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Lie derivative problem/error

  1. Sep 16, 2008 #1
    I'm trying to use an example to make sense out of the equation

    [tex]
    \mathcal{L}_X = d\circ i_X + i_X\circ d.
    [/tex]

    Some simple equations:

    [tex]
    \omega = \omega^1 dx_1 + \omega^2 dx_2
    [/tex]

    [tex]
    i_X\omega = X_1\omega^1 + X_2\omega^2
    [/tex]

    [tex]
    (d\omega)^{11} = (d\omega)^{22} = 0,\quad (d\omega)^{12} = \frac{1}{2}(\partial_1\omega^2 - \partial_2\omega^1) = - (d\omega)^{21}
    [/tex]


    [tex]
    \mu = \mu^{12} dx_1\wedge dx_2 + \mu^{21} dx_2\wedge dx_1
    [/tex]

    [tex]
    (i_X\mu)^1 = X_1 \mu^{11} + X_2 \mu^{21} = X_2 \mu^{21},\quad\quad (i_X\mu)^2 = X_1 \mu^{12}
    [/tex]


    [tex]
    \eta = \eta^0
    [/tex]

    [tex]
    (d\eta)^1 = \partial_1 \eta^0,\quad\quad (d\eta)^2 = \partial_2 \eta^0
    [/tex]

    Applying them:

    [tex]
    ((d\circ i_X)\omega)^1 = \partial_1 (i_X\omega) = \partial_1(X_1\omega^1 + X_2\omega^2)
    [/tex]

    [tex]
    ((d\circ i_X)\omega)^2 = \partial_2 (i_X\omega) = \partial_2(X_1\omega^1 + X_2\omega^2)
    [/tex]

    [tex]
    ((i_X\circ d)\omega)^1 = X_2 (d\omega)^{21} = -\frac{X_2}{2}(\partial_1\omega^2 - \partial_2\omega^1)
    [/tex]

    [tex]
    ((i_X\circ d)\omega)^2 = X_1 (d\omega)^{12} = \frac{X_1}{2}(\partial_1\omega^2 - \partial_2\omega^1)
    [/tex]

    [tex]
    ((d\circ i_X \;+\; i_X\circ d)\omega)^1 = (\partial_1 X_1)\omega^1 \;+\; (\partial_1 X_2)\omega^2 \;+\; X_1\partial_1\omega^1 \;+\; \frac{X_2}{2}\partial_1\omega^2 \;+\; \frac{X_2}{2}\partial_2\omega^1
    [/tex]

    [tex]
    ((d\circ i_X \;+\; i_X\circ d)\omega)^2 = (\partial_2 X_1)\omega^1 \;+\; (\partial_2 X_2)\omega^2 \;+\; \frac{X_1}{2}\partial_2 \omega^1 \;+\; X_2\partial_2\omega^2 \;+\; \frac{X_1}{2}\partial_1 \omega^2
    [/tex]

    There exists a following formula for the Lie derivative:

    [tex]
    (\mathcal{L}_X\omega)^{i_1,\ldots ,i_k} = X\cdot \omega^{i_1,\ldots ,i_k} \;+\; \sum_{\alpha = 1}^k (\partial_{i_{\alpha}} X_j) \omega^{i_1,\ldots ,i_{\alpha - 1},j, i_{\alpha + 1}, \ldots , i_k}
    [/tex]

    In this example it becomes

    [tex]
    (\mathcal{L}_X\omega)^1 = X_1\partial_1 \omega^1 \;+\; X_2\partial_2\omega^1 \;+\; (\partial_1 X_1) \omega^1 \;+\; (\partial_1 X_2) \omega^2
    [/tex]

    [tex]
    (\mathcal{L}_X\omega)^2 = X_1\partial_1 \omega^2 \;+\; X_2\partial_2 \omega^2 \;+\; (\partial_2 X_1)\omega^1 \;+\; (\partial_2 X_2) \omega^2
    [/tex]

    But this starts to look like

    [tex]
    \mathcal{L}_X \omega \neq (d\circ i_X + i_X\circ d)\omega
    [/tex]

    Where is this going wrong?
     
  2. jcsd
  3. Sep 17, 2008 #2

    haushofer

    User Avatar
    Science Advisor

    For your particular example of [itex]\omega[/itex] being a one-form and X the vector field, I get

    [tex]

    [di_{X}+i_{X}d ]\omega = d[\omega(X)] + d\omega (X) \\
    = d(\omega_{\mu}X^{\mu}) + 2[\partial_{[\mu}\omega_{\nu]}dx^{\mu}\wedge dx^{\nu}] (X)

    [/tex]

    This becomes

    [tex]
    [\omega_{\mu}\partial_{\nu}X^{\mu} + X^{\mu}\partial_{\nu}\omega_{\mu}]dx^{\nu} + X^{\mu}[\partial_{\mu}\omega_{\nu} - \partial_{\nu}\omega_{\mu}]dx^{\nu}
    [/tex]

    We see that two terms cancel and we obtain

    [tex]

    [\omega_{\mu}\partial_{\nu}X^{\mu} + X^{\mu}\partial_{\mu}\omega_{\nu}]dx^{\nu}

    [/tex]

    This looks like the Lie-derivative:

    [tex]

    \mathcal{L}_{X}\omega = [X^{\nu}\partial_{\nu}\omega_{\mu} + \partial_{\mu}X^{\nu}\omega_{\nu}]dx^{\mu}

    [/tex]

    If this doesn't clarify your problem I can look at all those explicit index-stuff you wrote down, but hopefully this helps :P
     
  4. Sep 20, 2008 #3
    The mistake was here. That was the same thing as

    [tex]
    i_X\mu = i_X\big(\mu^{ij} dx_i\wedge dx_j\big) = X_i \mu^{ij} dx_j
    [/tex]

    when it should have been

    [tex]
    i_X\mu = \frac{1}{2}i_X\big((\mu^{ij} - \mu^{ji})dx_i\otimes dx_j\big) = \frac{1}{2} X_i(\mu^{ij} - \mu^{ji})dx_j
    [/tex]

    I found that when following your calculation, and looking where it started going differently.
     
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