Proving ##\partial^{i} = g^{ik} \partial_{k}##

[SplinterCell]

Let $\varphi$ be some scalar field. In "The Classical Theory of Fields" by Landau it is claimed that
$$
\frac{\partial\varphi}{\partial x_i} = g^{ik} \frac{\partial \varphi}{\partial x^k}
$$
I wanted to prove this identity. Using the chain rule
$$
\frac{\partial}{\partial x_{i}}=\frac{\partial x^{k}}{\partial x_{i}}\frac{\partial}{\partial x^{k}}
$$
Differentiating $x^{k}=g^{ik}x_{i}$ with respect to $x_i$ yields
$$
\frac{\partial x^{k}}{\partial x_{i}}=g^{ik} \tag{*}
$$
which seems to prove the identity (?). However I'm not sure about the validity of the last step, especially because $(*)$ isn't mentioned in the relevant literature (either because it's too trivial or/and not useful or because it's wrong). I've only seen the metric tensor expressed as
$$
g_{ik}=\frac{\partial x^{l}}{\partial x^{\prime i}}\frac{\partial x^{l}}{\partial x^{\prime k}}
$$
where the primed coordinates refer to a different coordinate system we're transforming to.

 

[anuttarasammyak]

I intepret it by applying formula
$$A^i=g^{ij}A_j$$
for $A^i=dx^i$
$$dx^i=g^{ij}dx_j$$
$$\frac{\partial x^i}{\partial x_k}=g^{ik}$$
where partial derivatives are taken in condition that all the ohter components $x_j$ than $x_k$ do not change.

 

[PeroK]

Let $\varphi$ be some scalar field. In "The Classical Theory of Fields" by Landau it is claimed that
$$
\frac{\partial\varphi}{\partial x_i} = g^{ik} \frac{\partial \varphi}{\partial x^k}
$$
I wanted to prove this identity. Using the chain rule
$$
\frac{\partial}{\partial x_{i}}=\frac{\partial x^{k}}{\partial x_{i}}\frac{\partial}{\partial x^{k}}
$$
Differentiating $x^{k}=g^{ik}x_{i}$ with respect to $x_i$ yields
$$
\frac{\partial x^{k}}{\partial x_{i}}=g^{ik} \tag{*}
$$
which seems to prove the identity (?). However I'm not sure about the validity of the last step, especially because $(*)$ isn't mentioned in the relevant literature (either because it's too trivial or/and not useful or because it's wrong). I've only seen the metric tensor expressed as
$$
g_{ik}=\frac{\partial x^{l}}{\partial x^{\prime i}}\frac{\partial x^{l}}{\partial x^{\prime k}}
$$
where the primed coordinates refer to a different coordinate system we're transforming to.

There are a couple of conceptual issues here. First, the $x^i$ are the coordinates. They are not the components of a vector. Something like $x^{k}=g^{ik}x_{i}$ doesn't make sense.

We can, however, identify the "differentials" $dx^i$ as a set of basis one-forms; and $dx_i$ as a set of basis vectors. This is quite tricky to do formally, I recall.

The second issue, therefore, is what is meant by $\frac{\partial}{\partial x_i}$? As we have already noted that $x_i$ are not coordinates.

To resolve this we first note that the gradient of a scalar is a covector: $$\frac{\partial \varphi}{\partial x^i} = A_i$$This can be proved by considering the gradient under a coordinate transformation. There must, therefore, be a corresponding vector with components: $$A^i = g^{ij}A_j = g^{ij}\frac{\partial \varphi}{\partial x^j}$$We can identify $A^i$ with $\frac{\partial \varphi}{\partial x_i}$ in some sense. And, again, formalising this rigorously in terms of vectors and one-forms is quite tricky.

