A cylindrical Lienard-Wiechert potential calculation using multivector matrix products. Originally appeared at: http://sites.google.com/site/peeterjoot/math2011/matrixVectorPotentials.pdf Peeter Joot — [email protected] April 30, 2011

matrixVectorPotentials.tex

Contents 1

Motivation.

1

2

The goal of the calculation.

1

3

Calculating the potentials for an arbitrary cylindrical motion. 3.1 A motivation for a Hermitian like transposition operation. . . . . . . . . . . . . . . . 3.2 Back to the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 2 4

4

Doing this calculation with plain old cylindrical coordinates.

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5

Reflecting on two the calculation methods.

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1. Motivation. A while ago I worked the problem of determining the equations of motion for a chain like object [1]. This was idealized as a set of N interconnected spherical pendulums. One of the aspects of that problem that I found fun was that it allowed me to use a new construct, factoring vectors into multivector matrix products, multiplied using the Geometric (Clifford) product. It seemed at the time that this made the problem tractable, whereas a traditional formulation was much less so. Later I realized that a very similar factorization was possible with matrices directly [2]. This was a bit disappointing since I was enamored by my new calculation tool, and realized that the problem could be tackled with much less learning cost if the same factorization technique was applied using plain old matrices. I’ve now encountered a new use for this idea of factoring a vector into a product of multivector matrices. Namely, a calculation of the four vector Lienard-Wiechert potentials, given a general motion described in cylindrical coordinates. This I thought I’d try since we had a similar problem on our exam (with the motion of the charged particle additionally constrained to a circle). 2. The goal of the calculation. Our problem is to calculate q R∗ qvc A= cR∗

A0 =

1

(1) (2)

where xc (t) is the location of the charged particle, r is the point that the field is measured, and vc ·R c R2 = R2 = c2 ( t − tr )2

R∗ = R −

R = r − x c ( tr ) ∂xc vc = . ∂tr

(3) (4) (5) (6)

3. Calculating the potentials for an arbitrary cylindrical motion. Suppose that our charged particle has the trajectory xc (t) = h(t)e3 + a(t)e1 eiθ (t)

(7)

where i = e1 e2 , and we measure the field at the point r = ze3 + ρe1 eiφ

(8)

The vector separation between the two is R = r − xc

= (z − h)e3 + e1 (ρeiφ − aeiθ )   ρ  iφ  = e1 e −e1 eiθ e3  a  z−h Transposition does not change this at all, so the (squared) length of this vector difference is    e1 eiφ  ρ  R2 = ρ a (z − h) −e1 eiθ  e1 eiφ −e1 eiθ e3  a  z−h e3    iφ iφ iφ iθ e1 eiφ e3 ρ   e1 e iθe1 e iφ −e1 eiθ e1iθe = ρ a ( z − h )  − e1 e e1 e e1 e e1 e −e1 eiθ e3   a  z−h e3 e1 eiφ −e3 e1 eiθ e3 e3    i ( θ − φ ) iφ 1 −e e1 e e3 ρ   i ( φ − θ ) iθ    a  = ρ a (z − h) −e 1 − e1 e e3 iφ iθ z−h e3 e1 e − e3 e1 e 1 





3.1. A motivation for a Hermitian like transposition operation. There are a few things of note about this matrix. One of which is that it is not symmetric. This is a consequence of the non-commutative nature of the vector products. What we do have is a

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Hermitian transpose like symmetry. Observe that terms like the (1, 2) and the (2, 1) elements of the matrix are equal after all the vector products are reversed. Using tilde to denote this reversion, we have

(ei(θ −φ) )˜ = cos(θ − φ) + (e1 e2 )˜ sin(θ − φ) = cos(θ − φ) + e2 e1 sin(θ − φ) = cos(θ − φ) − e1 e2 sin(θ − φ) = e −i ( θ − φ ) . The fact that all the elements of this matrix, if non-scalar, have their reversed value in the transposed position, is sufficient to show that the end result is a scalar as expected. Consider a general quadratic form where the matrix has scalar and bivector grades as above, where there is reversion in all the transposed positions. That is bT Ab

