Difference and meaning of index the derivative operator

Question

I'm a beginner in this type of math, we are just starting to study it, but I need some clarifications about the meaning and the difference of when we write

$$\partial_i \qquad \text{and}\qquad \partial^i.$$

Also, is $$\dfrac{\partial}{\partial x^i} = \partial_i \quad \text{or}\quad \partial^i?$$

and why? What is the difference about the position of the indexes? I have troubles in understanding the difference of high and low index. We are working in $\mathbb{R}^3$ only, no higher dimensions.

Answer 1

You can eventually (if you need to) learn a more rigorous treatment later, so let me instead provide a cookbook approach:

1. An object with an open index means that its value changes when the observer changes orientation: $x^i$ is a vector, and I might measure it as $(0,0,1)$ while you are measuring it as $(0,0,-1)$ if I am upside down with respect to you. It is similar for other objects, e.g. $T^{ij}$, $A_i$, $B_i^{\;jk}$, and so on.
2. Only an object without an open index can be orientation-independent: my mass is same independent of whether I face east or north, so it should be denoted with an object without an open index, i.e. $m$.
3. You can only reduce the number of open indices by contraction (i.e. summing over them), and you can only contract indices if one is up and other is down. For instance, length of a road should be orientation-independent, so if we view road as a vector $l^i$, its length $L$ is given by $L=\sqrt{l_i l^i}$. Here, we are using Einstein's summation convention: if two indices are same (and one up one down), they are summed over.
4. The open indices (and their positions) should match in an equation: $T^{ij}=x^i y^j$ is fine, but $B_{ij}=x^iy_j$ is NOT. As long as they match, their position is irrelevant: $A_{ij}=x_i y_j$ and $A^{ij}=x^iy^j$ are same equations.

With these rules, start with the differential operator: $x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}$. You can explicitly compute by going to the spherical coordinates that this operator is equal to $r\frac{\partial}{\partial r}$; in other-words, it is angle-independent. As it is orientation independent, it should be a scalar, thus we write it as $x^i\partial_i$. We could have also written it as $x^i\left(\frac{\partial}{\partial x}\right)_i$, or as $x^i\frac{\partial}{\partial x^i}$.

In short, the equality $\partial_i=\frac{\partial}{\partial x^i}$ is rather a convention than an actual equality. Although it is clear that $x^i$ refers to an array of values $\{x^1,x^2,x^3\}$, it is not a priori clear what $\frac{\partial}{\partial x^i}$ refers to; in fact, one could naively think it refers to a derivation with respect to a vector, i.e. $\frac{\partial}{\partial \{x^1,x^2,x^3\}}$. By $\frac{\partial}{\partial x^i}$, we instead mean the array $\{\frac{\partial}{\partial x^1},\frac{\partial}{\partial x^2},\frac{\partial}{\partial x^3}\}$; and, in addition to this convention, if we further define the convention that $\partial_1=\frac{\partial}{\partial ^1}$ and so on, $\partial_i$ behaves like a proper one-index object: $a^i\partial_i$ is a scalar, $a_i\partial^i$ and $a^i\partial_i$ are same, and so on. With these two conventions, we end up with consistent equalities: $\partial_i=\frac{\partial}{\partial x^i}$, $\partial^i=\frac{\partial}{\partial x_i}$.

Answer 2

Here is a brief summary:

1. In (relativistic) physics, it is standard to adorn a local coordinate $x^{\mu}$ with a superindex.

2. We define a shorthand notation for the partial derivative $\partial_{\mu}:=\frac{\partial}{\partial x^{\mu}}$.

3. Indices are lowered (raised) with the metric tensor (inverse metric tensor), respectively. E.g. $\partial^{\mu}:=g^{\mu\nu}\partial_{\nu}$.

4. Einstein notation: There's an implicitly written summation over repeated indices.

As to why, check out Is it foolish to distinguish between covariant and contravariant vectors? and links therein.