Vector potential ##\vec A## in terms of magnetic field ##\vec B##

  • #1
154
11

Homework Statement:

We know that the magnetic field ##\vec B## can be expressed as a curl of the vector potential ##\vec A##. Invert this equation to express the vector potential in terms of the magnetic field.

Relevant Equations:

We have the magnetic field : ##\vec B = \vec \nabla \times \vec A##.

We can also show that if the linear velocity ##\vec v## of a point on a rigid body with position velocity ##\vec r## can be expressed in terms of the angular velocity ##\vec \omega## as ##\vec v = \vec \omega \times \vec r##, then ##\vec \omega = \frac{1}{2} \vec \nabla \times \vec v##.
My solution is making an analogy of the ##\text{Relevant equations}## as shown above, starting from the equation ##\vec \omega = \frac{1}{2} \vec \nabla \times \vec v##.

We have ##\vec B = \vec \nabla \times \vec A = \frac{1}{2} \vec \nabla \times 2\vec A \Rightarrow 2\vec A = \vec B \times \vec r##, whereupon I make the analogy with velocity (##\vec v \rightarrow 2\vec A) ##and angular velocity (##\vec \omega \rightarrow \vec B)## vectors as stated above.

Hence we have ##\mathbf{\boxed{\vec A = \frac{1}{2} \vec B \times \vec r}}## as the desired equation.

Is this correct?
 
Last edited:

Answers and Replies

  • #2
strangerep
Science Advisor
3,130
952
Well, you should work out whether taking the curl of the ##A## in your boxed equation gives back ##B##. (Hint: use the so-called "BAC-CAB" rule for a triple cross product.)
 
  • #3
154
11
Yes, thank you... I actually had that in mind, except for the fact that the BAC-CAB rule for the cross product of two space dependent vectors is an involved one.

##\nabla \times \left( {{\bf{A}} \times {\bf{B}}} \right)= {\bf{B}} \cdot \nabla {\bf{A}} - \left( {\nabla \cdot {\bf{A}}} \right){\bf{B}} + \left( {\nabla \cdot {\bf{B}}} \right){\bf{A}} - {\bf{A}} \cdot \nabla {\bf{B}}##.

Using it and my boxed equation above, we get :

##\begin{align*}
\vec \nabla \times \vec A &= \frac{1}{2} \vec \nabla \times \left( \vec B \times \vec r \right)\\ &=\frac{1}{2}\left[(\vec r \cdot \vec \nabla) \vec B - (\xcancel{\vec \nabla \cdot \vec B}) \vec r + (\vec \nabla \cdot \vec r) \vec B - (\vec B \cdot \vec \nabla) \vec r \right]\\ &= \frac{1}{2} \left[ x_i \partial_i \left(B_j \hat e_j \right) + 3 B_j \hat e_j - B_i \partial_i \left(x_j \hat e_j \right) \right]
\\&= \frac{1}{2} \left[ x_i \partial_i \left(B_j \hat e_j \right) + 2 \vec B \right]
\end{align*}##.

I am having trouble with the first term : ## x_i \partial_i \left(B_j \hat e_j \right)##. However, if I kept ##\vec \nabla \cdot \vec B \neq 0##, I find that this term exactly cancels with the first one and I am left with ##\vec \nabla \times \vec A = \vec B##, as desired. However, we know from elsewhere (magnetism) that ##\vec \nabla \cdot \vec B = 0##!

Can you help?
 
  • #4
strangerep
Science Advisor
3,130
952
I am having trouble with the first term : ## x_i \partial_i \left(B_j \hat e_j \right)##. However, if I kept ##\vec \nabla \cdot \vec B \neq 0##, I find that this term exactly cancels with the first one
You're sure about that, are you? [Translation: check it more carefully.]

However, we know from elsewhere (magnetism) that ##\vec \nabla \cdot \vec B = 0##!
Yes. So what does that tell you? (I.e., the answer to your original question is staring you in the face.)
 
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Is ##\vec{B}=\text{const}##?
 
  • #6
154
11
Is ##\vec{B}=\text{const}##?
No, the magnetic field ##\vec B = \vec B(\vec r)##, a function of position.
 
  • #7
154
11
You're sure about that, are you? [Translation: check it more carefully.]


