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Both my textbooks (Schaums Tensor calculus and Neuenschwander's Tensor calc for physics) define a contravariant tensor of order one (which they also state is synonymous with a simple vector) as anything who's coordinates obey the law of transformation $$A'^i=\frac{\partial x'^i}{\partial x^j}A^j$$ where $A'^i$ are the coordinates of the vector in the primed frame and $A^j$ are the coordinates in the unprimed frame. This definition is supposed to apply to any coordinate transformation. But there is something severely wrong with either this formula or my understanding. For suppose we are looking at the vector $\vec{v}=(x,y)$ in $R^2$ so that $A^1=x$ and $A^2=y$. This is definitely a contravariant tensor of order one since it is simply a vector in $R^2$. Now let us use a coordinate transformation to polar coordinates to see if $\vec{v}$ actually is a tensor by determining whether it obeys the transformation rule above. For the polar transformation, we have that $$x'^1=r=\sqrt{x^2+y^2}\,\,\,and\,\,\,y'^2=\theta = \arctan{(y/x)}$$ According to the transformation law above,, we should then have

$$A'^1=\frac{x}{\sqrt{x^2+y^2}}A^1+\frac{y}{\sqrt{x^2+y^2}}A^2\,\,\,\,and\,\,\,A'^2=\frac{-y}{x^2+y^2}A^1+\frac{1}{x+\frac{y^2}{x}}A^2$$ Now substituting $A^1=x$ and $A^2=y$ into the above we get that the polar coordinate components of the vector $\vec{v}=(x,y)$ are $$A'^1=\frac{x^2}{\sqrt{x^2+y^2}}+\frac{y^2}{\sqrt{x^2+y^2}}\,\,\,\,and\,\,\,A'^2=\frac{-yx}{x^2+y^2}+\frac{y}{x+\frac{y^2}{x}}$$ But this obviously can't be because if we suppose that $\vec{v}=(x,y)=(2,2)$ then the polar components of $\vec{v}$ are simply $r=\sqrt{8}$ and $\theta=\pi/4$ which is obvious from the definiton of polar coordinates. But according to the above, we get that the polar components of $\vec{v}=(2,2)$ are infact $A'^1=r=\frac{8}{\sqrt{8}}$ and that $A'^2=\theta =0(!!)$. These components of obviously incorrect! So where am I going wrong?

Any help on this issue would be most appreciated as its been driving me insane!

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    $\begingroup$ No, this is correct and the position is indeed not a vector. $\endgroup$
    – jacob1729
    Mar 19 at 13:52
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    $\begingroup$ Does this answer your question? Does spacetime position not form a four-vector? $\endgroup$
    – jacob1729
    Mar 19 at 13:53
  • $\begingroup$ @jacob1729 kind of but not fully. If $\vec{v}=(x,y)$ is not a vector then what is an example of a vector (preferably a simple example) that will obey that restrictive transformation law? I realize that $\vec{v}=(x,y)$ will obey the transformation law only in the event that the coordinate transformation is linear. Can you provide an example of something which will obey it for any arbitrary coordinate transformation? $\endgroup$ Mar 19 at 14:07
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    $\begingroup$ Velocities are vectors. That is, if you take a parametrised curve $\vec{r}(t)$ and differentiate it, the result will be a vector, although the points in the curve themselves are not. $\endgroup$
    – jacob1729
    Mar 19 at 14:12
  • $\begingroup$ Okay so i take a unit circle trajectory, derive it to get $r'(t)=[-\sin(t),\cos(t)]$. I then plug this into my transformation equation from my original question and i get that $r=A'^1=1$ and $\theta = A'^2=0$. All correct? My issue now is that originally the velocity was changing with time (experiencing acceleration) in $R^2$. Now in polar coordinates, the velocity is constant with time and so the force has vanished? The fact that $r=A'^1=1$ makes sense to me as the velocity in R^2 involves a unit circle by why does $\theta=0$ now? $\endgroup$ Mar 19 at 14:32
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Exactly as @Jacob1729 says. Strange though it may seem, positions are not vectors. Positioning in this or that coordinate system is just tagging points on a manifold. The definition of vectors essentially depends on a concept of tangency. And for that you need a derivative. Stop worrying about $x^i$ and $x'^i$. Those are just parametrisations of your manifold: $$ \psi^{i}:\underset{p}{\mathscr{\mathcal{M}}}\underset{\mapsto}{\longrightarrow}\underset{\psi^{i}\left(p\right)}{\mathbb{R}^{n}} $$ $$ \psi^{i}\left(p\right)=\left(x^{1}\left(p\right),\cdots,x^{n}\left(p\right)\right) $$ This is called a "local chart". Redo your calculations with, $$ \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right)=A^{i} $$ $$ \left(\frac{\partial}{\partial r},\frac{\partial}{\partial\theta}\right)=A'^{i} $$ From where, $$ \frac{\partial}{\partial r}=\frac{x}{\sqrt{x^{2}+y^{2}}}\frac{\partial}{\partial x}+\frac{y}{\sqrt{x^{2}+y^{2}}}\frac{\partial}{\partial y} $$ $$ \frac{\partial}{\partial\theta}=\frac{-y/x^{2}}{\sqrt{1+\left(y/x\right)^{2}}}\frac{\partial}{\partial x}+\frac{1/x}{\sqrt{1+\left(y/x\right)^{2}}}\frac{\partial}{\partial y} $$ And you'll see the light at the end of the tunnel. Those are the ones that must be vectors. And if they aren't, we'll write the definitive book on differential geometry.

