Vectors transforming under change of coordinates

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?

• Link to lecture? Minute? – Qmechanic Nov 14 '18 at 23:16

1 Answer

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:

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

• 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. – Aaron Stevens Nov 15 '18 at 1:18
• 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. – G. Smith Nov 15 '18 at 1:22
• 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. – Aaron Stevens Nov 15 '18 at 1:28
• If I have to be pedantic, the vectors living at the point $p=(x,y)$ are $\partial_x$ and $\partial_y$. – Gonenc Nov 15 '18 at 1:38
• I think you are using mathematicians’ definition of contravariance and covariance, and I am using physicists’. – G. Smith Nov 15 '18 at 1:51