Massless Scalar Field: Taking $ \partial / ( \partial( \partial_\mu \phi ) ) $

Question

The Lagrangian density for a massless scalar field is $ \mathcal{L} = \frac{1}{2} ( \partial_\mu \phi ) ( \partial^\mu \phi ) $. In order to derive the equations of motion for such a Lagrangian, we plug it into the Euler-Lagrange equation. In particular we need to compute the quantity $ \partial_\mu \frac{ \partial \mathcal{L} }{ \partial( \partial_\mu \phi ) } $. I have two questions:

1. Is the following derivation correct (ignoring the $1/2$) :

$$ \begin{align*} % \partial_\mu \frac{ \partial}{ \partial( \partial_\mu \phi ) } [ ( \partial_\nu \phi ) ( \partial^\nu \phi ) ] % &= \partial_\mu \frac{ \partial}{ \partial( \partial_\mu \phi ) } [ ( \partial_\nu \phi ) g^{\nu\alpha} ( \partial_\alpha \phi ) ] % \\ &= \partial_\mu g^{ \nu\alpha } [ ( \partial_\alpha \phi )\delta_\nu^\mu + ( \partial_\nu \phi ) \delta_\alpha^\mu ] \quad \text{(Using product rule)} % \\ &= \partial_\mu [ (\partial^\nu \phi ) \delta_\nu^\mu + ( \partial^\alpha \phi ) \delta_\alpha^\mu ] \quad \text{(Evaluating metric on both terms)} % \\ &= 2 \partial_\mu \partial^\mu \phi \quad \text{(Evaluating delta sum)} % \end{align*} $$

2. If it was correct, was such a derivation necessary, or is there a shorter way? Using the notation $ \mathcal{L} = ( \partial_\mu \phi )^2 $ makes it look very tempting to just apply the "power rule" to get $2 \partial_\mu \phi $ when we take $ \partial / \partial ( \partial_\mu \phi) $, but it just doesn't feel right... is that legitimate math?

Answer 1

I'll present a very explicit approach, which makes it obvious it's the correct answer. Notice we can express the kinetic term as,

$$T \equiv\partial_\mu \phi \partial^\mu \phi = (\partial_t\phi)^2 - \sum_{i=1}^{d-1}(\partial_i \phi)^2$$

since the metric $\eta = \mathrm{diag}(1,-1,\dots,-1)$. We see that,

$$\frac{\partial T}{\partial (\partial_t \phi)} = 2\partial_t\phi, \quad \frac{\partial T}{\partial(\partial_i \phi)} = -2\partial_i \phi$$

and noting that $\partial^\mu = (\partial_t,-\nabla)$ it immediately follows that,

$$\frac{\partial}{\partial(\partial_\mu \phi)} \partial_\nu \phi \partial^\nu \phi = 2 \partial^\mu \phi.$$

This should elucidate why we can quickly jot down the answer using an analogous 'power rule' though one should be careful when dealing with new terms, and double check explicitly.

Answer 2

One way could be to consider that this is the Lagrangian density of a massless real scalar field and therefore the equation will be Dalambertian phi equal to zero; Comparing terms with the Euler Lagrange equation you get the same without having to derive