Addition of vectors in different coordinate systems

Question

I've been working with vectors in einstein notation for a while and I believe I have a good intuition on how indices and transformations work in this notation. But for some reason, I can't figure out how to add 2 vectors in this notation.

Normally, if I have vectors: $\vec{v}$ and $\vec{u}$

Their sum is:

$$\vec{v} + \vec{u} = v^i \vec{e_i} + u^i\vec{e_i} = (v^i+u^i)\vec{e_i} = (v+u)^i \vec{e_i}$$

so $(v^i + u^i) = (v+u)^i$

but that's not how we add vectors in, say, polar coordinates. I.e: the vector $[r,\theta]^T+ [r',\theta']^T \neq [r+r', \theta + \theta']$.

What am I doing wrong?

Answer 1

Expanding on Ted Shifrin's comment:

It is shown in this answer that the relationship between the components of vector fields $V$ and $U$ in Cartesian and polar coordinates are:

\begin{align} V&=v^x\partial_x+v^y\partial_y\\ &=\underbrace{(v^x\cos\color{red}\varphi\,+v^y\sin\color{red}\varphi)}_{\displaystyle =:v^r}\,\partial_r+\underbrace{\frac{-v^x\sin\color{red}\varphi+v^y\cos\color{red}\varphi}{\color{red}r}}_{\displaystyle=:v^\varphi}\,\partial_\varphi\,,\\ U&=u^x\partial_x+u^y\partial_y\\ &=\underbrace{(u^x\cos\color{red}\varphi\,+u^y\sin\color{red}\varphi)}_{\displaystyle =:u^r}\,\partial_r+\underbrace{\frac{-u^x\sin\color{red}\varphi+u^y\cos\color{red}\varphi}{\color{red}r}}_{\displaystyle=:u^\varphi}\,\partial_\varphi\,. \end{align}

• $v^x,v^y,u^x,u^y$ are the Cartesian components.

• $v^r,v^\varphi,u^r,u^\varphi$ are the polar components.

• In both systems the upper indices are just labels.

• In contrast, the highlighted $\color{red}{r,\varphi}$ above are the coordinates of the point $\color{red}p$ on the manifold $\mathbb R^2$ at which both vectors $V,U$ must reside in order to be additible.

Then, \begin{align} V+U&=(v^x+u^x)\,\partial_x+(v^y+u^y)\,\partial_x\,\\[2mm] &=(v^r+u^r)\,\partial_r+(v^\varphi+u^\varphi)\,\partial_\varphi\,,\\[2mm] v^r+u^r&=(v^x+u^x)\cos\color{red}\varphi+(v^y+u^y)\sin\color{red}\varphi\,,\\[2mm] v^\varphi+u^\varphi&=\frac{-(v^x+u^x)\sin\color{red}\varphi+(v^y+u^y)\cos\color{red}\varphi}{\color{red}r}\,. \end{align}

• Regardless in which coordinate system we are: addition of vectors at the same point $\color{red}p$ is equivalent to addition of components. Anything else would be a nightmare.

Since there seems a disagreement with the answer by TurlocTheRed I elaborate further.

• The origin does not have polar coordinates so we cannot consider vectors that reside at $\color{red}p=(0,0)\,.$ Instead I shall consider $$ \color{red}p=(1,1)=(\sqrt{2}\,;\pi/4) $$ using the convention that the comma separates cartesian coordinates, and the semicolon separates polar coordinates. The same convention shall apply to vector components. This point has polar coordinates

\begin{align} \color{red}r&=\sqrt{2},\quad\color{red}\varphi=\pi/4\,,&\cos\color{red}\varphi &=\sin\color{red}\varphi=\frac{1}{\sqrt{2}}\,. \end{align} With the formulas above one can easily check that in both coordinates systems components add when we add the vectors: \begin{align} V_\color{red}p&=(1,1)=(\sqrt{2}\,;0)\,,&U_\color{red}p&=(0,1)=\textstyle(\frac{1}{\sqrt{2}};\frac12)\,,&V_\color{red}p+U_\color{red}p&=(1,2)=\textstyle(\frac{3}{\sqrt{2}};\frac12)\,. \end{align}

• The polar components of these vectors could be a bit puzzling at first but note that $V_\color{red}p$ points from $(1,1)$ to $(2,2)\,.$ This means we don't go in an angular direction and only in the radial direction. That's why the polar components of $V_\color{red}p$ are $(\sqrt{2}\,;0)\,.$

Answer 2

Basis vectors and their components mean different things in different coordinate systems. Further, the basis vectors vary from one point to another.

In Euclidean coordiantes, $\vec{i}(x_1,y_1,z_1)$ is the same $\vec{i}(x_2,y_2,z_2)$ It's the same basis vector anywhere in the coordinate plane.

In polar coordinates, $\vec{e_r}=(x/r)\hat{i}+(y/r)\hat{j}+(z/r)\hat{k}$ and $\vec{e_\theta}=(-y/r)\hat{i} + (x/r)\hat{j}$

There are different basis vectors at different positions, so you need different coordinates at different positions.

So $(1,1)$ in Cartesian coordinates is $(\sqrt{2}, \pi/4)$ in polar.

$(0,1)$ in Cartesian Coordinates is $(1, \pi/2)$ in polar.

$<1,1>+<0,1>=<1,2> $ in Cartesian. In polar that is $(\sqrt{5}, \arctan(2))$

The formula for adding cartesian coordinates is $<x_1,y_1>+<x_2,y_2> = <x_1+x_2, y_1+y_2>$

For polar, it's $<r_1,\theta_1> + <r_2, \theta_2> = <r,\theta>$

$r=\sqrt{r_1^2+r_2^2+2r_1r_2 \cos (\theta_1-\theta_2)}$

$\theta = \arctan (\frac{r_1\sin \theta_1+r_2\sin \theta_2}{r_1\cos\theta_1+ r_2\cos\theta_2})$