Differential forms & Maxwell equations

[Jay-qu]

Well why dont you give it a try, I am interested.

[Ben]

Ummm, let's see.

 

Recall I defined a space of p-vectors. This space arises as a result of "choosing", from a set of [math]n[/math] basis vectors, [math]p[/math] at a time, and calling them the basis for a new space called [math]\Lambda^p(V_n)[/math]. The eagle-eyed will notice these guys are not too dissimilar from tensors.

 

Now these p-vectors come equipped with a unique operation called the exterior (or wedge) product which is defined as follows:

 

[math]\wedge: \Lambda^p(V_n) \times \Lambda^q(V_n) \to \Lambda^{p+q}(V_n)[/math] where I do not insist that [math]p \ne q[/math].

 

So that, if say, [math]\alpha,\,\,\beta[/math] are 1-vectors, then [math]\alpha \wedge \beta = \gamma[/math] is a 2-vector.

 

This operation on 1-vectors has the property that [math]\alpha \wedge \beta = -\beta \wedge \alpha[/math] ; notice this mandates that [math]\alpha \wedge \alpha = 0[/math]. This doesn't quite generalize:

 

First, from the foregoing, it should be obvious that any p-vector can be "built" by the wedge product of p 1-vectors. If p is even, one says that our p-vector is of even degree, of odd degree otherwise. So we take as a definition that when [math]p,\,\,q[/math] are odd, then for [math]\alpha \in \Lambda^p (V_n),\,\, \beta \in \Lambda^q(V_n)[/math] that [math]\alpha \wedge \beta = - \beta \wedge \alpha[/math]. If [math]p,\,\,q [/math] is even, then [math]\alpha \wedge \beta = \beta \wedge \alpha[/math], and if just one of these guys is of even degree, then this last equality holds.

 

So for any [math] \alpha \in \Lambda^p(V_n),\,\,\beta \in \Lambda^q(V_n)[/math] one writes [math] \alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha[/math] (this follows from the elementary arithmetic fact that the product of odd numbers is odd, the product of an even number with any other number, odd or even, is even and [math]-1^2 = 1,\,\, -1^3 = -1[/math] and so on.

 

So finally we understand our notation; the existence of products implies the existence of powers, so [math]\Lambda^p(V_n)[/math] is the p-th exterior power on [math]V_n[/math]

[Jay-qu]

Ok I follow thus far. Where to next

[Qfwfq]

I see what you are saying, but I dont see the link to Maxwell's equations.

If you want to see the direction in which he is heading, I'll cheat :evil: and say that [imath]A_{\mu}dx^{\mu}[/imath] is a differential form and that [imath]A_{\mu}[/imath] is a pure gauge if this form is exact (equal to the gradient of something). So, is it closed? Is the domain simply connected?

 

Knowing Ben, of course, he's gotta formalize it to the maximum possible extent. :)

[Ben]

Ya well, someone has to play the pedant, and, as it is a game I play rather well, it may as well be me!

 

Anyway, so far I have been talking about a general class of objects called p-vectors. I now want to specialize somewhat with what might, at first sight, seem like a ghastly grinding of gears. So.....

 

Suppose that [math]f(x,y,z)[/math] is a function in 3 variables on an open subset of [math]R^3[/math]. Further let [math]x,\,y,\,z[/math] be continuously parametized by some [math]t[/math]. Then the chain rule from calculus gives me [math]\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{d x}{d t}+ \frac{\partial f}{\partial y}\frac{d y}{d t}+ \frac{\partial f}{\partial z}\frac{d z}{d t} [/math].

 

Then, independent of [math]t[/math], with a flourishing hand-wave, I "cancel" the [math]d t[/math] to get [math]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}d y + \frac{\partial f}{\partial z}d z[/math] which our calculus text calls the total differential of the function [math]f[/math]. (Don't worry, it's just for illustration!). We will call this a differential 1-form, or simply a 1-form.

