Covariant and Contravariant Transformation - Example of Polar Coordinates

Question

I would like to see the fact that the components of a vector transform differently (controvariant transformation) than the unit bases vectors (covariant transformation) for the specific case of cartesian to polar coordinate transformation.

The polar unit vectors $\hat{r}$ and $\hat{\theta}$ can be expressed in terms of cartesian unit vectors, $\hat{x}$ and $\hat{y}$, as the following \begin{equation} \hat{r}= \text{cos}\phi \ \hat{x} + \text{sin}\phi \ \hat{y} \\ \hat{\theta}= -\text{sin}\phi \ \hat{x} + \text{cos}\phi \ \hat{y} \tag{1} \end{equation}

Any vector, $\vec{V}$, can be expressed in the cartesian coordinate system as $\vec{V}=V_x \ \hat{x} + V_y \ \hat{y}$. The same vector can be expressed in polar coordinates as $\vec{V}=V_r \ \hat{r} + V_\theta \ \hat{\theta}$. We then have \begin{equation} V_x \ \hat{x} + V_y \ \hat{y}=V_r \ \hat{r} + V_\theta \ \hat{\theta}. \tag{2} \end{equation} I then project both sides of (2) once onto $\hat{r}$, and once onto $\hat{\theta}$. Using (1) and (2) we get \begin{equation} V_r= \text{cos}\phi \ V_x+\text{sin}\phi \ V_y \\ V_\theta= -\text{sin}\phi \ V_x+\text{cos}\phi \ V_y \tag{3} \end{equation}

Comparing (1) and (3), both the unit vectors and the components of a vector are transforming with the same rule, which is a contradiction! What am I missing here?

Answer 1

As @TedShifrin pointed out, the correct transformation for the dual is $(A^{-1})^T$.

For instance, in special relativity contravariant vectors transform as

$ V^{\mu '} = \Lambda ^{\mu '} _{\,\nu} V^{\nu}$,

where $ \Lambda ^{\mu '} _{\,\nu} $ is the Lorentz transformation taking component from the unprimed frame to the primed frame. For covariant components we have to use the inverse of the Lorentz transformation:

$ V_{\mu '} = \big( \Lambda^{-1} \big)^{\nu } _{\,\mu '} V_{\nu}$.

If we assume that $V'$ and $V$ are column vectors representing, $ V_{\mu '}$ and $V_{\nu}$, respectively. Then in matrix format we have:

$ V' = \big( \Lambda^{-1} \big)^T V$.

This can also be found here: https://en.wikipedia.org/wiki/Lorentz_transformation

Of course, one can get to the same conclusion in the context of group theory. If the vector $x$ transforms according to $x \rightarrow x'=Ax$ where $A$ is the group member. Then the dual $\tilde{x}$ transforms as $\tilde{x} \rightarrow \tilde{x} '= \big(A^{-1} \big)^T\tilde{x}$ so that $\tilde{x}^T x$ is an invariant in all frames.

Answer 2

I think that your answer is unnecessarily complicated for this question. In matrix notation equation (1) in the question is

\begin{equation} \begin{pmatrix} \hat{r} && \hat{\theta} \end{pmatrix} =\begin{pmatrix} \hat{x} && \hat{y} \end{pmatrix} \begin{pmatrix} cos\phi && -sin\phi \\ sin\phi && cos\phi \end{pmatrix} \triangleq \begin{pmatrix} \hat{x} && \hat{y} \end{pmatrix} M \end{equation}

and equation (3) expresses the transformation rule for a row (covariant) vector

\begin{equation} \begin{pmatrix} v_r && v_{\theta} \end{pmatrix}= \begin{pmatrix} v_x && v_y \end{pmatrix} M \end{equation}

so the transformation for a column (thus contravariant) vector is

\begin{equation} \begin{pmatrix} v_r \\ v_{\theta} \end{pmatrix} =M^T \begin{pmatrix}v_x\\ v_y\end{pmatrix} \end{equation}

Which is indeed the inverse of $M$ as the latter is orthogonal.