The Stress Tensor of the Electromagnetic Field

Generating a Symmetric 2-Tensor Using Quaternions

I will outline a way to generate the terms of the symmetric 2-rank stress-momentum tensor of an electromagnetic field using quaternions.  This method may provide some insight into what information the stress tensor contains.

Any equation written with 4-vectors can be rewritten with quaternions.  A straight translation of terms could probably be automated with a computer program.  What is more interesting is when an equation is generated by the product of operators acting on quaternion fields.  I have found that generator equations often yield useful insights.

A tensor is a bookkeeping device designed to keep together elements that transform in a similar way.  People can choose alternative bookkeeping systems, so long as the tensor behaves the same way under transformations.  Using the terms as defined in "The classical theory of fields" by Landau and Lifshitz, the antisymmetric 2-rank field tensor F is used to generate the stress tensor T

T sup ik = (- F sup iL F sub L sup k +delta(i,k) F sub LM F sup LM) over 4pi

I have a practical sense of an E field (the stuff that makes my hair stand on end) and a B field (the invisible hand directing a compass), but have little sense of the field tensor F, a particular combination of the other two.  Therefore, express the stress tensor T in terms of the E and B fields only:

T sup ik =  matrix((W, Sx, Sy, Sz), (Sx, mxx, mxy, myz), (Sy, myx, myy, myz), (Sz, mzx, mzy, mzz))

W = (E squared +  B squared ) over 8pi

Sa = (E Cross B) over 4pi

mab = (- Ea Eb - Ba Bb + 0.5 delta(a,b)(E squared +  B squared )) over 4 pi

Together, the energy density(W), Poynting's vector (Sa) and the Maxwell stress tensor (m_ab) are all the components of the stress tensor of the electromagnetic field.

Generating a Symmetric 2-Tensor Using Quaternions

How should one rationally go about to find a generator equation that creates these terms instead of using the month-long hunt-and-peck technique actually used?  Everything is symmetric, so use the symmetric product:

even(q, q prime) = (q q prime + q prime q ) over 2= (t t prime - X dot X prime, t X + X t prime)

The fields E and B are kept separate except for the cross product in the Poynting vector.  Individual directions of a field can be selected by using a unit vector Ua:

{E, Ux} = (-Ex, 0)   where Ux = (0, 1, 0, 0)

The following double sum generates all the terms of the stress tensor:

T sup ik =  Sum of a from x thru z Sum of b from x thru z  of (({Ua,Ub} over 3 - 1) times ((0,E) squared + (0,B) squared ) over 2 -

- even(E, Ua) even(E, Ub) - even(B, Ua) even(B, Ub) -

- even(odd(E, B), Ua) - even(odd(E, B), Ub)) over 4pi =

= (-Ex Ey - Ex Ez - Ey Ez - Bx By - Bx Bz - By Bz + Ey Bz - Ez By + Ez Bx - Ex Bz + Ex By - Ey Bx, 0) over 2 pi

The first line generates the energy density W, and part of the +0.5 delta(a, b)(E^2 + B^2) term of the Maxwell stress tensor.  The rest of that tensor is generated by the second line.  The third line creates the Poynting vector.  Using quaternions, the net sum of these terms ends up in the scalar.

Does the generator equation have the correct properties?  Switching the order of Ua and Ub leaves T unchanged, so it is symmetric.  Check the trace, when Ua = Ub

trace(T sup ik) =

=Sum over x, y & z of 1 over 4 pi((even(Ua,Ua) over 3 - 1)((0,E) squared + (0,B) squared) over 2 - even(E, Ua) squared  - even(B, Ua) squared ) = 0

The trace equals zero, as it should.

The generator is composed of three parts that have different dependencies on the unit vectors: those terms that involve Ua and Ub, those that involve Ua or Ub, and those that involve neither.  These are the Maxwell stress tensor, the Poynting vector and the energy density respectively.  Changing the basis vectors Ua and Ub will effect these three components differently.


So what does the stress tensor represent?  It looks like every combination of the 3-vectors E and B that avoids quadratics (like Ex^2) and over-counting cross terms.  I like what I will call the "net" stress quaternion:

net(T sup ik) =

= (-Ex Ey - Ex Ez - Ey Ez - Bx By - Bx Bz - By Bz + Ey Bz - Ez By + Ez Bx - Ex Bz + Ex By - Ey Bx, 0) over 2 pi

This has the same properties as an stress tensor.  Since the vector is zero, it commutes with any other quaternion (this may be a reason it is so useful).  Switching x terms for y terms would flip the signs of the terms produced by the Poynting vector as required, but not the others.  There are no terms of the form Ex^2, which is equivalent to the statement that the trace of the tensor is zero.

On a personal note, I never thought I would understand what a symmetric 2-rank tensor was, even though I listen in on a discussion of the topic.  Yes, I could nod along with the algebra, but without any sense of F, it felt hollow.  Now that I have a generator and a net quaternion expression, it looks quite elegant and straightforward to me.

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