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

Question

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?

Answer 1

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

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