p. 97]:
a A r I a ( A r I) … (3.54a)
a b ( I 3b ) a a ( I 3b )
or a A r a ( A r I)I … (3.54b)
… (3.51)
or a A r a ( A r I)I … (3.55b)
a b ( I 3 a) b
… (3.52) Note that to avoid confusion with the commutator product, the 3D vector cross product is indicated by an over arrow.
a A r I a ( A r I) …(3.55a) A r B s I A r (B s I) … (3.56) where the subscripts r and s indicate the grade of the multivector blade.
Product is symmetric
(a b) (c d e) (a b) (c d)e (a b) (c e)d (a b) (d e)c a (b (c d e)) Multiplying by the unit pseudoscalar returns the orthogonal complement of the multivector. (The orthogonal complement of the pseudoscalar is a scalar.) Since I is a unit pseudoscalar, it doesn’t change the magnitude of what it operates on. Since the inner and outer products of a vector and a multivector blade involve the components of the vector parallel and orthogonal to the multivector blade, one would expect that the inner and outer products would be linked by a duality transformation. Proof [Hestenes]: aM r I (aM r )I a(M r I) (a M r a M r )I a (M r I) a (M r I) Now a M r has grade r 1 so
(a M r )I has grade n (r 1) which is the grade of a (M r I) In contrast, (a M r )I has grade n (r 1) which is the grade of a (MI ) so (a M r )I a (M r I) and
(a M r )I a (M r I) These results provide a way of relating the axial vectors Chapter 3. Inner, outer, commutator and geometric products
18
of Gibbsian vector algebra with the associated GA bivector and are useful for generalizing integral theorems. Symmetry relation [DL: p. 97]: IA r (1) r ( nr ) A r I … (3.57) where r is the grade of the multivector and n the number of dimensions.
Chapter 3. Inner, outer, commutator and geometric products
19
Table 4. Applications Application
Formula
Notes
Finding the component of a vector which lies in the plane of a bivector blade
a|| (a B)B / B 2 … (3.58)
a B produces a vector proportional to a|| which lies in the plane of B but rotated through 90o. Take another inner product with B to either rotate through another 90o or back again depending on the signature of B. This inner product reduces to a geometric product since a B lies in the plane of B. Divide by B2 to remove dilation by |B| from each inner product with B. the signature of B2 automatically gets direction right. (Proof via a B (a a|| ) B 0 .)
Simpler: a|| a|| .1 a||BB / B2 (a B)B / B2 (a B) B / B 2
Finding the component of a vector which is orthogonal to the plane of a bivector blade
a (a B) B / B 2 … (3.59)
Splitting a simple bivector B into parts relative to which a vector a is orthogonal or “parallel” (i.e. a lies in the plane spanned by B||).
B || (B a) a / a 2
a a .1 a BB / B 2 a BB / B 2 (a B) B / B 2 since (a B)B (a B) B (a B) B (a B) B
and (a B) B 0 by definition. Also, (a B) B 0 since it is a trivector while a is a vector.
a (a B) / a 2 a(a B) / a 2
Meaning: a lies in the plane defined by B|| and B B|| B with a B . … (3.60)
Derivation: (1) B|| B|| .1 B||aa / a 2
aB || a B
(2) B||a B|| a B a because B|| a 0 and B a 0
Alternative [ref. ?]:
(3) (B a)a (B a) a because B a a
aBa B|| 12 B 2 … (3.61) a
Derivation of alternative: (B a ) a
(B a)a because B a a and so (B a) a 0 12 (Ba aB )a 12 (Ba2 aBa) B || (B a) a / a 2 desired expression.
Chapter 3. Inner, outer, commutator and geometric products
20
B (B a) a / a 2 … (3.62) (a aB a (a B)) / a 2
=> B B
Derivation (1) B B .1 B aa / a 2
a (a B)
(2) B a B a B a because B a 0 and B|| a 0 .
a2
(3) (B a)a (B a) a because (B a) a 0 by definition.
B B B|| , and so using (3.61), one obtains: aBa B 12 B 2 … (3.63) a
Rotations in Euclidean Vector Spaces Through an angle in the plane and with the sense of circulation given by ˆ the unit bivector B
ˆ / 2)a exp(B ˆ / 2) a exp(B … (3.64) ˆ is a unit bivector in where B the plane of rotation with the same orientation as required for the rotation. Formula written in short as: ~ a RaR … (3.65) ~ where R is a “rotor” and R is its reverse. Alternative form If aˆ and bˆ are unit vectors which define the plane of rotation and bˆ is aˆ rotated through the angle in the right direction, then ˆ sin and aˆ bˆ B aˆ bˆ cos , from which it ˆ / 2) follows from R exp(B that
Chapter 3. Inner, outer, commutator and geometric products
The idea of rotating about an axis only works in 3D, because only in 3D is there a unique orthogonal plane to an axis. However, rotating an object in 3D about an axis means that directions which lie in the plane orthogonal to the axis get rotated in the plane, while directions orthogonal to the plane are unrotated. This idea generalizes to arbitrary numbers of dimensions. Now in 3D, if one takes the rotation to happen uniformly for some length of time, then dr / dt r … (3.68) The GA perspective is to replace the axial vector with the bivector I 3 … (3.69) where defines the plane and sense of rotation and | || | . [Note that if k , then I 3 ijkk i j , which has the right orientation by the right hand rule.] Now dr / dt r ( I 3 ) r r … (3.70) by (3.52) and (3.69), and since r can be defined in an arbitrary number of dimensions and has the right geometric meaning, we now have a dimension independent formula. To integrate this DE, replace the inner product with geometric products: dr / dt r 12 (r r ) , which, taking into account that in general the geometric product does not commute, implies that: r (t ) exp(t / 2)r (0) exp(t / 2)
ˆ ˆ Taking t to be the angle of rotation, then r ( ) exp( / 2)r (0) exp( / 2) … (3.71)
Note that if r lies in the plane of , then we can write r r and we can get a
21
1 bˆ aˆ R ... (3.66) 2(1 aˆ bˆ ) Using this expression for R, it is easy to show that ~ ~ RR RR 1 … (3.67) [unimodularity of R]
single sided formula: r (t ) exp(t )r (0) … (3.72). (See 3.22’) ******************************************************************** ˆ / 2 and B B ˆ / 2 , note that in general, Warning: Letting B1 B 2 2 2 1 1
R1 R2 exp( B1 ) exp( B 2 ) exp( B1 B 2 ) . Geometrically, this is because two finite rotations do not in general commute, and the last expression loses track of the order of the rotations. Mathematically it follows because exp( B1 ) exp( B 2 ) (1 B1 12 B12 ...)(1 B 2 12 B 22 ...) 1 (B1 B 2 ) 12 (B12 2B1B 2 B 22 ) ...
while exp( B1 B 2 ) 1 (B1 B 2 ) 12 (B1 B 2 ) 2 ... 1 (B1 B 2 ) 12 (B12 B1B 2 B 2 B1 B 22 ) ...
and in general, B1B 2 12 (B1B 2 B 2 B1 ) because in general the geometric product is not commutative. The one exception to this general result is when B 1 and B 2 lie in the same plane, and in particular, it is valid to write: ~ RR exp(B) exp(B) exp(B B) exp(0) 1 .
Rotations in Minkowski Spacetime are of two types: in a spacelike plane and in a timelike plane (and mixtures of these two).
As above for spacelike rotations.
Now B a gives a vector orthogonal to a in the plane of B regardless of the signature of the vector space, and so can continue to define a rate of change of a which leaves the length of a unchanged and hence can be used to define a pure rotation. Now provided a is not orthogonal to B, B can be written as a||b where b is orthogonal to a||. Doing this it follows that B a a||ba || a||2 b which is parallel or antiparallel to b depending on whether a|| is spacelike or timelike respectively and whether the spacelike or timelike convention is being used for the metric. In these notes, the signature of spacetime will be taken to be ( ) . Define B to be a spacelike bivector if it contains no timelike vectors, in which case a|| is spacelike and B a rotates a|| in the sense of B as for Euclidean vector spaces. … (3.73) Define B to be a timelike bivector if it includes timelike directions. When this is the case, one of a|| or b must be timelike and the other spacelike since they are taken to be orthogonal. If a|| is spacelike, then again B rotates a|| in the sense of B, while if a|| is
Chapter 3. Inner, outer, commutator and geometric products
22
timelike, then B rotates a|| in the sense opposite to B. … (3.74) (See p. 38.) Note that if B is spacelike, then B 2 a||ba ||b a||2 b 2 0 … (3.75) while if B is timelike, then B2 a||ba ||b a||2b 2 0 … (3.76) => ˆ / 2) cos( / 2) B ˆ sin( / 2) for a spacelike bivector … (3.77) and exp( B
exp( Bˆ / 2) cosh( / 2) Bˆ sinh( / 2) for a timelike bivector … (3.78). ~ ~ ~ Note that if dx RdxR , then using RR R R 1 , ~ ~ ~ ~ dx dx dxdx RdxRRdxR Rdx dxR dx dxRR dx dx which means rotations leave the lengths of displacement vectors invariant as required. These ideas will be explored in more detail in Ch. 6.
Chapter 3. Inner, outer, commutator and geometric products
23
4. LINEAR (TENSOR) ALGEBRA WITH GEOMETRIC ALGEBRA This chapter is largely based on [DL: Sec. 4.4]. The goal is to develop frameindpendent approaches to performing linear algebra calculations which will have uses in relativistic physics.
4.1 Basic Definitions As tensors can be conceived of as (multi-)linear functions of multivector arguments, a geometric algebra approach to linear algebra is of fundamental importance. The approach starts by defining a linear vector-valued function F to have the property F(a b) F(a) F(b) … (4.1) if and are scalars. Using this property, if a is expanded in terms of basis vectors, then
F(a) F(a j e j ) a j F(e j ) … (4.2), where the Einstein convention that repeated upper and lower indices are summed over is being used. Now the action of F on ej must be able to be written as a linear sum of basis vectors, with the linear sum depending on j. That is,
F(e j ) F i j e i … (4.3) for some set of scalars Fij (if this set of scalars is a smoothly varying function of position x, then F constitutes a tensor field). And thus, F (a) F i j a j e i … (4.4),
which is equivalent to matrix multiplication of the matrix [Fij] (i = row number, j = column number) by the column vector [aj]. Note that F can be written in a frame-independent way. For example, F(a) a B(x) for some bivector field B(x) if F is antisymmetric. The symmetric stress-energy tensor for a relativistic perfect fluid is given by T(a) ( P)a vv Pa , where v is the local four-velocity of the fluid, P Chapter 4. Linear (Tensor) Algebra with Geometric Algebra
the local pressure relative to the local inertial rest frame, and the local inertial rest frame energy density [DL: p. 444 with a sign change due to using a metric with the opposite signature]. In general though, since a j e j a , it follows that (4.4) can be rewritten as: F(a) F i j e i e j a , … (4.5) To extract the matrix components of a linear function with respect to some set of basis vectors, note that
e j kj e k F(e j ) F( kj e k ) kj F(e k ) kj F m k e m F m j e m Next take the inner product with ei and use e i e m mi to obtain:
F i j e i F(e j ) … (4.6). One could also expand F(e j ) in terms of reciprocal basis vectors as
F(e j ) Fkj e k , from which it follows that Fij e i F(e j ) … (4.7). But from above,
F(e j ) F k j e k , so Fij e i F(e j ) e i F k j e k g ik F k j … (4.8) where g ik e i e k , thus obtaining the index lowering property of the metric tensor familiar from tensor analysis. These results allow one to link to the tensor product ideas expounded in [MTW]. Specifically,
F(..,..) Fij ei e j and
F(.., a) Fijei e j a In terms of basis vectors then, the stress-energy tensor for a perfect fluid is given by
24
T(a) T(.., a) ( P) vv a Pa
F(b) a F(a) b ... (4.12)
( P)vi e v j e a Pai e ( P)vi e v j e a Pe ( g ij e a)
Proof :
and thus
a F(a) b
i
j
i
i
T(..,..) ( P)vi v j e e Pg ij e e i
j
i
j
i
j
j
( P) v v Pg(..,..) ( P) v v Pg.
e i lim
F(a e i ) b F(a) b
e i lim
F(a) b F(e i ) b F(a) b
0
The action of F on a multivector is usefully defined as follows: F(a b ... c) F(a) F(b) ... F(c) … (4.9).
