# Expressing the magnetic vector potential in terms of its curl

We have the 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 its curl B, but I'm not sure how where to start.

vanhees71
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marcusl and dextercioby
Check your favorite textbook on vector calculus for the Helmholtz theorem, according to which you can always decompose a vector field in a gradient and a solenoidal field. The former takes into account its sources the latter its curl.

However, this is not as useful as the formulae you already have written down:

https://iopscience.iop.org/article/10.1088/0143-0807/37/6/065204/metahttps://arxiv.org/pdf/1609.08149

thank you, that paper is pretty informative.

equation 30 is a great, neat expression of A in terms of B, but unfortunately, it is for instantaneous potentials.

Equation 35, for causal potentials, has a ∇'Φ.

Do you know how to modify equation 35 to get it solely in terms of B?

vanhees71
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No, (30) holds for all fields, no matter whether time-dependent or not. It only deals with spatial integrals and derivatives, which "don't care" for an additional dependence on time. It's of course useless to solve the Maxwell equations, because there the task is to get both the vector and scalar potentials (or better said the four-potential) of the electromagnetic field from the sources, which are ##\rho## and ##j##, and these involve retarded solutions.

marcusl
No, (30) holds for all fields, no matter whether time-dependent or not. It only deals with spatial integrals and derivatives, which "don't care" for an additional dependence on time.

I see, thanks, that makes sense.

I noticed that from (30) to (31), the curl operator applied to the fraction ends up being equivalent to applying it to the numerator only. Do you know why that's the case?

vanhees71
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Unfortunately many textbooks as well as the author of the paper introduce always abbreviations in the integrals obscuring a bit their meaning. What's written is
$$\vec{A}=\vec{\nabla} \times \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\vec{B}(\vec{r}')}{4 \pi |\vec{r}-\vec{r}'|},$$
i.e., the nabla operator in front of the integral acts on the variables of ##\vec{r}## and this means it only acts on the denominator, if put under the integral. The next step is that obviously
$$\vec{\nabla} \frac{1}{|\vec{r}-\vec{r}'|}=-\vec{\nabla}' \frac{1}{|\vec{r}-\vec{r}'|},$$
where ##\vec{\nabla}'## refers derivatives with respect to the components of ##\vec{r}'##.

Unfortunately many textbooks as well as the author of the paper introduce always abbreviations in the integrals obscuring a bit their meaning. What's written is
$$\vec{A}=\vec{\nabla} \times \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\vec{B}(\vec{r}')}{4 \pi |\vec{r}-\vec{r}'|},$$
i.e., the nabla operator in front of the integral acts on the variables of ##\vec{r}## and this means it only acts on the denominator, if put under the integral. The next step is that obviously
$$\vec{\nabla} \frac{1}{|\vec{r}-\vec{r}'|}=-\vec{\nabla}' \frac{1}{|\vec{r}-\vec{r}'|},$$
where ##\vec{\nabla}'## refers derivatives with respect to the components of ##\vec{r}'##.
sorry, I'm confused, I thought the nabla applies to the numerator only.

vanhees71
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In the numerator there's no ##\vec{r}## it could be acting on, i.e., for ##\vec{\nabla}## the numerator, ##\vec{B}(\vec{r'})## is simply a constant.

In the numerator there's no ##\vec{r}## it could be acting on, i.e., for ##\vec{\nabla}## the numerator, ##\vec{B}(\vec{r'})## is simply a constant.
but in (31), it seems like the author's applying the curl to the numerator B only.

vanhees71
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Ok let's follow Heras's argument in detail. The first goal is to express ##\vec{A}## in terms of ##\vec{B}## only. It starts with
$$\vec{\nabla} \cdot \vec{B}=0.$$
According to Helmholtz's theorem this implies that there exists (at least locally) a vector potential with
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Then ##\vec{A}## is determined only up to a gradient field, and we can impose one constraint. Heras chooses the so-called Coulomb gauge, i.e.,
$$\vec{\nabla} \cdot \vec{A}=0.$$
Now get ##\vec{A}## in terms of ##\vec{B}## just take the curl of the defining equation, i.e.,
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A}) = \vec{\nabla} (\vec{\nabla} \cdot \vec{A}) - \Delta \vec{A}=-\Delta \vec{A}=\vec{\nabla} \times \vec{B}.$$
Using the Green's function of the Laplace operator then yields Eq. (30) of the paper after some further manipulations:
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{\nabla}' \times \vec{B}(t,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{B}(t,\vec{x}') \times \vec{\nabla}' \frac{1}{4 \pi |\vec{x}-\vec{x}'|} = -\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{B}(t,\vec{x}') \times \vec{\nabla} \frac{1}{4 \pi |\vec{x}-\vec{x}'|} = +\vec{\nabla} \times \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{B}(t,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}.$$
Here we have integrated by parts and rewritten the gradient wrt. ##\vec{x}'## in the gradient wrt. ##\vec{x}##, which then finally can be put out of the integral.

