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

In summary: So$$\partial_\mu \partial^\mu = \partial_0 \partial^0 - \partial_1 \partial^1 - \partial_2 \partial^2 - \partial_3 \partial^3 = \partial^0 \partial^0 - \partial^1 \partial^1 - \partial^2 \partial^2 - \partial^3 \partial^3,$$and if you switch the ##\partial^1## and ##\partial^3## terms, you get a minus sign.
  • #1
saadhusayn
22
1
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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  • #2
saadhusayn said:
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?
 
  • #3
George Jones said:
What does "##\frac{1}{2}(\partial_{\mu} \phi)^2##" mean?

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

$$ \frac{1}{2}(\partial_{\mu} \phi)^2 = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) $$
 
  • #4
saadhusayn said:
$$ \frac{1}{2}(\partial_{\mu} \phi)^2 = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) $$

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.
 
  • #5
George Jones said:
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
$$ \partial_{\beta}\frac{\partial \mathcal{L}}{\partial (\partial_{\beta} \phi)} = \partial^{\mu}\partial_{\mu}$$
 
  • #6
I haven't had a chance to look really closely (I will though)

saadhusayn said:
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:

saadhusayn said:
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.
 
  • #7
George Jones said:
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?
 
  • #8
saadhusayn said:
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##.

saadhusayn said:
Shouldn't $$ \partial_{x}\partial^{x}$$ and $$ \partial_{t}\partial^{t}$$ have opposite signs, since we are working with four vectors?

No. Remember,
$$A_\mu A^\mu = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3 = A^0 A^0 - A^1 A^1 - A^2 A^2 - A^3 A^3.$$
 

What is the $\partial_{\mu}$ operator?

The $\partial_{\mu}$ operator, also known as the partial derivative operator, is a mathematical tool used to calculate how a function changes with respect to one of its variables while holding the other variables constant. It is commonly used in calculus and is denoted by the symbol $\partial_{\mu}$, where $\mu$ represents the variable with respect to which the derivative is being taken.

What is the difference between $\partial_{\mu}$ and $\frac{\partial}{\partial x}$?

Both $\partial_{\mu}$ and $\frac{\partial}{\partial x}$ represent partial derivative operators. The main difference between them is that $\partial_{\mu}$ is used in the context of multivariate functions, where there are multiple independent variables, while $\frac{\partial}{\partial x}$ is used for functions with a single independent variable. Additionally, $\partial_{\mu}$ is not limited to just the variable $x$, but can be used for any independent variable denoted by $\mu$.

What is the purpose of using the $\partial_{\mu}$ operator?

The $\partial_{\mu}$ operator is used in many areas of math and science to calculate rates of change and to find the slope or tangent of a multivariate function at a specific point. It is especially useful in fields such as physics, where functions often have multiple independent variables.

Can the $\partial_{\mu}$ operator be applied to any function?

Yes, the $\partial_{\mu}$ operator can be applied to any function that has multiple independent variables. It is important to note that the result of the partial derivative may differ depending on which variable is being held constant, so it is essential to specify the variable with respect to which the derivative is being taken.

What are some common mistakes when using the $\partial_{\mu}$ operator?

One common mistake when using the $\partial_{\mu}$ operator is forgetting to specify the variable with respect to which the derivative is being taken. This can lead to incorrect results, especially when dealing with functions with multiple independent variables. Another mistake is confusing the $\partial_{\mu}$ operator with the total derivative operator, which is denoted by $\frac{d}{dx}$ and calculates the overall rate of change of a function.

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