# Suppose $\phi(x)$ is a field, how should I interpret $\partial^\mu\phi$ and $\partial_\mu\phi$?

I am really confused by the sub and upperscript notation sometimes. It might be really trivial but I have a difficult time interpreting the following things in for example this Lagrangian

$$\mathscr{L}=\frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{m^2}{2}\phi^2$$

I know how the repeating of the index implies a dot product between two four vectors, but what if there is no repeating of index and the $\partial_\mu\phi$ is not multiplied with anything, what is $\partial_\mu\phi$? Is it a four vector? Or is it a scalar? And in the case of $\phi$ being a vector field, i.e $\phi=(\phi_1,...,\phi_n)$ how should I then interpret $\partial_\mu\phi$, is it $(\partial_\mu\phi_1,...,\partial_\mu\phi_n)$ and is then each component of the vector field now a four vector or a scalar? It is all really confusing for me...

For example in the following lagrangian

$$\mathscr{L}=\frac{1}{2}\sum^N_{a=1}\partial^\mu\phi_a\partial_\mu\phi_a-\frac{m^2}{2}\sum^N_{a=1}\phi_a^2$$

In this case $\phi_a$ is a scalar field, but we can define $\phi=(\phi_1,...,\phi_N)$ as a vector field, can I then rewrite the above lagrangian as (T means transpose):

$$\mathscr{L}=\frac{1}{2}(\partial_\mu\phi)^T(\partial^\mu\phi)-\frac{m^2}{2}\phi^T\phi$$

So in a nutshell, what does $\partial_\mu\phi$ on it's own mean?

• Why would you think $\partial_\mu \phi$ is a scalar? It has an explicit space-time index... – JamalS Mar 4 '15 at 15:35

The derivative $\mathrm{d}\phi$ is a proper (co)vector:

$$\left(\begin{matrix}\frac{\partial\phi}{\partial x^1} \\ \frac{\partial\phi}{\partial x^2} \\ \frac{\partial\phi}{\partial x^3} \\ \frac{\partial\phi}{\partial x^4}\end{matrix}\right)$$

with components $\partial_\mu \phi$ and $\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$.

This works also in other than four dimensions, naturally.

• Perhaps I'm reading too much into this, but I think OP also wants a physical interpretation of $\partial_\mu\phi$. – Ryan Unger Mar 4 '15 at 17:17
• @0celo7: OP is out of luck then, I ain't got one :) – ACuriousMind Mar 4 '15 at 17:25

Rule of thumb is that things with indices are tensors, the order being given by the number of free indices you have. By free index I mean any index for an object which is not repeated, hence not involved in a sum (or contraction). For example $$g_{\mu\nu}$$ indicates a covariant tensor of order (or rank) 2. If this represents a metric, its inverse is the contravariant tensor of order 2 $g^{\mu\nu}$. Indices can get mixed up, like in $T_\mu^\nu$, which is a tensor of rank (1,1) meaning that it is contravariant of order 1 in the superscript index and covariant of order 1 in the subscript index. Another example is the Riemann curvature tensor $${R^\mu}_{\nu\rho\sigma},$$ which is a $(3,1)$-tensor.

Remark Observe that the index notation might be misleading, as there are objects with space-time indices which are not tensors. Typical examples are the coordinates themselves $x^\mu$, which do not transform tensorially under generic coordinate changes, and the Christoffel symbols ${\Gamma^\mu}_{\nu\rho}$.

Concerning the OP, $\partial^\mu$, despite being a differential operator rather than a tensor, can be interpreted as one because if its transformation law (in flat-spacetime with vanishing connection). When applied to a generic tensor it will increase its contravariant rank by 1. $\partial_\mu$ does the same with the covariant rank, so that if you take a scalar (i.e. a (0,0) tensor) $\partial_\mu\phi$ is a covariant vector, or a (1,0)-tensor. Similarly $\partial^\mu\phi$ is a contravariant vector, i.e. a (0,1)-tensor. The expression $\partial_\mu\phi\partial^\mu\phi$ has no free index, like $\phi$ itself, therefore it is a scalar, i.e. a (0,0)-tensor.

Remark For geometrical reasons, the right thing to consider is $\partial_\mu$, which are the formal component of the exterior derivative $\text d$. You get $\partial^\mu$ by contracting with the metric tensor (cf. ACuriousMind's answer), where again you have only one free index, in accordance to what has been state above.

