I Lorentz gauge

Tags:
1. Oct 29, 2016

DavideGenoa

Hello, friends! My textbook, Gettys's Physics, says that the Lorenz gauge choice uses the magnetic vector potential $$\mathbf{A}(\mathbf{x},t):=\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|}d^3y$$and the electric potential $$V(\mathbf{x},t):=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|}d^3y$$which are such that $$\nabla^2\mathbf{A}(\mathbf{x},t)-\varepsilon_0\mu_0\frac{\partial^2 \mathbf{A}(\mathbf{x},t)}{\partial t^2}=-\mu_0\mathbf{J}(\mathbf{x},t)$$but the book does not prove how $\mathbf{A}$ satisfies this equality.
How can we prove that it satisfies $\nabla^2\mathbf{A}-\varepsilon_0\mu_0\frac{\partial^2 \mathbf{A}}{\partial t^2}=-\mu_0\mathbf{J}$, as well as Maxwell's equations such as $$\nabla\cdot(\nabla\times\mathbf{A})=0$$ $$\oint_{\partial^+\Sigma} \nabla V\cdot d\mathbf{x}=-\frac{d}{dt}\int_{\Sigma} (\nabla\times\mathbf{A})\cdot d\mathbf{S}$$ $$\int_{\partial^+ \Sigma}(\nabla\times\mathbf{A})\cdot d\mathbf{x}=\mu_0\int_{\Sigma} \mathbf{J}\cdot d\mathbf{S}+\mu_0\varepsilon_0\frac{d}{dt}\int_{\Sigma}\nabla V\cdot d\mathbf{S}?$$
I heartily thank you for any answer!

Last edited: Oct 29, 2016
2. Oct 29, 2016

Jonathan Scott

3. Nov 20, 2016

DavideGenoa

As to $\nabla\cdot(\nabla\times\mathbf{A})=0$, I have realised that it is a trivial vector identity for any $\mathbf{A}\in C^2$.
I have not been able to find anything proving the other three identities yet, though...

4. Nov 24, 2016

DavideGenoa

$\nabla V$ should be $\mathbf{E}=-\nabla V$, with the minus sign, in the original post.

5. Nov 24, 2016

Jonathan Scott

I haven't looked it up recently, but I think it's a fairly messy calculation which is covered in various text books, including Griffiths. I think you may find some useful pointers in the Wikipedia article on Retarded potential.

6. Nov 24, 2016

vanhees71

What you need is the Green's function of the D'Alembert operator
$$\Box=\frac{1}{c^2} \partial_t^2 - \Delta,$$
i.e.,
$$\Box G(t,\vec{x})=\delta(t) \delta^{(3)}(\vec{x})$$
with the constraint, defining the retarded Green's function from all other Green's functions,
$$G(t,\vec{x}) \equiv 0 \quad \text{for} \quad t<0.$$
The most simple way to get this Green's function is via the use of the socalled "Mills representation", i.e.,
$$G(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \exp(\mathrm{i} \vec{k} \cdot \vec{x}) \tilde{G}(t,\vec{k}).$$
The equation of motion then translates to
$$\left (\frac{1}{c^2} \partial_t^2 +\vec{k}^2 \right) \tilde{G}(t,\vec{k}) = \delta(t).$$
It's easy to see that it's solution is
$$\tilde{G}(t,\vec{k})=\frac{c^2}{\omega_k}\sin(\omega_k t) \Theta(t) \quad \text{with} \quad \omega_k=c |\vec{k}|.$$
One just has to take the time derivatives using $\dot{\Theta}(t)=\delta(t)$.

Now we can easily transform back to position space with the final result
$$G(t,\vec{x})=\frac{\Theta(t)}{4 \pi r} \delta \left (t-\frac{r}{c} \right), \quad \text{with} \quad r=|\vec{x}|.$$

7. May 4, 2017

DavideGenoa

@vanhees 71, thank you very much for your hint, but I have never studied Green's function and I have no idee how what you say is derived and how to apply it to my problem, which precisely is finding the steps through which we see, by differentiating twice the components of $\mathbf{A}$ according to each variable and adding the second derivatives together, that the result is $\nabla^2\mathbf{A}-\varepsilon_0\mu_0\frac{\partial^2 \mathbf{A}}{\partial t^2}=-\mu_0\mathbf{J}$...

@Jonathan Scott , thank you very much for your answer! Where does Griffith, or any other author, explain the steps of the calculation?