Note that e i e i a e i ai a,
e i e i a F(a) b
0
and so e i e i 1 .
e F(e i ) b i
4.2 Adjoint
e i Fki e k b
Recall F(b) Fik b k e i , and so
Now note that
e i Fki b k
F(e i ) e j F k i e k e j F k i g kj F ji and
a F(b) a i Fki b k F(a) b
ei F(e j ) F k j ei e k F k j g ik Fij
a i F(e i ) b a i Fki e k b
This result motivates defining the adjoint F of F as being determined by:
a F(b) F(a) b ... (4.9)
Some useful identities [DL: p. 107]: A r F(B s ) F(F( A r ) B s ) r s ... (4.13a)
b F(a) F(b) a
F( A r ) B s F( A r F(B s )) r s ... (4.13b)
F(b) Fki b k ei .
from which it follows that ad(F)(a) F(a) … (4.10). Furthermore,
a ad(F(G (b)) F(G (a)) b by definition G (a) F(b) treatin g G (a) as being like a a G (F(b) using definition of adjoint, now treating F(b) like b ad(F(G (b)) G (F(b)) ... (4.11)
A frame-independent approach to calculating the adjoint is given by:
A symmetric function equals its own adjoint (see just above (4.9)), and so satisfies a F(b) F(a) b … (4.14)
4.3 Determinant [DL: pp. 108-109] F (I ) F(e1 ... e n ) F(e1 ) ... F(e n ) VI ... (4.15)
Chapter 4. Linear (Tensor) Algebra with Geometric Algebra
25
since any n-vector must be a multiple of I and where V is the volume of the parallelepiped spanned by the F(e i ) . Now F(e j ) F i j e i which corresponds to a vector given by column j of the matrix Fij. Since one interpretation of the determinant of a matrix is that it is the volume of the parallelepiped spanned by the set of vectors given by the columns of the matrix, it therefore follows that V det(F) , or F(I ) VI det( F)I … (4.16) From (4.16) it follows simply that F(G (I)) F(det(G )I) det( G )F(I) det( G ) det( F)I Or in other words, det( FG ) det( F) det( G ) … (4.17)
4.5 Trace a F (a) e (e i a F (a)) i
a eiei a
and so a (...) e i (e i a ...) F ( a e ) F ( a ) i e i lim since e i a is a scalar. 0 F (a) F (e i ) F (a) e i lim by linearity of F 0 e i F (e i ) ei F j i e j ij F
j
i
F ii
tr(F ) a F (a) ... (4.20)
Also, since F (F 1 (I )) I
it follows that det( FF 1 ) 1 det( F) det( F 1 ) 1
1
det( F ) det( F) ... (4.18)
a F (a b) e i (e i a F (a b))
[DL: pp. 109-110]
F ((a e i ) b) F (a b) e i lim 0 F (a b) F (e i b) F (a b) e i lim by linearity of F 0
F 1 ( A) I F (I 1A) det( F ) 1 ... (4.19a)
e i F (e i b j e j )
F 1 ( A) I F (I 1A) det( F ) 1 ... (4.19b)
b j e i F (e i e j ) ...(4.21)
4.4 Inverse
If, for example, F(a b) is bivector valued, then
Chapter 4. Linear (Tensor) Algebra with Geometric Algebra
26
b j ei F(ei e j ) b j ei 12 F kl ije k el ( with F kl ij antisymmet ric in kl) 12 F kl ijb j ( ki el li e k ) 12 F il ijb j el 12 F ki ijb j e k F li ijb j el F il ijb j el F(b) Thus,
a F(a b) F(b) … (4.22) where F(b) F(b j e j ) b j F(e j ) b j F l j el and
F l j F il ij … (4.23)
Chapter 4. Linear (Tensor) Algebra with Geometric Algebra
27
5. GEOMETRIC CALCULUS 5.1 Gradient operator A coordinate/frame-independent directional derivative operator (“grad”) is defined as follows: M ( x a) M ( x) a M (x) lim … (5.1) 0
for sufficiently smoothly varying multivector fields M(x). In addition, for scalar fields (x) , (x) has the usual frame independent geometric meaning of being a vector orthogonal to the level curves of (x) which points in the direction of the maximum rate of change of (x) and whose magnitude is that maximum rate of change.
5.2 Coordinate and reciprocal bases and the metric tensor Introduce a set of coordinates {xi} which label the points of an n-dimensional vector space. Relative to some origin O, the relative position of the points of the vector space can be specified with the relation:
x x( x1 ,..., x n ) x( x i ) … (5.2). The coordinate basis {ei} is defined by: ei x / x i … (5.3); i.e. ei is the direction in which xi but no other coordinate varies. The reciprocal basis {ei} is defined by: e i x i (x) … (5.4); i.e. the coordinates xi can be envisaged as being scalar functions of the position vector x and ei is orthogonal to the contour lines/hypersurfaces of xi (i.e. of the hypersurfaces xi = constant) – see (5.7).
and so e i
x i
… (5.6).
1 if i j x j j Also, e i e e i x (x) i i … (5.7), x 0 if i j j
j
which is why {ei} is called the reciprocal basis of {ei}. Now if a a i ei then a e j a i ei e j a i i j a j … (5.8). With a j e j a , it follows that a a j e j a e j e j and thus, to form an expansion of ej in terms of the basis {ei}, form the sum: e j e j ei ei g ji ei … (5.9) where g ji e j e i are the components of the reciprocal or inverse metric tensor. Similarly, e j e j ei ei g ji ei … (5.10) and thus the metric tensor and its inverse can be used to raise and lower indices. E.g. a a i ei a i g ij e j a j e j a j g ij a i g ji a i … (5.11) since g ij ei e j e j ei g ji . Note also that: i j ei e j g ik e k e j g ik g kj and hence g kj ( g kj ) 1 … (5.12).
From the above, it follows that relative to a coordinate basis: ei i ei i … (5.13). x
Now, by (5.1), (x e i ) x x e i x lim ei i … (5.5), 0 x Chapter 5. Geometric Calculus
28
5.3 “Div” and “Curl”
(x) ei i e j j ei e j i j 0 since ei e j is antisymmetric
The interior derivative or divergence:
in i and j whereas i j is symmetric in i and j and so for example:
M(x) e i M(x) e ( i M(x)) … (5.14) i
i
since the scalar operator i can cross the inner product. A frame independent definition of the divergence has to await the divergence theorem. Using the geometric product, , , M (x) 12 M M … (5.15)
where +/– is determined by the grade of M and the check marks indicate the “scope” of the gradient operator so that ,
,
,
,
M M e i i ( i M)e i … (5.16).
The exterior derivative or curl: M(x) ei i M(x) ei ( i M(x)) … (5.17)
since the scalar operator i can cross the outer product. A frame independent definition of the curl has to await the generalisation of Stokes’ theorem.
5.4 Useful Differential Identities
e1 e 2 1 2 e 2 e11 2 e 2 e1 2 1 Thus it is true in general that: (x) 0 … (5.18) for scalar functions (x) . Note the particular result: x i (x) 0 e i 0 … (5.19) for an arbitrary reciprocal coordinate basis (i.e. (5.19) does not rely on the {ei} being orthonormal and constant for its validity). This result allows some calculations involving curls to be simplified if the multivectors are expanded in terms of reciprocal coordinate basis vectors. Also, ( M) 0 … (5.20) Proof: Again for simplicity, assume a rectangular coordinate system so derivatives of the reciprocal basis vectors are zero. Thus
( M) ei i (e j j M) ei (e j i j M) (ei e j ) i j M 0 for the same reasons as above. ______________________________
Having a table of identities both makes doing physics calculations more efficient and also suggests useful directions for calculations. Hence, the purpose of this section is to derive a number of results useful for physics.
( M) ( ) M M … (5.21) Proof: (M ) e i i (M ) e i i (M ) e i [( i )M i M ]
Since the inner and outer product and grad operator are all frame-independent, then a result evaluated with respect to some special frame when written in a frame-independent way must be true for all frames. Thus for simplicity, all proofs in this section will be made assuming the basis vectors are orthonormal, constant, rectangular / Cartesian vectors, unless explicitly stated otherwise.
( i ) e i M e i i M ( ) M M
Thus for example, relative to a rectangular frame for which the coordinate and reciprocal basis vectors are constant, Chapter 5. Geometric Calculus
Thus a (a i e i ) (a i ) e i a i e i (e j j a i ) e i a i e j j e i i a i a i e j j e i
since e j ei i j .
29
Now j e i must be some linear combination of the basis vectors, so can write it as j e i kji e k for some set of position dependent coefficients kji . j Thus e j j ei kji e j e k kji kj jij and so a i a i ji a i …(5.22)
It can be shown that (see Sec. 8.3.6): kji 12 g kl ( j g il i g jl l g ji ) ... (5.23) jij 12 g jl ( j g il i g jl l g ji ) 12 g jl i g jl ... (5.24) after swapping some dummy repeated indices.
It can be shown using the expansion of the determinant, that 1 g 1 | g | 1 V … (5.25), jij 12 g jl i g jl i 2 g x V x i | g | x i where g det( g ij ) . 1 (Va i ) … (5.26) V x i where e1 ... e n VI … (5.27)
Thus a
and if e i e j g ij , V | det( g ij ) | … (5.28) As a special case: 2 ( ) (ei i )
1 1 (V i ) (Vg ij j ) i V x V x i
… (5.29) where the index raising property of the metric tensor has been used in the last step. Note that in these notes, in analogy with vectors, 2 will be taken to mean rather than . Now since a vector can be expanded in terms of either coordinate basis vectors or reciprocal basis vectors, this allows us to write: a (ai e i ) (ai ) e i ai e i (e j j ai ) e i e j e i j ai e i e j j ai e i e j i a j
and thus a ( i a j j ai )e i e j … (5.30)
Chapter 5. Geometric Calculus
provided it is understood that the expansion is not to include any identical bivectors up to the sign of rotation. ______________________________ ( M) 0 … (5.31) where the grade of M 2. Proof: Again for simplicity assume a rectangular coordinate system so derivatives of the reciprocal basis vectors are zero. Thus
( M) ei i (e j j M) ei (e j i j M) (ei e j ) ( i j M) 0
because ei e j is antisymmetric in i and j whereas i j M is symmetric in i and j. (The last step follows from (3.27) if M is a bivector, and (??) if M is a trivector, but can be shown to be true in general [DL: p. 172].) ______________________________ Similarly, ( M) ( ) M M … (5.32) ______________________________ (a b) b a b a a b a b … (5.33) (In 3D (a b ) ) ______________________________ (a B) ( a) B a ( B) … (5.34) Proof:
(a B) e [( a) B a ( B)] e [( a) B a ( B)] [e ( a)] B] [e ( B)] a ( a) B ( B) a ______________________________
( A r B s ) ??? … (5.35) Proof: ( A r B s ) ei i ( A r B s ) ei ( i A r B s A r i B s ) ______________________________ 30
Grad and symmetric tensors ( M) M ( M) … (5.36) if the grade of M 2 (In 3D ( M ) ). Proof: j
i
j
j
i
2
j
,
,
,
,
,
,
,
,
i
,
,
( iT jk )e j ki T jk ( i e j ) ki T jk e j ( i e k ) e i T jk ,k e j T kj ,k e j
j
,
e e i j M e (e i j M) M e j (e i M) i
,
i T(e ) i T jk e j e k e i (by (4.5))
( M) e i (e j M) e (e i j M) i
,
T() T(e i i ) i T(e i ) by linearity of T and i being a scalar. Now,
i
,
T() T jk ,k e j ( k T jk )e j … (5.43a)
M ( M) If M is only a vector a, then ( a) a a … (5.37) ______________________________
if the basis vectors are orthonormal, constant Cartesian vectors.
(a b) a b b a a ( b) b ( a) … (5.38) Proof:
T
(a b) e i (a b) e ( i a) b e a i b i
Now,
i
i
( a) b (e i i a) b e i b i a i a be i
e i ( i a) b ( a) b b a b a b ( a) etc. etc.
