Expressing the magnetic vector potential A-field in terms of the B-field

  • Thread starter tade
  • Start date
  • #1
552
18
We have a retarded magnetic vector potential ##\mathbf{A}(\mathbf{r},t) = \dfrac{\mu_0}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|} \mathrm{d}^3 \mathbf{r}'##

And its curl, ##\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r}'##


I would like to express A in terms of B




In this paper, equation 35 looks like a promising step on the way there:


$$\mathbf{A}=\frac{1}{4 \pi c^{2}} \frac{\partial}{\partial t} \int \frac{\left[\boldsymbol{\nabla}^{\prime} \Phi+\partial \mathbf{A} / \partial t\right]}{R} d^{3} r^{\prime}+\boldsymbol{\nabla} \times \int \frac{[\mathbf{B}]}{4 \pi R} d^{3} r^{\prime}$$


Unfortunately, it has ∂A/∂t on the RHS, and it also has a ∇'Φ.
I'm not sure if the correct answer involves Φ


If its easier to express ∂A/∂t in terms of B, rather than A in terms of B, that'll do just fine as well.
 
Last edited:

Answers and Replies

  • #2
Dale
Mentor
Insights Author
2021 Award
32,390
9,382
Isn’t it just ##A=\nabla \times B##
 
  • #3
552
18
Isn’t it just ##A=\nabla \times B##
B is supposed to be the curl of A, and A the inverse curl of B
 
  • Like
Likes Dale and vanhees71
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,518
10,278
Argh, somehow my posting got completely distorted. Here it's again:

You can of course express ##\vec{A}## as a functional of ##\vec{B}##, but it's easier in the Coulomb gauge than in the Lorenz gauge. You simply impose ##\vec{\nabla} \cdot \vec{A}=0## (Coulomb-gauge condition) and then you have
$$\vec{\nabla} \times \vec{B} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A},$$
with the solution
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\vec{\nabla}' \times \vec{B}(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
Note however that this is useless except for the special case of magnetostatics, because in this case ##\partial_t \vec{E}=0## and thus the Ampere-Maxwell Law simplifies to the Ampere Law,
$$\vec{\nabla} \times \vec{B}(\vec{x})=\vec{j}(\vec{x}),$$
and you simply get Biot-Savart's formula for the Coulomb-gauge vector potential of magnetostatics,
$$\vec{B}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
This is, btw also the result of the retarded potential in this case, because since ##\partial_t A^0=0## the Lorenz gauge condition is ##\partial_{\mu} A^{\mu}=\vec{\nabla} \cdot \vec{A}=0## and thus it coincides with the Coulomb gauge in this special case.
 
  • Like
Likes marcusl and Dale
  • #5
552
18
Argh, somehow my posting got completely distorted. Here it's again:

You can of course express ##\vec{A}## as a functional of ##\vec{B}##, but it's easier in the Coulomb gauge than in the Lorenz gauge. You simply impose ##\vec{\nabla} \cdot \vec{A}=0## (Coulomb-gauge condition) and then you have
$$\vec{\nabla} \times \vec{B} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A},$$
with the solution
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\vec{\nabla}' \times \vec{B}(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
Note however that this is useless except for the special case of magnetostatics, because in this case ##\partial_t \vec{E}=0## and thus the Ampere-Maxwell Law simplifies to the Ampere Law,
$$\vec{\nabla} \times \vec{B}(\vec{x})=\vec{j}(\vec{x}),$$
and you simply get Biot-Savart's formula for the Coulomb-gauge vector potential of magnetostatics,
$$\vec{B}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
This is, btw also the result of the retarded potential in this case, because since ##\partial_t A^0=0## the Lorenz gauge condition is ##\partial_{\mu} A^{\mu}=\vec{\nabla} \cdot \vec{A}=0## and thus it coincides with the Coulomb gauge in this special case.