 

This construction tells us at least three interesting things, which I will come to soon.

 

For now, we will look in our calculus text and there find a vector field defined as [math] F = P\vec{i}+Q \vec{j}+R \vec{k}[/math] where the P, Q, R are each functions in 3 variables, and the vectors are unit vectors (i.e. orthonormal basis vectors) on [math]R^3[/math]. The one-to-one correspondence with the 1-form above is obvious, so I feel justified in calling [math]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}d y + \frac{\partial f}{\partial z}d z[/math] both a differential 1-form AND a 1-vector. (Ungrinding of gears, I hope!)

 

Moreover, this correspondence suggests the following: since the terms [math]\frac{\partial f}{\partial x^i}[/math] are clearly scalar, then I must insist that the terms [math]dx^i[/math] are basis vectors, and we have the general form [math] \omega =\sum\nolimits_i \alpha_i dx^i[/math].

 

Now every time we see a differential given as a basis vector, we think "ah, differentiable manifolds". We don't need to know too much about what these are, beyond that they look locally like some [math]R^n[/math]. Obviously [math]R^3[/math] (our current playground) is trivially a manifold.

 

Manifolds also admit of 2 kinds of vector space - the tangent space and the co-tangent space.

 

Since [math]R^3[/math] is "flat", then these spaces can simply be identified with [math]R^3[/math] itself (or a subset thereof).

 

Moreover, we know (for reasons too lengthy to describe) that the tangent space has as basis vectors the differential operators [math] \frac{\partial}{\partial x_i}[/math], then we will say that our 1-forms are elements in a different space called the co-tangent space.

 

Which we all knew already, right?

 

Post edited as kindly pointed out by Guadalupe

[Guadalupe]

 

Suppose that [math]f(x,y,z)[/math] is a function in 3 variables on an open subset of [math]R^3[/math]. Further let [math]x,\,y,\,z[/math] be continuously parametized by some [math]t[/math]. Then the chain rule from calculus gives me [math]\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{d x}{d t}+ \frac{\partial f}{\partial y}\frac{d y}{d t}+ \frac{\partial f}{\partial z}\frac{d z}{d t} [/math].

 

Then, independent of [math]t[/math], with a flourishing hand-wave, I "cancel" the [math]d t[/math] to get [math]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}d y + \frac{\partial f}{\partial x}d z[/math] which our calculus text calls the total differential of the function [math]f[/math]. (Don't worry, it's just for illustration!). We will call this a differential 1-form, or simply a 1-form.

 

This construction tells us at least three interesting things, which I will come to soon.

 

For now, we will look in our calculus text and there find a vector field defined as [math] F = P\vec{i}+Q \vec{j}+R \vec{k}[/math] where the P, Q, R are each functions in 3 variables, and the vectors are unit vectors (i.e. orthonormal basis vectors) on [math]R^3[/math]. The one-to-one correspondence with the 1-form above is obvious, so I feel justified in calling [math]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}d y + \frac{\partial f}{\partial x}d z[/math] both a differential 1-form AND a 1-vector. (Ungrinding of gears, I hope!)

 

Which we all knew already, right?

 

 

 

Hi! Ben :)

 

I highlighted in red the denominator in question. Isn’t it supposed to be z instead of x?

 

 

 

 

 

:cup:

[Ben]

Eek! Though your highlighting doesn't show on my browser, I see what you mean. Sorry - post edited with notes

[Jay-qu]

Yeah I did learn some of this a few years ago, but I am not familiar with the language of differential forms so it is nice to see you present it here.

[Ben]

Anyway, recall first I said that what algebraists call the total differential of the function [math]f(,x,y,z)[/math] we were going to call a 1-form.

 

I will be talking exclusively about Euclidean 3-space with the usual metric.

