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

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?

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

• $(\partial_\mu\phi)^2=\partial_\mu\phi\partial^\mu\phi$. Your derivative of the Lagrangian is wrong – OON Jul 24 '17 at 4:12
• Do I differentiate this as a product and use the raising operator to get rid of the $\frac{1}{2}$? – saad Jul 24 '17 at 4:16
• @OON That should have been an answer, not a comment. – David Z Jul 24 '17 at 4:33
• @DavidZ I regretfully had no opportunity to write anything longer than that and thought that as an answer that would be almost rude. – OON Jul 24 '17 at 12:11
• @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. – David Z Jul 24 '17 at 12:16

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

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

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