For arbitrary coordinate basis vectors, ,
i T
T T
, , jk
,
e j e k e i ki T
jk
,k e j
jk jk
,k
jk
,i e j
ki T jk ijm e m T jk e j ikm e m e i
T jk kjm e m T jk e j mmk
j km T m k mmkT
jk
e
j
;k e j
… (5.43b) ,
,
i.e. T() T
jk
;k e j
… (5.43c)
Similarly, (B1 B 2 ) ( B1 ) B 2 ( B 2 ) B1 (B1 ) B 2 (B 2 ) B1 … (5.39) and thus (F F) 2( F) F 2(F ) F … (5.40). _____________________________
Note also that by the symmetry of T ,
(a B) ( a)B (B ) a a B a ( B) … (5.41) _____________________________
T jk ; j ak T jk ;k a j … (5.44) and as would have been attained from taking the inner product of (5.43c) and a.
( ) M ( M) ( M) M … (5.42) _____________________________
Chapter 5. Geometric Calculus
,
,
,
,
T() a T(a) by (4.14) ,
T ,
, , jk
,
e j ek a
ei i T
, , jk
,
e j ek a
31
5.5 Integral Theorems 3D Divergence Theorem: f ndS fdV ... (5.45a) V
V
f ndSI fdVI V
V
f (ndSI ) fdV V
V
f dS fdV ... (5.45b) V
V
Let the volume V be a sphere of radius r centred on the point P0 and take the limit as r 0 . In this limit, f is essentially a constant over the volume and hence (4.44a) gives: 1 f ( P0 ) lim f ndS r 0 V V so that the geometric interpretation of the divergence is that it is the flux per unit volume of the field f at each point.
where the orientation of dS is defined to be such that ndS | dS | I dSI … (5.46), so that ndSI n 2dS dS if n 2 1. Alternatively, post multiplying (5.45b) by I-1 gives:
(f dS)I
V
1
n(dS ) n dS n dS n dS since n dS
fdV … (5.45c). V
The form of the LHS of (5.45c) makes sense when one considers that f dS uses the part of f orthogonal to dS as does f ndS , and I-1 is needed to turn the trivector result into a scalar. Alternatively, (f dS )I 1 (f dS )I f (dSI ) … (5.47) (Note from above that ndS = dSI-1.)
3D Stokes Theorem: f dr ( f ) ndS ... (5.48a) S
S
f dr [I ( f )] (dSI 1 ) S
12
S
{[I( f )](dSI
1
) (dSI 1 )[I ( f )]}
S
( f ) dS ... (5.48b) S
because I commutes with bivectors and II -1 1. Here, the integral around the boundary has the same sense of circulation as the area bivector dS. The above two results can be rewritten as ,
,
f dS f dV
V
Let the area S be a circle of radius r centred on the point P0 and take the limit as r 0 . In this limit, f is essentially a constant over the area and hence (4.47) gives: [I 1 f ( P0 )] n 1 f dr r 0 A S lim
so that the geometric interpretation of the curl is that it is the circulation per unit area of the field f around each point with the circulation greatest in the plane of f .
V
and ,
,
,
,
f dr (f ) dS (f )dS
S
S
S
because in 3D the outer product of two bivectors must be zero (more ,
,
generally, in nD, because dS is of grade n – 1 and f is at least of grade 2, ,
,
then the outer product of f and dS must be zero). According to DL, these formulas can be written for objects with the same dimensions and added to give the general result valid in a flat space of any dimension: , ,
fdS f dV … (5.49)
R
R
Where R denotes a region, dS is the appropriate boundary measure of the region (for a two dimensional surface, the boundary measure is dr), dV the Chapter 5. Geometric Calculus
32
appropriate “volume” measure (for a two dimensional surface, the “volume” measure is dS), and now f is a multivector of arbitrary grade. Bivectors etc. are antisymmetric tensors though, so it would be good to have a version of this integral theorem valid for symmetric or mixed tensors. Since tensors can be conceived as linear functions of multivector arguments, the desired theorem according to DL is: ,
,
V
where dS is of grade n – 1. An alternative form is:
L(ndS )
V
,
,
,
,
L(n)dS L( dV ) L()dV … (5.51)
V
V
0
Now,
B A 12 B jk e j e k A 12 B (e k e j )e j e k A (by (3.28))
L(dS ) L( dV ) … (5.50)
V
Suppose now that f is a function with a bivector argument. By analogy then, f ( A B) f ( A ) B A f ( A ) lim … (5.55)
V
A 12 e k e j e j e k A … (5.56). In general,
M n n1! el ek ... eiei ... ek el M n … (5.57).
where n is the outward pointing unit normal defined above, and dS and dV can be brought outside the argument of L because of the linearity of L .
for a grade n multivector Mn.
5.6 Multivector Calculus
The value in expanding vector derivatives as above is that they allow the divergence and curl to be defined. Specifically, a f (a)
Tensor fields are functions not only of position, x, but also of multivector arguments. Hence, one can consider differentiating with respect to not only position, but also with respect to the multivector argument, and this idea has great value in reformulating linear algebra in the language of geometric algebra (see Ch. 4). To make these ideas concrete, consider a position-dependent, function of a vector argument a, f (a; x) , which will be written as f (a) for short. In analogy with the vector derivative with respect to position, the vector derivative with respect to the vector argument a is defined as follows: f (a b) f (a) b a f (a) lim … (5.52) 0
Using this notation, x … (5.53). Now b a bi ei a b ei ei a
(e i e i a ) f (a) e i [e i a f (a)] e i a a scalar f (a e i ) f (a) e i lim …(5.58). 0 And thus, for example, (a e i ) a a a e i lim e i e i tr ( g ij ) , … (5.58’) 0 where g ij ei e j are the metric coefficients.
Similarly,
f (a ei ) f (a) a f (a) ei lim … (5.59) 0
a e i e i a …(5.54)
Chapter 5. Geometric Calculus
33
aa A r
Some Identities:
aa A r
e i e i a (a A r )
e e i a (a A r ) i
e i lim
(a e i ) A r a A r
0
e i lim
(a e i ) A r a A r
0
e ei A r i
e ei A r
(n r ) A r e i A r only has n - r non - zero terms.
rA r e i dot a grade - r multivecto r gives r terms of grade r - 1.
a A r a (1) r (n r ) A r ... (5.61)
i
Geometric product wi th e i brings grade back to r. a A r a (1) r 1 rA r (1) r rA r ... (5.60)
Chapter 5. Geometric Calculus
From previous two results, it follows that a A r a (1) r (n 2r ) A r … (5.62)
34
6. SPECIAL RELATIVITY USING GEOMETRIC ALGEBRA 6.1 Vectors in Minkowski spacetime Gravity-free spacetime is taken to be Minkowskian; i.e. it is taken to be a flat (zero Riemannian curvature), four dimensional manifold with a Lorentzian metric. Inertial reference frames are taken to be vector manifolds which have a rectangular, orthonormal set of basis vectors {e }, 0,1,2,3 such that
e e … (6.1) with diag (1,1,1,1) … (6.2). The reciprocal basis {e } is defined to be such that e e … (6.3).
Thus e 0 e 0 and e i e i , i = 1, 2, 3 … (6.4). ______________ Aside: [DL] and [H] take diag (1, 1, 1,1) . This convention has the advantage that v v 1 , where v is the four velocity of a particle, but the disadvantage that the inner product of spacelike vectors is negative, the opposite of what is the case in 3D and so one has to remember to add negatives to the formulas one remembers from Newtonian physics. [DL] and [H] get around this problem by introducing “relative vectors”, which are timelike bivectors of the form: i e i e 0 . These relative vectors satisfy j k jk and so do behave in some sense like the space vectors we are familiar with from 3D Newtonian physics . Unfortunately, when working with the { i } , sometimes one can treat them like vectors while other times they need to be treated as bivectors. To avoid this “complication”, these notes explore the value of using the alternative metric (6.2) and not introducing the idea of relative vectors. The use of a metric with a different signature in these notes though, does mean that some formulas will appear as the negative of what they appear in [DL] and [H]. ______________
Chapter 6. Special Relativity Using Geometric Algebra
Relative to an inertial reference frame, points in spacetime are labeled by the position vectors: x x e te 0 xe1 ye 2 ze3 te 0 x … (6.5), where t ct c , where c is the speed of light in a vacuum in conventional units and tc is time in conventional units (i.e. t is a “geometric time” with the units of length, so 1 m of geometric time is approximately equivalent to
0.333 10 8 s ). This is often referred to as “c = 1” units. Note that full 4-vectors will be written in bold face type, while the parts of 4vectors which would be 3-vectors in Newtonian mechanics are written with over arrows (note that in special relativity, these are not vectors by themselves). Line element dx dx e ds 2 dx 2 dx dx dx e dx e dx dx dt 2 dx 2
… (6.6) Some conversions from c = 1 units to conventional units are as follows: Velocity = velocity in conventional units/c Acceleration = acceleration in conventional units/c2 Momentum = momentum in conventional units/c Energy = energy in conventional units/c2 With the above conventions, four-vectors a come in three types as determined as follows: a 2 , if a is timelike a 2 a 2 , if a is spacelike … (6.7) 0, if a is null where a | a | .
35
Clock Hypothesis If the rate of an ideal clock is not affected by its acceleration, then along the worldline of a particle, relative to the inertial frame in which the particle is instantaneously at rest: dx d e 0 ( ) dx dx dx 2 d 2 . I.e. proper (“wristwatch”) time gives, to within a sign, the magnitude of the arclength of the worldline of a particle. … (6.8)
conventional units, p ct E c / c ), and m (or mcc2 in conventional units) be identified as the rest energy of the particle. Thus p 2 m 2 v v m 2 E02 E 2 p 2 or E 02 E 2 p 2 … (6.19)
Four-velocity Parameterise the worldline of a particle by the elapsed proper time; i.e. x x( ) . Then the four velocity dx dt dx dt v x e0 e 0 V … (6.9) d d d d i where x x e i … (6.10) and V dx / dt … (6.11).
(d / dt)(e 0 V ) 2 A, ... (6.21) where A dV / dt ... (6.22)
Four-acceleration a v ... (6.20) (e 0 V ) V
d / dt 3V A … (6.23),
so
a 4V Ae 0 2 A 4V AV … (6.24)
Now dx dx dx d v v v 1 … (6.12) 2 dt dt (1 V 2 ) 1 / 2 … (6.13) Thus ( 1 V 2 ) 1 d d and so v (e 0 V ) … (6.14), where is the “Lorentz factor”.
Note that relative to the instantaneous rest inertial frame for an accelerating particle, V is instantaneously zero and 1 , so a 0 A0 … (6.25).
Four momentum of a particle A particle has no internal structure, so: p mv … (6.15) where m is a Lorentz invariant scalar called the rest mass. Note that p mV … (6.16)
Space-time splits Relative to the rest inertial frame of an observer moving with a constant fourvelocity va, the worldline of the observer is along the time axis of the reference frame, i.e. v a e0 (rest inertial frame). The time and space components of a vector vb relative to the rest frame of va can be found as follows. v b v b .1 v b v 2a v b v a v a ( v b v a v b v a ) v a
2
2
2
while p t p 0 m m 1 12 V 2 83 V 4 ... … (6.17).
For non-relativistic velocities, | V | 1 , and hence p t m 12 mV 2 … (6.18) Since 12 mV 2 is the (normalized) nonrelativistic kinetic energy of the particle, this suggests identifying pt with the total relativistic energy of the particle (in Chapter 6. Special Relativity Using Geometric Algebra
Taking a proper time derivative of v v 1 , one finds v v v a 0 … (6.26) I.e. v and a are always Minkowski orthogonal.
(v v a ) v a ( v b|| v a v b v a ) v a . v b v a v a b v v b a Thus from (6.14), ba v b v a … (6.27) [Lorentz factor of b relative to a.] 36
and
Vba ( v b v a ) v a / v b v a
vb v a … (6.28a) ( v b v a )
Form (6.28) it follows that
( v v ) 2 1 | Vba | Vba Vba b a 2 (vb v a )
1/ 2
… (6.28b)
For the case of collinear motion, see (6.60).
