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Expressing general rotation in terms of tensors

  1. Feb 15, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A general rotation through angle ##a## about the axis ##\underline{n}##, where ##\underline{n}^2 = 1## is given by $$R(a,\underline{n}) = \exp(-ia\underline{n} \cdot \underline{T}),$$ where ##(T_k)_{ij} = -i\epsilon_{ijk}##. By expanding the exponential as a power series in ##a##, and explicitly summing the resulting series, show that $$R_{ij}(a,\underline{n}) = \delta_{ij}\cos a + n_i n_j (1-\cos a) - \epsilon_{ijk} n_k \sin a$$ using ##(\underline{n} \cdot \underline{T})_{ij}^2 = \delta_{ij} - n_i n_j## and ##(\underline{n} \cdot \underline{T})_{ij}^3 = (\underline{n} \cdot \underline{T})_{ij}##, (which I derived earlier)

    2. Relevant equations
    ##\exp(ix) = \cos x + i\sin x## and cos and sin power series.

    3. The attempt at a solution
    $$R(a,\underline{n})_{ij} = [\cos(a (\underline{n} \cdot \underline{T})) - i\sin(a (\underline{n} \cdot \underline{T}))]_{ij}$$ Reexpressing in terms of the Taylor series for sin and cos: $$\left[\sum_{k=0}^{\infty} (-1)^k \frac{(a (\underline{n} \cdot \underline{T}))^{2k}}{(2k)!} - i \sum_{k=0}^{\infty} (-1)^k \frac{(a (\underline{n} \cdot \underline{T}))^{2k+1}}{(2k+1)!}\right]_{ij}$$ Reorganising the even and odd terms gives $$\left[\sum_{k=0}^{\infty} (-1)^k \frac{(\delta_{ij} - n_i n_j)^k}{(2k)!} - i \sum_{k=0}^{\infty} (-1)^k\frac{[(\delta_{ij} - n_i n_j)^{2k} (-i\epsilon_{ijk}n_k)]}{(2k+1)!}\right]$$

    I am not really sure how to progress. Thanks for any help.
     
  2. jcsd
  3. Feb 15, 2014 #2

    tiny-tim

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    Hi CAF123! :smile:

    (btw, you missed out all the aks :wink:
    sooo you should have no powers of ##(\delta_{ij} - n_i n_j)## or of ##(\underline{n} \cdot \underline{T})_{ij}## above 1 ! :smile:
     
  4. Feb 16, 2014 #3

    CAF123

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    Hi tiny-tim,

    I was able to progress and obtained for ##k \geq 1##, ##(\underline{n} \cdot \underline{T})_{ij}^{2k} = \delta_{ij} - n_in_j## and for ##k \geq 0##, ##(\underline{n} \cdot \underline{T})_{ij}^{2k+1} = (\underline{n} \cdot \underline{T})_{ij}##.

    For the case ##k=0## in ##(\underline{n} \cdot \underline{T})_{ij}^{2k}##, I was wondering if this should be ##\delta_{ij}##. If so, then I have at the end $$\delta_{ij} + (\delta_{ij} - n_i n_j) (\cos a - 1) - \epsilon_{ijl} n_l \sin a = \text{result}$$ My reasoning being that for all other powers of ##k=0## the result is a matrix, so for ##k=0##, the result should also be a matrix, (= the identity, which is naturally ##\delta_{ij}##.) Thanks.
     
  5. Feb 16, 2014 #4

    tiny-tim

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    Hi CAF123! :wink:
    Yes, A0 for any matrix is the identity, I,

    so (A0)ij = δij :smile:
     
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