# Raised index of partial derivative

I am having a really hard time wrapping my head around component notation for tensor fields. For example, I do not know exactly what the following expression means $$\partial_\mu\partial^\nu \phi, \tag{\#}$$ where $$\phi$$ is a scalar field. On the one hand $$\partial^\nu=g^{\lambda\nu}\partial_\lambda$$ where $$g_{\mu\nu}$$ is the Minkowski metric, and hence we could write explicitly $$\partial_\mu\partial^\nu \phi=\sum_{\mu,\nu,\lambda}g^{\lambda\nu}\partial_\mu \partial_{\lambda}\phi=\sum_{\mu,\nu}\partial_{\mu}\partial_{\nu}\phi=\partial_\mu\partial_\nu\phi. \tag{*}$$ On the other hand, we may think of $$\partial_\mu\partial^\nu=g(\partial_\mu,\partial^\nu)=\delta_\mu^\nu;$$ so that $$\partial_\mu\partial^\nu\phi=\phi?$$ Maybe? I am actually not sure of what this would mean. I am really confused. Any help is appreciated.

Edit: To give context of where this expression comes from: I was computing the Lagrangian $$\mathcal L=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu \phi)$$ considering an infinitesimal spacetime translation $$x^ \mu\to x^\mu-\alpha a^\mu$$. The scalar field thus transforms like $$\phi(x)\to \phi(x)+\alpha(\partial_\mu\phi(x))a^\mu.$$ Plugging thins into the Lagrangian yields the term I am referring to.

Edit 2: The change in placement of indices are actually my doubts. I try to elaborate.

I do not have any background in using indices to talk about tensors. I am used to interpret the expressions $$\partial_\mu$$ as the local vector field defined in some chart (local coordinates). I think about vector fields $$X$$ as abstract section of the Tangent bundle, which restricted to local coordinates can be expressed as $$X=X^\mu\partial_\mu$$. In the context of QFT, as far as I understand, the symbol $$\partial_\mu$$ denotes $$(\partial_t,\nabla)$$ in the local coordinates $$(t,x,y,z)$$. So that $$\partial_\mu\phi=(\partial_t \phi,\partial_x \phi,\partial_y\phi,\partial_z\phi)$$. This was supposed to be my justification on why I wrote the summation on $$\mu$$ and $$\nu$$ in $$(*)$$, but now I note that this only applies when $$\mu$$ or $$\nu$$ appear twice, indicating the scalar product; which leads me to the last remark. I think of $$g_{\mu \nu}$$ as the component of the matrix $$g=\begin{pmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1\\ \end{pmatrix}$$ which represents the pseudo-Riemmanian metric, which by definition acts on tangent vectors, i.e. linear combinations of the $$\partial_\mu$$ applied to a point. This is where my doubt comes, in which was the right way to interpret the notation; in particular what is the expresion $$(\#)$$ in explicit coordinates?

• @G. Smith I made the mistake of assuming this represents "shorthand" for the scalar term in the Lagrangian for a scalar field...it is not. What does it describe physically? Sep 27 '20 at 5:15
• A bunch of second derivatives with no physical significance that comes to mind. Perhaps the OP meant to ask about the kinetic term $\partial_\mu\phi\partial^\mu\phi$. Sep 27 '20 at 5:18
• @Drjh Well you didn’t answer what the OP asked. And it certainly isn’t shorthand for the scalar. But it might be typo/confusion. If the OP clarifies, maybe you can undelete your answer and explain what the kinetic term is. But you seemed confused about contracted vs. uncontracted indices. To make the scalar kinetic term, the indices are contracted. They aren’t two free indices like you wrote in your answer. Sep 27 '20 at 5:21
• General tip: Let's not have posts look like revision histories. Sep 27 '20 at 11:23
• Equation (#) stays as it is written, there is no summation, so either $\partial_\mu \partial^\nu$ or $g^{\nu \rho} \partial_\mu \partial_\rho$ Sep 27 '20 at 13:07

$$\renewcommand{\lag}{\mathcal{L}}\renewcommand{\pd}{\partial}\renewcommand{\d}{\mathrm{d}}\pd^\mu$$ is defined as $$\pd^\mu := g^{\mu\nu}\pd_{\nu}$$, where I use the convention that all repeated indices are summed and $$g^{\mu\nu}$$ are the components of the inverse metric tensor. Thus your Lagrangian can be rewritten as $$\lag=\tfrac12g^{\mu\nu}(\pd_\mu\phi)(\pd_\nu\phi)\tag{1}$$ and also your expression $$(\#)$$ is equal to $$g^{\mu\sigma}\pd_\nu\pd_\sigma\phi$$.
To see where all this comes from from a differential geometry point of view, this Lagrangian can be written in a coordinate free form as the top-form $$\lag = \tfrac12 \d\phi\wedge\star\d\phi,\tag{2}$$ where $$\d$$ is the exterior derivative and $$\star$$ is the Hodge-star. It is an easy exercise to restrict to a local coordinate system, $$\d x^\mu$$, in which case $$\d\phi$$ becomes $$\frac{\pd\phi}{\pd x^\mu}\d x^\mu\equiv\pd_\mu\phi\,\d x^\mu$$. The Hodge star will contribute a factor of $$g^{\mu\nu}$$ and so (2) will fall back to (1).
Moreover, you can think of $$a^\mu\pd_\mu\phi(x)$$ in a more formal setting as $$\iota_a \d\phi$$, where $$\iota_a$$ is the interior product along the vector field $$a$$ with components $$a^\mu$$. So the transformation $$\phi(x)\mapsto\phi(x)+\alpha a^\mu \pd_\mu\phi(x)$$ is written as $$\phi(x)\mapsto \phi(x) + \alpha\,(\iota_a\d\phi)(x).$$
The relevant term in your expression ($$\#$$) comes from a term $$\alpha \d\phi\wedge\star\d\iota_a \d\phi$$ in the Lagrangian, basically it is just the $$\alpha \star\d\iota_a \d\phi$$ part. If we expand this in local coordinates $$\{\d x^\sigma\}$$, we get: $$\alpha \star\d\iota_a \d\phi = \alpha a^\mu \pd_\sigma\pd_\mu \phi\;\star\d x^\sigma = \alpha a^\mu \pd_\sigma\pd_\mu \phi\ g^{\nu\sigma} \varepsilon_{\nu\lambda\kappa\rho}\d x^\lambda\wedge\d x^\kappa\wedge\d x^\rho,$$ where in the second equality I used the definition of the Hodge star acting on the basis differentials. Stripping off numbers, $$\varepsilon$$-symbols and the differentials, all we're left with is $$g^{\nu\sigma}\pd_\sigma\pd_\mu\phi,\tag{\#'}$$ which is exactly what you would have found (with your much shorter route) as $$\pd^\nu\pd_\mu\phi \tag{#}.$$ Thus, $$(\#')=(\#)$$.
Of course the typical way to arrive there is to simply use the fact that for any object $$\bullet_\mu$$ with a downstairs leg we can lift it using the inverse metric, i.e. $$\bullet^\mu := g^{\mu\nu}\bullet_\nu$$. But since you had trouble understanding where does this stem from from a differential geometry perspective, I wanted to stick with the differential geometry picture all the way through, from the Lagrangian to the final result. Hope this helped and didn't confuse you more.