Yes. So what does that tell you? (I.e., the answer to your original question is staring you in the face.)
Let me summarise. I put my original question (condensed) in blue and my solution so far in red. Thank you for your patience.

I began by asking that since the magnetic field ##\vec B## can be written as a curl of a vector potential field ##\vec A## ##\boxed{\left(\vec \nabla \times \vec A \right)}##, is it correct to say that the vector potential can be written thus : ##\boxed{\vec A = \frac{1}{2} \vec B \times \vec r}##?

[I came upon this solution by making an analogy with the way velocity vector (##\vec v##) at a point of a rigid body (##\vec r##) is related to the angular velocity vector (##\vec \omega##). The two relations that I used as analogies to the two boxed equations above were : (1) ##\vec v = \omega \times \vec r## and (2) ##\omega = \frac{1}{2} \vec \nabla \times \vec v##].

One way to check whether my answer ##\left( \vec A = \frac{1}{2} \vec B \times \vec r \right)## is correct is to take the curl of both sides and see whether we recover the original equation ##\left(\vec B = \vec \nabla \times \vec A \right)##. I find that this is indeed the case provided in the calculation I don't assume that ##\vec \nabla \cdot \vec B = 0##. However, the relation ##\vec B = \vec \nabla \times \vec A## being correct implies that ##\vec \nabla \cdot \vec B = 0##.

This is my confusion at the moment. Is it that my original answer (from the analogy) of ##\vec A = \frac{1}{2} \vec B \times \vec r## is incorrect?
 
  • #8
strangerep
Science Advisor
3,130
952
This is my confusion at the moment. Is it that my original answer (from the analogy) of ##\vec A = \frac{1}{2} \vec B \times \vec r## is incorrect?
It's only correct if ##B## is constant, but incorrect for non-constant ##B##.
 
  • #9
154
11
It's only correct if ##B## is constant, but incorrect for non-constant ##B##.
Can you explain? In my derivation above, nowhere did I assume a constant ##\vec B## and yet I could derive ##\vec \nabla \times \vec A = \vec B##.
 
  • #10
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
From a totally different view point, the view point of Helmholtz decomposition theorem , specifying only the curl of A is not enough to uniquely determine A ,we must also know the divergence of A , as well as some vanishing conditions for A, in order to uniquely determine the vector field A.

Even if your solution is correct (which I don't think it is btw, I think it is correct only for B non spatially varying cause only in this case the "annoying" first term in your derivation vanishes) Helmholtz decomposition theorem tell us that is not the only solution that satisfies ##\nabla\times \mathbf{A}=\mathbf{B}##. There might be other solutions for A that have the same curl.
 
  • #11
strangerep
Science Advisor
3,130
952
In my derivation above, nowhere did I assume a constant ##\vec B## and yet I could derive ##\vec \nabla \times \vec A = \vec B##.
In the derivation in your post #3 you said you were "having trouble" with a term involving ##x^k \partial_k B^i##. In fact, there was nothing wrong with that term. Its presence just shows that if ##B## is non-constant then you don't get your original ##B## back again.
 
  • #12
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
Can you explain? In my derivation above, nowhere did I assume a constant ##\vec B## and yet I could derive ##\vec \nabla \times \vec A = \vec B##.
The term ##(\mathbf{r}\cdot\nabla)\mathbf{B}## does not get cancelled by the term ##(\nabla\cdot\mathbf{B})\mathbf{r}## even if ##\nabla\cdot\mathbf{B}\neq 0##. The first term contains "mixed" derivatives of the form $$\frac{\partial B_j\mathbf{e_j}}{\partial x_i}$$ which dont exist in the term ##(\nabla\cdot\mathbf{B})\mathbf{r}## and therefore cannot be cancelled.

I think your starting point involving ##\mathbf{\omega}## and ##\mathbf{v}## holds only if ##\mathbf{\omega}## is constant independent of the location (equivalently spatially non varying) as such is the case in a rotation of a rigid body where all points of the rigid body rotate with the same angular velocity.
 
Last edited:
  • #13
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
No, the magnetic field ##\vec B = \vec B(\vec r)##, a function of position.
Then it's clear a prioro that ##\vec{A}=\vec{r} \times \vec{B}## is not the correct solution! Just take the curl!