 

Recall that [math]f[/math] is defined to be a 0-form. This suggests the operator [math]d[/math] maps 0-forms onto 1-forms. In fact this generalizes; this operator - called the exterior derivative - maps p-forms onto (p+1)-forms, as I will shortly show.

 

First, though, a caveat. While the above is certainly true, it is NOT true that every q-form can be written as the exterior derivative of some (q-1)-form. Where it can, one says our q-form is exact.

 

Let's say that [math]f[/math] is an arbitrary 0-form. Then we earlier had [math]df=\frac{\partial f}{\partial x}dx +....+ \frac{\partial f}{\partial z}dz[/math] as our 1-form. Recasting this as [math]df=(\sum\nolimits\frac{\partial}{\partial x}_i \,\,dx_i) f[/math] reveals the correspondence [math] df \simeq \nabla f [/math]. In other words, the exact 1-form here is the gradient "vector" written in covector form.

 

So what about the exterior derivative applied to a 1-form? The result will certainly be a 2-form, and probably a differential operator. I will get to this when I have more time (2 -3 days); meanwhile, take stab........

[Jay-qu]

Thank you for getting me up to speed with the notation. What do you mean 'take stab'?

[Guadalupe]

Thank you for getting me up to speed with the notation. What do you mean 'take stab'?

 

 

 

Hi! Jay-qu :eek2:

 

I believe our friend Ben was saying, “Have a stab at" or "Give it a stab” in other words he’s giving us the go ahead. :)

 

 

 

:eek2:

[Ben]

Yes, that was what I had meant, but it was rather silly of me, as I hadn't given enough information on how the exterior derivative works.

 

Let's do that now. Suppose that [math]f,\,g,\,h[/math] are each functions in the 3 variables [math]x,\,y,\,z[/math].

 

Write the 1-form [math]f\,dx+g\,dy+g\,dz[/math]. I may evaluate the exterior derivative [math]d(f\,dx+g\,dy+g\,dz)[/math] by observing that, for any p-forms [math]\alpha,\,\beta[/math] that

 

[math]d(\alpha +\beta) = d\alpha + d\beta[/math] (linearity)

 

[math]d(\alpha \wedge \beta) = d\alpha \wedge \beta +(-1)^p \alpha\wedge d \beta[/math] (Law of Leibniz) Woops-a-daisy, there is an edit here

 

[math]d(dx)=d(dy)=d(dz)=0[/math] (Lemma of Poincare).

 

For ease of notation let me set [math]f_x\equiv \frac{\partial f}{\partial x}[/math] etc.

 

We will find that for the 1-form above [math]d(f\,dx+g\,dy+h\,dz)=df \wedge dx+dg \wedge dy+ dh \wedge dz[/math]

 

[math]= (f_x dx+f_y\, dy+f_z dz)\wedge dx+...+(h_x dx+h_y dy+h_z dz)\wedge dz[/math].

 

Working through the first term in parentheses for illustration we get

 

[math]f_x dx \wedge dx + f_y dy \wedge dx+f_z dz \wedge dx[/math]

 

By the rules above, I have that [math]dx \wedge dx = 0[/math] and that [math]dy \wedge dx = -(dx \wedge dy)[/math].

 

Putting this together and gathering terms and rationalizing according to their "wedges", I arrive at

 

[math] (g_x-f_y)dx \wedge dy + (h_y -g_z)dy\wedge dz + (f_z -h_x)dz \wedge dx[/math].

 

This is just the covariant form of the curl, which, in our notation, is a 2-form, as expected.

 

It may come as no surprise that the exterior derivative of a 2-form is a 3-form, and that this will be be the covariant version of the divergence.

 

Now we can see where Maxwell comes in, but first I want to describe another operator. Later for that

[Ben]

Ah well,I seem to be boring you. Maybe this will perk your interest, as it is just so much fun.

 

Recall that, at the get-go, I informally described the vector space [math]\Lambda^p(V_n)[/math] as that which arises from the choice of p basis vectors from the n-basis for [math]V_n[/math]. Let's call this "choice space".

