# problem with differential forms -example

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• May 6th 2011, 02:07 AM
rayman
problem with differential forms -example
Hello! My teacher told me that the description of this example is not 100% correct. Here is the example
http://i51.tinypic.com/34dihx5.jpg

but my teacher says that f is not a 2-form, it is a Hodge * -dual to 1-form and when we write that f is the rotation than it should mean that f=dw which is not true in this case.

Does anyone know who is right? and where is the eventual error?

thanks
• May 7th 2011, 05:41 PM
ojones
Not sure what your instructor's point is here. $f$ is a 2-form and the flux is given by integrating this over the surface.
• May 7th 2011, 05:51 PM
TheEmptySet
I am not 100% sure what you are asking but consider the zero form

$g(x,y,z)=\frac{-E}{4\pi}(x^2+y^2+z^2)^{-\frac{1}{2}}$

Now if D acts on g we get

$Dg=\frac{E}{4\pi}\left( \frac{x}{(x^2+y^2+z^2)^\frac{3}{2}}dx + \frac{y}{(x^2+y^2+z^2)^\frac{3}{2}}dy + \frac{z}{(x^2+y^2+z^2)^\frac{3}{2}}dz\right)$

Now if we take its Hodge dual we get

$*Dg=\frac{E}{4\pi}\left( \frac{x}{(x^2+y^2+z^2)^\frac{3}{2}}(dy \wedge dz) + \frac{y}{(x^2+y^2+z^2)^\frac{3}{2}}(dz \wedge dx) + \frac{z}{(x^2+y^2+z^2)^\frac{3}{2}}(dx \wedge dy)\right)$

So

$f = *Dg$
• May 11th 2011, 10:28 PM
rayman

By the way the author of these exaples wrote to me some comments about these examples

''Both examples give a form f where df = 0, but f != dw for any w. That is why they are representatives of a non-trivial cohomology class.

The "solar-flux" 2-form f is not a vector field, right. The Hodge dual of f is the vector field q = E/(4*pi*r^2) r^ , where r^ is the unit radial vector. This vector field q is like the picture of "sunbeam vectors" in the slides.

The second example is just showing a vector field (which is a 1-form) v where the curl of v is zero but v != grad phi for any scalar function phi. If v were the gradient of a potential phi, then the line integral of v around any path should vanish. If the path doesn't "go around" the hole, then the line integral does vanish, since curl(v)=0, but if the path goes around the hole then the integral is non-zero. Therefore there is no such potential phi''

Does it makes sense to you? I have had a bit hard to understand his point
• May 11th 2011, 11:16 PM
ojones
A 2-form is a form, not a vector. Also, the Hodge-star operator maps forms to forms and so you don't get a vector from it.

I'm not quite clear what the question is here. You originally said your instructor thought there was something wrong with the example. Why don't you ask him/her to clarify what it is. The original post didn't really make sense.
• May 12th 2011, 01:59 AM
rayman
I have just talked to him. Yes it is a 2-form. My teacher says:

1) f is not a rotation of a vector potential in M
2) if we write that f i a rotation then we mean that 2-form f is a differential of a 1-form w i.e $f=dw$ but this is wrong

here is the plan he suggested to correctly solve this problem
1) find the Hodge dual to f
2) compute df
3) is f exact?
4) is f closed?
5) compute $d^+f$ (I do not know what he means with this one)
6) integrate f over a sphere
7) conclusions? f is a harmonical non-trivial 2-form.

