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Added an explicit answer to the OPs question on Lengths, changed example of scalar function from $x^2$ to $e^x$ for readability, fixed some primes but others might be wrong still.

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edited Mar 31, 2019 at 20:02

jacob1729

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$f'(x'(p)) = f(x(p)) = f( \Lambda^{-1}(x(p))$$f'(x'(p)) = f(x(p)) = f( \Lambda^{-1}(x'(p))$

Consider the scalar function $f(x)=x^2$$f(x)=e^x$ and the vector function $\vec{v}(x)=x^2\vec{e}$$\vec{v}(x)=e^x\hat{x}$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $x\mapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=\frac{1}{a^2}x'^2$$f'(x')=e^{x'/a}$.

The vector function transforms to become $\vec{v}'(x') = \frac{1}{a^2}x'^2 \vec{e}$$\vec{v}'(x') = e^{x'/a} \hat{x}$. However as noted above we must also transform the basis vector and the rule amounts to $\vec{e} = a \vec{e}'$$\hat{x} = a \hat{x}'$ in this case. So the new vector function is $\vec{v}'(x') = \frac{1}{a}\vec{e}'$$\vec{v}'(x') = a e^{x'/a} \hat{x'}$.

The case of length

For your specific example of length, we can just consider the effect of a scaling of the coordinates. If we scale our coordinates by a factor of $2$ (i.e. measure in units corresponding to twice as large as previously) then the number representing the length halves. This is consistent with the scalar transformation law for the quantity $L(x)=x$ rather than the vector transformation law for $\vec{L}=x\hat{x}$ which would be different.

*The notations $\hat{x}$ and $\hat{x}'$ mean the basis vector appropriate for the $x$ and $x'$ coordinate systems respectively. I'm not sure how to simply explain why scaling affects the natural basis vector to use, but it does.

$f'(x'(p)) = f(x(p)) = f( \Lambda^{-1}(x(p))$

Consider the scalar function $f(x)=x^2$ and the vector function $\vec{v}(x)=x^2\vec{e}$. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $x\mapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=\frac{1}{a^2}x'^2$.

The vector function transforms to become $\vec{v}'(x') = \frac{1}{a^2}x'^2 \vec{e}$. However as noted above we must also transform the basis vector and the rule amounts to $\vec{e} = a \vec{e}'$ in this case. So the new vector function is $\vec{v}'(x') = \frac{1}{a}\vec{e}'$

$f'(x'(p)) = f(x(p)) = f( \Lambda^{-1}(x'(p))$

Consider the scalar function $f(x)=e^x$ and the vector function $\vec{v}(x)=e^x\hat{x}$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $x\mapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=e^{x'/a}$.

The vector function transforms to become $\vec{v}'(x') = e^{x'/a} \hat{x}$. However as noted above we must also transform the basis vector and the rule amounts to $\hat{x} = a \hat{x}'$ in this case. So the new vector function is $\vec{v}'(x') = a e^{x'/a} \hat{x'}$.

The case of length

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created Mar 31, 2019 at 16:19

jacob1729

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I can understand why you would think they are the same, but even in the 1D case I think the answer is: no they are not.

You might think that a scalar function is that that outputs a single number, whilst a vector function outputs a single vector - but a vector in 1D can be described by only one number and so they must be the same! In fact, no this is not true. You might have heard that the difference is that vectors transform in such a way that "their components mix together", but again it's 1D so maybe that can't happen. In fact, the 'mixing' terminology comes from rotations, but if we focus on scaling our coordinates then the difference becomes clearer. The rest of this answer is an explicit example, along with what we mean by a scalar and vector field.

Scalar field

We want to consider a space $M$ consisting of points $p$ that we give coordinates $x(p)$ to. Note that it's very important that we don't confuse the coordinates we give points, and the points themselves! The coordinates are numbers, whilst points are physical points in space. This distinction is often lost, but it really is key here. We can always pick a different set of coordinates $x'(p)$ but the underlying point $p$ is the same. A scalar function is any function that assigns to each $p\in M$ real number:

$f:p\mapsto f(p)$.

However, we often abuse notation and write the function as a function of the coordinates:

$f:x\mapsto f(x)$ and $f':x'\mapsto f'(x')$

the issue that whilst these are easier functions to work with, you have different functions for each coordinate system. The consistency requirement is that they at least have the same value at the same value of $p$:

$f'(x'(p)) = f(x(p)) = f( \Lambda^{-1}(x(p))$

where $\Lambda$ is the function that changes coordinates i.e. $x\mapsto x' = \Lambda (x)$.

Vector Field

There are two definitions of these that are equivalent - the physicist's and the differential geometer's. We'll only give the physicists as the other is a little abstract (it has the advantage of being shorter though). A physicist's vector field picks out a vector space $V_p$ at each point $p\in M$ of the same dimension as $M$ (so 1 in our case). If then maps each point $p$ in $M$ to a vector in the vector space at that point. Finally, it also needs to have a transformation rule $v^i\mapsto v^{i'} = J^{i'}_{i} v^i$ on the individual components of and vector $v\in V_p$. Here, $J$ is the Jacobian matrix of the transformation and changing the components of the vector can be thought of as adjusting for the fact that changing coordinates also changes the basis in your vector space.

Example