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Take some general vector $\mathbf{u}$ which can be expressed in terms of is components and basis as $\mathbf{u} = u^{a} e_a$ (summation notation).

Now, if we transform to some different, primed basis, then the contravariant vector components transform as,

$$u^{a'} = \frac{\partial x^{a'}}{\partial x^b} u^b$$

In this way we can transform from e.g. Cartesian to polar coordinates. That is all fine

Now my confusion comes where if we have some vector defined at the origin. For example we have a velocity vector $\mathbf{u}$ with components $(u^x, u^y)$ which 'exists' at $x=0, y=0$.

If we then attempt to go to polar coordinates then

$$u^r = \frac{\partial r}{\partial x} u^x + \frac{\partial r}{\partial y} u^y $$

$$ = \frac{x}{r} u^x + \frac{y}{r} u^y$$

$$ = 0$$

but then , intuitively, I feel like the vector $\mathbf{u}$ should have some component in the $r$-direction, since $u^x, u^y$ are non zero? What gives?

user1887919
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1 Answers1

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Assume ${\bf{u}} = u_x {\bf{e_x}} + u_y {\bf{e_y}}$ with the cartesian basis vectors ${\bf{e_x}}$ and ${\bf{e_y}}$ with $u_x\neq0$ and $u_y\neq 0$. Then ${\bf{u}} = u_r {\bf{e_r}} + u_\vartheta {\bf{e_\vartheta}}$ in polar coordinates.

Because ${\bf{e_r}}=\cos(\vartheta)\bf{e_x}+\sin(\vartheta){\bf{e_y}}$ and ${\bf{e_\theta}}=-\sin(\vartheta)\bf{e_x}+\cos(\vartheta)\bf{e_y}$ we can substitute in the polar representation and we get

$$\begin{align*}{\bf{u}} &=u_r {\bf{e_r}} + u_\vartheta {\bf{e_\vartheta}} \\&= u_r (\cos(\vartheta){\bf{e_x}}+\sin(\vartheta){\bf{e_y}}) + u_\vartheta (-\sin(\vartheta)\bf{e_x}+\cos(\vartheta)\bf{e_y}) \\&= (u_r \cos(\vartheta)-u_\vartheta \sin(\vartheta)){\bf{e_x}}+(u_r \sin(\vartheta) + u_\vartheta \cos(\vartheta)){\bf{e_y}}\\&=u_x {\bf{e_x}} + u_y {\bf{e_y}} \end{align*}$$

From the last two equations we can read the (contravariant) coordinate trasformation of the vector components:

$$\begin{align*}u_x &=u_r \cos(\vartheta)-u_\vartheta \sin(\vartheta) \\u_y &= u_r \sin(\vartheta) + u_\vartheta \cos(\vartheta) \end{align*}$$

or

$$\begin{align*}u_r &=u_x \cos(\vartheta)+u_y \sin(\vartheta) \\u_\vartheta &= -u_x\sin(\vartheta) + u_y \cos(\vartheta) \end{align*}$$

EDIT

if $\vartheta = \arctan(u_y/u_x)$, then $\cos(\vartheta) = \frac{1}{\sqrt{1+\left(u_y/u_x\right)^2}}=\frac{u_x}{\sqrt{u_x^2+u_y^2}}$ and $\sin(\vartheta) = \frac{u_y/u_x}{\sqrt{1+\left(u_y/u_x\right)^2}} = \frac{u_y}{\sqrt{u_x^2+u_y^2}}$ then

$$\begin{align*}u_r &= u_x \cos(\vartheta)+u_y \sin(\vartheta) \\&=u_x \frac{1}{\sqrt{1+\left(u_y/u_x\right)^2}}+u_y \frac{u_y/u_x}{\sqrt{1+\left(u_y/u_x\right)^2}} \\&=\sqrt{u_x^2+u_y^2} \end{align*}$$

Ahmed Hossam
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  • Does it make sense to have $\theta$ = something at $x,y=0$? – user1887919 Jul 29 '19 at 10:36
  • We could plot this function $u_r =u_x \cos(\vartheta)+u_y \sin(\vartheta)$ of $0 \leq \vartheta \leq 2\pi$. It's a linear combination of $\sin$ and $\cos$ functions so, most of the time $u_r \neq 0$. I think $\vartheta = \arctan(u_y/u_x)$. I can't see how you come to the conclusion that $u_r = 0$, if $u_x \neq 0$ and $u_y \neq 0$. – Ahmed Hossam Jul 29 '19 at 10:47
  • Sure, but I am specifically interested in the case where the vector $u$ is defined at the origin. In this case, $\theta$ is not well-defined. I don't think $\theta = \arctan(u_y,u_x)$ - that is $\theta$ as the angle of the vector. I want $\theta$ as the position of the vector c.f. general relativity, tangent spaces etc. – user1887919 Jul 29 '19 at 10:56
  • When talking about polar coordinates, $\vartheta$ and $r$ are defining the position of the vector – Ahmed Hossam Jul 29 '19 at 11:54
  • Yep, they define the position, but not the components. The coordinate $\theta$ is related to the coordinates $x,y$ as $\theta = \arctan(y/x)$. But the vector components do not satisfy the same relation – user1887919 Jul 29 '19 at 12:09
  • The vector components $u_x$ and $u_y$ are the coordinates $x$ and $y$ in the cartesian coordinate system. The vecor components $u_r$ and $u_\vartheta$ are the coordinates $r$ and $\vartheta$ in the polar coordinate system. https://www.edumedia-sciences.com/de/media/264-polarkoordinaten-eines-vektors – Ahmed Hossam Jul 29 '19 at 13:12
  • Components are different from coordinates in differential geometry. See first answer here https://physics.stackexchange.com/questions/326462/general-relativity-how-are-components-of-ds2-specified – user1887919 Jul 29 '19 at 13:43
  • You are right, but you don't need general relativity for this. This will also do the trick: https://math.stackexchange.com/a/429386/430756 – Ahmed Hossam Jul 29 '19 at 14:31