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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 $1/2$) :

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

  1. 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?
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  • $\begingroup$ Indeed point 2 makes sense since you contract two objects which are the same (up to the placement of index). Otherwise if you would have say $\mathcal{L} = A_\mu B^\mu$ and you want $\frac{\partial \mathcal{L}}{\partial A_\mu} = B^\mu$ for some $A,B$ then you need to be careful, but now $A=B$ so it can be done quickly using the product rule $\endgroup$ Commented Apr 9, 2022 at 14:16

2 Answers 2

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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.

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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

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  • $\begingroup$ This does not answer the question, which was about the validity of applying a "product rule" to a non-scalar object. (also, this question was asked and answered well 4 years ago - why bring it up again?) $\endgroup$ Commented Aug 19, 2021 at 5:54

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