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

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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#### George Jones

Staff Emeritus
Gold Member
I'm trying to derive the Klein Gordon equation from the Lagrangian:

$$\mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi)^2 - \frac{1}{2}m^2 \phi^2$$
What does "$\frac{1}{2}(\partial_{\mu} \phi)^2$" mean?

Also, why are you splitting things into time and space components?

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

Also, why are you splitting things into time and space components?
$$\frac{1}{2}(\partial_{\mu} \phi)^2 = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi)$$

#### George Jones

Staff Emeritus
Gold Member
$$\frac{1}{2}(\partial_{\mu} \phi)^2 = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi)$$
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.

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.
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
$$\partial_{\beta}\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)} = \partial^{\mu}\partial_{\mu}$$

#### George Jones

Staff Emeritus
Gold Member
I haven't had a chance to look really closely (I will though)

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
$$\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)} = \frac{1}{2}g^{\mu\alpha}(\partial_{\mu} \phi \delta^{\alpha}_{\beta} + \partial_{\alpha} \phi \delta^{\beta}_{\mu})$$
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:

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.
Now that I look more closely, I don't see a missing minus sign.

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.
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?

#### George Jones

Staff Emeritus
Gold Member
The $\mu$ is contravariant and the $\beta$ covariant, right?
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 $\delta$.

Shouldn't $$\partial_{x}\partial^{x}$$ and $$\partial_{t}\partial^{t}$$ have opposite signs, since we are working with four vectors?
No. Remember,
$$A_\mu A^\mu = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3 = A^0 A^0 - A^1 A^1 - A^2 A^2 - A^3 A^3.$$

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

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