Trying to understand how tensors transform under a coordinate system change

Question

I'm trying to understand how tensor components transform under a change of coordinate system, but something doesn't look right. Below is the derivation that I wrote out:

---

We start with a general example of the tensor transformation equation for a tensor $\mathbf{T}$ of rank greater than 1, in this case 3:

$$\begin{align} T^{\mu^\prime}{}_{\nu^\prime \lambda^\prime} = \frac{\partial x^{\mu^\prime}}{\partial x^{\mu}}\frac{\partial x^{\nu}}{\partial x^{\nu^\prime}}\frac{\partial x^{\lambda}}{\partial x^{\lambda^\prime}} T^{\mu}{}_{\nu\lambda} \end{align}$$

To test out this equation, we'll simplify it to the case of a rank-1 tensor, which can be thought of as a vector, and work in 2 dimensions, where an index value of $0$ represents the first coordinate and an index value of $1$ represents the second coordinate: $$T^{\mu^\prime} = \frac{\partial x^{\mu^\prime}}{\partial x^{\mu}}T^{\mu} = \frac{\partial x^{\mu^\prime}}{\partial x^{0}}T^{0} + \frac{\partial x^{\mu^\prime}}{\partial x^{1}}T^{1}$$

Let's use the familiar example of 2-dimensional polar coordinates, mapped from the Cartesian $x$ and $y$ as follows:

$$\begin{align} x^{0^\prime} = r = \sqrt{x^2 + y^2} \\\\[4pt] x^{1^\prime} = \theta = \arctan\left(\frac{y}{x}\right) \end{align}$$

Here's what I expect the results to be based on standard linear algebra: $$\begin{bmatrix}T^r \\\ T^\theta\end{bmatrix} = \begin{bmatrix}\cos\theta & \sin\theta \\\ -\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}T^x \\\ T^y\end{bmatrix}$$ Let's try to verify this in tensor form:

Using the above expression for $T^{\mu^\prime}$, we can write out the components $T^{0^\prime}$ and $T^{1^\prime}$ explicitly by setting $\mu^\prime = 0$ and then $\mu^\prime = 1$. These are the expected components of our rank-1 tensor (vector) when transforming to the new coordinate system:

$$\begin{align} T^{0^\prime} = \frac{\partial x^{0^\prime}}{\partial x^{\mu}}T^{\mu} = \frac{\partial x^{0^\prime}}{\partial x^{0}}T^{0} + \frac{\partial x^{0^\prime}}{\partial x^{1}}T^{1} \\\\[6pt] T^{1^\prime} = \frac{\partial x^{1^\prime}}{\partial x^{\mu}}T^{\mu} = \frac{\partial x^{1^\prime}}{\partial x^{0}}T^{0} + \frac{\partial x^{1^\prime}}{\partial x^{1}}T^{1} \end{align}$$

It looks like we have some partial differentiation to do, so let's take all 4 of these partial derivatives $\partial x^{\mu^\prime} / \partial x^\mu$:

$$\begin{align} \frac{\partial x^{0^\prime}}{\partial x^0} & = \frac{\partial r}{\partial x} = \frac{\partial}{\partial x}\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}} \\\\[6pt] \frac{\partial x^{0^\prime}}{\partial x^1} & = \frac{\partial r}{\partial y} = \frac{\partial}{\partial y}\sqrt{x^2 + y^2} = \frac{y}{\sqrt{x^2 + y^2}} \\\\[6pt] \frac{\partial x^{1^\prime}}{\partial x^0} & = \frac{\partial \theta}{\partial x} = \frac{\partial}{\partial x}\arctan\left(\frac{y}{x}\right) = -\frac{y}{x^2 + y^2} \\\\[6pt] \frac{\partial x^{1^\prime}}{\partial x^1} & = \frac{\partial \theta}{\partial y} = \frac{\partial}{\partial y}\arctan\left(\frac{y}{x}\right) = \frac{x}{x^2 + y^2} \end{align}$$

Now all that's left is to plug in our partial derivatives $\partial x^{\mu^\prime} / \partial x^\mu$ into our above expression for $T^{0^\prime}$ and $T^{1^\prime}$ to obtain the components of $\mathbf{T}$ when transforming to polar coordinates: $$\begin{align} T^{0^\prime} = T^r & = \frac{x}{\sqrt{x^2 + y^2}}T^0 + \frac{y}{\sqrt{x^2 + y^2}}T^1 \\\\[6pt] & = \frac{x}{r}T^0 + \frac{y}{r}T^1 \\\\[6pt] & = T^0\cos\theta + T^1\sin\theta \end{align}$$

