How to find vector magnetic potential given magnetic field?

[yungman]

If we are given B, how can we find A? I can fine the magnitude of A by:

\int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}

So given B and the surface area, you can get the magnitude of A.

But how do you get the direction information? All I know is A is orthogonal to B and \vec B = \nabla X \vec A. But still I cannot find a formula to nail down the direction.

Anyone can help?

 

[Meir Achuz]

It's a bit like long division. There is no general result. For a specific B, you try to find a vector A whose curl is that B. If you can write B in a coordinate system, then you can write partial differential equations whose solution give A.
You can also use divB=-del^2 A and solve that equation for A.

 

[RedX]

If we are given B, how can we find A? I can fine the magnitude of A by:

\int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}

So given B and the surface area, you can get the magnitude of A.

But how do you get the direction information? All I know is A is orthogonal to B and \vec B = \nabla X \vec A. But still I cannot find a formula to nail down the direction.

Anyone can help?

How is it true that A is orthogonal to B?

Also, I don't see how:

\int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}

gets you the magnitude of A.

 

[RedX]

It's a bit like long division. There is no general result. For a specific B, you try to find a vector A whose curl is that B. If you can write B in a coordinate system, then you can write partial differential equations whose solution give A.
You can also use divB=-del^2 A and solve that equation for A.

divB=0 is one of the Maxwell equations.

 

[yungman]

How is it true that A is orthogonal to B?

Also, I don't see how:

\int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}

gets you the magnitude of A.

The curl of a vector is always orthogonal to the original vector. eg.

\hat x \;X\; \hat y = \hat z

\Phi = \int_{s'} \vec B \cdot d\vec{s'} \;\Rightarrow \;\int_{C} \vec A \cdot d\vec{l}= \Phi

So If you are given the B and given the surface that the flux cut through. You get the A within that surface.

 

[RedX]

The curl of a vector is always orthogonal to the original vector. eg.

\hat x \;X\; \hat y = \hat z

\Phi = \int_{s'} \vec B \cdot d\vec{s'} \;\Rightarrow \;\int_{C} \vec A \cdot d\vec{l}= \Phi

So If you are given the B and given the surface that the flux cut through. You get the A within that surface.

The vector field (y,0,1) has a curl in the z-direction, but is not perpendicular to the z-direction.

As for your second equation, you can get the line integral of A around the perimeter of the surface, but I don't think you can get A at a particular point on the perimeter, or at the center point of the surface.

addendum: here's a way to visualize why you can't get A at a particular point on the perimeter. This is a tri-force taken from a popular classic video game:

http://www.google.com/imgres?imgurl...=U3F4TbqWO4qcgQftqO3TBQ&sqi=2&ved=0CB0Q9QEwAA

You know the line integral of the middle triangle. But you want the field on the left side of that middle triangle. So how do you isolate just that part? Well you know the line integral of the leftmost triangle, and the rightmost part of the leftmost triangle is what you're looking for. So by considering the leftmost triangle, you do get an extra equation involving the left side of the middle triangle, but unfortunately this equation causes you to have two more sides (the remaining sides of the leftmost triangle), so in general you don't have enough information to get A at all. I hope this is right - someone correct me if I'm wrong!

 

[Meir Achuz]

divB=0 is one of the Maxwell equations.

I think he meant curl B.

 

[Troels]

If we are given B, how can we find A?

The simplest way to do it, for a general B, is via a PDE problem:

\nabla^2 \vec A =-\mu_0\vec J

whilst remembering that

\nabla\times \vec B =\mu_0\vec J

This, togther with suitable boundary conditions, (eg. specifing B on the boundary) should give you the result you are looking for. Please note, that an analytical solution may not be obtainable.

The "Amperes Law for A" method;

\int_S \vec B\cdot d\vec a = \oint_{\partial S} \vec A\cdot d\vec l

That you suggested in your OP works only if you have some suitable symmetry in the problem (ie A and dl are parallel at all points along the integration curve). The only case I can think of off the top of my head is the magnetic potential inside an infinitely long solenoid

 

[yungman]

The vector field (y,0,1) has a curl in the z-direction, but is not perpendicular to the z-direction.

My bad, I keep thinking about A X B is always orthogonal to A or B.

I am still reading the materials in the book.

