0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You need to change base from the nonholonomic orthonormal ordered base to the standard basis I think, $ (\hat{r},\hat{\theta})\neq(\partial_r, \partial_\theta)$. (They use standard basis so often in proofs I don't know which equations apply where so can't be more specific, but I think the structure group coefficients could take care of it). $\endgroup$ – Emil Feb 1 '18 at 6:15
  • $\begingroup$ They might also be called commutation coefficients or connection coefficients... unless I'm wrong... $\endgroup$ – Emil Feb 1 '18 at 6:22
  • $\begingroup$ I think it is dependent on how the pairing between the dual space is made... if it is made in terms on the standard basis then the inverse metric is the usual one, otherwise it is different? $\endgroup$ – Emil Feb 1 '18 at 7:48
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I cant get your point how dimensionality remains correct. Another thing knocks me badly. The length-squared of the vector is ${\dot r}^2+r^2{\dot\theta}^2$ which should come from $v^iv_i$. But it comes up as $v^iv_i={\dot r}^2+r^4{\dot\theta}^2$. something is going wrong. $\endgroup$ – som Dec 30 '17 at 19:30
  • $\begingroup$ Oh, this is because $v^i=\dot{r}\frac{\partial}{\partial r}+\dot{\theta}\frac{\partial}{\partial \theta}$ (at this point this is just the change of coordinates so if you think of it in differential geometry language, you shouldn't add $r$ here) , therefore $v_i=\dot{r}dr+r^2\dot{\theta}d \theta$ and the multiplication is correct $\endgroup$ – Weicheng Ye Feb 1 '18 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.