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A Confusion regarding the $\partial_{\mu}$ operator

  1. Jul 24, 2017 #1
    I'm trying to derive the Klein Gordon equation from the Lagrangian:

    $$ \mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi)^2 - \frac{1}{2}m^2 \phi^2$$

    $$\partial_{\mu}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\Bigg) = \partial_{t}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{t} \phi)}\Bigg) + \partial_{x}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{x} \phi)}\Bigg)$$

    But if
    $$ \frac{\partial \mathcal{L}}{\partial (\partial_{t} \phi)} = \partial_{t} \phi = \partial^{t} \phi$$
    And
    $$ \frac{\partial \mathcal{L}}{\partial (\partial_{x} \phi)} = -\partial_{x} \phi = \partial^{x} \phi$$
    Then
    $$\partial_{\mu}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\Bigg) = \partial_{t} \partial^{t} \phi + \partial_{x} \partial^{x} \phi$$
    We seem to missing a minus sign here. Where's the mistake? I'm supposed to get

    $$ \partial_{\mu}\partial^{\mu}\phi$$ for this term.
     
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  3. Jul 24, 2017 #2

    George Jones

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    What does "##\frac{1}{2}(\partial_{\mu} \phi)^2##" mean?

    Also, why are you splitting things into time and space components?
     
  4. Jul 24, 2017 #3
    $$ \frac{1}{2}(\partial_{\mu} \phi)^2 = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) $$
     
  5. Jul 24, 2017 #4

    George Jones

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    and
    $$\left(\partial_{\mu} \phi)(\partial^{\mu} \phi\right) = g^{\mu \alpha} \left(\partial_{\mu} \phi)(\partial_{\alpha} \phi\right)$$

    Now, find
    $$\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)}.$$

    Note that I introduced a new index ##\beta##, because an index different than the dummy summation indices ##\mu## and ##\alpha## is needed.
     
  6. Jul 24, 2017 #5
    Thank you. Is this correct?
    $$ \mathcal{L} =\frac{1}{2}g^{\mu \alpha} (\partial_{\mu} \phi)(\partial_{\alpha} \phi) - \frac{1}{2}m^{2} \phi^{2}$$
    So
    $$\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)} = \frac{1}{2}g^{\mu\alpha}(\partial_{\mu} \phi \delta^{\alpha}_{\beta} + \partial_{\alpha} \phi \delta^{\beta}_{\mu})$$
    Hence,
    $$\partial_{\beta}\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)} = \frac{1}{2}\partial_{\beta}(g^{\mu \beta} \partial_{\mu} \phi + g^{\beta \alpha} \partial_{\alpha} \phi)$$
    $$\partial_{\beta}\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)} = \frac{1}{2}(\partial^{\mu}\partial_{\mu} + \partial^{\alpha}\partial_{\alpha}) $$
    Since the first and second terms are the same, we can get rid of the half. And thus
    $$ \partial_{\beta}\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)} = \partial^{\mu}\partial_{\mu}$$
     
  7. Jul 24, 2017 #6

    George Jones

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    I haven't had a chance to look really closely (I will though)

    There is a small mistake in the placement of indices in the first term on the right side of the second equation.

    From the original post:

    Now that I look more closely, I don't see a missing minus sign.
     
  8. Jul 24, 2017 #7
    The ##\mu## is contravariant and the ##\beta ## covariant, right?

    Shouldn't $$ \partial_{x}\partial^{x}$$ and $$ \partial_{t}\partial^{t}$$ have opposite signs, since we are working with four vectors?
     
  9. Jul 24, 2017 #8

    George Jones

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    In the first term on the right side, the ##\delta^\alpha_\beta## should be ##\delta^\beta_\alpha##. Since the summation index ##\alpha## is upstairs on the ##g^{\mu \alpha}## and downstairs on ##\partial_\alpha \phi##, it must be downstairs in the ##\delta##; roughly, since the the index ##\beta## is downstairs in the "denominator" of
    $$\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)},$$
    ##\beta## should be upstairs in the ##\delta##.

    No. Remember,
    $$A_\mu A^\mu = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3 = A^0 A^0 - A^1 A^1 - A^2 A^2 - A^3 A^3.$$
     
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