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

17
0
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.
 

George Jones

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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?
 
17
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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?
$$ \frac{1}{2}(\partial_{\mu} \phi)^2 = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) $$
 

George Jones

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$$ \frac{1}{2}(\partial_{\mu} \phi)^2 = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) $$
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.
 
17
0
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

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772
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.
 
17
0
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
Science Advisor
Gold Member
7,223
772
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,
$$A_\mu A^\mu = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3 = A^0 A^0 - A^1 A^1 - A^2 A^2 - A^3 A^3.$$
 

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