Find EM Potential from EM Tensor | Math Solutions

[denijane]

Hello,
so my question is, if for some metric, we have found (somehow) F_μν, and we know that:
F_μν=∂_μA_ν-∂_νA_μ, how do we find A_ν?

I tried solving the differential system after imposing the Lorentz gauge ∂^μA_μ=0
but still, without some initial guess about which components of A are zero, the system cannot be solved.

Since books give often the potential A instead of F, I suppose there is a way to find it, I just can't figure it out on my own. Is there another equation to make the system determined?

 

[PeterDonis]

Since books give often the potential A instead of F, I suppose there is a way to find it

Not really, no. Books often give the potential A instead of the field F because F can always be derived from A, but A cannot always be derived from F. A is generally considered more fundamental.

 

[denijane]

Hi and thank you for replying.
So then, the only equations for the potential are:
$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ and
$\partial_{\mu}A_{\mu}=0$
In this case, any solution of the differential system is a valid potential and then it's a question of interpretation of the physics the potential represents with respect to E and B. Ook.

 

[PeterDonis]

the only equations for the potential are:

$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ and

$\partial_{\mu}A_{\mu}=0$

If you assume Lorentz gauge, yes. (Also, strictly speaking, one of the indexes in the second equation should be raised, as you wrote it in the OP.)

In this case, any solution of the differential system is a valid potential and then it's a question of interpretation of the physics the potential represents with respect to E and B.

Yes.

 

[dextercioby]

It's Lorenz, people, not Lorentz.

 

[robphy]

It's Lorenz, people, not Lorentz.

http://scienceworld.wolfram.com/physics/LorenzGauge.html

 

[vanhees71]

In textbooks usually you find the non-covariant solution of this problem in terms of Helmholtz fundamental theorem of vector calculus: Given a vector field $\vec{V}$ that falls sufficiently quickly to 0 at spatial infinity, can be decomposed uniquely in terms of a curl-free gradient field and a div-free solenoidal part, i.e.,
$$\vec{V}=\vec{V}_1+\vec{V}_2$$
with
$$\vec{\nabla} \times \vec{V}_1=0, \quad \vec{\nabla} \cdot \vec{V}_2=0.$$
From Poincare's Lemma you know that locally (i.e., in any simply connected region of space) there exists a scalar field $\phi$ such that
$$\vec{V}_1=-\vec{\nabla} \phi$$
and a vector potential $\vec{A}$ such that
$$\vec{V}_2=\vec{\nabla} \times \vec{A}$$
The vector potential is only defined up to a gradient and thus you can impose the condition (in em. called Coulomb gauge)
$$\vec{\nabla} \cdot \vec{A}=0.$$
Then you get
$$\rho=\vec{\nabla} \cdot \vec{V}=\vec{\nabla} \cdot \vec{V}_1=-\Delta \phi,$$
and this can be easily solved, using the Green's function of the Laplace operator,
$$\phi(\vec{x})=\frac{1}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
For the curl you find together with the Coulomb gauge
$$\vec{\nabla} \times \vec{V}=\vec{\nabla} \times \vec{V}_2 = \vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A}=\vec{\omega}.$$
Again using the Green's function of the Laplace operator yields
$$\vec{A}(\vec{x})=\frac{1}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\vec{\omega}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
If your fields are additionally time dependent, it doesn't matter in this approach; time is just a parameter left constant at all steps. This can be used to any vector field, and it doesn't rely on the special properties of electric and magnetic field components of the electromagnetic field, and it's not manifestly covariant.

For electrodynamics, it's clear that $F_{\mu \nu}$ must fulfill the Maxwell equations,
$$\partial_{\mu} F^{\mu \nu} = j^{nu}, \quad \epsilon^{\mu \nu \rho \sigma} \partial_{\mu} F_{\rho \sigma}=0.$$
The latter set of equations (the homogeneous Maxwell equations) imply the existence of a four-potential, which is defined up to a four-gradient of a scalar field, which allows you to introduce the Lorenz-gauge condition (it's indeed the Danish physicist Lorenz who should get the credit, because he figured this particular nice gauge out long before the Dutsch physicist Lorentz), i.e.,
$$F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}, \quad \partial_{\mu} A^{\mu}=0.$$
Now the inhomogeneous equations read
$$\Box A^{\nu} = j^{\nu},$$
and you need a Green's function of the D'Alembert operator, which is however not unique. Usually one needs the retarded propagator in classical field theory, which is given by
$$G_{\text{ret}}(x)=\frac{1}{4 \pi r} \Theta(x^0) \delta(x^0-|\vec{x}|) = \frac{1}{2 \pi} \Theta(x^0) \delta(x^2),$$
where the latter form explicitly shows that it's a Lorentz scalar. This gives, by integrating out the Dirac $\delta$, the retarded solution
$$A_{\text{ret}}^{\mu} = \frac{1}{4 \pi} \int_{\mathbb{R}^3} \frac{j^{\mu}(x^0-|\vec{x}-\vec{x}'|,\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
For a given $F_{\mu \nu}$ you have of course to check, whether it's really a retarded field. If not, you have to add an appropriate solution of the free Maxwell equations!

 

[davidge]

vanhees71, is it ok to also use the approach below to find $A_\mu(x)$?

Let's suppose that $\partial_\mu A^\mu (x) = 0 \Rightarrow A^\mu = \epsilon^{\mu \nu}{}_{\rho} \partial_{\nu}B^{\rho}(x)$.
In vector notation: $\nabla \cdot \vec{A} = 0 \Rightarrow \vec{A} = \nabla \times \vec{B}$.

Now, returning to tensor notation, if it's possible to find a $B_\kappa(x)$ that is not a function of $x$ for all $\kappa$, then $\partial_{\nu}B^{\rho}(x) = B_\kappa \partial_{\nu}g^{\rho \kappa}(x)$. In this way, if we know the metric at the point $x$ we would be able to find all the $A_\lambda(x)$, and so solve for $F_{\sigma \lambda}$.