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Writing Tensor Equations in Matrix Form

  1. Feb 24, 2012 #1
    I'm trying to express the tensor equation [itex]F'^{\mu\nu}=\Lambda^{\mu}_{\sigma}\Lambda^{\nu}_{ \rho }F^{\sigma\rho}[/itex] in matrix form. Here the indices range from 0 to 3, so we need 4 by 4 matrices. Let F', F, and [itex]\Lambda[/itex] be the matrices associated with the tensors appearing in our equation. Which of the following is the correct matrix translation of the tensor equation?

    [itex]F'=\Lambda F \Lambda[/itex]

    [itex]F'=\Lambda \Lambda F[/itex]

    [itex]F'=\Lambda^{\top} F \Lambda[/itex]

    [itex]F'=\Lambda F \Lambda^{\top}[/itex]

    Or something else entirely?

    I tried testing some of these out on the actual four-by-four matrices, but the algebra got too cumbersome. Usually when I figure out what order to put things in and where to put the transposes, I'm in a situation where I'm dealing with matrices and vectors, so that if you put it in the wrong order then the numbers of rows and columns don't match up. But in this case everything is four-by-four, so there is plenty of room for error.

    Any help would be greatly appreciated.

    Thank You in Advance.
  2. jcsd
  3. Feb 24, 2012 #2

    Ben Niehoff

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    Gold Member

    It will be one of these two, depending on how you decide to map your tensors to matrices.

    You see, a tensor and a matrix are not quite the same thing. A matrix is most naturally thought of as a "mixed" tensor, with one index up and one down: [itex]M^a{}_b[/itex]. Then the matrix product is quite natural to write:

    [itex]M^a{}_c = K^a{}_b L^b{}_c[/itex]

    On the other hand, a tensor with two "up" indices is technically considered a column vector whose elements are column vectors. However, the multiplication algorithm will end up being the same as matrix multiplication. To translate it to matrices, you just need to follow the indices carefully:

    [tex]\Lambda^a{}_c \Lambda^b{}_d F^{cd} = \Lambda^a{}_c F^{cd} \Lambda^b{}_d = \Lambda^a{}_c F^{cd} (\Lambda^\top)_d{}^b[/tex]
    where in the last step, we take the transpose because we need to switch the order of b and d to make it look like a matrix product. So we can write

    [tex]F' = \Lambda F \Lambda^\top[/tex]
    provided we interpret the first index of F as a row index, and the second as a column index.

    However, you should be careful with this notation to be clear what you mean. If we had a mixed tensor [itex]G^a{}_b[/itex], then in matrix notation its transformation would be

    [tex]G' = \Lambda G \Lambda^{-1}[/tex]
    For the pure "up" tensor [itex]F^{ab}[/itex], the most mathematically-correct way to write its transformation law is

    [tex]F' = (\Lambda \otimes \Lambda) \cdot F[/tex]
    where now F is unfolded into a single column vector with [itex]n \times n[/itex] entries.
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