6.2 Bivectors in spacetime Spacelike versus timelike bivectors Let B = ab with a b for simplicity. Thus
0, if a, b both spacelike B abab a b 0, if a or b timelike with … (6.29) the other spacelike –1
2
2
2
This difference in sign is important for making “boosts” / “hyperbolic rotations”.
From (3.57), I commutes with even grade multivectors, anticommutes with odd grade multivectors. Acceleration Bivector The acceleration bivector has significance for Thomas precession and for defining uniform acceleration. Using the results above, it is given by: vv v v 3 e 0 A V A … (6.35) The magnitude of this bivector is the magnitude of the acceleration relative to the instantaneous rest inertial frame of the particle. This can be seen by noting that relative to the instantaneous rest inertial frame of the particle, V 0 and 1 and so v v e 0 A0 , and since A0 e 0 and | e 0 | 1 , it follows that | e 0 A0 || A0 | .
–1 Note that: 2 v v v vvv vv v ( v v ) v … (6.36) 0. ) (The last step uses v v
6.3 Lorentz Transformations A set of basis bivectors e 0i e 0 e i , i = 1, 2, 3 … (6.30a). e 23 e 2 e 3 Ie 01, e 31 e 3 e1 Ie 02 , e12 e1 e 2 Ie 03 … (6.30b) Note that the ordering of the vectors in the spacelike bivectors is such that Ie 0 e i Ie 0i e jk e j e k … (6.31a) Ie jk e 0i … (6.31b)
where i, j, k are a cyclic permutation of 1, 2, 3. Using (3.26) and (6.1), it is easily shown that: e 0i e 0 j ij … (6.32a)
e 0i e jk e 0i (Ie 0m ) 0 … (6.32b)
(Ie 0i ) (Ie 0 j ) ij … (6.32c)
Suppose the basis vectors {e } of an arbitrary inertial reference frame are rigidly rotated with the spacetime origin kept fixed. If the amount of rotation is smoothly parameterized by the continuous variable , then the instantaneous rate of change of the {e } with respect to must be orthogonal to their instantaneous orientation in order to keep their magnitudes constant. That is: de ( ) e ( ) … (6.37) d GA gives that this is automatically satisfied if de ( ) B e ( ) … (6.38) (see (3.70)) d for some bivector B.
Pseudoscalar: I = e0e1e2e3 … (6.33) so I2 = –1 … (6.34). Chapter 6. Special Relativity Using Geometric Algebra
37
where cosh( ) … (6.46) and V tanh( ) … (6.47), thus giving a definition of in terms of V.
Replacing the inner product with geometric products gives: de ( ) 1 2 (e ( )B Be ( )) … (6.39), d From which it follows that if B is a constant bivector, e ( ) exp(B / 2)e (0) exp(B / 2) … (6.40).
By reciprocity, e 0 [e0 Ve1 ] … (6.48)
ˆ 2 1 0 . Case 1: Let Bˆ be a unit timelike bivector so that B Thus 2
3
4
1 1 1 exp(Bˆ / 2) 1 Bˆ Bˆ 2 Bˆ 3 Bˆ 4 ... 2 2 2 3! 2 4! 2 1 2 1 4 1 3 1 ... Bˆ ... 4! 2 2 2 2 3! 2 cosh( / 2) sinh( / 2)Bˆ ... (6.41a)
Proceeding as above, e1 ( ) e1 sinh( )e 0 cosh( )e1
cosh( )tanh( )e 0 e1 Ve 0 e1 ... (6.50).
ˆ e e e . The action of R( ) on e0 is thus: For simplicity, consider B 10 1 0 e 0 ( ) e0 –1
cosh( )e 0 cosh( ) tanh( )e1
cosh( )e 0 tanh( )e1 e0 v e 0 Ve1 … (6.45)
Chapter 6. Special Relativity Using Geometric Algebra
e′0
Bˆ
e′1 e1
ˆ / 2) … (6.42) If we call R( ) exp(B ~ ˆ / 2) R a “rotor”, then evidently exp(B ( ) … (6.43) and from the expansions for the rotors given above, it is easy to show that the rotors are unimodular, i.e. ~ ~ RR RR 1 … (6.44).
cosh( )e 0 sinh( )e1
e0
Again by reciprocity, e1 [e1 Ve0 ] … (6.51)
Similarly, ˆ / 2) cosh( / 2) sinh( / 2)B ˆ … (6.41b) exp(B
[cosh( / 2) sinh( / 2)e1e 0 ]e 0 [cosh( / 2) sinh( / 2)e1e 0 ]
Note that e 0 e 0 1 … (6.49) as required (i.e. rigid rotation of axes leaves magnitude of basis vectors unchanged).
e 0 e1e 0 e1e 02 e1
Note that R( ) rotates the timelike and spacelike vectors in opposite ˆ cannot give the sense of directions, and hence the orientation of B orientation for both rotations. For the conventions chosen in the above ˆ gives the orientation of the rotation of the example, the orientation of B spacelike vector and has the opposite orientation of the rotation of the timelike vector. Furthermore, e 2 ( ) e2 e 2 , e 3 ( ) e3 e 3 … (6.52) as one would expect for a rotation in the e1 e 0 plane defined by Bˆ . It is also straightforward to show that e0 e1 0 so that the basis vectors in the “rotated frame” are still orthonormal as required. A position vector x can be expanded relative to either set of basis vectors so that 38
x x e x e … (6.53) from which it follows directly that x 2 x 2 , x 3 x 3 … (6.54).
To find x 0 t in terms of t and x , use the projection property of the inner product: t x e0 x e (e0 Ve1 ) (t Vx) … (6.55) where x1 x . Similarly, x x e1 x e (e1 Ve0 ) ( x Vt ) … (6.56). Formula for relative speeds Consider for simplicity, two particles labeled a and b moving along the x-axis of the inertial frame S. Thus: v a a (e 0 Va e1 ) … (6.57a) and v b b (e 0 Vb e1 ) … (6.57b). Now if v a is the four velocity of the frame S’, then the velocity of particle b relative to S’ is, using the Lorentz transformation results (6.48) and (6.51), given simply by
v b b a (e0 Va e1 ) Vb a (e1 Va e0 )
(e1 e 0 ) v a a (e1 e 0 ) (e 0 Va e1 ) a e1 aVa e 0 e1 and thus from (6.28), V Va Vba b e1 as required. 1 VaVb More generally, if v b b (e0 Vbx e1 Vby e 2 Vbz e3 ) … (6.61), then ba a b (1 VaVbx ) … (6.62) and Vby V x Va Vbz 1 2 Vba b e e e3 … (6.63). 1 VaVbx a (1 VaVbx ) a (1 VaVbx ) Action of R( ) on a bivector B a b ~ B R( )BR ( ) ~ R( )a bR ( ) ~ 12 R( )[ab ba]R ( ) ~ ~ ~ ~ 12 R( )[aR ( ) R( )b bR ( ) R( )a]R ( ) using R ( ) R( ) 1 12 [ab ba]
V Va a b (1 VaVb ) e0 b e1 ... (6.58) 1 VaVb ba a b (1 VaVb ) ... (6.59)
a b ... (6.64). Note that R( ) boosts multivectors, it doesn’t tell you what a multivector looks like relative to boosted frames (or at least, not directly).
and
Components of a bivector relative to a boosted frame: From (6.30), we can expand an arbitrary bivector F as follows (pre-empting Ch. 7 where E and B will be identified as describing the electric and magnetic fields):
Vba
V Va b ... (6.60). 1 VaVb
Note also from (6.27), that ba v b v a and from (6.28) that Vba ( v b v a ) v a / v b v a . Now, v b v a a b (Vb Va )e1 e 0 and since e1 e 0 is a unit bivector, it follows that | v b | a b | (Vb Va ) | . Dividing through by ba then gives (6.60) apart from the sign. However, Chapter 6. Special Relativity Using Geometric Algebra
F E i e 0 e i IB i e 0 e i E i e0 ei IB i e0 ei
Thus if a boost is along the x-axis:
39
1
01
1
0
E F F e e [by (3.28)] F e0 e1 [by (6.32) and by (6.4)] F (1 V )e 0 e1 E [by (6.32)] ... (6.65a) 2
2
1
E 2 F e0 e2 F (e 0 e 2 Ve1 e 2 ) E 2 VB 3 ... (6.65b)
E 3 F e0 e3 F (e 0 e 3 Ve1 e 3 ) E 3 VB 2 ... (6.65c)
e0 e
e0 e
e0e2 e
By (6.32c), B1 F (Ie0 e1 ) F (I 2 (1 V 2 )e 0 e1 ) B1 ... (6.65d)
B 2 F (Ie 0 e 2 ) F (Ie 0 e 2 IVe1 e 2 ) B 2 VE 3 ... (6.65e) B 3 F (Ie0 e3 ) F (Ie 0 e 3 IVe1 e 3 ) B 3 VE 2 ... (6.65f)
Alternatively, and perhaps more simply, simply replace the {e } in F by the
2e1e2 e 1e0e2 e e2, e2 e e1 e
Generalising: If E e 0 E, B e 0 B and V is in an arbitrary direction, then
E|| E|| ... (6.66a) B|| B|| ... (6.66b) E ( E V B ) ... (6.66c) B ( B V E ) ... (6.66d)
{e } using (6.45), (6.51) and (6.52) and collect terms. For example,
F E 2 e 0 e 2 E 2 (e0 Ve1 ) e2 E 2 e0 e2 VE 2 e1 e2 E 2e0 e2 B 3 Ie0 e3 E 2 E 2 and B 3 VE 2 consistent with (6.65b) and (6.65f).
Note from the following diagram [R: Fig. 3] how GA provides a beautiful geometric picture of how the hyperbolic rotation of axes means that a field which is purely timelike relative to one frame, and hence only has an electric field, has both timelike and spacelike components relative to a boosted frame, and hence has both a magnetic field component as well as an electric field component. This geometric picture is not available in the tensor formalism, and is considerably simpler than what is available from the differential forms perspective (e.g. [MTW: Ch. 4]).
Another approach to finding components relative to a boosted frame If the components of a multivector are known relative to a frame S, then the components relative to a frame S’ obtained by boosting S using the rotor R( ) can be obtained as follows. Consider a multivector in S’ with the same components as it has in S, then use the rotor R(– ) to rotate that multivector. For example: Suppose relative to frame S the vector x has the components: x te 0 xe1 ye 2 ze 3 , then the components of x relative to S’ are determined by: R( )[te0 xe1 ye2 ze3 ]R( ) t (e0 Ve1 ) x(e1 Ve0 ) ye2 ze3 (t Vx)e0 ( x Vt )e1 ye2 ze3 t (t Vx); x ( x Vt ); etc.
Similarly, if B B i e 0 e i IC i e 0 e i relative to S, then its components relative to S’ can be determined via: Chapter 6. Special Relativity Using Geometric Algebra
40
R( )[ B i e0 ei IC i e0 ei ]R( )
Now using
B1 (e0 Ve1 ) (e1 Ve0 ) B 2 (e0 Ve1 ) e2
cosh( / 2)
B 3 (e0 Ve1 ) e3 C 1e2 e3 C 2 e3 (e1 Ve0 )
cosh 1 2 and then multiplying by 1 cosh
C (e1 Ve0 ) e2 3
e0 e1 [ 2 2V 2 ]B1 e0 e2 [B 2 VC 3 ] e0 e3 [B 3 VC 2 ] e1 e2 [C 3 VB 2 ] e2 e3C 1 e3 e1 [C 2 VB 3 ] e0 e1 B1 e0 e2 ( B 2 VC 3 ) e0 e3 ( B 3 C 2 ) e1 e2 (C 3 VB 2 )
1 cosh , 2
sinh( / 2)
1 cosh one finds: 1 cosh | sinh | vˆ u L( ) 2(1 cosh ) 1 v u v u
e2 e3C 1 e3 e1 (C 2 VB 3 )
6.4 Alternative form of a pure boost
And thus
The alternative form for a pure boost derived in this section will be the more convenient form to use in a number of subsequent sections (from [DL: Subsection 5.4.4]).
L( )
2(1 v u)
1 vu 2(1 v u)
… (6.70).