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where A = Aij , a m × m matrix where Aij = A˜ji and contains scalar and bivector grades, and b = kbi k, a m × 1 column matrix of scalars. Then the product is

∑ bi Aij bj = ∑ bi Aij bj + ∑ bi Aij bj + ∑ bk Akk bk ij

i< j

j
k

= ∑ bi Aij b j + ∑ b j A ji bi + ∑ bk Akk bk i< j

i< j

k

= ∑ bk Akk bk + 2 ∑ bi ( Aij + A ji )b j i< j

k

= ∑ bk Akk bk + 2 ∑ bi ( Aij + A˜ij )b j i< j

k

The quantity in braces Aij + A˜ij is a scalar since any of the bivector grades in Aij cancel out. Consider a similar general product of a vector after the vector has been factored into a product of matrices of multivector elements   b1     b2  x = a1 a2 . . . a m  .  (10)  ..  bm The (squared) length of the vector is x2 = ( ai bi )( a j b j )

= ( ai bi )˜a j b j = b˜i a˜i a j b j = b˜i ( a˜i a j )b j . 3

It is clear that we want a transposition operation that includes reversal of its elements, so with a general factorization of a vector into matrices of multivectors x = Ab, it’s square will be x = b˜ T A˜ T Ab. As with purely complex valued matrices, it is convenient to use the dagger notation, and define A† = A˜ T

(11)

where A˜ contains the reversed elements of A. By extension, we can define dot and wedge products of vectors expressed as products of multivector matrices. Given x = Ab, a row vector and column vector product, and y = Cd, where each of the rows or columns has m elements, the dot and wedge products are D E x · y = d† C † Ab D E x ∧ y = d† C † Ab . 2

(12) (13)

In particular, if b and d are matrices of scalars we have E D C † A + A† C x · y = dT C † A b = dT b 2 E D C † A − A† C b. x ∧ y = dT C † A b = dT 2 2

(14) (15)

The dot product is seen as a generator of symmetric matrices, and the wedge product a generator of purely antisymmetric matrices. 3.2. Back to the problem Now, returning to the example above, where we want R2 . We’ve seen that we can drop any bivector terms from the matrix, so that the squared length can be reduced as   i (θ −φ) 0 − e ρ   R2 = ρ a ( z − h )  − e i ( φ − θ ) 1 0  a  z−h 0 0 1    1 − cos(θ − φ) 0 ρ   1 0  a  = ρ a (z − h) − cos(θ − φ) 0 0 1 z−h     ρ − a cos(θ − φ) = ρ a (z − h) −ρ cos(θ − φ) + a z−h 

1

So we have R2 = ρ2 + a2 + (z − h)2 − 2aρ cos(θ − φ) Now consider the velocity of the charged particle. We can write this as

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(16)

 h˙  a˙  aθ˙ 

 dxc = e3 e1 eiθ dt

e2 e

 iθ

(17)

To compute vc · R we have to extract scalar grades of the matrix product * e eiφ  1  −e1 eiθ  e3 e1 eiθ e3

+ e2 eiθ



+ * e eiφ  1   = −e1 eiθ  e3 e1 eiθ e2 eiθ e3  * e eiφ e e1 eiφ e1 eiθ e1 eiφ e2 eiθ + 3 1 = −e1 eiθ e3 −e1 eiθ e1 eiθ −e1 eiθ e2 eiθ  e3 e3 e3 e1 eiθ e3 e2 eiθ   0 cos(θ − φ) − sin(θ − φ) . −1 0 = 0 1 0 0

So the dot product is    h˙  0 cos(θ − φ) − sin(θ − φ)   a˙  −1 0 R · v = ρ a ( z − h ) 0 1 0 0 aθ˙   ˙   a˙ cos(θ − φ) − aθ sin(θ − φ)   − a˙ = ρ a (z − h) ˙h ˙ + ρ a˙ cos(θ − φ) − ρaθ˙ sin(θ − φ) = (z − h)h˙ − aa 