Ironically to get (31) Heras does these final steps backwards. We can just use the Ampere-Maxwell Law
$$\vec{\nabla} \times \vec{B}=\mu_0 \vec{j} + \mu_0 \epsilon_0 \partial_t \vec{E}$$
to get immediately (31)
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{\nabla}' \times \vec{B}(t,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(t,\vec{x}')+\epsilon_0 \partial_t \vec{E}(t,\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
As Heras himself says, all this is pretty useless in practice. You may ask, for which this paper is good for, but it answers your question ;-)).

thank you for the detailed analysis

Heras chooses the so-called Coulomb gauge, i.e.,
$$\vec{\nabla} \cdot \vec{A}=0.$$
If we start with the Lorenz gauge, we will eventually arrive at the same spot right?

Using the Green's function of the Laplace operator then yields Eq. (30) of the paper after some further manipulations:
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{\nabla}' \times \vec{B}(t,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{B}(t,\vec{x}') \times \vec{\nabla}' \frac{1}{4 \pi |\vec{x}-\vec{x}'|} = -\int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{B}(t,\vec{x}') \times \vec{\nabla} \frac{1}{4 \pi |\vec{x}-\vec{x}'|} = +\vec{\nabla} \times \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{B}(t,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}.$$
Here we have integrated by parts and rewritten the gradient wrt. ##\vec{x}'## in the gradient wrt. ##\vec{x}##, which then finally can be put out of the integral.
interesting, that's kinda tricky

As Heras himself says, all this is pretty useless in practice. You may ask, for which this paper is good for, but it answers your question ;-)).
Don't worry, its been pretty useful to me for understanding electromagnetism. :)

I'd just like to see one more step.

Is it correct to say that $$\frac{∂}{∂t}\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{B}(t,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\frac{∂\vec{B}(t,\vec{x}')}{∂t}}{4 \pi|\vec{x}-\vec{x}'|}$$?

vanhees71
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$$\Box \vec{A}=\vec{j}.$$
It's "physical solution" is the retarded rather than the instantaneous solution,
$$\vec{A}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(t-|\vec{x}-\vec{x}'|/c,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}.$$
Your formula concerning the time derivative is correct.

thanks!
It's "physical solution" is the retarded rather than the instantaneous solution,
$$\vec{A}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(t-|\vec{x}-\vec{x}'|/c,\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}.$$

But in terms of B, its the same right?

No, (30) holds for all fields, no matter whether time-dependent or not. It only deals with spatial integrals and derivatives, which "don't care" for an additional dependence on time.

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vanhees71
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No, if ##\vec{\nabla} \cdot \vec{A} \neq 0## the formula above doesn't hold, because for any vector field, ##\vec{V}## you have ##\vec{\nabla} \cdot (\vec{\nabla} \times \vec{V})=0.##

No, if ##\vec{\nabla} \cdot \vec{A} \neq 0## the formula above doesn't hold, because for any vector field, ##\vec{V}## you have ##\vec{\nabla} \cdot (\vec{\nabla} \times \vec{V})=0.##

ok, regardless, (30) is correct regarding the retarded vector fields described in the OP right? Just to confirm.

We have the 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 its curl B, but I'm not sure how where to start.

vanhees71
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No, it's not correct. As Heras states the formula holds in the Coulomb gauge. It obviously cannot hold in the Lorenz gauge, because for the Lorenz gauge ##\vec{\nabla} \cdot \vec{A}=-\frac{1}{c} \partial_t \Phi \neq 0##.

(writing in progress...)

No, it's not correct. As Heras states the formula holds in the Coulomb gauge. It obviously cannot hold in the Lorenz gauge, because for the Lorenz gauge ##\vec{\nabla} \cdot \vec{A}=-\frac{1}{c} \partial_t \Phi \neq 0##.
hmm, sadly, I think there might've been a misunderstanding

thank you, that paper is pretty informative.

equation 30 is a great, neat expression of A in terms of B, but unfortunately, it is for instantaneous potentials.