In short, you can't prove it simply using the chain rule because the $x_i$ need to be defined first - and that takes you into more advanced differential geometry. The simplest approach, I suggest, is to take the following as a definition of the differential operator $\frac{\partial}{\partial x_i}$: $$\frac{\partial}{\partial x_i} \equiv \partial^i \equiv g^{ij}\partial_j \equiv g^{ij}\frac{\partial}{\partial x^j}$$

 

[SplinterCell]

There are a couple of conceptual issues here. First, the $x^i$ are the coordinates. They are not the components of a vector. Something like $x^{k}=g^{ik}x_{i}$ doesn't make sense.

In "Classical Electrodynamics" by Jackson he writes (p. 542):

The covariant coordinate 4-vector $x_{\alpha}$ can be obtained from the contravariant $x^{\beta}$ by contraction with $g_{\alpha\beta}$, that is,
$$
x_{\alpha} = g_{\alpha\beta} x^{\beta}
$$
and its inverse,
$$
x^{\alpha} = g^{\alpha\beta}x_{\beta}
$$

In other words, $x^i$ can be treated as components of a (contravariant) coordinate vector, can't they?

Landau & Lifshitz also define $x^i$ and $x_i$ this way. In particular, they write
$$
x^{i}=\left(ct,\mathbf{r}\right),\quad x_{i}=\left(ct,-\mathbf{r}\right),\quad x^{i}x_{i}=c^{2}t^{2}-\mathbf{r}^{2}
$$
(where they use the signature $(+,-,-,-)$ which is common in special relativity).

 

[anuttarasammyak]

I do not have Jackson but assume that he says about SR where metric tensor components are constants and not functions of coordinates as they are in GR.
$$dx_i = g_{ik} dx^k$$
is integrated in SR as
$$\int dx_i = \int g_{ik} dx^k=g_{ik} \int dx^k$$
to be
$$x_i = g_{ik} x^k$$

 

[PeroK]

In "Classical Electrodynamics" by Jackson he writes (p. 542):

In other words, $x^i$ can be treated as components of a (contravariant) coordinate vector, can't they?

You have to be careful. In flat spacetime you have a position vector, whose components are often denoted by $x^{\alpha}$. But, in curved spacetime, vectors are local objects, defined (in the tangent space) at each point in spacetime. There's no generalisation to the coordinates themselves being the components of a (position) vector.

 

[PeroK]

Landau & Lifshitz also define $x^i$ and $x_i$ this way. In particular, they write
$$
x^{i}=\left(ct,\mathbf{r}\right),\quad x_{i}=\left(ct,-\mathbf{r}\right),\quad x^{i}x_{i}=c^{2}t^{2}-\mathbf{r}^{2}
$$
(where they use the signature $(+,-,-,-)$ which is common in special relativity).

In your OP, are you assuming $g^{ij} = \eta^{ij}$?

 

[SplinterCell]

In your OP, are you assuming $g^{ij} = \eta^{ij}$?

Indeed, the context here is special relativity, i.e. flat spacetime with Minkowski metric.

You have to be careful. In flat spacetime you have a position vector, whose components are often denoted by $x^{\alpha}$. But, in curved spacetime, vectors are local objects, defined (in the tangent space) at each point in spacetime. There's no generalisation to the coordinates themselves being the components of a (position) vector.

Ok, didn't know that (haven't studied general relativity yet). So does this mean that $x^{k}=g^{ik}x_{i}$ can make sense, but only if the spacetime is flat (like @anuttarasammyak showed)?

 

[PeroK]

Indeed, the context here is special relativity, i.e. flat spacetime with Minkowski metric.

Ok, didn't know that (haven't studied general relativity yet). So does this mean that $x^{k}=g^{ik}x_{i}$ can make sense, but only if the spacetime is flat (like @anuttarasammyak showed)?

A position vector only makes sense in flat spacetime. That equation is only valid for the Minkowski metric - or, perhaps more generally, when $g^{ij}$ is independent of the coordinates.

You could try out that identity in the case of spherical coordinates (in flat spacetime) and see what you get.