 

I define an operator [math]\ast: \Lambda^p(V_n) \to \Lambda^{n-p}(V_n)[/math] that maps "choice space" onto "remainder space". So that if I have [math]x \in \Lambda^1(V_3)[/math] as an element in my choice space, I will have [math]\ast(x)=(y,z) \in \Lambda^{3-1}(V_3)[/math] as an element in remainder space.

 

Likewise, if [math](x,y) \in \Lambda^2(V_3)[/math], then [math]\ast(x,y)=z \in \Lambda^{3-2}(V_3)[/math]

 

Well, that's a very loose way of putting it, but it will do for now. We are going to look for the covariant form of the Laplacian which, as is well known, is a second-order differential operator that maps scalar or vector fields onto scalar or vector fields, respectively. It is often written as [math]\nabla^2[/math], though some care is needing in interpreting the exponent here (perhaps a better notation is [math]\Delta[/math], but I'm old-fashioned).

 

Let's start with the arbitrary 0-form [math]f(x,y,z)[/math] in Euclidean 3-space (usual metric). Then we know that the exterior derivative [math]df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial

y}dy +\frac{\partial f}{\partial z}dz[/math].

 

We may think of this as "multiplying" our 0-form [math]f[/math] by the elements [math]\frac{\partial}{\partial x^i}dx^i[/math]

 

Then [math]\ast df =\frac{\partial f}{\partial x}dy\wedge dz+\frac{\partial f}{\partial y}dz \wedge dx+\frac{\partial f}{\partial z}dx \wedge dy[/math]. Obviously this is a (3 - 1 = 2)-form.

 

Applying the exterior derivative again I find a (2 + 1 =3)-form [math]d \ast df[/math], which since we are simply "multiplying trough by the [math]\frac{\partial}{\partial x^i}[/math] gives me

 

[math]d\ast df=(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2})dx\wedge dy\wedge dz= (\nabla^2 f)dx\wedge dy \wedge dz[/math], obviously a 3-form.

 

Then a final application of Hodge yields the (3 - 3 =0)-form [math]\ast d\ast df= \nabla^2 f[/math].

 

Now how cool is that?

[Ben]

Hey, you guys are no fun! I happen to think this stuff is fun - obviously you don't agree, so I shall bore you no further on this subject.

 

Ho hum..... anyway, I have given you all the tools you need to express Maxwell's equations in terms of differential forms.

 

Good luck with that!

 

PS. Or they can be found in any half-decent text

[Jay-qu]

Sorry Ben, I do find this fun and interesting on a deep level, but I feel my math is not strong enough..

 

I am aware that the magnetic field is actually not a vector but a 1-form, but I dont get how this changes Maxwell's equations.

[Qfwfq]

Ben, I think folks find the effort a bit heavy and would need a bit more clarity. Despite having seen this type of machinery back years ago, it takes me a bit of doing to follow you and get it straight. I really don't think I could have figured out:

This is just the covariant form of the curl, which, in our notation, is a 2-form, as expected.

 

It may come as no surprise that the exterior derivative of a 2-form is a 3-form, and that this will be be the covariant version of the divergence.

if I hadn't already been through certain courses back then, so I imagine how much harder it must be for many others here.

 

If I only had more time I could be expanding a bit, to make it easier for all.

[Ben]

Yeah. While I strongly dispute Jay's claim that his math isn't strong enough, I do take your point.

 

In fact, I may make the general point that this site is, after all, just a glorified chat-room - we come here for fun. It just so happens that, sad bastard that I am, this sort of math IS how I get my fun. But I ought not to have assumed that others are equally twisted.

 

Anyway, on a general note: Without casting doubt on the knowledge and ability of any poster here, I think that if any of our many lurkers come here looking for help with course-work (say), they are taking a huge risk - who is to say that I, for one, have any clue what I am talking about?