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I've been confused about the $\partial_{\mu}$ operator.

Peskin and Schroeder defines it as $\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$

For example, the Euler Lagrange equation of motion is

$$ \frac{\partial}{\partial x^{\mu}}\Big( \frac{\partial \mathcal{L}}{\partial (\partial \phi/\partial x^{\mu})}\Big) - \frac{\partial\mathcal{L}}{\partial \phi} = 0$$

If we apply this to the Lagrangian

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

Then

$$ \frac{\partial \mathcal{L}}{\partial (\partial \phi/\partial x^{\mu})} = \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} = \partial_{\mu}\phi$$

And thus the first term should be $$ \partial_{\mu} \partial_{\mu} \phi$$

But the first term in the Klein Gordon equation is

$$ \partial^{\mu} \partial_{\mu} \phi$$

Where am I going wrong here? How do contravariant and covariant tensors and the Minkowski metric tensor apply?

[Edit]

Follow up question:

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

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    $\begingroup$ $(\partial_\mu\phi)^2=\partial_\mu\phi\partial^\mu\phi$. Your derivative of the Lagrangian is wrong $\endgroup$ – OON Jul 24 '17 at 4:12
  • $\begingroup$ Do I differentiate this as a product and use the raising operator to get rid of the $\frac{1}{2}$? $\endgroup$ – saad Jul 24 '17 at 4:16
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    $\begingroup$ @OON That should have been an answer, not a comment. $\endgroup$ – David Z Jul 24 '17 at 4:33
  • $\begingroup$ @DavidZ I regretfully had no opportunity to write anything longer than that and thought that as an answer that would be almost rude. $\endgroup$ – OON Jul 24 '17 at 12:11
  • $\begingroup$ @OON No, it wouldn't be rude. Length is not a factor that distinguishes answers from comments; all that matters is whether the post answers the question or not. If you have something that answers the question, post it as an answer; or if for some reason you don't want to make it an answer, don't post it at all. It shouldn't be a comment though. $\endgroup$ – David Z Jul 24 '17 at 12:16
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Perhaps it is useful to just expand these terms without the fancy notation.

$$ \frac{1}{2} (\partial_\mu \phi)^2 = \frac{1}{2} ( \partial_\mu \phi) (\partial^\mu \phi) = \frac{1}{2} \left( (\partial_t \phi)^2 - (\partial_x \phi)^2 \right) $$

Therefore

$$ \frac{\partial \frac{1}{2} (\partial_\mu \phi)^2}{\partial (\partial_t \phi)} = \frac{\partial \frac{1}{2} (\partial_t \phi)^2}{\partial (\partial_t \phi)} = \partial_t \phi $$

$$ \frac{\partial \frac{1}{2} (\partial_\mu \phi)^2}{\partial (\partial_x \phi)} = \frac{\partial -\frac{1}{2} (\partial_x \phi)^2}{\partial (\partial_x \phi)} = -\partial_x \phi $$

These two things can be summarized as

$$ \frac{\partial \frac{1}{2} (\partial_\nu \phi)^2}{\partial (\partial_\mu \phi)} = \partial^\mu \phi $$

The operator $\partial^\mu$ is different from the operator $\partial_\mu$. Differential operators exist naturally with the lowered indices. Raising them involved multiplying them by the metric. So in the mostly minus convention $(+---)$ we have

$$ \partial_\mu = (\partial_t, \partial_x, \partial_y, \partial_z) $$ $$ \partial^\mu = (\partial_t, -\partial_x, -\partial_y, -\partial_z) $$

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  • $\begingroup$ That explains a lot! Can you please have a look at the edit though? $\endgroup$ – saad Jul 24 '17 at 15:04
  • $\begingroup$ $(\partial_t \partial^t + \partial_x \partial^x) \phi = (\partial_t^2 - \partial_x^2) \phi.$ I don't see what the problem is $\endgroup$ – user1379857 Jul 24 '17 at 21:46

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