How to Derive Equations of Motion from Lagrange Density?

[Markus Kahn]

Homework Statement

I'd like to derive the equations of motion for a system with Lagrange density
$$\mathcal{L}= \frac{1}{2}\partial_\mu\phi\partial^\mu\phi,$$
for $\phi:\mathcal{M}\to \mathbb{R}$ a real scalar field.

Homework Equations

$$\frac{\partial \mathcal{L}}{\partial\phi}-\partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}=0$$

The Attempt at a Solution

$$\begin{align*}\frac{\partial \mathcal{L}}{\partial\phi}-\partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}& = -\partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\\
&= - \frac{1}{2}\partial_\mu \frac{\partial}{\partial(\partial_\mu\phi)} \partial_\nu\phi\partial^\nu\phi\\
&=- \frac{1}{2}\partial_\mu \left( \partial^\nu \phi\frac{\partial}{\partial(\partial_\mu\phi)}\partial_\nu\phi + \partial_\nu\phi \frac{\partial}{\partial(\partial_\mu\phi)} \partial^\nu\phi\right)\\
&= -\frac{1}{2} \partial_\mu (\partial^\mu\phi+\partial_\mu\phi)\end{align*}$$
As far as I know I should get $\Box \phi =0$, but for this to be true I need to show that
$$\frac{1}{2}(\partial^\mu\phi+\partial_\mu \phi)= \partial^\mu\phi$$
holds, and I honestly don't know why this should be the case.

Can somebody help?

 

[stevendaryl]

First of all, the meaning of $\partial^\mu \phi$ is just $g^{\mu \nu} \partial_\nu \phi$, where $g$ is the metric tensor. So you can rewrite $\mathcal{L}$ in terms of $\partial_\nu \phi$:

$\mathcal{L} = \frac{1}{2} g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi$

The other thing is that in the above, $\mu$ is a dummy index. To make sure it doesn't get confused with the $\mu$ in $\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}$, you should replace $\mu$ by some other index, say $\lambda$:

$\mathcal{L} = \frac{1}{2} g^{\lambda \nu} \partial_\lambda \phi \partial_\nu \phi$

Now, the only thing you need to know is: What is $\frac{\partial (\partial_\lambda \phi)}{\partial (\partial_\mu \phi)}$? It's zero if $\mu \neq \lambda$, and it's 1 if $\mu = \lambda$. That can be summarized by the kronecker delta:

$\frac{\partial (\partial_\lambda \phi)}{\partial (\partial_\mu \phi)} = \delta^\mu_\lambda$

You can similarly figure out $\frac{\partial (\partial_\nu \phi)}{\partial (\partial_\mu \phi)}$

See if that helps.

 

[bjnartowt]

As stevendaryl indicated, be careful to distinguish equation-indices with those indices that are summed over. In the following, I put a "prime" on indices that I sum over:
$0 = \frac{{\partial {\cal L}}}{{\partial \phi }} - {\partial _\mu }\frac{{\partial {\cal L}}}{{\partial \left( {{\partial _\mu }\phi } \right)}} = 0 - {\partial _\mu }\frac{\partial }{{\partial \left( {{\partial _\mu }\phi } \right)}}\left( {\frac{1}{2}{\partial _{\mu '}}\phi {\partial ^{\mu '}}\phi } \right) = - {\partial _\mu }\frac{1}{2}\left( {{g^\mu }_{\mu '} \cdot {\partial ^{\mu '}}\phi + {\partial _{\mu '}}\phi \cdot {g_{\mu '}}^\mu } \right) = - {\partial _\mu }\frac{1}{2}\left( {{\partial ^\mu }\phi + {\partial ^\mu }\phi } \right) = - {\partial _\mu }{\partial ^\mu }\phi$

I believe this is what you're looking for.

Afterword: note that this demonstrates that ${\partial _{{\partial _\mu }\phi }}$ is a contravariant vector (i.e., its action upon a scalar gives a vector with an upper index) while $x_\mu$ (having a lower index) is an example of a covariant vector.

 

[Markus Kahn]

First of all, the meaning of $\partial^\mu \phi$ is just $g^{\mu \nu} \partial_\nu \phi$, where $g$ is the metric tensor. So you can rewrite $\mathcal{L}$ in terms of $\partial_\nu\phi$

This was the trick that solved my problem... Didn't think that would simplify my issue this much, so thank you for that!

I believe this is what you're looking for.

This is exactly what I was looking for, but I must admit that I find your notation rather confusing... It's the first time for me seeing the metric tensor with both, indices and up and down at the same time. Since I'm not sure how I would evaluate this, I would need to transform it back to something I'm more used (aka $g_{\nu}{}^\mu = g^{\rho\mu}g_{\rho\nu}$)...

 

[stevendaryl]

This was the trick that solved my problem... Didn't think that would simplify my issue this much, so thank you for that!This is exactly what I was looking for, but I must admit that I find your notation rather confusing... It's the first time for me seeing the metric tensor with both, indices and up and down at the same time. Since I'm not sure how I would evaluate this, I would need to transform it back to something I'm more used (aka $g_{\nu}{}^\mu = g^{\rho\mu}g_{\rho\nu}$)...

With one index up and one index down, the metric tensor is the same as the Kronecker delta: $g_\nu^\mu = 0$ if $\nu \neq \mu$ and $g_\nu^\mu = 1$ if $\nu = \mu$.