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I was watching a lecture on tensors and the professor said that a defining feature of a vector $v$ is that it transforms under a coordinate transformation $x^{\mu} \rightarrow x^{\mu'}$ as

$$v^{\mu'}(x^{\mu'}) \equiv \frac{\partial x^{\mu'} }{\partial x^\mu} v^{\mu}(x^{\mu}(x^{\mu'})) $$

Basically the last last term on the right hand side is the $x^{\mu}$ coordinates expressed in terms of $x^{\mu'}$ coordinates.

I am trying to understand this in $\mathbb R^2$ for the Cartesian and Polar coordinates. If the $x^{\mu'} \equiv (x,y)$ and $x^{\mu} \equiv (r,\theta)$, then I get

$$\bigg(\begin{matrix}x\\y \end{matrix}\bigg) \equiv \bigg[ \begin{matrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta \end{matrix}\bigg] \bigg(\begin{matrix}r\\\theta \end{matrix}\bigg)$$

which doesn't work out to be true. Can you please tell me what I'm missing here?

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    $\begingroup$ Link to lecture? Minute? $\endgroup$
    – Qmechanic
    Commented Nov 14, 2018 at 23:16

1 Answer 1

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This transformation equation applies to a vector $v$ that "lives at" at the point $x$ (i.e., is in the tangent space at $x$). The point $x$ is described by the coordinates $(x, y)$ or $(r, \theta)$, but these coordinate tuples are not vectors at $x$. Examples of vectors at $x$ are $(dx, dy)$ and $(dr, d\theta)$, and your matrix equation applies to them:

$$\bigg(\begin{matrix}dx\\dy \end{matrix}\bigg) = \bigg[ \begin{matrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta \end{matrix}\bigg] \bigg(\begin{matrix}dr\\d\theta \end{matrix}\bigg)$$

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  • $\begingroup$ This is right. So many times I see people thinking that coordinates are vectors, and it leads to so much confusion when we have a vector space defined over those coordinates. $\endgroup$ Commented Nov 15, 2018 at 1:18
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    $\begingroup$ For a general manifold, as opposed to flat space, why do you have to think of vectors and tensors as living at a point? Well, think about the curved surface of the Earth. A vector pointing north in New York is different from a vector pointing north in Tokyo. A lot of differential geometry is about how to compare vectors at different points. $\endgroup$
    – G. Smith
    Commented Nov 15, 2018 at 1:22
  • $\begingroup$ Right. I see a similar confusion in introductory EM in expressing vectors using the unit vectors $\hat r$, $\hat\theta$, and $\hat\phi$ depends on where the vector is located. $\endgroup$ Commented Nov 15, 2018 at 1:28
  • $\begingroup$ If I have to be pedantic, the vectors living at the point $p=(x,y)$ are $\partial_x$ and $\partial_y$. $\endgroup$
    – Gonenc
    Commented Nov 15, 2018 at 1:38
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    $\begingroup$ I think you are using mathematicians’ definition of contravariance and covariance, and I am using physicists’. $\endgroup$
    – G. Smith
    Commented Nov 15, 2018 at 1:51

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