Hint: You can just solve the equation ##\vec{\nabla} \times \vec{A}=\vec{B}## component wise in Cartesian coordinates. Note that of course you must have ##\vec{\nabla} \cdot \vec{B}=0## and that ##\vec{A}## is only determined up to a gradient of a scalar field!
 
  • #14
154
11
Thank you all for your comments. The matter was fairly tricky and am relieved to have resolved it.
 
  • Like
Likes Delta2
  • #15
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Would you like to share your solution? I find it nice for other people who may encounter the same problem.

My idea is the following. I don't know, whether it's in any textbook, but it's very simple, so maybe the textbook authors don't find it worth to present it:

First of all the equation
$$\vec{\nabla} \times \vec{A}=\vec{B}$$
with ##\vec{B}## a given vector field implies the integrability constraint
$$\vec{\nabla} \cdot \vec{B}=0.$$
Further ##\vec{A}## is only determined up to a gradient of an arbitrary scalar field, and thus I've one free choice of a "gauge constraint condition". Here I choose the axial gauge condition
$$A_3=0.$$
Then the equation reads, split in Cartesian components,
$$\vec{\nabla} \times \begin{pmatrix} A_1 \\ A_2 \\ 0 \end{pmatrix} = \begin{pmatrix} -\partial_3 A_2 \\ \partial_3 A_1 \\ \partial_1 A_2-\partial_2 A_1 \end{pmatrix}.$$
Of course we use the first two components to find ##A_1## and ##A_2##. From the 2nd component we get
$$\partial_3 A_1=B_2 \; \Rightarrow \; A_1(\vec{x})=\int_0^{x_3} \mathrm{d} x_3' B_2(x_1,x_2,x_3')+\tilde{A}_1(x_1,x_2).$$
From the first component we get
$$-\partial_3 A_2=B_1 \; \Rightarrow \; A_2(\vec{x})=-\int_0^{x_3} \mathrm{d} x_3' B_1(x_1,x_2,x_3')+\tilde{A}_2(x_1,x_2).$$
To get also ##A_3## we note that the gauge constraint is not determining the vector potential completely. We can just regauge with an arbitrary scalar field ##\chi=\chi(x_1,x_2)## without destroying the condition ##A_3=0##. Thus we can additionally make
$$\tilde{A}_2=0.$$
Then the 3rd component reads
$$B_3(\vec{x})=\partial_1 A_2 - \partial_2 A_1 = -\int_0^{x_3} \mathrm{d} x_3' [\partial_1 B_1(x_1,x_2,x_3') + \partial_2 B_2(x_1,x_2,x_3') -\partial_2 \tilde{A}_1(x_1,x_2).$$
Because of ##\vec{\nabla} \cdot \vec{B}=0## this reads
$$B_3(\vec{x})=+\int_0^{x_3} \mathrm{d} x_3' \partial_3 B_3(x_1,x_2,x_3') - \partial_2 \tilde{A}_1(x_1,x_2) = B_3(x_1,x_2,x_3) - B_3(x_1,x_2,0) - \partial_2 \tilde{A}_1(x_1,x_2).$$
Thus finally setting
$$\partial_2 \tilde{A}_1(x_1,x_2)=-B_3(x_1,x_2,0) \; \Rightarrow \; \tilde{A}_1(x_1,x_2)=-\int_0^{x_1} \mathrm{d} x_1' B_3(x_1',x_2,0).$$
This shows that there's always a vector potential for ##\vec{B}## provided ##\vec{\nabla} \cdot \vec{B}=0##.
 
  • Like
Likes PhDeezNutz and etotheipi
  • #16
rude man
Homework Helper
Insights Author
Gold Member
7,786
767
Kind of late to be joining, but Feynman and many others derive the relationship
$$ \bf A = (1/(4\pi) \int \frac {\nabla \times \bf B} {r} dv $$

similar to the scalar potential expression.

based on ## \nabla \times \bf B = \nabla \times (\nabla \times \bf A) = -\nabla^2 \bf A ##
if ## \nabla \cdot \bf A = 0 ## is chosen which can be shown to be OK working back from the result..

The solution is the above integral. Don't know how that compares with the previous posters' work.
 
  • #17
strangerep
Science Advisor
3,130
952
Kind of late to be joining, but Feynman and many others derive the relationship
$$ \bf A = (1/(4\pi) \int \frac {\nabla \times \bf B} {r} dv $$The solution is the above integral. [...]
But what about the case ##\;{\bf B} = ## const?
 