[cf. (3.66)]
A pure boost changes the velocity of an inertial frame, but includes no spatial rotation. The bivector which boosts a frame with velocity u to velocity v with no additional spatial rotation is obviously: v u vu Bˆ vˆ u … (6.67) [Note that since a rotation of a | u v | | u || v | timelike vector, orientation of Bˆ is opposite to the required rotation.]
Check
Calling the rotor for a pure Lorentz boost L, ˆ sinh( / 2) cosh( / 2) sinh( / 2) vˆ u … (6.68) L( ) cosh( / 2) B where cosh vu v u … (6.69).
6.5 Fermi Transport
Note that |u| = 1 and | v | ba | Vba | cosh( ) | tanh( ) || sinh( ) | and hence | v u || sinh | … (6.67’), vˆ u v u / | sinh | … (6.67”).
Chapter 6. Special Relativity Using Geometric Algebra
(1 u v v u)u(1 u v u v) ~ L( )uL ( ) 2(1 u v ) (u v)(1 u v u v) 2(1 u v) 2 v 2u vv v 2(1 u v) as required.
Let a particle undergoing arbitrary (but smooth) accelerated motion carry an orthonormal set of basis vectors around with it. If the basis vectors vary from 41
instant to instant via pure Lorentz boosts, the frame (“gyros”) is said to be Fermi transported. If the four velocity of the particle at time is v ( ) and at an instant later it is v( ) v( ) v ( ) , then the pure Lorentz boost needed to achieve that change in velocity from (6.70) is: 1 [ v v ]v 2 v v L 1 12 v v 1 12 vv … (6.71) 2 2[1 v ( v v )] using v 2 1 and v v 0 , and vv v v is recognized as the acceleration bivector defined above. Now if the rotor which takes the frame carried by the particle to its current configuration is called R(), then ~ R( ) R( ) R ( ) [1 R ( ) R ( ) ]R( ) ... (6.72) ~ using RR 1. Now R( ) L( ) R( ) , … (6.73) so a comparison of (6.71) and (6.72) gives: ~ R R 12 v v or R 12 v vR 12 vv R … (6.74).
6.6 Fermi derivative Relative to the instantaneous Fermi transported frame of a particle, a vector b can be expanded in terms of the basis vectors of the frame as: b b e ( ) b b e ( ) b e ( )
Now ~ e ( ) R( )e R ( ) where e e (0) ~ ~ e ( ) R ( )e R ( ) R( )e R ( ) ~ ~ ~ ~ ~ ~ Since RR 1, R R RR 0 R RR R
Thus, ~ ~ ~ ~ e ( ) R R Re R Re R R R ~ ~ R R e ( ) e ( ) R R ~ 2( R R ) e ( ) ( v v ) e ( )
Consequently, b b e ( ) ( v v ) b … (6.75). Now define the Fermi derivative to be the rate of change of b relative to the Fermi frame of the particle. This derivative is thus given by: Db b e ( ) b ( v v ) b … (6.76). D If b is fixed relative to a Fermi frame, then Db / D 0 b (v v ) b . … (6.77). In particular, if b = ei, then Fermi transported basis vectors vary with time according to de i ( v v ) e i ( ) … (6.78). d and thus appears to rotate relative to a fixed inertial frame. This is referred to as Thomas precession.
6.7 Thomas precession (Based on [DL: Subsection 5.5.2].) Even though from instant to instant the frame carried by the particle undergoes only pure Lorentz boosts, relative to a single inertial frame the axes are seen to precess. This means that the rotor R() can be decomposed into a pure spatial rotation of the axes followed by a pure Lorentz boost. I.e. R( ) L( )U ( ) ... (6.79) R L U LU ... (6.80) (Note that this is not the same L given above.)
Chapter 6. Special Relativity Using Geometric Algebra
42
Thus from (6.74) and (6.79), R L U LU 12 vv LU
L
LU 12 vv LU L U ~ ~ ~ L U 12 L vv LU L L U
U 12 T U ... (6.81) ~ ~ where T 2 L L L vv L ... (6.82) is a bivector which describes the rate of precession. This can be seen as follows. Since ˆ sin( / 2), … (6.83) U ( ) cos( / 2) B where is the angle of rotation of the axes, not the amount the particle has rotated about the origin by, then
U 12 sin( / 2) 12 Bˆ cos( / 2) Bˆ sin( / 2) 12 Bˆ U Bˆ sin( / 2) using Bˆ 2 1 ~ ~ 12 Bˆ 2 sin( / 2)Bˆ U U using UU 1 12 Bˆ sin( )Bˆ 1 cos( ) Bˆ Bˆ U ... (6.84) ~ by substituti ng for U using (6.83). Now since U describes a spatial rotation in 3D space and in 3D any sum of bivectors can in principle be written as a single bivector blade, it follows that U 12 T U for some bivector T which gives the plane and rate of precession of the axes. This is obvious for the case when the plane of ˆ precession is constant so that Bˆ 0 , and thus (6.84) implies that T B … (6.85).
~ ~ To calculate T 2L L L vv L , use 1 ve0 1 e0 v ~ L L 2(1 ) 2(1 ) from (6.70), which implies that Chapter 6. Special Relativity Using Geometric Algebra
v e 0 2(1 ) v e 0 2(1 )
(1 ve0 ) [2(1 )]3 / 2
L 2(1 )
From the above equations, it follows that (1 e 0 v) v e 0 ~ ~ , using L L 1 . 2 L L 1 In addition, ~ L vv L ~ ~ ~ L vLL v L using LL 1 ~ e 0 L v L e v vv e 0 v ve0 v e 0 0 using v v -vv and v 2 1 2(1 ) Using v e 0 e 0 v 2 v e 0 2 , it then follows that: ~ ~ 2 L L L vv L
1 vv e 0 vv e 0 2(1 )
e 0 vv e 0 1 vv 2(1 ) e 02
1 ( vv ) (1 ) where ( vv ) is the part of vv e 0 [see (3.63)], which from (6.35) is
3V A .
Thus, in general, 3A V … (6.86), T 1
43
and the rate of precession relative to a reference inertial frame using coordinate time is T 2 A V … (6.87). 1
For uniform circular motion, | V | r0 , | A | r0 2 | V | , and V A , so it ˆ where Bˆ is a unit bivector with the same follows that T ( 1) B orientation as A V . Relative to frame S, the rate of precession is just 1/ this result, i.e. ( 1) in a sense opposite to the sense of rotation of the particle.
Uniform Acceleration Since the four acceleration a is Minkowski orthogonal to the four velocity v, and v constantly changes along the worldline of an accelerating particle, it must also be that a necessarily constantly changes along the worldline of an accelerating particle, even if the particle is moving in a straight line relative to some inertial frame. Consequently, uniform acceleration cannot be defined as a being a constant. However, it seems sensible to define a uniformly accelerating frame as being one in which an observer at rest in the frame feels a constant inertial force. This means that at each point along a particle’s worldline, the instantaneous acceleration relative to the instantaneous rest inertial frame, A0 , is a constant.
6.8 Rotor equation of motion From ~ e ( ) 2( R R ) e ( ) it follows that ~ v 2( R R ) v … (6.88) since v e0 ( ) . As will be shown in the next chapter, the equation of motion of a charged particle is mv qF v and thus
~ 2 R R (q / m)F
or R 12 (q / m)FR ... (6.89) and thus the dynamics can be put into a rotor equation of motion.
6.9 Rindler Frame The Rindler frame is the rest frame of a uniformly accelerating particle. This example is of interest for its pedagogical value in exploring the physical consequences of Einstein’s Equivalence principle in General Relativity. Chapter 6. Special Relativity Using Geometric Algebra
But A0 a constant is not frame invariant, and so not exactly what we want. But, this is where the fact that bivectors don’t have any particular shape comes to the rescue. From (6.25) we know that relative to the instantaneous rest inertial frame A0 a0 , so relative to this inertial frame, the acceleration
bivector, (6.35), is given by e0 A0 . Relative to an arbitrary inertial frame,
e0 A0 v a , and so the fact that e0 A0 is a constant along the
worldline also means that v a is a constant along the worldline of the particle. … (6.90)
Let | A0 | g … (6.91),
then ( v a) ( v a) vava av 2a a2 | A0 |2 g 2 … (6.92). Suppose a particle is accelerating uniformly along the x-axis of the inertial frame S. If at time t 0 the particle is instantaneously at rest relative to S, then at that instant, its acceleration bivector is given by ge 0e1 . Since the acceleration bivector is a constant for uniform acceleration, it follows that v a va ge0e1 … (6.93) along the entire worldline of the particle.
44
Since v a , the order of v and a can be swapped in a geometric product with a change in sign. Thus from (6.93), av ge 0e1 . Right multiplying this expression by –v gives:
a v ge 0e1 v v exp( ge 0e1 ) v(0) (a single sided rotor is possible because the rotation of v is in a single plane : see (3.72))
cosh( g ) e 0e1 sinh( g ) v(0) ... (6.94).
cosh( gT )[tanh( gT )e 0 e1 ] –1
If the particle starts from rest so that v(0) = e0, then
v( ) cosh( g )e 0 sinh( g )e1 ... (6.95)
cosh( g ) ... (6.96) and V 1 V tanh( g ) ... (6.97).
sinh( gT )e 0 cosh( gT )e1 ... (6.104).
e 0 e1e 0 e1e 02 e1
If x(0) = 0, then integrating v() gives the worldline of the particle to be: x( ) g 1 sinh( g )e 0 g 1[cosh( g ) 1]e1 … (6.98). Thus from the definition of v, dt / d cosh( g ) t g 1 sinh( g ) ... (6.99) and dx / d sinh( g ) x g 1[cosh( g ) 1] ... (6.100) ( x g 1 ) 2 t 2 g 2 ... (6.101) is the equation for the worldline . Returning factors of c, the equation of the worldline is: ( x c 2 / g ) 2 c 2 t 2 (c 2 / g ) 2
Rindler Frame The Rindler frame, A, is the rest frame of the uniformly accelerating observer constructed by using the instantaneous rest inertial frames at each instant to determine the hypersurfaces of simultaneity for A.
Chapter 6. Special Relativity Using Geometric Algebra
Using capital letters to label the coordinates relative to A, a spacetime point is given by: x te 0 x x 0 (T ) Xe X (T ) Ye Y Ze Z … (6.101) where from (6.98) and defining T … (6.102), x 0 (T ) g 1 sinh( gT )e 0 g 1[cosh( gT ) 1]e1 … (6.103) is the location of the spatial origin of A at time T, and from (6.50), e X (T ) (T )[V (T )e 0 e1 ]
Thus x te 0 x ( X g 1 ) sinh( gT )e 0 [( X g 1 ) cosh( gT ) g 1 ]e1 Ye 2 Ze 3 g 1 (1 gX ) sinh( gT )e 0 g 1[(1 gX ) cosh( gT ) 1]e1 Ye 2 Ze 3 ... (6.105)
From which it follows that t ( X X h ) sinh( T / X h ) ... (6.106a) x ( X X h ) cosh(T / X h ) X h ... (6.106b) y Y ... (6.106c) z Z ... (6.106d)
where X h g 1 ( c 2 / g c ) … (6.107) is the distance to the horizon in the Rindler frame. Because cosh2 sinh 2 1 , it follows from (6.106a) and (6.106b) that the curves of constant X relative to an inertial frame (IF) S are given by the hyperbolae: ( x X h ) 2 t 2 ( X X h ) 2 … (6.108a) or, by multiplying through by g2, (1 gx ) 2 g 2 t 2 (1 gX ) 2 … (6.108b) while the curves of constant T are given by the straight lines: t tanh( gT )x X h … (6.109a) 45
or gt tanh( gT )1 gx … (6.109b). The curvilinear coordinates of the Rindler frame A are illustrated in the following diagram where the horizon distance, g-1, has been normalized to one.