Equation 35, for causal potentials, has a ∇'Φ.

Do you know how to modify equation 35 to get it solely in terms of B?
No, (30) holds for all fields, no matter whether time-dependent or not. It only deals with spatial integrals and derivatives, which "don't care" for an additional dependence on time.

but anyway, equation 35:
$$\mathbf{A}=\frac{1}{4 \pi c^{2}} \frac{\partial}{\partial t} \int \frac{\left[\boldsymbol{\nabla}^{\prime} \Phi+\partial \mathbf{A} / \partial t\right]}{R} d^{3} r^{\prime}+\boldsymbol{\nabla} \times \int \frac{[\mathbf{B}]}{4 \pi R} d^{3} r^{\prime}$$

I'm no longer concerned about getting rid of ∇'Φ, but I need to solve for A.

What should I do to get all the A out of the RHS and to be only on the LHS?

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vanhees71
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It depends on the gauge, as I said. But still I don't understand for what all these effords should be good for. The only gauge-independent formulation which makes sense are Jefimenko's equations, defining the (components of) the electromagnetic field in terms of (the components of) its sources, i.e., ##\rho## and ##\vec{j}##. These are "retarded integrals" as is natural in view of the causality principle.

It depends on the gauge, as I said.
well now we're at (35).

But still I don't understand for what all these effords should be good for.
I'm trying to express the electric field in terms of the "magnetic displacement current".

vanhees71
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As I said, Heras's paper is utterly confusing. He's basically going in circles, and also his notation is unclear.

The logic in classical electrodynamics is as follows:

The fundamental laws are Maxwell's equations, which are expressed in completely in gauge-invariant terms. In (1+3)-dimensional notation in Heaviside-Lorentz units they read
$$\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
These homogeneous Maxwell equations are basically constraint equations for the various field components. Then there are the inhomogeneous Maxwell equations, which relate the field components to the sources:
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
The homogeneous equations imply that the 6 field components can be expressed in terms of a scalar and a vector potential,
$$\vec{E}=-\vec{\nabla} \Phi -\frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
The potentials are defined only up to a gauge transformation, i.e., if ##\Phi## and ##\vec{A}## describe a physical situation then with any scalar field ##\chi##, also the potentials
$$\Phi'=\Phi+\frac{1}{c} \partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi$$
describe the very same physical situation, indeed evaluating the fields from the new potentials you get
$$-\vec{\nabla} \Phi'-\frac{1}{c} \partial_t \vec{A}' = -\vec{\nabla} (\Phi+\frac{1}{c} \partial_t \chi) -\frac{1}{c} \partial_t (\vec{A}-\vec{\nabla} \chi)=-\vec{\nabla} \Phi-\frac{1}{c} \partial_t \vec{A} =\vec{E}$$
and
$$\vec{\nabla} \times \vec{A}'=\vec{\nabla} \times (\vec{A}-\frac{1}{c} \vec{\nabla} \chi) = \vec{\nabla} \times \vec{A}=\vec{B}.$$
Now we can use this "gauge freedom" to impose one "gauge condition" to (at least) partially fix the potentials. This can be done to make life as easy as possible.