  • #18
rude man
Homework Helper
Insights Author
Gold Member
7,786
767
But what about the case ##\;{\bf B} = ## const?
Any B field must be set up by a current.
You need that current-B field relationship (as given by Maxwell's ## \nabla \times \bf H = \bf j ##.
 
  • #19
rude man
Homework Helper
Insights Author
Gold Member
7,786
767
OK that was probably not the answer you were looking for.

If B = constant that is an infinite field and so ## \nabla \times \bf B = 0 ## and the A vector would be zero also. That is an impossible situation. I don't know.

Let's say your B field is a right circular cylinder ##B_0## of radius ##a## extending from - infinity to + infinity along the z axis. Then ## \nabla \times \bf B = -B_0~\delta(r-a) \hat{\bf \theta} ## in cylindrical coordinates. The integration would be difficult to set up IMO.
 
  • #20
rude man
Homework Helper
Insights Author
Gold Member
7,786
767
OK that was probably not the answer you were looking for.

If B = constant that is an infinite field and so ## \nabla \times \bf B = 0 ## and the A vector would be zero also. That is an impossible situation. I don't know.

Let's say your B field is a right circular cylinder ##B_0## of radius ##a## extending from - infinity to + infinity along the z axis. Then ## \nabla \times \bf B = -B_0~\delta(r-a) \hat{\bf \theta} ## in cylindrical coordinates. The integration would be difficult to set up IMO.
Maybe ## \bf A = (-B_0/4\pi)( 2\pi a) \int_{-\infty}^{+\infty} \frac {dz} {(a^2 + z^2)^{1/2}}. ##
Not sure about this. At least the units check. I do know that the A vector can be limitless but if you replace the limits of integration by say ## +/- b ## and compute the A vector as A(b), the curl of A will be limited even as ## b \to \infty. ##

I tried this by computing the electrical vector potential of such a B field so ##\nabla \times \bf E = -\partial \bf B/\partial t ## to verify Faraday's law & that's what happened.
 
Last edited:
  • #21
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
Kind of late to be joining, but Feynman and many others derive the relationship
$$ \bf A = (1/(4\pi) \int \frac {\nabla \times \bf B} {r} dv $$

similar to the scalar potential expression.

based on ## \nabla \times \bf B = \nabla \times (\nabla \times \bf A) = -\nabla^2 \bf A ##
if ## \nabla \cdot \bf A = 0 ## is chosen which can be shown to be OK working back from the result..

The solution is the above integral. Don't know how that compares with the previous posters' work.
Are you sure this is the equation derived by Feynman? This equation obviously fails if ##\mathbf{B}=constant\neq 0## cause it gives ##\nabla\times \mathbf{B}=0## and hence ##\mathbf{A}=0##.
I have in mind a slightly different equation which i derive from Helmholtz decomposition theorem which is $$\mathbf{A}=\frac{1}{4\pi}\nabla\times\int \frac{\mathbf{B(r')}}{|\mathbf{r-r'}|}d^3\mathbf{r'}$$ where the curl operator is with respect to ##\mathbf{r}##, while the integration is with respect to ##\mathbf{r'}##.
This is also with the assumption that ##\nabla\cdot\mathbf{A}=0##.
 
Last edited:
  • #22
rude man
Homework Helper
Insights Author
Gold Member
7,786
767
Are you sure this is the equation derived by Feynman? This equation obviously fails if ##\mathbf{B}=constant\neq 0## cause it gives ##\nabla\times \mathbf{B}=0## and hence ##\mathbf{A}=0##.
I have in mind a slightly different equation which i derive from Helmholtz decomposition theorem which is $$\mathbf{A}=\frac{1}{4\pi}\nabla\times\int \frac{\mathbf{B(r')}}{|\mathbf{r-r'}|}d^3\mathbf{r'}$$ where the curl operator is with respect to ##\mathbf{r}##, while the integration is with respect to ##\mathbf{r'}##.
This is also with the assumption that ##\nabla\cdot\mathbf{A}=0##.
Yours may be correct also but certainly mine is:
## \nabla \times \bf H = \bf j ## (absent time-varying fields).
Put that in my integral and you get ##\bf A = (1/4\pi) \int \bf j/r dv ##
which is the recognized integral for the ## \bf A ## magnetic vector potential
and which gives the correct B field and agrees with Ampere's law - easily shown.

