Inverse Rotation Function Question

In summary: The theorem of Helmholtz says that if a vector field satisfies properties of continuity and differentiability on an open domain from R^n,then the equations:\nabla\cdot\vec{A}=f(\vec{r}) \nabla\times\vec{A}=\vec{g}(\vec{r}) +bondary conditions will uniquely determine it...That's it.
  • #1
TheDestroyer
402
1
Hi guys, I'm a student in the physics college, and do need to ask math people a difficult question, and hope you can give me the answer as SIMPLE as you can,

if we have a known vector B defined as B = rot(A)

(In physics, B is a magnetic field, A is a Vector Voltage of the field)

How can we get A? I know this will create a 3x3 system of a partial differential equation, and the answer will have a gradient for scalar function as a constant for integration, I've studied differential equations, and parial differential equations, but i didn't study yet the systems of partial differential equations

if anyone would answer, please try giving me the answer step by step after what I studied, and thanks

TheDestroyer
 
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  • #2
Do how we physicists do when solving Maxwell's equations.Search for a particular solution,namely one which verifies:
[tex] \nabla\cdot\vec{A}=0 [/tex]

Then:
[tex] \nabla\times\vec{B}=\nabla\times\nabla\times\vec{A}=\nabla (\nabla\cdot\vec{A})-\Delta\vec{A}=-\Delta\vec{A} [/tex]

,which can be projected onto a system of independent axis and solved each scalar equation like a regular Poisson one.

In the general case,it's very difficult to integrate that system.


Daniel.
 
  • #3
I'm very sorry, I don't know maxwells equations, would you give me a push with an example :)?
 
  • #4
For a magnetostatic field in vacuum,Maxwell's equations are:

[tex] \nabla\times\vec{B}=\vec{j} [/tex] (*)

[tex] \nabla\cdot\vec{B}=0 [/tex]

,which are solved by the method of potential as i examplified above...

Daniel.

------------------------------------------------------
(*)-Heviside-Lorentz units used.
 
  • #5
AAAAAAAAAAAAAAAAAAAAAAAAAh things is going more complicated, I'm just a second year boy !

can you give me solution to this as an example then?
B = B(x/r^3, y/r^3, z/r^3)

r=(x^2+y^2+z^2)^(1/2)

and sorry for my heavy understanding mind
 
  • #6
You've given a scalar and you've asked a solution to a vector equation...

Give a vector field.

Daniel.
 
  • #7
Oh god ! sorry if i wasn't clear

B(x,y,z) = x/r^3 (i) + y/r^3 (j) + z/r^3 (k)

(i),(j),(k) unit vectors for axes x,y,z

r=(x^2+y^2+z^2)^(1/2)
 
  • #8
The solution to your equation:

[tex] \frac{\vec{r}}{r^{3}}=\nabla\times\vec{A} [/tex]

is obtained,if you come up with certain conditions that the vector field A must obey.As you have figured out,the general solution is:

[tex] \vec{A}=\vec{B}+\nabla f(\vec{r}) [/tex]

,where "f" is a solution of the Laplace equation...

The particular solution (B) is found by specifying the divergence of "A"...
That way,we can use the theorem of Helmholtz...

Daniel.
 
Last edited:
  • #9
Daniel, Pleeeease, Spread a little more, I'm not english, I can't understand you, you gave me the solution, please give me step by step !

and thank you very much
 
  • #10
The theorem of Helmholtz says that if a vector field [itex] \vec{A} [/itex] satisfies properties of continuity and differentiability on an open domain from R^n,then the equations:

[tex] \nabla\cdot\vec{A}=f(\vec{r}) [/tex]

[tex] \nabla\times\vec{A}=\vec{g}(\vec{r}) [/tex]

+bondary conditions will uniquely determine it...That's it.

Daniel.
 
  • #11
What are bondary conditions? and what's f,g?
 
  • #12
f & g are functions given.They are so-called nonhomogeneity functions.And for a complete treatment of boundary conditions in PDE-s,i'll infer you to a book on PDE-s...

Daniel.
 
  • #13
TheDestroyer said:
(In physics, B is a magnetic field, A is a Vector Voltage of the field)

And one more thing:A is VECTOR MAGNETIC POTENTIAL and its SI-mKgs unit is not "volt".

Daniel.
 
  • #14
Great? Can you guide me to a pdf partial differential equations books?
 
  • #15
I told you man I'm not english, it's just a damn translation lol, thanks for the info.
 
  • #16
TheDestroyer said:
Great? Can you guide me to a pdf partial differential equations books?

I don't know of any.Maybe someone else does.I'm sure u'll find a ton at the closest university library,though...

Daniel.
 
  • #17
(where "f" is a solution of the Laplace equation...

The particular solution (B) is found by specifying the divergence of "A"...
That way,we can use the theorem of Helmholtz...

Daniel.)

Can you tell me which laplace equation? and how did you get A to have it's divergence? And tell me theorem of helmholtz step by step please, 1 then 2 then 3 then 4 and so...

and thanks...
 

FAQ: Inverse Rotation Function Question

What is an inverse rotation function?

An inverse rotation function is a mathematical function that reverses the direction of a rotation. It takes in a rotated object or coordinate and outputs the original object or coordinate before the rotation took place.

Why is the inverse rotation function important?

The inverse rotation function is important because it allows us to undo a rotation and return to the original orientation of an object or coordinate. This is especially useful in applications such as computer graphics, robotics, and navigation.

How is the inverse rotation function calculated?

The inverse rotation function is calculated by taking the inverse of the rotation matrix or by using the inverse trigonometric functions (such as arctan or arccos) on the rotation angle.

Can the inverse rotation function be used for any type of rotation?

Yes, the inverse rotation function can be used for any type of rotation, including rotations in 2D or 3D space. It can also be used for rotations in any direction, such as clockwise or counterclockwise.

Are there any limitations to the inverse rotation function?

One limitation of the inverse rotation function is that it assumes a perfect rotation with no translation or scaling. Additionally, the inverse rotation function may not be well-defined for certain types of rotations, such as reflections or skew transformations.

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