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

In summary, the conversation discusses the expression of a retarded magnetic vector potential and its curl, and how to express one in terms of the other. It is mentioned that this can be done in the Coulomb gauge for magnetostatics, but in the general case it is a cumbersome and non-local expression. It is also mentioned that in the magnetostatic case, it is possible to express the inverse curl of B in terms of A. However, it is not possible to express A in terms of B for the given A and B.
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 ΦIf its easier to express ∂A/∂t in terms of B, rather than A in terms of B, that'll do just fine as well.

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Isn’t it just ##A=\nabla \times B##

Dale said:
Isn’t it just ##A=\nabla \times B##
B is supposed to be the curl of A, and A the inverse curl of B

Dale and vanhees71
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.

marcusl and Dale
vanhees71 said:
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

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!

B is supposed to be the curl of A, and A the inverse curl of B
D’oh! Yes, of course. Sorry

vanhees71 said:
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

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Dale said:
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

Dale
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.

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## 1. How is the magnetic vector potential A-field related to the B-field?

The magnetic vector potential A-field is a mathematical construct that is used to describe the behavior of the B-field. It is defined as the curl of the A-field is equal to the B-field, making them directly related to each other.

## 2. Why is it useful to express the A-field in terms of the B-field?

Expressing the A-field in terms of the B-field allows for a more simplified and efficient way to analyze and understand the behavior of magnetic fields. It also allows for easier calculations and predictions of the behavior of the B-field.

## 3. What is the mathematical equation for expressing the A-field in terms of the B-field?

The mathematical equation is ∇ x A = B, where ∇ is the del operator and x represents the cross product. This equation is known as the vector potential equation and is used to express the A-field in terms of the B-field.

## 4. Can the A-field be expressed in terms of the B-field in all situations?

No, there are certain situations where it is not possible to express the A-field in terms of the B-field. These situations include when the B-field is changing rapidly or when there are certain boundary conditions present.

## 5. How does expressing the A-field in terms of the B-field relate to Maxwell's equations?

Expressing the A-field in terms of the B-field is a result of one of Maxwell's equations, specifically the curl of the magnetic field equation. This equation is an essential part of understanding the behavior of electromagnetic fields and is used in many applications of electromagnetism.

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