When you have something like $\partial_\mu\phi_a$, you have mixed indices which refer to different spaces. Greek letters are usually used to denote tensor fields in the (co)tangent bundle to Minkowksi space-time, while other indices usually refer to internal degrees of freedom and leave in other fibre bundles over space-time (e.g. a spinor bundle for spinors). Their behaviour follow pretty much the same rules as the other indices, except that they indicate that the object lives in many different fibres over space-time.

I'm going to go a bit overboard here and give you the sketch of how vectors are geometrically constructed, since I think it's helpful to know. While writing this I found I was phrasing things very carefully, which means: you may need to reread parts of this in a quiet corner if it doesn't all make sense at first.

Suppose that you have a set of scalar fields $\mathbf{S} \subseteq (\mathcal{M} \rightarrow \mathbb R)$ on an underlying set of "points" $\mathcal{M}$. This set S contains every scalar field that you are interested in. It's helpful to define some jargon: let an $n$-functor be a smooth functions in $\text C^\infty(\mathbb R^n \rightarrow \mathbb R)$. We can of course apply them to real values (which I'll denote with parentheses as $\Phi(\dots)$, but also we can use them to combine scalar fields, which I'll denote with square brackets as $\Phi[a, b, \dots]$ which is the function from any $p \in \mathcal M$ to $\Phi(a(p), b(p), \dots)$. Henceforth I will just call scalar fields scalars, since we're talking about field theory.

When we close S under $n$-functors for all $n$, we get a lot of stuff: first off we can add and multiply scalars since + and * are 2-functors; but we can even define a topology on $\mathcal M$ with those scalars which makes all of them continuous (the kernel topology; $s \subseteq \mathcal M$ is closed when $s = \operatorname{ker} \sigma$ for some $\sigma \in \mathbf S$), and we can use this topology to define whether $\mathcal M$ is connected or not, etc. If you go down this rabbit hole far enough, you get to specify criteria by which $\mathcal M$ is a "manifold". But let's get more concrete and specific.

Now in special relativity in particular, there are special fields $w, x, y, z$ which we use as global coordinates, meaning that every scalar $\phi$ in $\mathbf{S}$ can be written as a 4-functor $\Phi(x_1, x_2, x_3, x_4)$ applied to those scalar fields; $\phi = \Phi[w, x, y, z].$ (In general relativity, you still have coordinates, but you will have to use the topology to define "neighborhoods" of a point and then say that for every point there is a neighborhood enclosing that point with a set of $d$ coordinates which distinguish points in that neighborhood.)

We can take your intuitive 3D notion of "vector fields" and associate them with derivations $D_v = v^\mu \partial_\mu$, so that we can define vectors (again dropping the word "field") in a purely geometric sense as the set of linear maps $V \in (\mathbf S \rightarrow \mathbf S)$ satisfying the chain rule: given a $k$-functor $\Phi(x_1\dots x_k)$ with partial derivatives $\Phi_{(i)} = \partial \Phi/\partial x_i$, then $$V [\Phi[s_1\dots s_k]] = \sum_{i=1}^k ~ V[s_i] ~ \Phi_{(i)}[s_1\dots s_k].$$ We can then identify the components of any vector field by applying it to our coordinates, so you see there is a two-way correspondence between these abstract mathematical operations on S and the lists-of-coordinates that you're used to.

We can write this space of derivations on $\mathbf S$ as $\mathbf S^\bullet$, and make independent copies of it for every symbol that we want to stick up top: this is known as "abstract index notation". We write elements of this set as symbols with a corresponding superscript; so $v^\alpha \in \mathbf S^\alpha$.

Then there are covectors, like the general gradient of a field, which are linear maps from vectors to scalars. For example, given a scalar $\phi$ there is a covector $\nabla\phi$ defined as $$\nabla\phi~(V) = V[\phi].$$ These we can denote by $\mathbf S_\bullet$ and we copy the space for every symbol. Now you know what $\partial_\mu \phi$ means; it is the copy of $\nabla \phi$ that lives in the space $\mathbf S_\mu$. It is a covector field. (Somewhat important: you don't quite know what $\partial_\mu v^\nu$ means yet. More complicated derivatives of this kind -- derivatives of vectors rather than scalars -- require some added geometric structure called a "connection" on the manifold.)

Finally we have the metric, which is a linear map from a pair of vectors to a scalar; this can be used to convert covectors to vectors.

• Overkill, but beautiful overkill. One might mention that this is known in differential geometry as the statement that the vector fields on a manifold are derivations, if I see it correctly. – ACuriousMind Mar 4 '15 at 18:06