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A Killing vectors corresponding to the Lorentz transformations

  1. Feb 13, 2019 at 7:00 AM #1
    Hi everyone! I have a problem with one thing.

    Let's consider the Lorentz group and the vicinity of the unit matrix. For each ##\hat{L}##
    from such vicinity one can prove that there exists only one matrix ##\hat{\epsilon}## such that ##\hat{L}=exp[\hat{\epsilon}]##. If we take ##\epsilon^{μν}##, μ<ν, as the parameters on the Lorentz group in a vicinity of the unit matrix, then we can compute the corresponding Killing vectors as ##\xi ^μ_{αβ}=\frac{∂x′^μ}{∂ϵ^{αβ}}(\hat{ϵ}=0)## where ##x′^μ=L^μ_νx^ν##. Here is my problem: during the computations there is one line that I do not get, namely: ##\cfrac{∂L^{μν}}{∂ϵ^{αβ}}(\hat{ϵ}=0)=δ^μ_αδ^ν_β−δ^μ_βδ^ν_α##. The ## (\epsilon^{μν}) ## matrix is antisymmetric, so that is why we get the difference of the Kronecer delta?

    Thank you in advance!

    MW
     
  2. jcsd
  3. Feb 13, 2019 at 10:31 AM #2

    haushofer

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    Yes. That partial derivative is defined such that the (a)symmetries of alpha and beta are the same on the left and right hand side. You can also check explicitly by taking e.g. alpha=0, beta=1.
     
  4. Feb 13, 2019 at 11:24 AM #3
  5. Feb 13, 2019 at 12:13 PM #4

    Orodruin

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    Because that page treats a general matrix where the components are all independent, this is not the case for the Lorentz group generators.
     
  6. Feb 13, 2019 at 3:45 PM #5

    haushofer

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  7. Feb 15, 2019 at 7:00 AM #6

    haushofer

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    So, to extend a little bit on Orodruin's remark: in this case, the antisymmetry in alpha and beta indicates that not all the components are independent. So if you then define the functional derivative of a tensor with respect to that epsilon, you get an expression containing again indices alpha and beta. By definition, these components are then still considered to be dependent. This means the (a)symmetry carries over in the functional derivative.

    Or, rephrased: the (a)symmetry has to be respected on the left and right hand side in the definition of the functional derivative.
     
  8. Feb 15, 2019 at 7:07 AM #7

    Orodruin

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    To expand a little on the little expansion, just with a hands-on example, consider the Lorentz transformations in 1+1D Minkowski space, where you would have
    $$
    \hat \epsilon = \begin{pmatrix}0 & \epsilon_{01} \\ - \epsilon_{01} & 0 \end{pmatrix}.
    $$
    Differentiating ##\hat L = \exp(\hat\epsilon)## with respect to ##\epsilon_{01}## would give you
    $$
    \frac{\partial \hat L}{\partial \epsilon_{01}} = \frac{\partial \hat \epsilon}{\partial \epsilon_{01}} \hat L,
    $$
    which evaluated in ##\epsilon = 0## leads to
    $$
    \left. \frac{\partial \hat L}{\partial \epsilon_{01}}\right|_{\epsilon = 0} = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}.
    $$
    You will find that this satisfies your relation.
     
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