Can one construct a covariant velocity from the contravariant one using the metric tensor in plane polar coordinate system?

Question

In plane polar coordinate we know $ds^2=dr^2+r^2d\theta^2$ from which we can write the components of the covariant metric tensor $g_{11}=1$, $g_{22}=r^2$, $g_{12}=g_{21}=0$.

Upto this point I believe I'm correct. But after this I may go wrong and need your help.

The velocity ${\vec v}=\dot r\hat{r} + r\dot\theta\hat{\theta}$ is a contravariant vector with components $v^1=\dot r$ and $v^2=r\dot \theta$. Correspondingly, atleast as per definition, the components of covariant velocity should be obtained by $v_i=g_{ij}v^j$. Thus $v_1=\dot r$ and $v_2=r^3\dot\theta$. So the covariant velocity is ${\vec v}=\dot r\hat{r} + r^3\dot\theta\hat{\theta}$. This is clearly just bogus because dimensionality does not hold right. Where is the whole thing became faulty?

Is it that a scale factor removes the dimensionality problem and leaves the rest correct?

Answer 1

Your assumption for the velocity vector to be contravariant is kind of wrong. You can define any vector in the following manner, $$\vec{V}=V^i\mathbf{e}_i=V_i\mathbf{e}^i $$ In case we want to assume contravariant components, we also need to define set of covariant basis. We can do that through the covariant transformation, which is, $${V'}_i=\frac{\partial{x'}_j}{\partial x_i}V_j $$ In this case we shall get the two covariant basis as, $$\mathbf{e}_r=\frac{\partial x}{\partial r}\mathbf{e}_x+\frac{\partial y}{\partial r}\mathbf{e}_y $$ $$\mathbf{e}_{\theta}=\frac{\partial x}{\partial \theta}\mathbf{e}_x+\frac{\partial y}{\partial\theta}\mathbf{e}_y $$ That would give us, $$\mathbf{e}_r=\cos\theta\, \mathbf{e}_x+\sin\theta\, \mathbf{e}_y $$ $$\mathbf{e}_{\theta}=-r\sin\theta\, \mathbf{e}_x+r\cos\theta\, \mathbf{e}_y $$ Note that these aren't unit basis vector, which is the one generally introduced at the pre-tensor level. If you consider the inner product of each of the basis in the Cartesian coordinate then we shall get, $$\mathbf{e}_r\cdot\mathbf{e}_r=1 $$ $$\mathbf{e}_{\theta}\cdot\mathbf{e}_{\theta}=r^2 $$ Which also means that the tangential basis, is not a unit basis. The position vector in this case will be, $$\vec{r}=r\, \mathbf{e}_r $$ Which makes the velocity vector, $$\vec{v}=\frac{d\vec{r}}{dt}=\dot{r}\, \mathbf{e}_r+r\left(\frac{\dot{\theta}}{r}\right) \, \mathbf{e}_{\theta}$$ $$\Rightarrow \vec{v}=\dot{r}\, \mathbf{e}_r+\dot{\theta} \, \mathbf{e}_{\theta}$$ Note that the dimension of this vector is that of velocity. The tangential basis has the dimension of length which makes the total expression having a dimension of velocity. And now you can use the metric tensor while taking the dot product to see the invariant scalar which will be, $$\vec{v}\cdot\vec{v}=\eta_{\alpha\beta}v^{\alpha}v^{\beta}=\dot{r}^2+r^2\dot{\theta}^2 $$ When you normalize the covariant basis (in this case also called vector basis, while the contravariant basis will be called one form basis) you are basically changing the basis definition from a coordinate basis to a noncoordinate basis. Noncoordinate basis cannot be obtained from coordinate transformation, which is why it isn't used in most of the cases of tensor application.

Answer 2

Your answer is correct, but your way of writing things is at least not standard and makes you wonder it could be wrong

For a contravariant tensor, a better way of writing it is similar to how you write a metric (because they transform in the same way), namely $$\overrightarrow{V}=\dot{r}dr+r^3 \dot{\theta}d\theta$$ And you can check that it has the correct dimensionality manifestly

Any good textbook on differential geometry will explain why this is the standard expression