Dividing (6.110a) into (6.110b) for fixed X gives dx / dt tanh( gT ) , a result independent of X. Thus, all points in A have the same speed relative to the inertial frame S at the same A frame coordinate time T as one would expect of a rigid frame. Note though, that different values of X take differing amounts of S frame coordinate time t to get to that common speed, indicating that relative to S, the points in A have X dependent rates of acceleration, even though the A frame is “rigid”. Note that from (6.109a), dx gt gt … (6.112) ( X ) tanh( gT ) dt 1 gx [ g 2t 2 (1 gX ) 2 ]1/ 2 A( X )
d 2x dt 2
V ( X 0)
Rindler metric: dt dX sinh( T / X h ) ( X X h ) cosh(T / X h )dT / X h ... (6.110a)
dx dX cosh(T / X h ) ( X X h ) sinh( T / X h )dT / X h ... (6.110b) Thus ds 2 dt 2 dx2 dy 2 dz 2 (1 X / X h ) 2 dT 2 dX 2 dY 2 dZ 2 (1 gX ) dT dX dY dZ ... (6.111) Note that when X = –Xh, g TT 0 , indicating the presence of a horizon. 2
2
2
2
2
gTT (1 gX ) 2 implies that a standard (not coordinate) clock at height X has a rate 1 gX times that of the standard clock at X = 0, illustrating “gravitational” time dilation: clocks deeper in the “gravitational field” seem to run slower than clocks higher in the “gravitational field”.
Chapter 6. Special Relativity Using Geometric Algebra
(X )
g (1 gX ) 2 [ g 2 t 2 (1 gX ) 2 ]3 / 2
… (6.113)
gt
… (6.114) [1 g 2t 2 ]1/ 2 g and A( X 0) … (6.115) 2 2 3/ 2 [1 g t ] Note that relative to S, the observer does not have a constant acceleration, and in fact A( X 0) 0 as t . This is necessary because relative to S, the observer’s speed can’t exceed the speed of light.
6.10 Turntable Frame A uniformly rotating disk is of interest in SR for “Ehrenfest’s paradox” and in GR because Einstein used it to heuristically motivate the need for a nonEuclidean spatial geometry in GR. As shown in the figure below, for the purposes of this subsection, the centre of the disk will be taken to be located at (r0 ,0,0) relative to the lab frame, L. A Fermi frame, F, for an observer on the periphery of the disk will be taken for simplicity to have a spacetime origin coincident with L’s spacetime origin. Coordinates relative to L are in lower case letters while the coordinates of points relative to F will be given in capital letters.
46
Using (6.116) and (6.119) with a constant, it follows that
eT (e 0 r0 sin( t )e1 r0 cos( t )e 2 )
[e 0 r0 sin( T )e1 r0 cos( T )e 2 ] ... (6.120) Thus
a deT / dT [ r0 2 cos(T )e1 r0 2 sin( T )e 2 ] ... (6.121) e X cos(T )e1 sin( T )e 2 ... (6.122). From (6.120), it follows that
eY [r0 e 0 sin( T )e1 cos(T )e 2 ] ... (6.123)
as required to make it orthogonal to both e T and e X . Relative to L, the worldline of the origin of F is given by:
t t x r (cos t 1) 0 … (6.116) y r0 sin t 0 z
If P(T) denotes the spacetime position of P when its clock reads an elapsed proper time of T, then an event E which has coordinates (t, x, y, z) relative to L has coordinates (T, X, Y, Z) relative to F if the following vector relation is satisfied:
OE OP(T ) Xe X (T ) YeY (T ) ZeZ (T ) … (6.124),
and thus its speed is given by V r0 … (6.117) and the Lorentz factor relative to L is given by (1 r0 2 2 ) 1/ 2 … (6.118). By time dilation, the proper time T of the observer at the origin of F is related to lab time t by T t / … (6.119) and this defines coordinate time relative to F.
where O is the spacetime origin of L. Thus,
Let the point P be at the spatial origin of F. An orthonormal set of basis vectors {eT , e X , eY , e Z } carried by P can be determined as functions of the basis vectors of L as follows: e T is given by the four-velocity and e X by
x (r0 X ) cos(T ) Y sin( T ) r0 ... (6.125b)
( 2 r0 2 ) 1 times the four-acceleration of P respectively; by symmetry,
e Z e z , where e z is the unit vector in the direction of the z-axis of L; and e Y is taken to be orthogonal to both e X and e Z with a direction such that e X , e Y and e Z make up a right handed set of axes.
Chapter 6. Special Relativity Using Geometric Algebra
t e0 x e x y e y z e z Te 0 r0 (cos(T ) 1)e1 r0 sin( T )e 2 X [cos(T )e1 sin( T )e 2 ] Y [r0 e 0 sin( T )e1 cos(T )e 2 ] Ze Z t (T r0Y ) ... (6.125a) y (r0 X ) sin( T ) Y cos(T ) ... (6.125c) z Z ... (6.125d) The metric of F is then found to be:
47
ds 2 dt 2 dx 2 dy 2 dz 2 2 (dT r0dY ) 2 [cos(T )dX sin( T )dY (r0 X ) sin( T )dT
and c2 r0 0.034 . Thus with c / c , the turntable frame on the surface of the earth is essentially a Minkowskian frame.
2Y cos(T )dT ]2 [sin( T )dX cos(T )dY (r0 X ) cos(T )dT 2Y sin( T )dT ]2 dZ 2 [1 2 2 ( X 2 2Y 2 2r0 X )]dT 2 dX 2 dY 2 dZ 2 2 2 ( XdY YdX )dT ... (6.126). Note that at the spatial origin of F, the metric reduces to diag(-1, 1, 1,1), but the first derivatives of the metric coefficients do not vanish there, indicating that the frame is an accelerating, not inertial one, and so while the speed of light is locally isotropic, it is globally anisotropic. (The fact that at each point on a cur`ved line the line is locally straight does not imply that the line has to be globally straight.) Selleri has argued that the rotating frame reveals an inconsistency in special relativity because one should get an inertial frame in the limit as the centripetal acceleration goes to zero but the velocity remains finite. This is achieved by letting r0 and 0 in such a way that r0 stays fixed, which then implies that the magnitude of the centripetal acceleration 2 r0 0 . But in this limit, (6.126) gives exactly the required result provided X and Y remain finite, which is at it must be, because different points on the disk are travelling in different directions, so one cannot have a global inertial frame, only a local one.
6.11 Length and orientation of a moving rod Let S’ be an inertial reference frame with axes parallel to frame S moving with speed V along the x-axis of frame S. Consider a rod fixed in frame S’. The worldlines of the ends of the rod relative to S’ are: x A t 'e0 … (6.127a) and x B t ' e0 L cos( i )ei … (6.127b) where the cos( i) are direction cosines relative to S’.
Relative to S, the worldlines are then: x A t ' (e 0 Ve1 ) te 0 Vte1 … (6.128a) and x B t ' (e 0 Ve1 ) L cos(1 )(Ve 0 e1 ) L cos( 2 )e 2 L cos( 3 )e 3 t VL cos(1 ) e 0 Vt L cos(1 ) e1 L cos( 2 )e 2 L cos( 3 )e 3
L' te 0 Vt cos(1 ) e1 L cos( 2 )e 2 L cos( 3 )e 3 ... (6.128b) using t (t 'VL' cos 1 ) to replace t’ by t.
L (x B (t ) x A (t )) 2
1/ 2
L cos 2 (1) / 2 cos 2 2 cos 2 3
1/ 2
... (6.129)
with Consider a lab on the equator of Earth ignoring gravity.
r0 6.38 10 m 6
c 7.29 10 5 rad/s r0c 465 m/s and 1 to a very high degree of accuracy Chapter 6. Special Relativity Using Geometric Algebra
L' cos1 , … (6.130a) L L ' cos 2 cos 2 … (6.130b) etc. L cos1
48
For a rod lying in the x-y plane, cos 2 sin 1 etc. tan 1 tan 1 … (6.131).
0 / ... (6.135c) 0 due to time dilation.
6.12 Relativistic Doppler effect
6.13 Aberration of light
Consider a source moving along the positive x-axis with speed V so that its velocity relative to the observation frame is v (e 0 Ve1 ) Consider a photon received at the spatial origin of the observer frame which makes an angle relative to the positive x-axis. The energy-momentum of this photon relative to the observer frame is thus p e 0 k cos e1 k sin e 2 … (6.132) The angular frequency of the photon relative to the rest frame of the source is simply
Consider two frames S and S’ in standard configuration. At the instant their spatial origins coincide, suppose a ray of light from some source is received at the spatial origins. If relative to S the ray makes an angle to the positive xaxis and the ray lies in the xy-plane, then relative to S, photons in the ray have energy-momentum p e 0 cos e1 sin e 2 … (6.136). Using (6.48) and (6.51), the energy-momentum relative to S’ is therefore
0 p v / (e 0 k cos e1 k sin e 2 ) (e 0 Ve1 ) (1 V cos ) ... (6.133) using k ... (6.134) in c 1 units.
Thus
0 1 V cos … (6.134) 1/ 2 1V 2
p / (e0 Ve1 ) cos (e1 Ve0 ) sin e2 (1 V cos )e0 (V cos )e1 sin e2 e0
Thus (1 V cos ) … (6.138),
cos
Special cases: = 0 (source receding along line of sight)
and
sin
0
... (6.135a)
(1 V ) 0 = (source approaching along line of sight) 0 ... (6.135b) (1 V ) 0 = (transverse Doppler effect)
Chapter 6. Special Relativity Using Geometric Algebra
(V cos ) sin e1 e2 ... (6.137). (1 V cos ) (1 V cos )
(cos V ) ... (6.139a) (1 V cos ) sin ... (6.139b). (1 V cos )
With a bit of trigonometry, it can be shown that tan
1 1V 2
1 V
1/ 2
tan 12 … (6.140).
6.14 Boost in an arbitrary direction ˆ / 2) with I.e. let R( ) exp( B
49
ˆ cos( )e e … (6.141) B i i 0 (summation over i implied, i a direction cosine)
e0 [cosh( / 2) Bˆ sinh( / 2)]e 0 [cosh( / 2) Bˆ sinh( / 2)] [cosh( ) Bˆ sinh( )]e since e lies in the plane of rotation. 0
0
[1 Bˆ V ]e 0 [e 0 V cos( i )e i ] ...(6.141), with the final result following from cos( i )ei e 0e 0 cos( i )ei .
Conservation of energy-momentum is encapsulated by the local equation (this part mirrors [MTW: Sec. 22.3]: ,
,
0 T() ,
,
, ,
,
,
,
( P) v v P [ v ( P)]v ( P) v v ( P)( v) v P ... (6.147) The v component of this equation gives ,
,
0 T() v v ( P) ( P)( v v) v ( P)( v) v P, and thus
More interesting is what the boost does to the spatial basis vectors: ˆ sinh( / 2)]e [cosh( / 2) B ˆ sinh( / 2)] … e1 [cosh( / 2) B 1 (6.142) Now, the expansion of Bˆ consists of a bivector in which e1 lies and bivectors which e1 is orthogonal to. [To be completed.]
6.15 Stress-energy tensor for a perfect fluid T(a) ( P) vv a Pa … (6.143) where 𝜌 and 𝑃 are the energy density and pressure relative to the local instantaneous rest inertial frame and v is the local bulk four velocity of the fluid. From (6.143), T(v) ( P)vv v Pv ( P)v Pv v , … (6.144) the negative of the four momentum density of the fluid. Consequently, v T(v) ( P)v vv v Pv v P P , … (6.145) the proper energy density of the fluid.
0 v ( P) v ... (6.148)
which is the relativistic generalization of conservation of mass. In obtaining (6.148), ( v v) v v v 0 has ben used. The part of (6.147) orthogonal to v is given by ,
,
,
,
,
,
,
,
0 T() T() vv / v 2 T() T() vv ( P) v v ( P) v [P ( v P) v], ... (6.149) which is the relativistic generalization of Euler’s equation of fluid motion. Note that P ( v P) v P (P) vv / v 2 , and so is the part of P orthogonal to v (i.e. the spatial part of P in the local rest inertial frame of the fluid).