8. May 5, 2017

vanhees71

You find the derivation in the attached pdf. It's about the quasistationary limit of electrodynamics, but a wide part is about the retarded solution. The Green's function (which is a subject every physicist must know!).

Attached Files:

• retarded-potential.pdf
File size:
138.8 KB
Views:
36
9. May 27, 2017

DavideGenoa

Thank you so much! The problem is that I do not understand how the desired identities are derived in the .pdf.
I mean, if $\mathbf{A}$ does not depend upon $t$, provided, as I think we can do in physics, that $\mathbf{J}$ is compactly supported and of class $C^2(\mathbb{R}^3)$, by using this result we can prove, as explained here, that $\nabla^2\mathbf{A}=-\mu_0\mathbf{J}$. Under the same assumptions we can prove that $\int_{\partial^+ \Sigma}(\nabla\times\mathbf{A})\cdot d\mathbf{x}$$=\mu_0 \int_{\Sigma}\mathbf{J}\cdot d\mathbf{S}$, by differentiating under the integral sign when that can be done and using other techniques in other cases, as shown here by using an argument explained by @Hawkeye18 , whom I thank again.
In this case, how can we proceed? To calculate the Laplacian of $\mathbf{A}$, for example, I need start with calculating $\frac{\partial \mathbf{A}}{\partial x_1}$, but how do we calculate $$\frac{\partial}{\partial x_1}\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|}d^3y ?$$Can we differentiate under the integral sign and, if we can, why can we?
I heartily thank anybody who will answer me!

10. May 27, 2017

vanhees71

How do you come to the conclusion that $\vec{A}$ is time-independent? That's only the case in magnetostatics, i.e., with stationary charge-current densities.

In physics we usually assume that we can differentiate under the integral sign. Of course, you are right, if you want to do the entire thing with mathematical rigour you have to prove that you can do that. If you need this, have a look in textbooks on analysis, and don't bother yourself with oldfashioned Riemann integrals but use Lebesgue ones. They are much closer to the robust way physicists use integrals and derivatives :-).

11. May 27, 2017

DavideGenoa

Thank you so much again! I am not saying that $\mathbf{A}$ is time independent in general. I mean: if
$$\mathbf{A}(\mathbf{x}):=\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$
and $\mathbf{J}\in C^2(\mathbb{R}^3)$ is compactly supported, by using this result we can prove, as explained here step by step, that $\nabla^2\mathbf{A}=-\mu_0\mathbf{J}$. If we use the same time independent $\mathbf{A}$, we can prove that $\int_{\partial^+ \Sigma}(\nabla\times\mathbf{A})\cdot d\mathbf{x}$$=\mu_0 \int_{\Sigma}\mathbf{J}\cdot d\mathbf{S}$ as shown here step by step.
In this case, with
$$\mathbf{A}(\mathbf{x},t):=\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}},$$
I do not know how to proceed to calculate the Laplacian of $\mathbf{A}$ and, yes, I would be interested in knowing what mathematical assumptions and mathematically rigourous steps we have to take and very grateful to anybody helping me in finding a resources containing them or explaining them here...

12. May 27, 2017

vanhees71

Minkowski and modern theory of generalized functions is your friend in answering your question. What's derived in my pdf is that the retarded Green's function of the D'Alembert operator is given by
$$G_{\text{ret}}(x)=\frac{1}{4 \pi |\vec{x}|} \Theta(x^0) \delta(x^0-|\vec{x}|),$$
i.e.,
$$\Box_x G_{\text{ret}}(x-y)=\delta^{(4)}(x-y).$$
Now we have
$$\Box A^{\mu}=\frac{1}{c} j^{\mu}$$
and thus
$$A^{\mu}(x)=\int_{\mathbb{R}^4} \mathrm{d}^4 x' G_{\text{ret}}(x-y) \frac{1}{c} j^{\mu}(y).$$
As I said before, for a mathematically strict proof you have to show that you can commute integration with differentiation, but provided you can do it, applying the d'Alembert operator leads immediately to the result due to the property of the Green's function.

13. Jun 1, 2017

samalkhaiat

14. Jun 15, 2017

DavideGenoa

Thank you both for your answers! Since I still do not understand the steps necessary to prove the desired result shown in your links, nor the steps I have found in Griffiths' Introduction to Electrodynamics, § 10.2.1, I am going to ask about my problems with these last ones, which are the clearest for my present (low) level of preparation, here.