7
$\begingroup$

I am an engineer trying to understand the mathematical definitions and physical significance of vectors, dual vectors, and dual spaces.

I understand how we take dot products and the covector "eats" the vector and produces a number. But those are numbers, not ideas. And they are just columns and rows.

I understand that some people say "a row is a covector and a column is a vector," but a column and a row should have nothing to do with ideas or physics.

I'd like to see an explanation of the form: "Work is a scalar. We get Work by the dot product of force and a displacement. Force is a one form. And the duality of covector/vector is related to force/displacement.

But how does one transform between force and displacement the way I see people transforming from vector to covector? Kinetic energy is a scalar, right? Mathematicians say that understanding vectors and covectors reveals a lot about the underlying algebra, but I just don't see any revelation here.

Trying to interpret what I read leads into seemingly contradictory language: I get kinetic energy by dotting the velocity vector into a velocity vector. So which velocity is the vector and which is the co-vector? I know that in a Cartesian space, the metric is the identity matrix and the distinctions are lost, but, still I just can't see the distinction in a way that makes sense to me.

Could someone please explain what a dual space is?

Why is it worth my time?

What are the physical meanings of it all?


Editor's note: This substantial revision is an after-the-fact attempt to paraphrase the OP's question, retaining as much of the original wording as possible.

$\endgroup$
  • 1
    $\begingroup$ Just a reminder: I acknowlege that my frustration is the covector of my ignorance. So please forgive me, but I want to express it that way to reveal my frustration. $\endgroup$ – thomas Oct 28 '15 at 14:32
  • 2
    $\begingroup$ This is the first rant I've seen on this site. Impressive. $\endgroup$ – PepperSausage Oct 28 '15 at 14:39
  • 1
    $\begingroup$ Have a look at the wiki site on this: en.wikipedia.org/wiki/Dual_space And related questions: math.stackexchange.com/questions/3749/… math.stackexchange.com/questions/170481/… And some notes: math.ubc.ca/~feldman/m425/dual.pdf $\endgroup$ – PepperSausage Oct 28 '15 at 14:43
  • 1
    $\begingroup$ For a discussion of force and work in the DG context see Hubbard&Hubbard's "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach" $\endgroup$ – rych Oct 29 '15 at 10:12
  • 2
    $\begingroup$ Another book you might like is Bamberg/Sternberg's A Course in Mathematics for Students of Physics. $\endgroup$ – Ted Shifrin Oct 30 '15 at 20:40
7
$\begingroup$

$\newcommand{\dd}{\partial}$In the sense that you're asking, velocity of a point particle is a vector, and any object that "measures a velocity and returns a scalar" is a covector.

Though there's a meaningful mathematical distinction between a vector space $V$ and its dual space $V^{*}$, it's probably easier to see the distinction between a vector-valued function and a covector-valued function, because (as lists of component functions) they transform differently under change of coordinates.

For simplicity, assume we're in a plane region, equipped with coordinates $(x_{1}, x_{2})$. Rather than writing vectors as column matrices and covectors as row matrices, introduce the coordinate vector fields $$ \frac{\dd}{\dd x_{1}},\qquad \frac{\dd}{\dd x_{2}}, \tag{1a} $$ and the coordinate one-forms $$ dx_{1},\qquad dx_{2}. \tag{1b} $$

A smooth vector field is a "linear combination" of the coordinate fields with smooth functions coefficients, e.g. $$ X = X^{1}\, \frac{\dd}{\dd x_{1}} + X^{2}\, \frac{\dd}{\dd x_{2}}. \tag{1c} $$ A smooth covector field (or smooth one-form) is a "linear combination" of the coordinate one-forms with smooth functions coefficients, e.g. $$ \xi = \xi_{1}\, dx_{1} + \xi_{2}\, dx_{2}. \tag{1d} $$

If $(y_{1}, y_{2})$ is another coordinate system, the chain rule gives \begin{align*} \frac{\dd}{\dd x_{1}} &= \frac{\dd y_{1}}{\dd x_{1}}\, \frac{\dd}{\dd y_{1}} + \frac{\dd y_{2}}{\dd x_{1}}\, \frac{\dd}{\dd y_{2}}, & \frac{\dd}{\dd x_{2}} &= \frac{\dd y_{1}}{\dd x_{2}}\, \frac{\dd}{\dd y_{1}} + \frac{\dd y_{2}}{\dd x_{2}}\, \frac{\dd}{\dd y_{2}}; \tag{2a} \\ dx_{1} &= \frac{\dd x_{1}}{\dd y_{1}}\, dy_{1} + \frac{\dd x_{1}}{\dd y_{2}}\, dy_{2}, & dx_{2} &= \frac{\dd x_{2}}{\dd y_{1}}\, dy_{1} + \frac{\dd x_{2}}{\dd y_{2}}\, dy_{2}. \tag{2b} \end{align*}

