Given a vector $\vec{p}$ = $<p^1,p^2>$ show $<ap^1,bp^2>$ for $a \ne b$ is not a vector.

Question

In the textbook Einstein Gravity in a nutshell by Zee, on page 43, it claims that given a vector $\vec{p}$ = $<p^1,p^2>$ then for any 2 real numbers $a\ne b$ then the object $<ap^1,bp^2>$ is definitely not a vector. Supposedly it's magnitude is not rotationally invariant? I do not understand firstly why it is not rotationally invariant (how would I show this?) I tried using the rotation matrix but I got the same magnitude as before [i.e. I got |R($\theta)(ap^1,bp^2)^T$| = |$(ap^1,bp^2)^T$|, where $R(\theta)$ is the 2D rotation matrix]! Also I'm guessing we can treat these vectors as being in $A^2$ rather than ${R}^2$. But secondly is there any difference between this definition of a vector and the usual definition given in an elementary physics course as something having both a magnitude and direction?

Answer 1

In the context of gravity, a "vector" is defined by objects satisfying the transformation rule under coordinate transformation, $$ p'^{\mu} = R^{\mu}_{\;\nu}p^{\nu} \quad \mbox{repeated indices are summed}$$

For an object like $q^{\mu} = a(\mu)p^{\mu}$, given the transfromation property of $p^{\mu}$, $$ \mbox{in general}\quad q'^{\mu} \neq R^{\mu}_{\;\nu}q^{\nu} $$

For instance, if $\vec{p}=(p^1,p^2)$ is a vector, then $\vec{p}' =(R^{1}_{\;1}p^1+R^{1}_{\;2}p^2,R^{2}_{\;1}p^1+R^{2}_{\;2}p^2) $. However, for $\vec{q} =(ap^1,bp^2) $, \begin{align} \vec{q}' &=(R^{1}_{\;1}\color{red}{a}p^1+R^{1}_{\;2}\color{red}{b}p^2,R^{2}_{\;1}\color{red}{a}p^1+R^{2}_{\;2}\color{red}{b}p^2) \\ &\neq (\color{red}{a}(R^{1}_{\;1}p^1+R^{1}_{\;2}p^2),\color{red}{b}(R^{2}_{\;1}p^1+R^{2}_{\;2}p^2)) \end{align} which does not match $\vec{q}'=(ap'^1,bp'^2)$ in the new coordinate system.