 

[yungman]

The simplest way to do it, for a general B, is via a PDE problem:

\nabla^2 \vec A =-\mu_0\vec J

whilst remembering that

\nabla\times \vec B =\mu_0\vec J

This, togther with suitable boundary conditions, (eg. specifing B on the boundary) should give you the result you are looking for. Please note, that an analytical solution may not be obtainable.

The "Amperes Law for A" method;

\int_S \vec B\cdot d\vec a = \oint_{\partial S} \vec A\cdot d\vec l

That you suggested in your OP works only if you have some suitable symmetry in the problem (ie A and dl are parallel at all points along the integration curve). The only case I can think of off the top of my head is the magnetic potential inside an infinitely long solenoid

So

\int_S \vec B\cdot d\vec a = \oint_{C} \vec A\cdot d\vec l

can only be true with certain symetry like a long wire carrying current or a long solenoid?

Yes this is from a problem of a long solenoid.

 

[yungman]

I read Griffiths page 234 in Magnetic Vector Potential in STATIC condition where \nabla X \vec B = \mu_0 \vec J.

\nabla X \vec B= \nabla X \nabla X \vec A = \nabla(\nabla \cdot \vec A ) - \nabla^2 \vac A = \mu_0 \vec J

We invoke gauge where \nabla \cdot \vec A = 0 [/tex]<br /> <br /> \Rightarrow\; \nabla^2 \vec A =-\mu_0 \vec J<br /> <br /> Does this imply in all cases of static condition, <b>A</b> and <b>J</b> are in the same direction in all static case?<br /> <br /> <br /> <br /> <br /> On the other note, for time varying condition where:<br /> <br /> \nabla X \vec B = \mu_0 \vec J + \mu_0 \epsilon_0 \frac {\partial {\vec E}}{ \partial t}<br /> <br /> <b>A</b> and <b>J</b> are not in the same direction.<br /> <br /> <br /> Am I correct?

 

[Meir Achuz]

Del^2A need not be in the same direction as A.

 

[yungman]

Del^2A need not be in the same direction as A.

I skipped a step to show that

\nabla^2 \vec A = -\mu_0 \vec J \;\Rightarrow\; \vec A(\vec r) = \frac {\mu_0}{4\pi}\int_{v&#039;} \frac {\vec J(\vec r)}{|\vec r|} dv&#039;

And

\vec A = \frac {\mu_0}{4\pi}\oint_{C&#039;} \frac {\vec I}{|\vec r|} dl&#039;


That shows A is same direction of J. Is this right?

 

[Troels]

So

\int_S \vec B\cdot d\vec a = \oint_{C} \vec A\cdot d\vec l

can only be true with certain symetry like a long wire carrying current or a long solenoid?

Don't get me wrong. The equation - like amperes law - is always true, but - likewise like amperes law - it is only useful in finding the magnetude of A along the integration curve - you have to dream up the direction from a symmetry argument.

To address your other question, no. In general you cannot assume that A and J are parallel at all points, but this is where symmetry comes in. In case of a a solenoid, you have current running in the \hat \phi-direction only and thus the PDE becomes:
\nabla^2 A_r = 0
\nabla^2 A_\phi = J_{\phi}
\nabla^2 A_z = 0

A trial solution of A_r=A_z=0 satisfies the first and last equation and thus, by the uniqueness theorem, are the solutions, thus A has only a phi-component, in this case.

 

[Meir Achuz]

I skipped a step to show that

\nabla^2 \vec A = -\mu_0 \vec J \;\Rightarrow\; \vec A(\vec r) = \frac {\mu_0}{4\pi}\int_{v&#039;} \frac {\vec J(\vec r)}{|\vec r|} dv&#039;

And

\vec A = \frac {\mu_0}{4\pi}\oint_{C&#039;} \frac {\vec I}{|\vec r|} dl&#039;


That shows A is same direction of J. Is this right?

Still NO.

 

[yungman]

Still NO.

Can you explain? There is only one vector on the left and only one on the right. How can they not be in the same direction. In fact it is common to pull the unit vector out and put them in front of the equation.