Writing $$ X = Y^{1}\, \frac{\dd}{\dd y_{1}} + Y^{2}\, \frac{\dd}{\dd y_{2}}, $$ substituting (2a) into (1c), and equating components gives the transformation rule for vector fields: $$ Y^{1} = X^{1}\, \frac{\dd y_{1}}{\dd x_{1}} + X^{2}\, \frac{\dd y_{1}}{\dd x_{2}},\quad Y^{2} = X^{1}\, \frac{\dd y_{2}}{\dd x_{1}} + X^{2}\, \frac{\dd y_{2}}{\dd x_{2}}. \tag{3a} $$

Similarly, writing $$ \xi = \eta_{1}\, dy_{1} + \eta_{2}\, dy_{2}, $$ substituting (2b) into (1d), and equating components gives the transformation rule for one-forms: $$ \eta_{1} = \xi_{1}\, \frac{\dd x_{1}}{\dd y_{1}} + \xi_{2}\, \frac{\dd x_{2}}{\dd y_{1}},\qquad \eta_{2} = \xi_{1}\, \frac{\dd x_{1}}{\dd y_{2}} + \xi_{2}\, \frac{\dd x_{2}}{\dd y_{2}}. \tag{3b} $$

Classically, a vector field is a collection of functions associated to a coordinate system that transform like (3a), and a one-form is a collection of functions that transforms like (3b). And that, in a sense, is the distinction between vectors and covectors.

Admittedly, the coordinate vector field and coordinate one-form notation is merely a formalism, just like column and row matrices. The tie-in with the chain rule, however, may make this formalism appealing, even compelling.

The connection with matrix notation comes from expressing (3a) and (3b) in terms of matrix products. For brevity, write $$ a_{j}^{i} = \frac{\dd y_{i}}{\dd x_{j}},\qquad b_{j}^{i} = \frac{\dd x_{i}}{\dd y_{j}}, $$ so that the matrices $$ A = \left[\begin{array}{@{}cc@{}} a_{1}^{1} & a_{2}^{1} \\ a_{1}^{2} & a_{2}^{2} \\ \end{array}\right] = \frac{\dd(y_{1}, y_{2})}{\dd(x_{1}, x_{2})} $$ and $$ B = \left[\begin{array}{@{}cc@{}} b_{1}^{1} & b_{2}^{1} \\ b_{1}^{2} & b_{2}^{2} \\ \end{array}\right] = \frac{\dd(x_{1}, x_{2})}{\dd(y_{1}, y_{2})}, $$ the Jacobian of the change of coordinates from $x$ to $y$ and the Jacobian from $y$ to $x$, are inverses.

If we identify $$ \frac{\dd}{\dd x_{1}} = \left[\begin{array}{@{}c@{}} 1 \\ 0 \\ \end{array}\right],\qquad \frac{\dd}{\dd x_{2}} = \left[\begin{array}{@{}c@{}} 0 \\ 1 \\ \end{array}\right],\qquad X = \left[\begin{array}{@{}c@{}} X^{1} \\ X^{2} \\ \end{array}\right], $$ and $$ dx_{1} = \left[\begin{array}{@{}cc@{}} 1 & 0 \\ \end{array}\right],\qquad dx_{2} = \left[\begin{array}{@{}cc@{}} 0 & 1 \\ \end{array}\right],\qquad \xi = \left[\begin{array}{@{}cc@{}} \xi_{1} & \xi_{2} \\ \end{array}\right], $$ then (3a) and (3b) read, respectively, $$ \left[\begin{array}{@{}c@{}} Y^{1} \\ Y^{2} \\ \end{array}\right] = A \left[\begin{array}{@{}c@{}} X^{1} \\ X^{2} \\ \end{array}\right] $$ and $$ \left[\begin{array}{@{}cc@{}} \eta_{1} & \eta_{2} \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} \xi_{1} & \xi_{2} \\ \end{array}\right] B. $$

Particularly, $$ \xi(X) = \xi_{1} X^{1} + \xi_{2} X^{2} = \eta_{1} Y^{1} + \eta_{2} Y^{2}; $$ the pairing between a one-form and a vector field is independent of the coordinate system, and consequently represents a "physical" or "geometric" concept.

$\endgroup$
  • 2
    $\begingroup$ The author of the question is long gone, was only seen on October 28. Please feel free to edit their question so that it makes sense and matches your answer. As is, it looks like a rant and may well get deleted. $\endgroup$ – user147263 Nov 8 '15 at 2:38
0
$\begingroup$

I'd like to answer the following part of the question:

I get kinetic energy by dotting the velocity vector into a velocity vector. So which velocity is the vector and which is the co-vector?

Rather than thinking of it as the dot product of two velocity vectors, think of it as momentum covector and velocity vector.

The momentum, $\mathbf{p} = m\mathbf{v}$, is a covector that takes the velocity vector $\mathbf{v}$ and gives double the kinetic energy $\mathbf{p}^T\mathbf{v}$ = $m\mathbf{v}^T\mathbf{v}$.

This derivation of kinetic energy might also shed some light on the role of momentum as a covector, and where the factor $\frac{1}{2}$ comes from.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.