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  • $\begingroup$ Thanks for the response! Okay so I have gathered from your response as well as from jacob that positions are not vectors. But velocities are vectors. For example, the velocity of a clockwise circular trajectory is $r'(t)=[-\sin(t),\cos(t)]$. This is a vector. Yet when I use the transformation law in the form as shown in my question, i get that $ r=A′^1=1$ and $θ=A′^2=0$. My issue now is that originally the velocity was changing with time (experiencing centripetal acceleration) in $R^2$. Now in polar coordinates, the velocity is constant in time and so the force has vanished? $\endgroup$ Mar 19 at 18:24
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    $\begingroup$ I think you're conflating the polar and the Cartesian description. You must be consistent and drop the Cartesian description altogether. In polar coordinates $\ddot{\boldsymbol{r}}\left(t\right)=r\omega\dot{\boldsymbol{e}}_{\theta}=-r\omega^{2}\boldsymbol{e}_{r}\left(t\right)$, and $\boldsymbol{F}\left(r\right)=F_{0}\boldsymbol{e}_{r}\left(t\right)$, so that $F_{0}=-mr\omega^{2}$. Velocity is not constant, because $\boldsymbol{e}_{\theta}\left(t\right)$ is changing. $\endgroup$
    – joigus
    Mar 19 at 19:18
  • $\begingroup$ Here let me show you: i.imgur.com/DmhMmCs.jpg . In the link, I have provided all working out showing that a velocity vector for a circular trajectory loses all time dependence in the polar coordinate system if we apply the transformation rule. I must be doing something wrong but I'm not sure what $\endgroup$ Mar 20 at 6:00
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    $\begingroup$ You did your calculations fine. You're just confused. In this case you're in a flat manifold, and you can compare vectors everywhere. It's just that you've changed to a curvilinear coord. system. $$\boldsymbol{A}=A^{r}\boldsymbol{e}_{r}+A^{\theta}\boldsymbol{e}_{\theta}=A^{x}\boldsymbol{e}_{x}+A^{y}\boldsymbol{e}_{y}$$ $$A^{r}\boldsymbol{e}_{r}\left(t\right)=A^{x}\left(t\right)\boldsymbol{e}_{x}+A^{y}\left(t\right)\boldsymbol{e}_{y}$$ In Cartesian coords., the basis vectors are fixed; in polar coords., it's the coordinates that are fixed for your particular motion. $\endgroup$
    – joigus
    Mar 20 at 9:23
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    $\begingroup$ Thanks, okay I think I'm getting it now. So I'm correct that the velocity $r'(t)=[-sin(t),cos(t)]$ has components of $r=1$ and $\theta=0$ in polar coordinates and that these polar components do not change in time? It is actually the polar basis vectors themselves that are changing in time and so the force has been "hidden" by the fact that they are changing which allows the components to remain constant despite the presence of a force. $\endgroup$ Mar 20 at 10:36

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