Inverse Rotation Function Question

[TheDestroyer]

I've asked in the calculus and analysis this and dextercioboy (thanked) has gave a not specific (or i was not able to understand him) for the particular solution of the following equation,

If B was a known Vector defined as:

B=rot(A)

How can i get A?

As I know, we will get a system of partial differential equations, I want some one to give me the way to get the general solution of that system, and please to be specific,

And Thanks

 

[mathwonk]

lets see, "rot" was maxwells notation in the 19th century for what we usually call the "curl", so you have a differential equation of form curl(A) = B, and you want to solve it.

as i recall, curl, which also is more modernly known as exterior derivative of a one form,

would be d(pdx + qdy) =
(round d q/(round d x - (round d p/(round d y)dx^dy, i hope. so you have some function H and an equation of form

(round d q/round d x - round d p/round d y) = H and you want to solve for p and q.


well of course the curl of any gradient is zero, so this solution is not unique by a long shot.

it seems there are many solutions to this, and you can probably pick p or q, to be almost anything and solve for the other.

sorry for not more details. i am not too motivated by these calculations.

this is sophomore differential equations. so any book, even a calculus book should help.

but at least i hope i got you started and shared the correct terminology so someone better versed in diff eq can take over.

 

[Dr Transport]

$$\vec{A} = \pm\frac{\vec{B} \times \vec{r}}{2}$$ I forget the correct sign...check it yourself.

 

[TheDestroyer]

Sorry, Mr. Mathwonk, I didn't understand, do you mind using tex script?

 

[Galileo]

$$\vec B = \nabla \times \vec A$$
is very general. I don't know how to find the general solution.
In any case, the divergence of B is zero since the divergence of a curl (or rot) is always zero.
A vector field is uniquely determined by its divergence, curl and boundary conditions (Helmholtz theorem), so we only need the rotation, but it's pretty complicated:

$$\nabla \times \vec B = \nabla \times (\nabla \times \vec A)=\nabla(\nabla \cdot \vec A)-\nabla^2 \vec A$$

If you add the requirement that $\nabla \cdot \vec A=0$, then you can solve it:

$$\vec A = \frac{1}{4\pi}\int \frac{\vec B \times \vec r}{r^2}d^3\vec r$$.

 

[TheDestroyer]

LOL, How can i solve the integration with $$d^3 \vec{r}$$?

 

[dextercioby]

It's nothing funny.Galileo's notation is rather sloppy.For a comprehensive treatment you may want to check the web for "Biot-Savart-Laplace",or a treatise on electromagnetism...

The integration is generally on R^{3}.

Daniel.

 

[TheDestroyer]

I've studied them, in my first semester, but the way our university is teaching is very bad, can you believe they teach us mathematical formulas in physics while we don't know how to proof it, and while we don't know what it means ! math and physics has no relations in our study, and i know this is wrong, ... what can i do, I'm asking here in this forum to learn more, i can't do anything else !

For example, Vector potential is defined as:

$$\vec{B} = \vec{curl}\vec{A}$$

And the solution is:

$$A_{x} = \frac{\mu_{0}}{4\pi}\int\int\int_{V} \frac{j_{x}}{r}dV$$

(I don't remember if it's like that, i forgot because i don't know the proof)

But why, no one knows, this way of teaching is killing me !

 

[TheDestroyer]

And What if Rot B = 0??

 

[dextercioby]

We can't teach you the way you'd probably want to.That's why universities have been created.You can go to the library and pick a decent book,as Griffiths,or Jackson.There's a lotta explaining behind the mathematical fomulas you've just written...We can't do that,sorry.

Daniel.

 

[dextercioby]

And What if Rot B = 0??

Then it would be a simple Laplace equation:
$$\Delta\vec{A}=\vec{0}$$
,which can be integrated only under certain conditions,as a typical boundary problem may ask.

Daniel.

 

[TheDestroyer]

Our university library is very old, and doesn't have the required books, believe it?

 

[dextercioby]

Yes...I'm sorry to hear that,that is really bad...

Daniel.