# The size of a vector

## Summary:

How to calculate the size of a vector, confusion with basis vectors.

## Main Question or Discussion Point

I'm stumbling on something rather basic here, will explain with an example. (Pardon the LaTeX problems, trying to fix..)

Suppose I have a plane, and in the plane I put the familiar (x,y) Cartesian coordinate system, and the metric is the usual Euclidean metric with $ds^2 = dx ^2 + dy^2$.

Now suppose I add into this another coordinate system defined by

u=x+2y
v=x-y

and so it follows that

x=(1/3)(u + 2v)
y=(1/3)(u-v).

The basis vectors for u and v are
$\vec{e}_u = < \frac {\partial x}{\partial u},\frac {\partial y}{\partial u} > ,= < 1/3 , 1/3 >$
$\vec{e}_v = < \frac {\partial x}{\partial v},\frac {\partial y}{\partial v} > ,= < 2/3 , -1/3 >$
and the corresponding covectors are

$\bar{e}^u = < \frac {\partial u}{\partial x},\frac {\partial u}{\partial y} > ,= < 1 , 2 >$
$\bar{e}^v = < \frac {\partial v}{\partial x},\frac {\partial v}{\partial y} > ,= < 1 , -1 >$.

The inner products of the basis vectors and covectors are
$\vec{e}_u \cdot \bar{e}^u = < 1/3 , 1/3 > \cdot < 1 , 2 > =1$
$\vec{e}_v \cdot \bar{e}^v = < 2/3 , -1/3 > \cdot < 1 , -1 > =1$

. . as one would expect.

My confusion is this: Isn't the magnitude of a vector equal to the square root of its inner product with itself?

$\vec {v} \cdot \vec{v} = v^a v^b g_{ab} = v^a (v^b g_{ab}) = v^a v_a$

This would imply that $\vec{e}_u$, for example, has a magnitude of 1, when clearly it's the square root of $(1/3)^2 + (1/3)^2 = 2/9$.

Last edited:

Related Linear and Abstract Algebra News on Phys.org
fresh_42
Mentor
The inner product changes, too, or better: its calculation changes! You cannot turn to ##(u,v)## coordinates and still compute angles and lengths in ##(x,y)## coordinates. Angles, lengths and the quadratic form might be invariant, its matrix is not.

I should add that I thought this rule was good for a tensor of any rank — just use index gymnastics to create the "opposite" tensor and stick them together.

$$R^{a}_{bcd} R ^{bcd}_{a} =$$
the square of the size, or magnitude of the fourth rank tensor $\bf{R}$, although perhaps in cases like this there is no clear meaning to this quantity.

So would you say that the size of $\vec {e}_u$ is 1 or the square root of 2/9? I'm guessing the latter since, as you say, vector lengths must be invariant.

Yes, I'm mixing up (x,y) and (u,v) here - since I guess in (u,v) $\vec{e}_u$ is simply (1,0). Perhaps that's the root of my confusion.

fresh_42
Mentor
What you have is ##(x,y) \longmapsto x^\tau Q y##. So we have for ##v=Sx, w=Sy##
$$x^\tau Q y= v^\tau (S^{-1})^\tau Q S^{-1}w = v^\tau \left( \underbrace{(S^{-1})^\tau Q S^{-1}}_{=Q'} \right)w$$
and the matrix changed from ##Q## to ##Q'##.

Thanks fresh_42, that's fascinating. Will see if I can compute the inner products properly in (u,v)..

DrClaude
Mentor
@Orodruin can probably help here.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
My confusion is this: Isn't the magnitude of a vector equal to the square root of its inner product with itself?

$\vec {v} \cdot \vec{v} = v^a v^b g_{ab} = v^a (v^b g_{ab}) = v^a v_a$

This would imply that $\vec{e}_u$, for example, has a magnitude of 1, when clearly it's the square root of $(1/3)^2 + (1/3)^2 = 2/9$.
It is ##\sqrt{2/9}##. ##\vec e_u \cdot \vec e^u## is not the inner product of ##\vec e_u## with itself, it is the inner product between ##\vec e_u## and ##\vec e^u##, which are different vectors.

I should add, what led me here was a YouTube video of a man deriving the Christoffel symbols for the surface of a sphere. He took the inner products of the basis vector derivatives with the basis covectors, and I've been trying to understand why he bothered to make sure they were covectors. Will post a link to the video when I find it..

(In that case there was again the complication of using two coordinate systems simultaneously - the phi and theta for the sphere and the (x,y,z) of the Cartesian 3-space.)

Orodruin
Staff Emeritus
I agree. My interest was to show by example that the covariant derivative of a metric tensor is zero, even in a non-flat continuum. So I wanted the Christoffel symbols for the surface of a sphere. I know one can produce them using that formula with the partial derivatives of the components of the metric, but of course that assumes $\nabla g = 0$ in the first place.