Question about retarded scalar potential and Coulomb Gauge.

Main Question or Discussion Point

This is exact copy from Griffiths Introduction to Electrodynamics 3rd edition page 421. This is regarding to information travel in space. In time varying situation, E depend not only on V, but on A also.

There is a peculiar thing about the scalar potential in the Coulomb gauge: it is determined by the distribution of charge right now. If I move an electron in my laboratory, the potential V on the moon immediately records this change. That sounds particularly odd in the light of special relativity, which allows no message to travel faster than the speed of light. The point is that V by itself is not a physically measurable quantity-all the man in the moon can measure is E, and that involves A as well. Somehow it is built into the vector potential, in the Coulomb gauge, that whereas V instantaneously reflects all changes in $\rho$, the combination:

$$\vec E = -\nabra V - \frac{\partial \vec A}{\partial t}$$

does not; E will change only after sufficient time has elapsed for the "news" to arrive.

[ End Quote]

These are my questions:

1) If the information of moving the electron in the lab can only transmit as E to the moon, and it takes time for E to travel to the moon, why then the man on the moon can immediately records the change?

2) Why is potential not a measurable quantity?

3) In the next section about retarded potential:

$$V_{(\vec r , t)} =\frac 1 {4\pi \epsilon_0} \int_{v} \frac { \rho(\vec r’\hbox{,} t_r)}{|\vec r - \vec r'|} d\tau$$

This obvious show that V is time dependent. Is this because the information travel through space as E and this potential is obtained at the field point ( the moon in the current case) by calculating

$$V_{(\vec r , t)} = -\int \vec E \cdot d\vec l$$

Thanks

Alan

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Anyone?

I thought it is a very simple concept. Please tell me whether I have the correct idea:

1) Potential information cannot be transmitted.

2) Only E propagates, and it propagates at speed of light in free space.

3) Since potential is not transmitted, potential at the field point in distance can only be deduced from E that travel from the source to the field point.

4) Therefore the potential deduced from E at the field point at time t is actually what happened at tr=(t-r/c). Where r is the distance and c is the speed of light. Therefore the potential you see at the field point at t is due to the potential happened at tr at the source.

I thought this is very simple until I read the book!!! Please correct if I am wrong in this post.

Thanks

Born2bwire
Gold Member
The electric and magnetic fields are the measurable quantities because they are the force fields. They represent the forces exerted on charges due to other charges. The scalar and vector potentials are another set of variables that we can use to decompose the fields but obviously they aren't measurable since they are not the force fields.

The problem with the potentials is that they are non-unique since we can apply a suitable translation operator to either one and still result in the same fields. For ease of communication then, we choose a set of gauge conditions that will ensure unique results that can be readily compared. The Coulomb gauge is one that has the interesting consequence that the scalar potential is an instantaneous potential. However, since the scalar potential is not the measurable quantity then we find that the electric field is still retarded due to the effects of the vector potential.

If you want, look into the Lorenz gauge which decouples the scalar and vector potentials. The resulting equations is that the scalar and vector potentials are now wave equations and their solutions are retarded potentials.

The confusion probably arises due to your use of the scalar potential in electrostatics. In electrostatics, there is no gauge confusion because the scalar and vector potentials are time-independent. Thus, the electric field is only dependent upon the scalar potential. We do not talk about gauge conditions explicitly here but it is still a problem because we discuss the problem of defining the zero potential point. But this is such a trivial concept at that point that it is not discussed in detail with relation to gauge theorem.

The electric and magnetic fields are the measurable quantities because they are the force fields. They represent the forces exerted on charges due to other charges. The scalar and vector potentials are another set of variables that we can use to decompose the fields but obviously they aren't measurable since they are not the force fields.
I thought voltage can be measured!!