And BTW it is also derivable from the Helmholtz Decomposition theorem:
http://farside.ph.utexas.edu/teaching/em/lectures/node37.html
which BTW is a writeup I really like. It helped me a lot buttressing my split-field blog.
 
  • Informative
  • Like
Likes vanhees71 and Delta2
  • #23
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
Yours may be correct also but certainly mine is:
## \nabla \times \bf H = \bf j ## (absent time-varying fields).
Put that in my integral and you get ##\bf A = (1/4\pi) \int \bf j/r dv ##
which is the recognized integral for the ## \bf A ## magnetic vector potential
and which gives the correct B field and agrees with Ampere's law - easily shown.

And BTW it is also derivable from the Helmholtz Decomposition theorem:
http://farside.ph.utexas.edu/teaching/em/lectures/node37.html
which BTW is a writeup I really like. It helped me a lot buttressing my split-field blog.
I have to look it up abit more thoroughsly (give me time to think heh ) but we can agree for now that your equation is correct in the case of non time varying currents.
It doesnt seem to be correct in the case of time varying currents cause then the well known expression for A contains the retarded time inside the current density J, which retarded (or advanced ) time doesnt appear anywhere in your equation.

After reading the following post #24 by vanhees71 i think your equation is correct at all cases, forget my comment about retarded or advanced times.
 
Last edited:
  • #24
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,371
6,754
Kind of late to be joining, but Feynman and many others derive the relationship
$$ \bf A = (1/(4\pi) \int \frac {\nabla \times \bf B} {r} dv $$

similar to the scalar potential expression.

based on ## \nabla \times \bf B = \nabla \times (\nabla \times \bf A) = -\nabla^2 \bf A ##
if ## \nabla \cdot \bf A = 0 ## is chosen which can be shown to be OK working back from the result..

The solution is the above integral. Don't know how that compares with the previous posters' work.
That's also a standard solution of the problem, but it needs the curl of ##\vec{B}##. The idea behind this is to use the ansatz
$$\vec{B}=\vec{\nabla} \times \vec{A},$$
because of
$$\vec{\nabla} \cdot \vec{B}=0.$$
Then
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}.$$
Since ##\vec{A}## is only determined up to a gradient field, one can impose one gauge condition. Here it's most convenient to use the Coulomb-gauge condition,
$$\vec{\nabla} \cdot \vec{A}=0,$$
and then you get
$$\vec{\nabla} \times \vec{B}=-\Delta \vec{A},$$
and with the Green's function of the Laplace operator, which you know from electrostatics, you get the said equation, which in more detail reads
$$\vec{A}(t,\vec{r})=\int_{\mathbb{R}} \mathrm{d}^3 \vec{r}' = \frac{\vec{\nabla}' \times \vec{B}(t,\vec{r}')}{4 \pi |\vec{r}-\vec{r}'|},$$
which is of course gauge equivalent to the solution, I've given above in "axial gauge".
 
  • Like
Likes Delta2
  • #25
rude man
Homework Helper
Insights Author
Gold Member
7,786
767
I have to look it up abit more thoroughsly (give me time to think heh ) but we can agree for now that your equation is correct in the case of non time varying currents.
It doesnt seem to be correct in the case of time varying currents cause then the well known expression for A contains the retarded time inside the current density J, which retarded (or advanced ) time doesnt appear anywhere in your equation.

After reading the following post #24 by vanhees71 i think your equation is correct at all cases, forget my comment about retarded or advanced times.
OK.
My main motivation was to prove that in my r-2r ring (see my last blog) the E fields are made up of both non-conservative and conservative fields. The Helmholtz Decomposition theorem helped me derive the electrical vector potential which determines the non-conservative E field at every point along the ring, thus proving beyond doubt the split-field nature of the E fields.

I will look further into the time-varying situation if I have the motivation. Perhaps it's OK already but I have to convince myself.
 
Last edited:
  • Like
Likes Delta2

Related Threads on Vector potential ##\vec A## in terms of magnetic field ##\vec B##

Replies
4
Views
7K
Replies
3
Views
1K
  • Last Post
Replies
10
Views
17K
Replies
1
Views
6K
Replies
1
Views
878
Replies
5
Views
643
Replies
5
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
7
Views
11K
Top