This looks great. Likewise, for the angular component, $$\begin{align} T^{1^\prime} = T^\theta & = -\frac{y}{x^2 + y^2}T^0 + \frac{x}{x^2 + y^2}T^1 \\\\[6pt] & = \frac{1}{\sqrt{x^2 + y^2}}\left[-\frac{y}{\sqrt{x^2 + y^2}}T^0 + \frac{x}{\sqrt{x^2 + y^2}}T^1\right] \\\\[6pt] & = \frac{1}{r}\left[-\frac{y}{r}T^0 + \frac{x}{r}T^1\right] \\\\[6pt] & = \frac{1}{r}\left[T^1\cos\theta - T^0\sin\theta \right] \\\\[6pt] \end{align}$$ Now, the bracketed term is what I expect, but I'm obtaining a $1/r$ term in front, which seems strange to me. I'm not sure whether my mistake is simple algebra or if I'm fundamentally misunderstanding something here. Any ideas as to where I went wrong?

Thanks!

Answer 1

To every coordinate system $(x^1,\dots,x^n)$, there is a naturally associated (I will call it canonical) basis $\partial_{x^1},\dots,\partial_{x^n}$. Tensor equations such as $$T^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^\mu} T^\mu$$ assume you are using these canonical bases everywhere, with respect to the both the old and the new coordinates: $$ T = T^\mu \partial_{x^\mu} = T^{\mu'}\partial_{x^{\mu'}}. $$ Your calculations are correct, so the matrix you obtained $$ \begin{bmatrix} \frac{\partial x^{0^\prime}}{\partial x^0} & \frac{\partial x^{0^\prime}}{\partial x^1} \\ \frac{\partial x^{1^\prime}}{\partial x^0} & \frac{\partial x^{1^\prime}}{\partial x^1} \end{bmatrix} = \begin{bmatrix} \frac{x}{\sqrt{x^2 + y^2}} & \frac{y}{\sqrt{x^2 + y^2}} \\ -\frac{y}{x^2 + y^2} & \frac{x}{x^2 + y^2} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\frac{1}{r}\sin\theta & \frac{1}{r}\cos\theta \end{bmatrix} $$ correctly transforms components with respect to the canonical basis of cartesian coordinates: $\partial_x,\partial_y$ into components with respect to the canonical basis of polar coordinates: $\partial_r,\partial_\theta$: $$\begin{align} T &= T^x \partial_x + T^y\partial_y \\ &= T^{r}\partial_r + T^{\theta}\partial_\theta \end{align}$$ where $$ \begin{bmatrix} T^{r} \\ T^{\theta} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\frac{1}{r}\sin\theta & \frac{1}{r}\cos\theta \end{bmatrix} \begin{bmatrix} T^x \\ T^y \end{bmatrix} $$ are the components of $T$ with respect to the basis $\partial_x,\partial_y$ and the basis $\partial_r,\partial_\theta$.

The mistake is that the equation

$$ \begin{bmatrix} T^{r} \\ T^{\theta} \end{bmatrix} \overset{?}{=} \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} T^{x} \\ T^{y} \end{bmatrix} \tag 1 $$ assumes a different basis. This last equation $(1)$ supposes you are using the canonical basis $\partial_x,\partial_y$ in Cartesian coordinates, but you are using the orthonormal basis $e_r,e_\theta$ in polar coordinates. Actually, let me denote the components of $T$ with respect to the orthonormal basis $e_r,e_\theta$ as $\hat T^r,\hat T^\theta$. Now the correct version of $(1)$ is

$$ \begin{bmatrix} \hat T^{r} \\ \hat T^{\theta} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} T^{x} \\ T^{y} \end{bmatrix} \tag 2 $$

The relation between the bases $\partial_r, \partial_\theta$ and $e_r,e_\theta$ is \begin{align} \partial_r &= e_r \tag 3\\ \partial_\theta &= r e_\theta. \tag 4 \end{align}

That is: the only difference between both bases is that the canonical angular basis vector $\partial_r$ has length $r$ but the orthonormal angular basis vector $e_\theta$ has length $1$.

To get the expression you want (the one in orthonormal coordinates) all you have to do is to take the (correct) expression you got (the one in canonical coordinates): \begin{align} T &= T^r\partial_r + T^\theta \partial_\theta \\ &= [T^x\cos\theta + T^y\sin\theta]\partial_r +\frac{1}{r} [-T^x\sin\theta + T^y\cos\theta]\partial_\theta. \end{align} and substitute with equations $(3)$ and $(4)$; that is $\partial_r=e_r$ and $\partial_\theta = r e_\theta$.

Personally I wouldn't reccomend it. If you wanted to further change coordinates from polar to, say, some slanted affine coordinate system $x^{\mu''}$, you would need to undo this in order to apply the tensor formula $$T^{\mu''} = \frac{\partial x^{\mu''}}{\partial x^{\mu'}} T^{\mu'}.$$