The problem with the potentials is that they are non-unique since we can apply a suitable translation operator to either one and still result in the same fields. For ease of communication then, we choose a set of gauge conditions that will ensure unique results that can be readily compared. The Coulomb gauge is one that has the interesting consequence that the scalar potential is an instantaneous potential.
Do you mean by instantaneous potential is because using Coulomb gauge, you get rid of the time dependent A part so there is no time information anymore?
However, since the scalar potential is not the measurable quantity then we find that the electric field is still retarded due to the effects of the vector potential.
Is this the part where the potential is not propagate from the source to the field point, only the E propagate and it take time to reach the field point. The potential established at the field point can only be coming from the E. Therefore the potential at the field point has a delay and are the retarded potentials?
If you want, look into the Lorenz gauge which decouples the scalar and vector potentials. The resulting equations is that the scalar and vector potentials are now wave equations and their solutions are retarded potentials.

The confusion probably arises due to your use of the scalar potential in electrostatics. In electrostatics, there is no gauge confusion because the scalar and vector potentials are time-independent. Thus, the electric field is only dependent upon the scalar potential. We do not talk about gauge conditions explicitly here but it is still a problem because we discuss the problem of defining the zero potential point. But this is such a trivial concept at that point that it is not discussed in detail with relation to gauge theorem.
When you said Lorentz gauge decouple the scalar and vector potential, do you mean using Lorentz gauge, we get:

$$\nabla ^2 V -\mu_0\epsilon_0 \frac {\partial ^2 V}{\partial t^2}= -\frac {\rho}{\epsilon_0} \;\hbox { and } \nabla ^2 \vec A -\mu_0\epsilon_0 \frac {\partial ^2 \vec A}{\partial t^2}= -\mu\vec J$$

Where each equation contain only one potential (either V or A).

Thanks for your time. It is very confusing. Can you comment on my post #2 whether I understand correctly?

Thanks

Alan

vanhees71
Gold Member
2019 Award
In classical electromagnetics the four-potential is an unobservable auxilliary quantity. It is only determined up to a gauge transformation.

Let's consider the Maxwell equations in vacuum, i.e., without polarizable media. In Heaviside-Lorentz units with $$c=1$$ the homogeneous equations read

$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$

These equations are identically fulfilled, if one introduces the four-potential, such that

$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} A^0, \quad \vec{B} = \vec{\nabla} \times \vec{A}.$$

For a given em. Field $$(\vec{E},\vec{B})$$ also any other four-potential with

$$\vec{A}'=\vec{A}+\vec{\nabla} \chi, \quad A^0={A'}^0-\partial_t \chi$$

for any scalar field, $$\chi$$ leads to the same em. field. Thus, one can impose an arbitrary auxilliary condition to fix the gauge potential. The most common choices are the Lorenz gauge (which is manifestly Poincare covariant),

$$\partial_t A^0+\vec{\nabla} \cdot \vec{A}=0$$,

or the Coulomb gauge, where the three-vector part of the four-potential is transverse,

$$\vec{\nabla} \cdot \vec{A}=0.$$

To determine the four-potential, subject to the one or the other gauge condition, we need the inhomogeneous Maxwell Eqs.

$$\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}-\partial_t \vec{E}=\vec{j}.$$

Introducing the four-potentials for the fields yields

$$-\partial_t \vec{\nabla} \cdot \vec{A}-\Delta A^0=\rho, \quad \vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}+\partial_t^2 \vec{A}+\vec{\nabla} (\partial_t A^0)=\vec{j}.$$

For the Lorenz gauge this simplifies to the set of decoupled wave equations for the four-momentum components

$$\Box A^0=\rho, \quad \Box \vec{A}=\vec{j} (\text{LG})$$

and for the Coulomb gauge

$$-\Delta {A'}^0=\rho, \quad \Box \vec{A}'+\vec{\nabla} (\partial_t {A'}^0)=\vec{j}.$$

In both cases, we can solve these equations with Green's function methods. For that we define the retarded propagator of the wave equation by

$$\Box_x G_{\text{ret}}(x,x')=\delta^{(4)}(x-x'), \quad G_{\text{ret}}(x,x') \propto \Theta(t-t').$$

The latter condition defines the retarded propagator as the one, where only information about the sources for $$t'<t$$ are relevant, i.e., there are causal actions only from the past not from the future (causality).

