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Reading Zee's new "Einstein Gravity in a Nutshell" I just don't get it what exactly is meant by "transforms like a ...". He explicitly writes that not any odd column of numbers is a vector, so I am trying to figure out what exactly is meant. Here is what I think I did understand so far:

- "transforms" refers to the transformation from one coordinate systems to another.

- Specifically a Vector [itex]V^\nu[/itex] shall transform like [itex]dx^\nu[/itex] in [itex]dx'^\mu = \frac{∂x'^\mu}{∂x^\nu}dx^\nu[/itex], in particular

[tex]V'^\mu(\vec{x}') = \frac{∂x'^\mu}{∂x^\nu}V^\nu(\vec{x})[/tex]

If I take the equality above as thedefinitionof [itex]V'^\nu[/itex], the statement "transforms like ..." does not have any deeper meaning. The "transforms like ..." only has a deeper semantic meaning, if [itex]V'^\nu[/itex] is defined independently of the equality above and we can then say, "Hey, this equality does not hold for any odd column of numbers but only for those which we want to call a vector".

Now I ask: what is this independent definition of [itex]V'^\nu[/itex]?

To be more specific: suppose [itex]V^\mu(\vec{x})\in R^2[/itex] is some physical quantity with a length and a direction and let ourunprimedcoordinate system be polar coordinates. For the sake of argument let [itex]V^\mu(r,\theta) = (l,\xi)[/itex], where [itex](l,\xi)[/itex] is the physical quantity written in polar coordinates.

Let the primed coordinate system be cartesian coordiinates, meaning [itex]x'=r \cos(\theta)[/itex] and [itex]y'=r \sin(\theta)[/itex].

Assume that [itex](l,\xi)[/itex] is indeed a vector, then I can tranform it by [itex]\frac{∂x'^\mu}{∂x^\nu}[/itex] into a [itex]V'(r\cos(\theta),r\sin(\theta))=(x_0',y_0')[/itex] of the primed coordinate system. (Still correct?)

My hunch would be that "transforms like a vector" means that [itex]x_0' = l\cos(\xi)[/itex] and [itex]y_0'=l\sin(\xi)[/itex], where the right hand sides are the "independent definition" I was looking for. Is that true?

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# What does it mean: transforms like a vector

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