\vec A = \frac {\mu_0}{4\pi}\oint_{C&#039;} \frac {\vec I}{|\vec r|} dl&#039;

\Rightarrow\; (\hat A)\; A \;=\; (\hat I)\;[\frac {\mu_0}{4\pi}\oint_{C&#039;} \frac {I}{|\vec r|} dl&#039;]

\Rightarrow\;\hat A =\hat I

 

[kcdodd]

Lookup Helmholtz decomposition for vector fields, which I think is what you're tiptoeing around.

http://en.wikipedia.org/wiki/Helmholtz_decomposition

It will give you the vector potential from the field, or field from the potential.

 

[Meir Achuz]

The unit vector I CAN NOT be pulled out of the integral.
It can, and commonly does, vary in direction during the integration.

 

[yungman]

The unit vector I CAN NOT be pulled out of the integral.
It can, and commonly does, vary in direction during the integration.

I was thinking about this and I started another post to concentrate on this alone. Can yhou answer that one?

 

[Meir Achuz]

Can you explain? There is only one vector on the left and only one on the right. How can they not be in the same direction. In fact it is common to pull the unit vector out and put them in front of the equation.

\vec A = \frac {\mu_0}{4\pi}\oint_{C&#039;} \frac {\vec I}{|\vec r|} dl&#039;

\Rightarrow\; (\hat A)\; A \;=\; (\hat I)\;[\frac {\mu_0}{4\pi}\oint_{C&#039;} \frac {I}{|\vec r|} dl&#039;]

\Rightarrow\;\hat A =\hat I

The integral is of I around a closed circuit. It should be obvious, without any algebra, that I has to change direction to go around a closed path. Try walking in a circle without changing direction.

 

[yungman]

Lookup Helmholtz decomposition for vector fields, which I think is what you're tiptoeing around.

http://en.wikipedia.org/wiki/Helmholtz_decomposition

It will give you the vector potential from the field, or field from the potential.

I looked at the Helmholtz that decompose any vector into the sum of irrotational and solenoidal vectors. But I don't see the relation of this to my question.

I think I understand why the differentiation or the integration of a vector is not necessary the same direction. as in this case, the integration of vector J/B] is not the same direction as J/B] so I cannot count on A/B] is the same direction as J/B].

Thanks for your time.

 

[RedX]

I looked at the Helmholtz that decompose any vector into the sum of irrotational and solenoidal vectors. But I don't see the relation of this to my question.

The second equation in kcdodd's link gives an explicit formula for the potential in terms of the magnetic field (just replace F with B in the equation for A). You know the magnetic field then a nasty looking equation gives you the potential.

 

[Troels]

\vec A = \frac {\mu_0}{4\pi}\oint_{C&#039;} \frac {\vec I}{|\vec r|} dl&#039;

It's important to note that integral is not a universial solution to the poisson equation for A. Most importantly, the current distrubiton has to be localized, among a few other restrictions

YOu are almost always off to safer grounds by starting directly from the poisson equation and deduce what you can from that. See my previous post for the example of the infinite solenoid

 

[mishuo]

The answer to your question is that you cannot get a unique form of a vector potential from given magnetic field B. The the set of vector potential to a corresponding set of their magnetic field is not one-to-one but multiple-to-one.

So normally we have to choose a special form that is simple to calculate, or we can choose a gauge form (e.g. Lorentz gauge, Coulomb gauge, etc ).

You can refer to some books of classical electrodynamics. Here I suggest you read "Mathematic methods for Physicist", 3rd edition (International Edition), written by George Arfken, page 69-72


By the way, the anti-question of yours, i.e. given a vector potential a magnetic field B is determined, is true. In this way, you can simply use B=\nabla\times A, by making partial derivative over vector A, you can get the unique form a B

 

[vanhees71]

I've given an explicit answer, using a "spacial gauge" in the following thread

https://www.physicsforums.com/showthread.php?t=531128&highlight=vector+potential

An alternative is to use the Debye potentials, which unfortunately is not treated in textbooks. A nice introduction can be found here

C. Gray, B. Nickel, Debye potential representation of vector fields, American Journal of Physics 46, 735 (1978).
http://dx.doi.org/10.1119/1.11111

For those, who understand German, here's the link to my "Mathematical Methods for Physicists" manuscript:

http://fias.uni-frankfurt.de/~hees/publ/maphy.pdf

where some examples for their application are given (multipole expansion for magneto static and dynamical em. fields).