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I'm reading Einstein Gravity in a Nutshell (by Zee) and here he defines a vector as an object which is invariant under coordinate representation; concretely, if in one coordinate representation, $V$, $p_V=(p^1,p^2)$ then when we transform it via a rotation $p_W = R(\theta)p_V$ then we do not violate any physical laws.

To motivate the question further, I understand that individual terms after a transformation are not preserved but the law as a whole is: For example if $ma_V = F_V$ then $ma_W=F_W$ and although $F_W \neq F_V$ it is the case that the scalar (tensor) is preserved $$ma_W-F_w=0=ma_V-F_V$$

The question that motivated this post is the following: Prove that if $p_V$ is a vector then $p'=(ap^1,bp^2)$ cannot be a vector unless $a \equiv b$. This seems easy enough to show

$$Rp'= (p_1a\cos(\theta)-p_2b\sin(\theta), p_1 a\sin(\theta) + p_2b\cos(\theta))^t$$

And clearly, we cannot just factor out the $a$ and $b$ unless $a\equiv b$. But what does this really mean? The converse must be true, namely if $p'$ is a vector then surely $p_V$ cannot be a vector.

I suspect if I keep reading there will be ample examples and it will click better (for example the force-mass-acceleration equation makes sense in how it transforms). I suppose my confusion rests in the idea that "well of course a vector is just a tuple and if rotate it that's another tuple and that old-new tuple pair was prescribed by a rotation so of course all tuples are vectors!"

EDIT: Here's another example to make the point (this is an actual exercise): Suppose we are given two vectors ${p}$ and $q$ in ordinary 3-dimensional space. Consider this array of three numbers: $(p^2q^3,p^3q^1,p^1q^2)^t$. Prove that is not a vector, even though it looks like a vector. (Check how it transforms under rotation!) In contrast, $(p^2q^3-p^3q^2, p^3q^1-p^1q^3,p^1q^2-p^2q^1)^t$ does transform like a vector (so it's a vector). It is in fact the vector cross product $p \times q$.

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  • $\begingroup$ It might in fact be the case that I'm not in fact clearly understanding what a scalar is.... if that's the case, I hope it shows above in the first centred equation (namely, I'm under the impression that neither $F$ nor $a$ are scalars but rather the "scalar" is the quantity $\chi = ma-F$) $\endgroup$ – Squirtle Jun 29 '14 at 0:43
  • $\begingroup$ I don't understand "the question that motivated this post..." Is $p'$ supposed to be the result of some transformation? Is it merely supposed to be some other vector-like quantity that may or may not obey the correct transformation laws? $\endgroup$ – Muphrid Jun 29 '14 at 1:11
  • $\begingroup$ Well... I suppose that's the point! The question was given exactly as I described... I think the lack of a more concrete example caused my confusion. $\endgroup$ – Squirtle Jun 29 '14 at 1:16
  • $\begingroup$ Perhaps they mean that $p'$ cannot have that form in every basis, which would be true. $\endgroup$ – Muphrid Jun 29 '14 at 1:18
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Let's do this systematically: (I must admit that the part of your question with "prove that $p'$ is not a vector if $p$ is" did not really make sense to me, so I decided to show you how vectors in GR are really defined and how transformations really work. Feel free to tell me to mind my own business if this is not what you are looking for ;) )

Let $\mathcal{M}$ be our spacetime manifold (think $\mathbb{R}^4$). Let $p \in \mathcal{M}$ be a point and $x,y : \mathcal{M} \rightarrow \mathbb{R}^4$ be coordinates on $\mathcal{M}$, i.e. the 4-tupel $(x^0(p),x^1(p),x^2(p),x^3(p))$ and $(y^0(p),y^1(p),y^2(p),y^3(p))$ represent the same point. At the point $p$, we define the tangent vector space $T_p\mathcal{M}$ to be spanned by the partial derivaties w.r.t. to the coordinates, i.e we have the space of linear combinations

$$ T_p\mathcal{M}[x] := \left\{\sum_{\mu = 0}^3 c^\mu \frac{\partial}{\partial x^\mu} | c^\mu \in \mathbb{R}\right\}$$

and the coefficients $c^\mu$ are what we ordinarily call our vector components. If we say that some tupel $(v^0,v^1,v^2,v^3)$ is a vector, we mean by definition that it represents the vector $\sum_\mu v^\mu \frac{\partial}{\partial x^\mu}$ at some (often tacitly) understood point.

A coordinate transformation is now given by $y \circ x^{-1}$, often denoted $y(x)$ by abuse of notation. Now, our vectors must be expressed in derivatives w.r.t to the new coordinates $y^\mu$. By the chain rule, $\frac{\partial}{\partial x^\mu} = \sum_{\nu = 0}^3 \frac{\partial y^\nu}{\partial x^\mu}\frac{\partial}{\partial y^\nu}$, so the transformed coefficients are

$$c'^\mu = \sum_{\nu = 0}^3 \frac{\partial y^\mu}{\partial x^\nu} c^\nu$$

This is the transformation law for vectors. If your coordinate transformation is just a constant matrix like a rotation or a Lorentz transformation, $\frac{\partial y^\mu}{\partial x^\nu}$ are just its matrix components. Since our vectors at a point live by definition in the tangent space, they transform by definition like this.

