# Proving Lorenz Gauge Choice: A Getty's Physics Exploration

• DavideGenoa
In summary, the Lorenz gauge choice uses the magnetic vector potential and electric potential to satisfy Maxwell's equations, including the identity of ##\nabla^2\mathbf{A}-\varepsilon_0\mu_0\frac{\partial^2 \mathbf{A}}{\partial t^2}=-\mu_0\mathbf{J}##. To prove this, one can use the Green's function of the D'Alembert operator and the Mills representation to transform the time-dependent potential into a position-dependent one. However, to fully understand the calculation, one would need to have a deeper understanding of Green's functions and Lebesgue integrals.
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!

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DavideGenoa
As to ##\nabla\cdot(\nabla\times\mathbf{A})=0##, I have realized 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...

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

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

DavideGenoa
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
$$G(t,\vec{x})=\frac{\Theta(t)}{4 \pi r} \delta \left (t-\frac{r}{c} \right), \quad \text{with} \quad r=|\vec{x}|.$$

@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?

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!).

#### Attachments

• retarded-potential.pdf
138.8 KB · Views: 211
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!

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 :-).

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...

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.

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.

## 1. What is the Lorenz gauge choice and why is it important in physics?

The Lorenz gauge choice is a mathematical condition used in the study of electromagnetic fields. It ensures that the equations of motion for the fields are consistent with the laws of electromagnetism. This gauge choice is important because it simplifies the equations and makes them easier to solve, leading to a better understanding of electromagnetic phenomena.

## 2. How does Proving Lorenz Gauge Choice contribute to the field of physics?

Proving Lorenz Gauge Choice provides a rigorous mathematical proof of the validity of this gauge choice. This contributes to the field of physics by giving scientists and researchers a deeper understanding of the fundamental principles of electromagnetism and providing a solid foundation for future studies and applications in this area.

## 3. What methods were used in the exploration of Proving Lorenz Gauge Choice?

The Getty's Physics Exploration used a combination of mathematical analysis and computer simulations to explore the properties of the Lorenz gauge choice. This involved creating and solving equations, as well as testing the results with various numerical methods to ensure accuracy.

## 4. Are there any limitations to the findings of Proving Lorenz Gauge Choice?

As with any scientific study, there are always limitations to consider. In this case, the exploration was focused on a specific set of conditions and assumptions, so the results may not apply to all situations. Additionally, new discoveries and advancements in the field of physics may lead to further refinement of the Lorenz gauge choice in the future.

## 5. How can the findings of Proving Lorenz Gauge Choice be applied in real-world scenarios?

The Lorenz gauge choice has applications in many areas of physics, including electromagnetism, quantum mechanics, and relativity. By providing a solid proof of its validity, the findings of Proving Lorenz Gauge Choice can be used to guide and inform future research in these fields. Additionally, the simplified equations resulting from this gauge choice can be applied to practical problems in engineering and technology.

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