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I am reading A. Zee's book Einstein Gravity in a Nutshell and encountered this definition of a vector on page 43:

An array of two numbers $p = \begin{pmatrix} p^1 \\ p^2 \end{pmatrix} $ is a vector if it transforms according to $p' = R(\theta)p$

and then he proceeds to give an example that $\begin{pmatrix} ap^1 \\ bp^2 \end{pmatrix}$ is not a vector if $a \neq b$.

I am utterly confused by this definition. Can anyone explain what does he mean in this case?

Qmechanic
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Percy
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2 Answers2

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Mr. Zee seems to be saying that, since we shall define the array $\vec{p}$ as a vector, we have

$$\begin{align} p_1'&=R_{11}p_1+R_{12}p_2\\ p_2'&=R_{21}p_1+R_{22}p_2\\ \end{align}$$

Let us define a new array (not vector) $\vec q$ as

$$\vec q \equiv \begin{pmatrix} ap_1\\ bp_2 \end{pmatrix}$$

If $\vec q$ as an array transforms naively under the transformation $R$, we get

$$\begin{align} q_1'&=ap_1'\\ &=R_{11}ap_1+\frac{a}{b}R_{12}bp_2\\ &=R_{11}q_1+\frac{a}{b}R_{12}q_2\\ q_2'&=bp_2'\\ &=\frac{b}{a}R_{21}ap_1+R_{22}bp_2\\ &=\frac{b}{a}R_{21}q_1+R_{22}q_2\\ \end{align}$$

On the other hand, if $\vec q$ is indeed a vector, it must transform under $R$ as simply

$$\begin{align} q_1'&=R_{11}q_1+R_{12}q_2\\ q_2'&=R_{21}q_1+R_{22}q_2\\ \end{align}$$

If these two transformation rules — the "naive" and "true" ones — are to be the same, then we must have $a=b$. However if $\vec q$ is simply defined to be a vector from the start, and therefore transform under the second transformation rule, then there is no such restriction necessary.

Ultimately I do not see the point in this at all.

Arturo don Juan
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  • Thanks! This seems to be a rather round about way to define a "vector"... – Percy Jul 09 '20 at 14:04
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    @Percy The definition of a vector is very straightforward and is essentially the second definition I gave. A vector is an array $v^{\mu}$ which transforms under a general coordinate transformation $x^{\mu}\rightarrow y^{\mu}(x)$ according to the rule $v'^{\mu}=\frac{\partial y^{\mu}}{\partial x^{\nu}} v^{\nu}$. Equivalently, a coordinate-free vector $\mathbf{v}=v^{\mu}e_{\mu}$ where $e_{\mu}$ are basis (co-)vectors, should not physically change under coordinate transformations - e.g. a pencil is the same pencil irrespective of your choice of coordinates. – Arturo don Juan Jul 09 '20 at 17:26
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The components of a vector $\vec{p}$ are $x$ and $y$

$$\vec{p}=\begin{bmatrix} x \\ y \\ \end{bmatrix}$$

where $x,y\in\Re$

thus the magnitude of $\vec{p}$ is $$|\vec{p}|=\sqrt{x^2+y^2}$$ if $\vec{p}\mapsto \vec{p}'$ the magnitude must be preserve: $$|\vec{p}'|=|\vec{p}|$$

with: $$\vec{p}'=R\,\vec{p}\quad,\Rightarrow |\vec{p}'|=|\vec{p}^T\,R^T\,R\,\vec{p}|= |\vec{p}|$$ where R is a orthogonal transformation matrix .

so the magnitude preservation is fulfilled with any $x,y$ components!, this is also the case if $x\mapsto a\,x$ and $y\mapsto b\,y$ where a and b are real numbers

Eli
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