A scalar is any quantity that does not transform under coordinate transformation in any way, or, equivalently, on which any transformation takes the form of the identity. Obviously, vectors are not scalars. The "space of scalars" at a point is just $\mathbb{R}$ (or $\mathbb{C}$, depending on your preference), i.e numbers which do not change under coordinate transformations.

Now, how do we build scalars out of vectors? We define the dual vector space or cotangent space to be spanned by the differentials $\mathrm{d}x$, i.e.

$$T^*_p\mathcal{M}[x] := \left\{\sum_{\mu = 0}^3 \omega_\mu \mathrm{d}x^\mu\right\}$$

and, again by the chain rule, these now transform exactly inversely to vectors, i.e.

$$ \omega'_\mu = \sum_{\nu = 0}^3 \frac{\partial x^\nu}{\partial y^\mu}\omega_\nu$$

Now, in GR, there is a metric on spacetime given by coefficients $g_{\mu\nu}$. Given a vector $c^\mu$, we define its dual vector to be $c_\mu := \sum_\nu g_{\mu\nu}c^\nu$. Now, given any tuple as a vector $c^\mu$, we may form the scalar product

$$(c,c) = \sum_\mu c^\mu c_\mu$$

For any given coordinate transformation, you may easily check that the derivaties incurred from the transforms of the vector and the covector cancel, and that this thing is indeed invariant under coordinate transformations, and thus is a scalar.

EDIT: To consider now a general class of the examples you seem to be asked to look at, observe that, for vectors $p^\mu,q^\nu$ in the above sense tupels like $(p^0 q^0, p^1,p^2,p^3)$ cannot be vectors, since the first components would transform twice w.r.t to the coordinate transformation, while the others only transform once, but the law of vectors says all components transform the same, and exaclty in the way given above.

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  • $\begingroup$ Hi. Is it possible to give some insight for the reason we define a vector on a point? Is it correct to say that the reason is the curvature in a sense that when the space is curved we cannot compare any two vectors on different points? What's the mathematical motivation of defining a vector on a point? Thank you. $\endgroup$ – Constantine Black Mar 25 '16 at 17:40
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I just wanted to comment on what I think the question from Zee's book is getting at. I don't think that this question from Zee is a particularly good question, and I think ACuriousMind's answer is excellent and gets at the real points that you should be trying to learn. So this is being written purely for the sake of trying to clarify Zee's book, rather than trying to give a competing answer to this question.

Let's say we have a vector $\vec{p}$ which is expressed in some basis (let's call it the 'unprimed basis') as $\vec{p}=p_1 \hat{e}_1 + p_2 \hat{e}_2$. We can express $\vec{p}$ in a new basis formed by rotating by an angle $\theta$ (this is a 'passive' view of rotations) \begin{equation} \vec{p} = p_1'\hat{e}_1' + p_2' \hat{e}_2' = \left(p_1\cos\theta+ p_2 \sin \theta \right)\hat{e}_1' + \left(-p_1 \sin \theta + p_2 \cos \theta \right) \hat{e}_2' \end{equation} So far, so good.

Now let's define an object $\vec{q}$, which has the following components in the unprime basis: \begin{equation} \vec{q}=q_1 \hat{e}_1 + q_2 \hat{e}_2 = a p_1 \hat{e}_1 + b p_2 \hat{e}_2 \end{equation}

Now in order for $\vec{q}$ to be a vector it must be that expressed in the prime basis, it has the form

\begin{equation} \vec{q} = \left(a p_1 \cos \theta + b p_2 \sin \theta\right)\hat{e}_1' + \left(- a p_1 \sin \theta + b p_2 \cos \theta \right)\hat{e}_2' \end{equation}

So far there is no problem for any value of $a$ or $b$. In a fixed basis, the components of a vector can have any value they want, so I might as well just call the components of $\vec{q}$ in the unprimed basis $a p_1$ and $b p_2$. It's just a strange looking parameterization of an arbitrary vector in the unprimed basis. So long as, after a rotation, $\vec{q}$ has the form above, then everything is fine.

I think the point that Zee is trying to get at with his question can be phrased like this: In the prime basis, \begin{equation} \vec{q} \neq a p_1' \hat{e}_1' + bp_2' \hat{e}_2' \end{equation} unless $a=b$.

In other words, even though in the original frame it is true that $q_1 = a p_1$ and $q_2= b p_2$, in the rotated frame it is certainly not true that $q_1' = a p_1'$ and $q_2' = b p_2'$.

Of course this is completely obvious it you just rotate $\vec{q}$, then you clearly see that $q_1' = ap_1 \cos \theta + b p_2 \sin \theta \neq a (p_1 \cos \theta + p_2 \sin \theta) = a p_1'$.

I think the larger point that Zee is trying to get at is that you should only trust vector relationships, not relationships between components: $q_1 = a p_1$ is not a "good" relationship because it is only true in a specific frame.

The only exception to this is if $a=b$; in other words, if it turns out that if all the components of two vectors are proportional in some frame, then you actually do have a vector statement $\vec{q}=a\vec{p}$ which will then be true in any frame. This fact is used frequently in GR--to show two tensors are equal you only need to show they are equal in some frame, so you often choose the simplest possible frame in which to evaluate the tensors.

However in my opinion, this is a bad question (or at least poorly phrased). There is nothing wrong with $\vec{q}$ being a vector for any values of $a$ and $b$, the real problem is that $q_1=ap_1$ is not a covariant statement, and that was pretty unclear from the original question.

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