The solution is

$$G_{\text{ret}}(x,x')=\frac{1}{4 \pi} \Theta(t-t') \delta[(t-t')-|\vec{x}-\vec{x}'|],$$

which is a Lorentz invariant quantity, i.e., it behaves as a scalar field under proper orthochronous Lorentz transformations as to be expected from the manifest covariance of the Lorenz-gauge condition and Maxwell's Equations. The solution for the Lorenz gauge then reads

$$A^{\mu}(x)=\int_{\mathbb{R}^4} \mathrm{d}^4 x' G_{\text{ret}}(x,x') j^{\mu}(t',x')=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{j^{\mu}(t-|\vec{x}-\vec{x}'|,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$

For the Coulomb gauge we first solve for $$A^0$$. Here we need the Green's function of the Laplace operator

$$\Delta g(\vec{x},\vec{x}')=-\delta^{(3)}(\vec{x}-\vec{x}')$$

with the solution

$$g(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|}$$

From this we find the instantaneous scalar potential

$${A'}^0(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' g(\vec{x},\vec{x}') \rho(t,\vec{x}').$$

Plugging this solution in the equation for $$\vec{A}'$$ leads to

$$\Box \vec{A}'(x)=\vec{j}_{\perp}(x):=\vec{j}(x)-\vec{\nabla} A^0(x).$$

Using the conservation of the electric current which must hold due to Maxwell's Equations

$$\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0$$,

we find

$$\vec{j}_{\perp}(t,\vec{x})=\vec{j}(t,\vec{x})+\vec{\nabla}_x \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x'} \frac{\vec{\nabla}_{x'} \cdot \vec{j}(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}$$.

The equation for $$\vec{A}'$$ itself must again be solved with the retarded propagator due to the causality constraint,

$$\vec{A}'(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^4 x' G_{\text{ret}}(x,x') \vec{j}_{\perp}(x')$$.

However, one should note that $$\vec{j}_{\perp}$$ contains an instantaneous (non-local) part, which precisely cancels the contribution from the instantaneous Coulomb scalar potential $${A'}^0$$ in the electric field.

One can show after some somewhat tedious vector manipulations that the fields from the Coulomb and the Lorenz gauge give indeed the same field $$(\vec{E},\vec{B})$$, i.e., the retarded solutions of Maxwell's equations, as it must be. For the details see the very illuminating article,

J. D. Jackson, Am. Jour. Phys. 70, 917 (2002).

Born2bwire
Gold Member
When you said Lorentz gauge decouple the scalar and vector potential, do you mean using Lorentz gauge, we get:

$$\nabla ^2 V -\mu_0\epsilon_0 \frac {\partial ^2 V}{\partial t^2}= -\frac {\rho}{\epsilon_0} \;\hbox { and } \nabla ^2 \vec A -\mu_0\epsilon_0 \frac {\partial ^2 \vec A}{\partial t^2}= -\mu\vec J$$

Where each equation contain only one potential (either V or A).

Thanks for your time. It is very confusing. Can you comment on my post #2 whether I understand correctly?

Thanks

Alan
Yeah, and you can easily see that those are wave equations and thus the resulting equations for the potentials will be retarded and propagate at the speed of light.

As for potential information, using the Coulomb gauge the scalar potential changes instantaneously. That is why the man in the moon (smug jerk) can measure the change in the scalar potential instantly. Except the potential is not an observable, the electric field is. And thus since the electric field is still dependent upon the vector potential, the vector potential must change in such a way as to retain the time retardation of the electric field.

So yeah, I would say your post is correct except that I would abandon the idea of deriving what the potentials are (although I guess you could figure them out if you know both of the fields).

Yeah, and you can easily see that those are wave equations and thus the resulting equations for the potentials will be retarded and propagate at the speed of light.