To see what gauge choice makes life as easy as possible, we express the fields in the inhomogeneous Maxwell equations, which remain to be solved, in terms of the potentials:
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})+\frac{1}{c} \partial_t (\vec{\nabla} \Phi+\frac{1}{c} \partial_t \vec{A}) = \frac{1}{c}\vec{j}.$$
This you can rearrange a bit further to get
$$(\frac{1}{c^2} \partial_t^2 -\Delta) \vec{A} + \vec{\nabla} (\vec{\nabla} \cdot \vec{A} + \frac{1}{c} \partial_t \Phi)=\Box \vec{A} + \vec{\nabla} (\vec{\nabla} \cdot \vec{A} + \frac{1}{c} \partial_t \Phi)=\frac{1}{c} \vec{j}.$$
The other inhomogeneous equation becomes
$$\vec{\nabla} \cdot (\vec{\nabla} \Phi+\frac{1}{c} \partial_t \vec{A})=-\rho$$
or
$$\frac{1}{c} \partial_t (\vec{\nabla} \cdot \vec{A}) +\Delta \Phi=-\rho.$$
Now there are two choices where these equations become "convenient". In some sense the most easy choice is the Lorenz gauge, defined by the contraint
$$\frac{1}{c} \partial_t \Phi + \vec{\nabla} \cdot \vec{A}=0.$$
Then in the equation with ##\vec{j}## on the right-hand side the second term vanishes, and you get simply three separate wave equations for the components,
$$\Box \vec{A}=\frac{1}{c} \vec{j}.$$
In the equation with ##\rho## on the right-hand side you use the Lorenz gauge condition to get rid of ##\vec{A}## in this equation, also yielding the simple wave equation for the scalar potential,
$$\Box \Phi=\rho.$$
Not he box operator has no unique "inverse", but there's a most convenient choice, the "retarded solutions", which give the potentials at times ##t## in terms of the sources at time arguments ##t'<t## only, i.e., you evaluate the potentials at time ##t## only from the past history of the sources' evolution. This matches the causality principle of physics: All quantities at present time depend only on the past (or the present) state of the system but never on its future. This is manifest in Lorenz gauge for the potentials, i.e.,
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{j(t-|\vec{x}-\vec{x}'|/c,\vec{x}')}{4 \pi c|\vec{x}-\vec{x}'|}, \quad \Phi(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(t-|\vec{x}-\vec{x}'|/c,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
This makes even the somewhat stricter relativistic sense of the causality principle manifest: Nothing physical can spread faster than the speed of light. Indeed here the potentials depend on sources at times earlier by the time light needs to travel from the source point ##\vec{x}'## and the point of observation ##\vec{x}##. Since the fields are given by derivatives ##\partial_t## and ##\vec{\nabla}##, this holds true for them too, and only for them it always must hold true, because they are the only really observable fields.

This becomes clear if you choose the Coulomb gauge condition for the potentials, which often has also its advantages. There the idea is to get rid of ##\vec{A}'## (we label the potentials in the Coulomb gauge with a prime to dinstinguish them from those in the Lorenz gauge above) in the equation with ##\rho## on the right-hand side in the most simple way by demanding the Coulomb-gauge condition,
$$\vec{\nabla} \cdot \vec{A}'=0.$$
Then the said equation simply becomes
$$-\Delta \Phi'=\rho.$$
This is solved as in electrostatics with the "instantaneous integral"
$$\Phi'(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
Now the equation with ##\vec{j}## on the right-hand side, gets somewhat more complicated, but can be written as a wave equation with the source containing not only ##\vec{j}## but also the just calculated scalar potential:
$$\Box \vec{A}'=\frac{1}{c} (\vec{j} + \partial_t \vec{\nabla} \Phi')=\frac{1}{c} \vec{j}_{\perp}.$$
This you can solve again with help of the retareded integral,
$$\vec{A}'(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}_{\perp}(t-|\vec{x}-\vec{x}'|/c,\vec{x}')}{4 \pi c |\vec{x}-\vec{x}'|}.$$
You can easily check that the Coulomb-gauge potentials, ##\Phi'## and ##\vec{A}'## are related to the Lorenz-gauge potential ##\Phi## and ##\vec{A}## by a gauge transformation. It's a good exercise to determine the corresponding gauge field ##\chi##.

The upshot is that with the Coulomb gauge, despite its mixture of "instantaneous" and "retarded" integrals leads to precisely the same electromagnetic field which is entirely given by "retarded" integrals, as it should be since the Maxwell equations fulfill the stricter notion of causality according to relativity, because the Maxwell equations are in fact a relativistic field theory of a "massless vector field", although that's hidden in using the (1+3)-notation. Everything becomes much more lucid when writing it in the more appropriate Minkowski-space 4-vector notation, which is the natural mathematical language of everything related to relativity.

SiennaTheGr8
hmm, sadly, I think there might've been a misunderstanding

but anyway, equation 35:
$$\mathbf{A}=\frac{1}{4 \pi c^{2}} \frac{\partial}{\partial t} \int \frac{\left[\boldsymbol{\nabla}^{\prime} \Phi+\partial \mathbf{A} / \partial t\right]}{R} d^{3} r^{\prime}+\boldsymbol{\nabla} \times \int \frac{[\mathbf{B}]}{4 \pi R} d^{3} r^{\prime}$$

I'm no longer concerned about getting rid of ∇'Φ, but I need to solve for A.

What should I do to get all the A out of the RHS and to be only on the LHS?

@vanhees71 thanks, but I just wanna find A in terms of ∇'Φ and B.