In contrast, if n is a unit spacelike vector orthogonal to v, then n T(n) ( P)n vv n Pn n P , … (6.146) the local fluid pressure. Chapter 6. Special Relativity Using Geometric Algebra
50
7. CLASSICAL ELECTRODYNAMICS
7.2 Covariant Form of Maxwell’s Equations
7.1 Units and Conventions
Now a b I 3a b so E I 3 E … (7.7)
The spacetime metric will be taken to have the spacelike signature (-1, 1, 1, 1).
where e i i e i i , i = 1..3 … (7.8) x is the spatial part of the of the spacetime gradient operator e e , = 0..3 … (7.9). x Note that Latin sub- and superscripts will be taken to run over the three space dimensions, while Greek sub- and superscripts will be taken to run over all four spacetime dimensions.
Units such that “ c 0 0 1 ” will be used. What this means is as follows. Using a subscript c to indicate quantities with conventional units, the above convention means the energy density of an electromagnetic field in vaccuo is given by: c / c 2 12 0 Ec2 / c 2 12 Bc2 /( 0 c 2 ) 12 ( E 2 B 2 ) … (7.1) Thus, E 0 Ec / c … (7.2a) and B Bc /(c 0 ) 0 Bc … (7.2b) In S.I. units, Maxwell’s equations away from polarisable and magnetic media are: (1) Ec Bc / tc … (7.3a) Multiply by 0 / c to obtain: E B / t … (7.3b) (2) Bc / 0 J c 0 Ec / tc … (7.4a)
Multiply by 1/(c 2 0 ) to obtain: c vc B E / t J E / t … (7.4b) c 0 c
(3) Bc 0 obviously just becomes B 0 … (7.5) (4) Ec c / 0 … (7.6a)
Multiply by
0 / c to obtain: E c /(c 0 ) … (7.6b)
Chapter 7. Classical Electrodynamics
Inserting the constant e 02 1 in the lhs of (7.3b) and noting that as a constant e0 can cross the derivatives, it follows that I 3 E I E … (7.10) –1 –1 –1 where E e 0 E e 0 E … (7.11). I 3e0 e1e 2e3e0 I Likewise letting B e 0 B e 0 B … (7.12) and using e 0 e 0 e 0 e 0 1 in the rhs of (7.3b), (7.3b) as a whole becomes I E e 0 B / t … (7.13) Finally, multiplying by I and using the fact that since I is a constant, it follows that: (1’) e0IB / t E 0 … (7.14) ____________________________________ Aside: In the expressions for E and B, why put e0 first? (Note that Hestenes, and Doran and Lasenby, put it second, but use a metric with negative trace). The reason is to make formulas consistent with the usual electromagnetic expressions. For example, B A I 3 A … (7.15) Left multiply by e0, then left multiply by I e 0 I 3 to find IB A … (7.16) 51
with B e 0 B . Choosing to put e0 second would introduce a negative into this expression. ______________________________________ Similarly, (7.4b) gives I B J e0E / t … (7.17). From (a M r )I a (M r I) [see (3.54a)] it follows that I B (IB) and so we get: (2’) e 0 E / t (IB) J … (7.18). From (7.6a) and (a b) b a b a a b a b , we get: (4’) E e 0 E e 0 … (7.19). Similarly, from (7.5) we get: B 0 which implies ( B)I 0 and from (a M r )I a (M r I) we get
Consequently, since ( F) 0 is an identity [see (4.30)], we get charge conservation: J e 0 / t ( e 0 ) J / t J 0 … (7.27) automatically from F J . Also, because A 0 is also an identity [see (4.19)], it follows from F 0 that there exists a four potential A such that F A … (7.28) and thus F J implies, using (4.35), that ( A) 2 A ( A) J … (7.29) which is an inhomogeneous wave equation for A if the Lorentz gauge A 0 is taken. The d’Alembertian
2 2 … (7.30). t 2
(3’) (IB) 0 … (7.20).
2
Adding (7.14) to (7.19) and using E E E etc., one gets: (E IB) e 0 (E IB) / t e 0 J … (7.21) which can be rewritten as: F J … (7.22) where e e 0 / t … (7.9), J e 0 J ( / ) v 0 v … (7.23) [Note is measured relative to the frame of reference, not the rest inertial frame as 0 is.] and F = E + IB … (7.24).
Note that in these notes, the convention is that 2 , while the geometric product of the gradient operator will be written as .
Since F F F and F is a trivector while J is a vector, it must be therefore from (7.22), that the covariant form of Maxwell’s equations are: F J … (7.25) and F 0 … (7.26).
… (7.31)
Chapter 7. Classical Electrodynamics
___________________________________ Comparison with tensor formulation F A e A e A e e 12 F e e F 2 A F 2 A F 12 ( F F ) A A
52
In components, F E i e 0i 12 i jk B i e jk ... (7.32a) E e 0i IB e 0i ... (7.32b) [see (6.31a)] where e e e , … (7.33) i
i
Proof: ,
Ei and Bi are the usual 3D components of the electric and magnetic fields, and i jk is the permutation tensor. Thus if
F 12 F e e , … (7.34) F 0 j F
j0
,
,
,
F(e ) ,
,
,
,
F(e ) F e F , e v F , e ... (7.39)
E j … (7.35a)
and F jk i jk B i [no sum over i] … (7.35b) and consequently, F e 12 F (e e ) 12 F , ( e e ) F , e So F J F , e J e … (7.36).
Note that it is the divergence being on the first index which makes the rhs –J [see S. Weinberg (1972), Gravitation and Cosmology]. Using the antisymmetry of F, one obtains
F , e J e … (7.37) If F is treated as a linear function of a vector argument [see Ch. 4], then this last expression can be rewritten as: ,
,
F()
,
F() J … (7.38) where the check marks indicate the scope of the gradient operator (see Eqs. (5.43)).
Chapter 7. Classical Electrodynamics
7.3 Covariant form of Lorentz Force Law Start from: dp q( E V B) dt dp q(E V B) d q[e 02 E (V ) ( I 3 B)] q[E (e 0 ) (V ) (IB)] q[E (e 0 ) (IB) (V )] qF v qE (V ) since (IB) e 0 0
Now noting that qE (V ) qE Ve 0 gives the proper time rate at which the energy of the particle increases, it follows from the last result above that if p (energy of charged particle )e 0 p , then p dp / d qF v … (7.40) where p = mv … (7.41). _______________________________
53
Comparison with tensor formula qF v
A ( x)
G (x, x' ) J t ,V
( x' ) d 4 x ' … (7.43)
[| x x ' | (t t ' )] 4 | x x ' |
qF (e e ) v e
with G(x, x' )
12 qF v (e e e e e e )
[As a wave equation for waves with speed c = 1, would expect G to represent a point impulse travelling with the same speed, hence the numerator, but also falling off as 1/r so energy is conserved, hence the denominator.]
1 2
12 qF v ( e e ) ... qF v e ... (7.42). _____________________________ Axiomatic Derivation of the Lorentz Force Law If there exist structureless charged particles whose rest masses are conserved (i.e. no heating), then the state of motion of the particle can be described simply by its four-velocity v. Since v has a constant magnitude, the fouracceleration v must be orthogonal to v (this follows from taking the proper time derivative of v v 1 which gives v v 0 ). To get a force law which necessarily produces a four-acceleration orthogonal to the particle’s four velocity, the force law must be a function of v. The simplest possibility is a linear function of v, and we know from previously that the inner product of a vector and a bivector produces a vector in the plane of the bivector which is orthogonal to the original vector. Hence, the simplest candidate law for the action of an electromagnetic field on a charged particle is mv qF v for some bivector field F and q is the charge on the particle (actually, there should also be a proportionality constant too, but it has been assumed that the units have been chosen to make this unity). Considering F Ee 0 e E Be a e b and expanding out the inner product, we find that the timelike component behaves like an electric field in the direction e E and the spacelike bivector behaves like a magnetic field in the direction e a e b .
7.4 Lienard-Wiechert Potentials The wave equation, (7.29), can be solved using Green’s function methods:
Chapter 7. Classical Electrodynamics
… (7.44)
Note that because of the Lorentzian metric, | x x | is zero not only when x x , but also when t t | x x | . For a point particle, J(x' ) (x' )[e 0 V (t ' )] ( / ) v 1q (3) ( x ' x p (t ' ))v(t ' ) … (7.45) and thus
A ( x)
t ,V
qv(t ' ) (3) (| x x ' | (t t ' )) (x ' x p (t ' ))dV ' dt' 4 | x x ' |
qv(t ' ) (| x x p (t ' ) | (t t ' ))dt' 4 | x x p (t ' ) | t
… (7.46)
Now ( g (t ' )) j
1 (t 't j ) … (7.47) | g ' (t j ) |
where the t j are the zeros of g (t ) .
In the above equation, g (t ) | ( x x p (t )) | (t t ) … (7.48) so we need the derivative wrt t which is: 1 ( x x p (t )) (dx p dt ) / | x x p (t ) | … (7.49) (t t ) ( x x p (t )) dx p dt / | x x p (t ) |
Hence A(t , x )
qv(t ) … (7.50) 4 | (t t 0 ) [ x x p (t 0 )] dx (t 0 ) / dt | with t 0 t | x x p (t 0 ) | … (7.51) (an implicit equation). 54
The requirement t t ' | x x ' | means that the source point x' t ' e 0 x p (t ' ) and the field point x te 0 x are joined by a null vector X = x – x’ which satisfies X X 0 . With X v 0 , it then follows that A ( x)
v (t 0 ) q v (t 0 ) q qv … (7.52) 4 X v (t 0 ) 4 (x x) v (t 0 ) 4Rv
where x and t 0 are implicitly functions of x, and Rv is the spatial distance from the source point to the field point relative to the inertial frame in which the particle is momentarily at rest at the time at which the particle is at the source point.
7.5 Fields of an arbitrarily moving charge Let the position and velocity of an arbitrarily moving charged particle be given as functions of the elapsed proper time along the worldline of the particle. This means that the potential can be written as q v ( ) A ( x) … (7.53) 4 X v ( ) with X x x p ( ) such that X X 0 and x p ( ) is the position of the particle at time Since A(x) is of the form v where is a scalar function, F A (v) ( ) v v . (7.54) Now 0 X X (x x p ( )) (x x p ( )) …. (7.55) provides an implicit equation for (x) , which means that d and hence e ( (x))d / d … (7.56) d x x To find an expression for (x) , take the gradient of (7.54) to get: 0 ( X X)
e ( X) X 0
X ( x e ) x p ( ) e ( ) v( )
… (7.57)
From this equation it follows that X . … (7.58) Xv X v Thus v v … (7.59) Xv To determine , we need ( X v) 1 ( X v) 2 ( X v) … (7.60) Now ( X v ) e ( X) v e X ( v ) … (7.61) X X v X v Xv Xv Collecting all the terms together and doing a bit of rearranging, it can be shown that q v X X [ X ( v v )] F , … (7.62) 3 4 ( X v ) ( X v) 3 which can be seen to consist of a retarded Coulomb (velocity) field q vX … (7.63) Fb 4 ( X v) 3 And a radiation (acceleration) field q X [ X ( v v )] . … (7.64) Fr 4 ( X v) 3 Note that since X ( v v ) produces a vector orthogonal to X lying in the v v plane, X [X ( v v )] is a bivector consisting of X and a vector orthogonal to X, and hence represents a TEM wave emanating from the source point.
e ( X) X e X ( X) Chapter 7. Classical Electrodynamics
55
7.6 Stress-Energy-Momentum Tensor for a Field in Vacuo
Thus e 0 P 12 (E 2 B 2 )e 0 (e 0 F) F E 2 e 0
The field energy density of an em field in vacuo is: 12 ( E 2 B2 ) … (7.65) Now
(e 0 F) F 12 F Fe 0
Sign issue Conventionally, the stress-energy tensor is such that T00 T 00 , but since T00 e 0 T(e 0 ) and e 0 e 0 1 , this means that to be consistent requires the e0 component of T(e 0 ) to equal . Thus we write the above result as a linear function of e0 as T(e 0 ) 12 F Fe 0 (e 0 F) F … (7.76)
E 2 E j E k (e 0 e j ) (e 0 ek ) E j E k [e 0 ek e j e 0 e 0 e 0 e j ek ] E 2 …(7.66)
therefore 12 (E 2 B 2 ) … (7.67) Now F F (E IB) (E IB) E E (IB) (IB) … (7.68) Since E is a timelike bivector while IB is spacelike, E (IB) 0 . … (7.69) Since IB lies in a single plane, (IB) (IB) IBIB I B B … (7.70) 2
2
2
so F F E B . … (7.71) 2
… (7.75)
2
Since the choice of frame e0 was arbitrary, we can write in general that the energy-momentum tensor of the electromagnetic field in vacuo is given by: T( v) 12 F Fv ( v F) F … (7.77) which gives the negative of the energy-momentum density as measured by an observer with 4-velocity v (see MTW pp. 131-132) .