I would appreciate if someone could help me with these steps.
Finding a Hodge dual is to operate with the operator * on f, right?
$*f=\frac{E}{4\pi}(\frac{x}{(x^2+y^2+z^2)^{3/2}}dy\wedge dz+\frac{y}{(x^2+y^2+z^2)^{3/2}}dz\wedge dx+\frac{x}{(x^2+y^2+z^2)^{3/2}}dx\wedge dy)= \frac{E}{4\pi}}(\frac{x}{(x^2+y^2+z^2)^{3/2}}dx+ \frac{y}{(x^2+y^2+z^2)^{3/2}}dy+\frac{z}{(x^2+y^2+z^2)^{3/2}}dz$ so we have showed that our 2-form is represented now by the dual 1-form.
this can be written in a shorter way
$*f=\frac{E}{\pi}\frac{\hat{r}}{r^2}$

2) find df , should we simply differentiate f with respect to x,y,z?
• May 12th 2011, 11:41 PM
ojones
OK, I think I understand what's going on. Your teacher is correct in that the 2-form in the example can't be exact. This is because if it were, the flux through any closed surface would have to be zero. However, we know from Gauss's Law that the flux through any closed surface containing the origin must be E (compare your field with the electric field generated by a point charge at the origin). The exampe is not wrong either. The author doesn't claim that a vector potential for the field exists globally. He considers small open balls (which presumably don't include the origin). He states clearly that these potentials can't be patched together to give a global one.

I'll need to think a little bit more about his suggestion as to how to show f is not exact.
• May 12th 2011, 11:53 PM
rayman
Quote:

Originally Posted by ojones
OK, I think I understand what's going on. Your teacher is correct in that the 2-form in the example can't be exact. This is because if it were, the flux through any closed surface would have to be zero. However, we know from Gauss's Law that the flux through any closed surface containing the origin must be E (compare your field with the electric field generated by a point charge at the origin). The exampe is not wrong either. The author doesn't claim that a vector potential for the field exists globally. He considers small open balls (which presumably don't include the origin). He states clearly that these potentials can't be patched together to give a global one.

I'll need to think a little bit more about his suggestion as to how to show f is not exact.

Yes now I understand that too. I have been trying to do some more calculations but I am getting nowhere.

Refering to the differential forms
if $df=0$ then it is a closed form
if $f=dw$ then it is exact
if f is an exact form then it is closed. If we modulate our topological manifold (we suppose that we can shrink it to a point- Poincare theorem) then we can even say that every closed form is also exact, but that is not a case in this problem

I found also a formula for calculating df for our 2-form
$df=(\frac{\partial f_{yz}}{\partial x}+\frac{\partial f_{zx}}{\partial y}+\frac{\partial f_{xy}}{\partial z})dx\wedge dy\wedge dz$ and I managed to calculate it and I get zero, so our 2-form is a closed form.
My teacher said that this form is not exact but I have no idea how to show it.
Second thing how do we inegrate r-forms over some manifold (the spehere in our case)?
• May 13th 2011, 02:04 PM
ojones
I don't know what you mean by $f_{yz}$, ... etc. These should be $\frac{E}{4\pi}\frac{x}{r^3}$, .... But anyway, it's the right approach to show $df=0$.

The easiest way to show the form is not exact is to show that the flux though the sphere of radius 1 isn't zero.

Your teacher seems to be making heavy weather of this and I still don't see his ultimate point. There's no need to introduce the Hodge dual or do any of that other stuff. Also, the example isn't wrong! The form is locally exact which is all he claimed.

The integral of the 2-form will be the same as $\int_S\mathbf{F}\cdot \mathbf{n}\, dS$ where $\mathbf{F}$ is the associated vector field.
• May 13th 2011, 11:25 PM
rayman
the formula comes from the book I use, and finally df gives 0 like it should be. I guess my teacher wants me to do all this calculations so I can get some practise with differential forms and see how they behave and what opperations can be done with them.

I have been thinking about this integral and I am sure I need to use Gauss-Ostrogradski theorem, which is more less what you have mentioned. I will struggle with it today, but I am guessing that if our divergence of the field gave zero than the integral should also give zero....but this definitely has to be checked.
• May 14th 2011, 03:45 PM
ojones
By $f_{yz}$ do you mean $\partial ^2 f/\partial z\partial y$ or the $dy\wedge dz$ component of the 2-form? If it's the latter, then we're in agreement.

You can't use Gauss's theorem in this case for surfaces that include the origin because the field is not defined there. In any event, you don't need it. The surface integral is a trivial calculation for radial fields.
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