Can I see an example of transforming a tensor from polar to cartesian coordinates?

Question

I have been learning about tensors, and I understand about creating the Jacobian matrix to obtain the coordinate transformation for infinitesimals, so that we have $$d\bar{x}^j=\frac{\partial \bar{x}^j }{\partial x^h}dx^h$$

So the transformation from polar coordinates to Cartesian is: $$ dx = dr \cos \theta - \sin \theta d\theta $$ $$ dy=dr\sin\theta +r\sin\theta d\theta $$

I can understand that (I think!), but my problem is that several of the books I have read say things like "a vector transforms like coordinate differentials", and include statements such as $$\bar{V}^j=\frac{\partial \bar{x}^j }{\partial x^h}V^h$$

and then they usually show examples of transforming a vector from one Cartesian to a different Cartesian, where it's easy because the basis vectors are constant throughout space (eg the standard Lorentz transformation). But I have never seen any examples showing the transformation for Polar to Cartesian. What's more, it doesn't make sense to me that vectors would transform like this for Polars unless they were infinitesimals anyway, because the transformation isn't linear.

Could someone please provide an example of transforming a vector from Polar coordinates to Cartesian coordinates using the above two equations? Or tell me where my thought processes have let me down?

Edit: I'm sorry, but I haven't been clear. I'm currently reading Schutz General Relativity 3rd edition, and he quotes the polar-to-cartesian transformation I use above (admittedly he uses the inverse), and he explicitly says the equations are "valid to first order". He then goes on to use the tensor expression I have quoted above and says, "a vector can be defined as an object whose components transform according to [this expression]". So my naive reading of that was that the transformation only applies to first order. But forget about that. I feel that if my request (below) is answered then I will understand.

I understand how the expressions that @KurtG provides are derived, and I have moved on to the curvature tensor, the Einstein field equations, etc, but someone asked me about transformations and I realised I couldn't explain this part. I need to see an actual example of a transformation using actual numbers. Take a vector in polar coordinates, tell me its components, show it in a diagram, and then transform it to Cartesian using the expressions above, so that I can see the vector is the same but the components have changed.

Answer 1

Is seems like you are familiar with the transformations of basis one forms $dr,d\theta$ to $dx,dy\,.$ The similar formulas for the basis vector fields $\partial_r,\partial_\theta,\partial_x,\partial_y$ you can find here. I advertise this because the derivation method (chain rule) is hard to forget. Now to tensors: we know that a tensor is a linear combination, say, $$ \mathbf{T}={T^r}_\theta\;\partial_r\otimes d\theta\,. $$ To transform it to Cartesian plug in the transformations of $\partial_r$ and $d\theta$ and collect the $dx$ and $dy$ terms. Hard to make a mistake. Same method works for all other tensors. Exercise: transform the metric tensor from $$ g=dx\otimes dx+dy\otimes dy $$ to $$ g=dr\otimes dr+r^2\,d\theta\otimes d\theta\,. $$ In short:

• When $\bar{z}$ denotes the polar coordinates $r,\theta$ and $z$ denotes the Cartesian coordinates $x,y$ the Jacobian of $\bar{z}\mapsto z\,,$ that is, $$ (r,\theta)\mapsto (r\cos\theta,r\sin\theta) $$ and its inverse are -as we know- \begin{align} \frac{\partial z_h}{\partial\bar{z}_j}&= \begin{pmatrix} \cos\theta & -r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix}\,, & \frac{\partial\bar{z}_j}{\partial z_h}= \begin{pmatrix} \cos\theta & \sin\theta\\ -\frac1r\sin\theta&\frac1r\cos\theta\end{pmatrix}\,. \end{align} In the following I use the transposed Jacobian and the transposed inverse which is more compatible with the fact that $\partial_x,\partial_y,\partial_r,\partial_\theta$ are vectors and $dx,dy,dr,d\theta$ and the components of vectors are covectors. But using the transpose is just a matter of taste. What is more important is to keep track of Jacobian vs. its inverse. Then,

• the basis one-forms transform as $$\tag{1} \begin{pmatrix}dr & d\theta\end{pmatrix}=\begin{pmatrix}dx & dy\end{pmatrix} \begin{pmatrix} \cos\theta & -\frac1r\sin\theta\\\sin\theta&\frac1r\cos\theta\end{pmatrix} \,.$$

• the basis vector fields transform as

$$\tag{2} \begin{pmatrix}\partial_r\\\partial_\theta\end{pmatrix}= \begin{pmatrix} \cos\theta & \sin\theta\\ -r\sin\theta&r\cos\theta\end{pmatrix} \begin{pmatrix}\partial_x\\\partial_y\end{pmatrix}\,. $$

• the components of a vector $$ \boldsymbol{V}=V^x\partial_x+V^y\partial_y=V^r\partial_r+V^\theta\partial_\theta $$ transform as

$$\tag{3} \begin{pmatrix}V^r &V^\theta\end{pmatrix} =\begin{pmatrix}V^x & V^y\end{pmatrix} \begin{pmatrix} \cos\theta & -\frac1r\sin\theta\\ \sin\theta&\frac1r\cos\theta\end{pmatrix}\,. $$

• (1) and (3) are contravariant (use the transpose of the inverse Jacobian), and (2) is covariant (uses the transpose of the Jacobian).

• I do not understand why anything should be valid only to "first order". The transformation rule $$ \bar{V}^j=\frac{\partial\bar{z}_j}{\partial z_h}V^h $$ is valid to any order since it follows from the chain rule. Like in the Cartesian/polar example that rule is dictated by the invariance of the vector $$ \boldsymbol{V}=V^h\partial_{z_h}=\bar{V}^j\partial_{\bar{z}_j} $$ under arbitrary coordinate transformations.

• The Cartesian point $\color{red}p=(1,0)$ has polar coordinates $(r,\theta)=(1;0)$ where I am using the coloring and notation from this highly recommended answer. Thus, the Jacobian in (3) at this $\color{red}p$ is indeed $$\begin{pmatrix}1&0\\0&1\end{pmatrix}\,.$$ Congrats!

• The vector $\boldsymbol{V}=\partial_x+\partial_y$ that points at this $\color{red}p$ into the Cartesian direction $(1,1)$ has polar components $(V^r;V^\theta)=(1;1)\,.$ I admit that this may look confusing but lets keep struggling.

• This vector points from $\color{red}p$ to $\color{red}q=(2,1)=(\sqrt{5};\arctan(1/2))\,.$ Now you can use plotting of polar vectors in Geogebra to convince yourself that nothing is wrong.

• The reason for $\boldsymbol{V}_{\color{red}p}$ having identical components in both coordinate systems is that at $\color{red}p$ the basis vectors happen to coincide which is not that puzzling when one thinks about it.