So, if ##\mathbf{A}(\mathbf{r},t) = \dfrac{\mu_0}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|} \mathrm{d}^3 \mathbf{r}'##

And ##\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r}'##

is it correct to say that $$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\vec{\nabla}' \times \vec{B}(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$


from your explanation, i'm not sure which parts exactly the Coulomb gauge/magnetostatics/Lorenz guage/retarded time etc. apply to, sorry
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,518
10,278
No in the general case it's not correct the same, because your ##\vec{A}## is in Lorenz gauge and mine is in Coulomb gauge. As I said in this most general case the given integral is useless. The reason for that is that (I'm using Heaviside Lorentz units rather than SI units; so in my case ##\epsilon_0=\mu_0=1##, and ##c## is written explicitly) in this case you have the full Ampere-Maxwell Law, i.e.,
$$\vec{\nabla} \times \vec{B}=\vec{j} +\frac{1}{c} \partial_t \vec{E}.$$
This would make my formula above
$$\vec{A}(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{1}{4 \pi |\vec{x}-\vec{x}'|} \left [\vec{j}(t,\vec{x}')+\frac{1}{c} \partial_t \vec{E}(t,\vec{x}') \right],$$
and this is a very cumbersome non-local expression when you want to write everything in terms of the sources ##\rho## and ##\vec{j}##.

Only in the magnetostatic case it's of practical use!
 
  • #7
Dale
Mentor
Insights Author
2021 Award
32,390
9,382
B is supposed to be the curl of A, and A the inverse curl of B
D’oh! Yes, of course. Sorry
 
  • Like
  • Haha
Likes tade and vanhees71
  • #8
552
18
No in the general case it's not correct the same, because your ##\vec{A}## is in Lorenz gauge and mine is in Coulomb gauge. As I said in this most general case the given integral is useless. The reason for that is that (I'm using Heaviside Lorentz units rather than SI units; so in my case ##\epsilon_0=\mu_0=1##, and ##c## is written explicitly) in this case you have the full Ampere-Maxwell Law, i.e.,
$$\vec{\nabla} \times \vec{B}=\vec{j} +\frac{1}{c} \partial_t \vec{E}.$$
This would make my formula above
$$\vec{A}(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{1}{4 \pi |\vec{x}-\vec{x}'|} \left [\vec{j}(t,\vec{x}')+\frac{1}{c} \partial_t \vec{E}(t,\vec{x}') \right],$$
and this is a very cumbersome non-local expression when you want to write everything in terms of the sources ##\rho## and ##\vec{j}##.

Only in the magnetostatic case it's of practical use!
i see, cool

though could you help me find A in terms of B for my given A and B?

you could also do ∂A/∂t in terms of B if its easier
 
Last edited:
  • #9
552
18
D’oh! Yes, of course. Sorry
D'oh!

On a side note, its really cool how so many physicists and mathematicians have written for the Simpsons, and inserted many formulas into scenes
 
  • #10
552
18
i see, cool

though could you help me find A in terms of B for my given A and B?

you could also do ∂A/∂t in terms of B if its easier


@vanhees71 from eqn. 35, we can re-arrange:

$$\mathbf{A}+\frac{1}{4 \pi c^{2}} \frac{\partial}{\partial t} \int \frac{[\partial \mathbf{A} / \partial t]}{R} d^{3} r^{\prime}=\frac{1}{4 \pi c^{2}} \frac{\partial}{\partial t} \int \frac{[\boldsymbol{\nabla}^{\prime} \Phi]}{R} d^{3} r^{\prime}+\boldsymbol{\nabla} \times \int \frac{[\mathbf{B}]}{4 \pi R} d^{3} r^{\prime}$$

and then we need to sort it properly for A or ∂A/∂t.
 
Last edited:

Related Threads on Expressing the magnetic vector potential A-field in terms of the B-field

Replies
21
Views
923
Replies
4
Views
2K
Replies
1
Views
2K
Replies
24
Views
32K
Replies
2
Views
1K
Replies
1
Views
3K
Replies
3
Views
1K
Replies
2
Views
3K
Top