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Dyadic Product and Index of Refraction

  1. Nov 22, 2008 #1
    Dear readers,

    In my recent study of dyadic products I found out that physical quantities expressend as 2nd rank tensors can also be expressed as a dyadic product of two vectors. Similarly 2nd rank tensor fields can be expressed as pointwise dyadic products of two vector fields.

    One such quantity seems to be the index of refraction. I am used to solving geometrical optics problems, say in 2 dimensions, using calculus of variations and a (Riemannian) metric tensor with elements n^2 on the diameter and 0's elsewhere. But I've been at a loss for a while now regarding how to understand that metric tensor in terms of a dyadic product.

    I'll be grateful if someone explains this to me or points me to some works (preferably online) wherein this is explained. Please note that I'm primarily concerned with real indices of refraction and transparent media.

    Thanks in advance.
  2. jcsd
  3. Nov 22, 2008 #2


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    To be more precise second rank tensors may be expressed as linear combinations of dyadic products. The dyadic/n-adic product is just alternative notation for the tensor product.
    [tex] \hat{x}\hat{y} \equiv \hat{x}\otimes\hat{y}[/tex]
    In this case note that [tex]\hat{x}\hat{y}[/tex] and [tex]\hat{y}\hat{x}[/tex] are distinct and independent rank 2 tensors.

    Be careful as often (and especially with differentials) the implied product is not the tensor product but a outer ( = anti-symmetric = Grassmann) product.
    [tex] dx dy \equiv dx\wedge\dy = dx\otimes dy - dy\otimes dx[/tex]

    If you are representing a general 2x2 tensor then the dyadic representation is:
    [tex] \left( \begin{array}{cc} A & B \\ C & D \end{array}\right) = A\hat{x}\hat{x} + B \hat{x}\hat{y} + C\hat{y}\hat{x} + D\hat{y}\hat{y}[/tex]

    Where [tex]\{\hat{x}, \hat{y}\}[/tex] is the basis and I've rewritten a rank (2,0) tensor as a matrix { rank (1,1) } for compactness.

    Usually dyadics are used when there is a (Euclidean) metric and thus one can transform between covariant and contravariant. The metric is then simply represented by the dot product: [tex] \hat{x}\bullet \hat{y} = g_{xy}[/tex]

    You can extend the inner product to the various dyadics and invoke multiple times. Example:
    [tex] pqrs\bullet \bullet uvw = (r\bullet v)(s\bullet u)pqrw = g_{su}g_{rv}\cdot pqw[/tex]

    Now if you must distinguish between covariant and contravariant vectors then you could use a star notation or simply use raised and lower indices.

    The metric then on a space with basis [tex]\{\mathbf{e}_1, \mathbf{e}_2,\cdots\}[/tex] would be expressed in dyadic form using dyadics of the dual basis, [tex] \{ \mathbf{e}^k\} : \mathbf{e}^k ( e_j) = \delta^k_j[/tex]

    The metric (or any other rank (0,2) tensor) is then expressed as:
    [tex]\mathbf{g} = \sum_{i,j} g_{ij} \mathbf{e}^i \mathbf{e}^j[/tex]

    The main problem with dyadic notation is that it is difficult to express all the possible index contracting products since the order the vectors are written matters. Ultimately one drops the dyadic notation altogether except in introductory definitions. Simply work with the indexed coefficients themselves. This with Einstein's summation convention makes expressing all tensor math straightforward and compact.
  4. Nov 22, 2008 #3
    Thanks a lot, jambaugh. Very nice explanation. Though, I am rather familiar with tensor analysis in coordinates. Perhaps I haven't stated clearly where my understanding fails.

    Assume you have a Riemannian metric g, e.g. the one induced by the existence of a transparent medium, you can simply compute the Christoffel symbols of second kind, write down the geodesic equations, and attempt to solve them. I know how to do that.

    However, upon being introduced to dyads/dyadics I was told that a dyad understood as an operator on the vector space (in this case the tangent space at each point) could be interpreted as the physical entity it represents, for example the transparent medium, acting on other physical things, light in the example I'm trying to comprehend. The medium acts on light, a vector, and changes its direction and magnitude.

    Now if I have the refraction index at every point in a space, say R^2, I can create a simple metric out of it and solve the problem as I described in the second paragraph. Nonetheless, for the purpose of demonstration I am interested in finding the dyad corresponding to the scalar refraction index. I thought this could be done from the Riemannian metric much in the same way you explained above. But it doesn't work out:

    Let u = a i + b j, v = c i + d j be the vectors that form the sought-for dyad, then we have

    uv = ac ii + ad ij + bc ji + bd jj

    g11 = n^2, g12 = g21 = 0, g22 = n^2 => ac = n^2, ad = 0 = bc, bd = n^2

    These equations don't yield a result for n not equal to zero, not even for a, b, c, d complex numbers, i.e. a vector space over the complex field. And I don't know what else to do. I have been warned that the multiplicands of a dyadic product need not themselves correspond to physical quantities even if their product is indeed such a quantity but anyway there must be a way of obtaining them for mathematical purposes.

    I presume the multiplicands of the dyad in question should somehow be derived from an application of Snell's law but the procedure fails me.
  5. Nov 24, 2008 #4


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    I'm short on time and need time to understand what you're trying to do. Just to be sure though you do understand that not all dyad's i.e. rank 2 tensors are going to be obtained from the dyadic (tensor) product of a single pair of vectors e.g. uv.

    Rather you must take linear combinations of such paired products to get the general case, at most I believe a number of pairs equal to your dimension. uv + rs.

    I don't know if that helps but class is starting and I've got to go. I'll look at this some more this evening.
  6. Nov 25, 2008 #5
    Thanks a lot for caring, jambaugh. Further thinking and a look into Pedrottis' Introduction to Optics solved the problem. It turned out what I sought after was the usual refraction matrix: a11 = 1, a12 = a21 = 0, a22 = n1/n2. This constitudes the tensorial form of Snell's law. I even managed to write a little raytracing Maple code with it :-)

    Your note about second-rank tensors being linear combinations of more than one dyadic products also explains why my uninformed approach failed. For the case of aij described above I have to suffice to the linear combination: 1 * ii + n1/n2 * jj. There are no two vectors u and v whose dyadic product give this combination so I have to think of it in terms of the dyadic product of i by i combined with the dyadic product of, say, n1/n2 * j by j.
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