so P Pc / c E B … (7.73)
An alternative derivation proceeds as follows. Energy-momentum balance requires the four divergence of the em field energy momentum tensor to equal the four force density of free charges and currents on the field. By N3, this is just the negative of the four force density of the field on free charges and currents. Thus would expect something like Tem F J … (7.78). However, Tem is a symmetric tensor and hence can’t be written as a bivector and hence (7.78) can’t be the correct form. But we discovered previously that when treated as linear functions of multivector
(e 0 E) (IB) e 0 E e 0 E since e 0 E 0 E contains e 0
arguments, T() gives the divergence of T(a) on its second index, so expect the balance equation for field energy-momentum would have the form
The Poynting vector in vacuo is given by Pc 0 Ec Bc … (7.72)
E B E ( I 3 B) from (3.51) e 02 E ( I 3e 02 B) inserting two factors of e 02 1 (e 0 E) (IB) using I 3e 0 I (e 0 F) (IB) since e 0 (IB) 0
,
,
(e 0 F) F (e 0 E) E (e 0 F) F E Ee 0 E (E e 0 ) To see this, expand E (E e 0 ) (e 0 F) F E 2 e 0 since E e 0 0 … (7.74) Chapter 7. Classical Electrodynamics
56
,
,
T() F J …(7.79) F ( F) from the Maxwell equation F J ( F) F the inner product of a vector and bivector is antisymmet ric
7.7 Separate conservation of the radiation and velocity field energies away from a charge Let the stress-energy tensor of the radiation field be given by T r (a) 12 Fr Fr a (a Fr ) Fr . … (7.82)
,
,
,
,
,
,
,
,
,
T r () 0 . … (7.83) Let us explicitly prove that this is in fact the case.
,
T r () 12 (Fr Fr ) ( F r ) F r
( F) F (e F) F ( F) F (F ) F ( F) F 12 (F F) ( F) F by a vector identity ,
,
( F) F 12 (F F) since F 0 from Maxwell' s equns.
Away from the worldline of the charge producing the radiation field, conservation of energy in the radiation field is given by ,
,
,
,
,
,
,
e ( Fr ) Fr (e Fr ) Fr (e Fr ) Fr
… (7.79’)
e ( Fr ) Fr (e Fr ) Fr e ( Fr ) Fr ( Fr ) Fr
It then follows that
using (a B1 ) B 2 B1 B 2 a (a B 2 ) B1 Thus
T( v)
1 2
F Fv ( v F) F … (7.80) as above.
_______________________________ Aside ,
,
T() ,
,
T(e ) ,
,
T(e ) T e if basis vectors are constant vectors e 0 ( 0T 00 1T 01 2T 02 3T 03 ) e1 (...) ... … (7.81) And thus T00 is the density of mass-energy and T0k is the k component of energy flux as required. ____________________________________
,
,
,
,
T r () ( Fr ) Fr ( Fr ) Fr Now F J F Fb Fr 0 away from the charge. This means that either Fb and Fr are separately zero or Fb Fr which means that Fb and Fr can differ only by a sign and the addition of a constant and the divergence of a trivector field. Since Fb and Fr are in fact functionally very different, it must be that Fb and Fr are separately zero. By similar reasoning, F 0 Fr 0 . Thus T r () ( Fr ) Fr ( Fr ) Fr 0 as required to imply that the amount of energy in the radiation field is conserved away from the source. Consider now the stress-energy tensor for the total field: T(a) 12 (Fb Fr ) (Fb Fr )a (a (Fb Fr )) (Fb Fr )
T r (a) 12 (Fb Fb 2Fb Fr )a (a Fb ) Fb (a Fb ) Fr (a Fr ) Fb and thus T b (a) 12 (Fb Fb 2Fb Fr )a (a Fb ) Fb (a Fb ) Fr (a Fr ) Fb
Chapter 7. Classical Electrodynamics
57
the energy of which is also conserved away from the source since energy as a whole is conserved (that total energy is conserved away from the source ,
,
follows from (7.79) which reduces to T() 0 when J = 0).
7.8 Back reaction on an accelerating charge See [R] for more details and references. 7.8.1. The Lorentz-Abraham-Dirac equation of motion Since the four-velocity v of a point particle has unit magnitude, its proper time rate of change must always be orthogonal to v. By (3.19), this means that the equation of motion of a point charge incorporating radiation reaction is necessarily of the form v (e self ) v , where e and self are bivectors which together with v determine the external force per unit mass and selfforce per unit mass respectively. For an object as simple as a point charge, we might expect that the self-force bivector depends only on the magnitude of the charge carried by the particle, the properties of spacetime (because these would determine the rate at which radiation is carried away from the particle), and the state of motion of the charge; that is, on its velocity v, its acceleration v , and (with the benefit of hindsight), its hyperacceleration v . Ignoring for the moment the constants of proportionality and taking into account that the outer product of a vector with itself is zero, the possibilities for self are proportional to v v , v v , and v v . Checking each of these in turn and using (3.21), it is easily shown that the Lorentz-Abraham-Dirac equation, mv qFe v m 0 (v v 2 v) , is given by
mv qFe v m 0 ( v v) v , … (7.84) where Fe is the external electromagnetic field causing the charge to accelerate, and 0 c 0,conv q 2 /(60 mc 2 ) in SI units. That Eq. (7.84) is the required form could have been predicted as follows. First, the possibilities v v and v v when dotted with v yield terms proportional to v , and hence are inertial terms. The first of these is the selfforce arising from the fact that the bound field of the charge needs to be “pushed” along with the bare mass and is incorporated into the physically Chapter 7. Classical Electrodynamics
observed rest mass. The second describes an inertial mass which is acceleration dependent and thus can be ruled out on experimental grounds. By using (3.21) to convert (7.84) into its standard form, it can be argued that m 0 v 2 v must be the term that describes the rate at which energymomentum is radiated from the particle because radiation describes the loss of energy, and this term is the only one whose time component is necessarily negative. These arguments can be used to simply motivate the form of the Lorentz-Abraham-Dirac equation up to the constants of proportionality. (In a course where the Liénard-Wiechert potentials are derived, a procedure developed by Barut [Phys. Rev. D 10(10), 3335-3336 (1974)] can be put into geometric algebra form and used to make a complete derivation of the Lorentz-Abraham-Dirac equation.) Apart from the fact that geometric algebra makes the form of the LorentzAbraham-Dirac equation readily predictable, is there any value in expressing the self-force as m 0 ( v v) v rather than in the conventional form
m 0 (v v 2 v) ? The answer is yes, because from (3.8) and the fact that v is equivalent to e0 for the instantaneous rest inertial frame of the particle, it is evident that m 0 ( v v) v is the spatial part of the hyperacceleration relative to the instantaneous rest inertial frame of the particle, which is more obviously the covariant generalization of the self-force in the non-relativistic equation of motion, mdV / dt Fe m 0 d 2V / dt 2 , than is, m 0 (v v 2 v) . That the spatial part of the hyperacceleration relative to the instantaneous rest inertial frame of the particle, which is d 2V / dt 2 relative to that frame, should play such a pivotal role in determining the self-force follows from Einstein’s equivalence principle. Because the rest frame of a uniformly accelerating charge is equivalent to a uniform gravitational field, and a charge at rest in a uniform gravitational field has no net work done on it, then relative to that frame of reference there can be no radiation and hence no radiation reaction and the self-force should be just what is required to support the effective mass of the bound part of the charge’s field (this self-force is incorporated in the definition of the mass m). Consequently, for uniform acceleration, the selfforce term incorporating radiation reaction should vanish. Now since v v 0 , it follows that 58
m 0 ( v v) v m 0
d v v v . … (7.85) d
Equation (7.85) thus shows that the self-force vanishes when d ( v v ) / d 0 , or in other words, when the acceleration bivector v v is constant in magnitude and direction, which corresponds to uniform acceleration as required. 7.8.2. Modified Lorentz-Abraham-Dirac equation The Lorentz-Abraham-Dirac equation has several difficulties, such as noncausal pre-acceleration and runaway solutions. A consequence of the analysis above is that geometric algebra shows that the Lorentz-Abraham-Dirac equation is necessarily the simplest possible equation of motion of a charged particle incorporating radiation reaction, thus providing support for arguments that the problem must be with the assumption of a structureless point charge rather than with the non-rigorous methods with which the Lorentz-AbrahamDirac equation is derived. If, classically speaking, charged “particles” really must have a finite size and structure, we would expect that the equation of motion of such an object would be an integro-differential equation rather than a differential equation. Although this view is correct, if the external force varies “sufficiently slowly” (the meaning of which is defined precisely in the discussion surrounding (7.89)) and the finite sized structure can be considered to be quasi-rigid, then even an object with structure can appear point-like and have its motion given by a differential equation. When the conditions for a point-like description of motion are met, then from derivations of the equation of motion of an accelerating charge based on classical mechanics and classical electromagnetic theory and quantum mechanics after the classical limit is taken, the resulting modified Lorentz-Abraham-Dirac equation is found to be mv fe 0 fe fe v v , … (7.86)
where fe is the external force, fe v fe (x( )) , e / x , and
e e . The non-relativistic limit of (7.86) is mdV / dt Fe 0 dFe / dt . … (7.87)
Chapter 7. Classical Electrodynamics
Recall that for a pure force (that is, one that doesn’t affect the rest mass of the forced particle) fe v 0 . Hence, it follows from taking a proper time derivative that f e v fe v . In addition, again using 1 v v and Eq. (8) in reverse, it follows that (7.86) can be expressed as mv f e 0 v fe v . … (7.88) Using (3.23) with b replced by v, (7.88) shows that the particle behaves like a neutral particle when the spatial component of the external force in the instantaneous rest frame of the particle, the normal three force, is a constant. This statement is closely related to the implication of the Lorentz-AbrahamDirac equation that a charged particle responds as if neutral when the spatial component of the particle’s acceleration in the instantaneous rest frame of the particle is a constant (and so can be used to “predict” (7.88)). As mentioned, both Eqs. (7.84) and (7.88) can be understood qualitatively in terms of Einstein’s equivalence principle. Note though, that although the LorentzAbraham-Dirac equation is a third order equation for particle position, (7.86) is only second order, which solves the problems of pre-acceleration and runaway motion. Thus it has been argued that Eqs. (7.86) and (7.87) are more physically valid equations of motion than the Lorentz-Abraham-Dirac equation. (Interestingly, (7.86) also equates to the first-order perturbation solution to the Lorentz-Abraham-Dirac equation obtained by using the zeroth order solution, v f e / m , to estimate v , though the physical status of (7.86) has been changed by considering it to be the point-like limit of the equation of motion of a quasi-rigid finite size charge.)
Although the modified Lorentz-Abraham-Dirac equation doesn’t exhibit preacceleration and runaway motion, it is an approximation that is valid only when the external force varies sufficiently slowly. In particular, it gives unphysical results (energy-momentum is not conserved) if a rapidly changing field is approximated as a step discontinuity. What does “sufficiently slowly” mean? For the external force to “not see” any structure in the charge, it must vary negligibly in the time it takes light to travel across the radius of the charge. For an electron, the time of travel of light across the charge is of order 0, in which time the external force varies by approximately | 0 dFe / dt | . Thus, “sufficiently slowly” means that 59
| 0 dFe / dt | / | Fe | 1. … (7.89)
mv qFe v m 0 ( v v) v … (7.90)
Note that this condition is not satisfied by forces with a step discontinuity, so this simplifying approximation is not available. If we consider rectilinear motion for simplicity and note that dFe/dt = (dFe/dx)V and Vmax = 1 (assuming motion in the positive x direction and an increasing force), it follows that (dFe/dt)max = dFe/dx. Using this result and setting the right-hand side of (7.89) equal to with