This is the last of what we needed for the potentials, so we have A0 = p

q ˙ + a a/c ˙ ˙ + ρ cos(θ − φ) a/c ˙ − ρa sin(θ − φ)θ/c ρ2 + a2 + (z − h)2 − 2aρ cos(θ − φ) − (z − h)h/c (18)

A=

˙ 3 + ( ae ˙ 2 ˙ 1 + aθe he c

)eiθ

A0 ,

(19)

where all the time dependent terms in the potentials are evaluated at the retarded time tr , defined implicitly by the messy relationship q c(t − tr ) = (ρ(tr ))2 + ( a(tr ))2 + (z − h(tr ))2 − 2a(tr )ρ cos(θ (tr ) − φ). (20) 4. Doing this calculation with plain old cylindrical coordinates. It’s worth trying this same calculation without any geometric algebra to contrast it. I’d expect that the same sort of factorization could also be performed. Let’s try it

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

 a cos θ xc =  a sin θ  h   ρ cos φ r =  ρ sin φ  z

(21)

(22)

R = r − xc   ρ cos φ − a cos θ =  ρ sin φ − a sin θ  z−h    cos φ − cos θ 0 ρ =  sin φ − sin θ 0  a  0 0 1 z−h So for R2 we really just need to multiply out two matrices    cos φ sin φ 0 cos φ − cos θ 0 cos2 φ + sin2 φ − cos θ − sin θ 0  sin φ − sin θ 0 = −(cos φ cos θ + sin θ sin φ) 0 0 1 0 0 1 0  1 − cos(φ − θ )  1 = − cos(φ − θ ) 0 0 

 −(cos φ cos φ + sin φ sin θ ) 0 cos2 θ + sin2 θ 0 0 1  0 0 1

So for R2 we have 

  1 − cos ( φ − θ ) 0 ρ   1 0  a  R2 = ρ a (z − h) − cos(φ − θ ) 0 0 1 z−h     ρ − a cos(φ − θ ) = ρ a (z − h) −ρ cos(φ − θ ) + a z−h

= (z − h)2 + ρ2 + a2 − 2aρ cos(φ − θ ) We get the same result this way, as expected. The matrices of multivector products provide a small computational savings, since we don’t have to look up the cos φ cos φ + sin φ sin θ = cos(φ − θ ) identity, but other than that minor detail, we get the same result. For the particle velocity we have

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 a˙ cos θ − aθ˙ sin θ vc =  a˙ sin θ + aθ˙ cos θ  h˙    a˙ cos θ − sin θ 0 =  sin θ cos θ 0  aθ˙  0 0 1 h˙ 

So the dot product is 

vc · R =

=

= =

   cos θ sin θ 0 cos φ − cos θ 0 ρ   a˙ aθ˙ h˙ − sin θ cos θ 0  sin φ − sin θ 0  a  0 0 1 0 0 1 z−h    2 2 ρ   cos θ cos φ + sin θ sin φ − cos θ − sin θ 0  a  a˙ aθ˙ h˙ − cos φ sin θ + cos θ sin φ 0 z−h 0 0 1    ρ   cos(φ − θ ) −1 0 a˙ aθ˙ h˙  sin(φ − θ ) 0 0  a  0 0 1 z−h ˙h(z − h) − aa ˙ ˙ + ρ a˙ cos(φ − θ ) + ρaθ sin(φ − θ )

5. Reflecting on two the calculation methods. With a learning curve to both Geometric Algebra, and overhead required for this new multivector matrix formalism, it is definitely not a clear winner as a calculation method. Having worked a couple examples now this way, the first being the N spherical pendulum problem, and now this potentials problem, I’ll keep my eye out for new opportunities. If nothing else this can be a useful private calculation tool, and the translation into more pedestrian matrix methods has been seen in both cases to not be too difficult. References [1] Peeter Joot. Spherical polar pendulum for one and multiple masses (Take II) [online]. Available from: http://sites.google.com/site/peeterjoot/math2009/multiPendulumSpherical2. pdf. 1 [2] Peeter Joot. Lagrangian and Euler-Lagrange equation evaluation for the spherical Npendulum problem [online]. Available from: http://sites.google.com/site/peeterjoot/ math2009/multiPendulumSphericalMatrix.pdf. 1

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