Undergrad Basis Vectors & Inner Product: A No-Nonsense Introduction

[kent davidge]

I read from this page https://properphysics.wordpress.com...se-introduction-to-special-relativity-part-6/

that the basis vectors are the canonical basis vectors in any coordinate system. This seems to be wrong, because if that was the case the metric would be the identity matrix in any coordinate system, and we know that, for example, in spherical coordinates the metric has components $\{1, r^2, r^2 \sin^2 \varphi \}$. So should I conclude that what is said on that page is wrong?

On the other hand, what is said on there seems to be more consistent with the inner product properties, because then a vector $V = Ae_1 + Be_2$ would have norm $\sqrt{A^2 + B^2}$ in any coordinate system.

So I'm not sure what conclusion I should draw from that page.

 

[sweet springs]

It seems this lecture deals with special relativity. You can take it like Cartesian coordinates. In general relativity curvilinear coordinates or no orthogonal coordinate are used. You will learn mathematics for such a complex geometry at that time.

 

[fresh_42]

There should have been an additional adjective: Cartesian or orthonormal and eventually finite-dimensional, too. Given any basis $v_1,v_2,\cdot ,v_n$ the coordinates of these vectors according to this basis, i.e. according to themselves, are always $(1,0,\ldots ,0)\; , \;(0,1,0,\ldots ,0)\; , \; etc.$.

A metric is something else. You can have any positive definite, symmetric bilinear form to define a inner (scalar) product for a metric and a norm, or define a metric otherwise. The standard inner product is simply the one, that belongs to the identity matrix as bilinear form.

 

[stevendaryl]

I read from this page https://properphysics.wordpress.com...se-introduction-to-special-relativity-part-6/

that the basis vectors are the canonical basis vectors in any coordinate system. This seems to be wrong, because if that was the case the metric would be the identity matrix in any coordinate system, and we know that, for example, in spherical coordinates the metric has components $\{1, r^2, r^2 \sin^2 \varphi \}$. So should I conclude that what is said on that page is wrong?

On the other hand, what is said on there seems to be more consistent with the inner product properties, because then a vector $V = Ae_1 + Be_2$ would have norm $\sqrt{A^2 + B^2}$ in any coordinate system.

So I'm not sure what conclusion I should draw from that page.

People mean different things by the basis vectors. If you have a path through space $\mathcal{P}(s)$ as a function of a path parameter, $s$, then the corresponding tangent vector is just: $V = \frac{d \mathcal{P}}{ds}$. If you have a coordinate system $x^\mu$, then the most straight-forward way to define the "components" of $V$ are using the coordinates: $V^\mu \equiv \frac{dx^\mu}{ds}$. That is, $V^\mu$ is just the rate of change in the coordinate $x^\mu$ as you move along the path $\mathcal{P}(s)$.

So let's take the example of 2D polar coordinates $r, \theta$. Then the components of velocity are given by: $V^r = \frac{dr}{ds}$ and $V^\theta = \frac{d\theta}{ds}$. These are related to the rectangular coordinates $(x,y)$ through:

$V^x = \frac{\partial x}{\partial r} V^r + \frac{\partial x}{\partial \theta} V^\theta$
$V^y = \frac{\partial y}{\partial r} V^r + \frac{\partial y}{\partial \theta} V^\theta$

Using $x = r\ cos(\theta)$ and $y = r\ sin(\theta)$, this means:

$V^x = cos(\theta) V^r - r sin(\theta) V^\theta$
$V^y = sin(\theta) V^r + r cos(\theta) V^\theta$

Now, we have: $|V|^2 = (V^x)^2 + (V^y)^2 = cos^2(\theta) (V^r)^2 - 2 r sin(\theta) cos(\theta) V^r V^\theta + r^2 sin^2(\theta) (V^\theta)^2 + sin^2(\theta) (V^r)^2 + 2 r sin(\theta) cos(\theta) V^r V^\theta + r^2 cos^2(\theta) (V^\theta)^2 = (V^r)^2 + r^2 (V^\theta)^2$

So if you want $|V|^2$ to be independent of what coordinate system you are using, then you have to use the metric tensor:

$|V|^2 = \sum_{\alpha \beta} g_{\alpha \beta} V^\alpha V^\beta$

In cartesian coordinates: $g_{xx} = g_{yy} = 1$. In polar coordinates, $g_{rr} = 1$ but $g_{\theta \theta} = r^2$

In terms of basis vectors, this means choosing $e_r = cos(\theta) e_x + sin(\theta) e_y$ and $e_\theta = - r sin(\theta) e_x + r cos(\theta) e_y$. In this basis, the magnitudes of the basis vectors are not 1. $|e_\theta| = r$.

Now, in a lot of introductory books about vector calculus, they try to keep the metric tensor as the simple form $g_{rr} = g_{\theta \theta} = 1$. To do this, they have to absorb the $r^2$ into the definition of $V^\theta$: They use $\tilde{V}^\theta \equiv r V^\theta$. This means choosing different basis vectors:

$\hat{r} \equiv e_r$
$\hat{\theta} \equiv \frac{1}{r} e_\theta$

This is a non-coordinate basis, in that $\tilde{V}^\theta$ is unequal to $\frac{d\theta}{ds}$.

 

[kent davidge]

People mean different things by the basis vectors. If you have a path through space $\mathcal{P}(s)$ as a function of a path parameter, $s$, [...]

Hmm, interesting. So we can build all the vectors of our theories (special and general relativity) in terms of curves through out appropriate manifolds? Does this mean we can associate, say, a four-momentum, or a four-velocity, to a path in some manifold? i.e. not just "position in space and time" but also any other vector quantity?

 

[stevendaryl]

Hmm, interesting. So we can build all the vectors of our theories (special and general relativity) in terms of curves through out appropriate manifolds? Does that mean we can associate, say, the four-momentum, or a four-velocity, to a path in some manifold?

Given a parametrized path through the manifold, you have an associated vector, namely the "velocity" (but with position viewed as a function of the parameter, not time). But a vector doesn't uniquely define a path, though. If you have a vector field $V(\mathcal{P})$, that is, a vector at every point in space, then a vector field plus a starting location $\mathcal{P}_0$ uniquely determines a path: There is only one path $\mathcal{P}(s)$ satisfying $\mathcal{P}(0) = \mathcal{P}_0$ and $\frac{d\mathcal{P}}{ds} = V(\mathcal{P}(s))$

 

[martinbn]

As stevendaryl says different curves can give rise to the same tangent vector, but you can define tangent vectors at a point as equivalent classes of curves.

 

[kent davidge]

As stevendaryl says different curves can give rise to the same tangent vector, but you can define tangent vectors at a point as equivalent classes of curves.

Oh yea, I have read about that

 

[fresh_42]

Oh yea, I have read about that

I hope here: https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/