As for potential information, using the Coulomb gauge the scalar potential changes instantaneously. That is why the man in the moon (smug jerk) can measure the change in the scalar potential instantly. Except the potential is not an observable, the electric field is. And thus since the electric field is still dependent upon the vector potential, the vector potential must change in such a way as to retain the time retardation of the electric field.

So yeah, I would say your post is correct except that I would abandon the idea of deriving what the potentials are (although I guess you could figure them out if you know both of the fields).
Thank you for taking the time to explain.

One more question about the highlighted sentence, why shouldn't I derive the potential ( on the moon) from the E and B field arrive at the moon? I thought those are the retarded potentials that are derived from the EM field propagate from the lab on earth.

Thanks

Alan

In classical electromagnetics the four-potential is an unobservable auxilliary quantity. It is only determined up to a gauge transformation.
what is the 4 potential. I only see $A^0$ which is the scalar potential V and the A which is the vector magnetic potential.

Let's consider the Maxwell equations in vacuum, i.e., without polarizable media. In Heaviside-Lorentz units with $$c=1$$ the homogeneous equations read

$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$

These equations are identically fulfilled, if one introduces the four-potential, such that

$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} A^0, \quad \vec{B} = \vec{\nabla} \times \vec{A}.$$

For a given em. Field $$(\vec{E},\vec{B})$$ also any other four-potential with

$$\vec{A}'=\vec{A}+\vec{\nabla} \chi, \quad A^0={A'}^0-\partial_t \chi$$

for any scalar field, $$\chi$$ leads to the same em. field. Thus, one can impose an arbitrary auxilliary condition to fix the gauge potential. The most common choices are the Lorenz gauge (which is manifestly Poincare covariant),

$$\partial_t A^0+\vec{\nabla} \cdot \vec{A}=0$$,

or the Coulomb gauge, where the three-vector part of the four-potential is transverse,

$$\vec{\nabla} \cdot \vec{A}=0.$$

To determine the four-potential, subject to the one or the other gauge condition, we need the inhomogeneous Maxwell Eqs.

$$\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}-\partial_t \vec{E}=\vec{j}.$$

Introducing the four-potentials for the fields yields

$$-\partial_t \vec{\nabla} \cdot \vec{A}-\Delta A^0=\rho, \quad \vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}+\partial_t^2 \vec{A}+\vec{\nabla} (\partial_t A^0)=\vec{j}.$$
Is $$\Delta A^0 = \nabla^2 A^0$$?

For the Lorenz gauge this simplifies to the set of decoupled wave equations for the four-momentum components

$$\Box A^0=\rho, \quad \Box \vec{A}=\vec{j} (\text{LG})$$

and for the Coulomb gauge

$$-\Delta {A'}^0=\rho, \quad \Box \vec{A}'+\vec{\nabla} (\partial_t {A'}^0)=\vec{j}.$$
Do you consider $$\nabla \chi$$ and $$\frac {\partial \chi}{\partial t}$$ the third and forth potential after V and A?

Thanks

Alan

vanhees71
Gold Member
2019 Award
Perhaps I should have explained my naming scheme better. For me electromagnetism is relativistic, and thus $$A^0=\Phi$$ and $$\vec{A}$$ together build a four-vector in Minkowski space $$(A^{\mu})=(A^0,\vec{A})$$.

Perhaps I should have explained my naming scheme better. For me electromagnetism is relativistic, and thus $$A^0=\Phi$$ and $$\vec{A}$$ together build a four-vector in Minkowski space $$(A^{\mu})=(A^0,\vec{A})$$.
Sorry, the book I use is still an undergrad book, I don't know what you refer to. I never learn Minkowski space.

vanhees71
Gold Member
2019 Award
Then forget about this at the moment. Then usually one talks about the scalar and the vector potential $$\Phi$$ and $$\vec{A}$$.

Born2bwire
Gold Member
Thank you for taking the time to explain.

One more question about the highlighted sentence, why shouldn't I derive the potential ( on the moon) from the E and B field arrive at the moon? I thought those are the retarded potentials that are derived from the EM field propagate from the lab on earth.

Thanks

Alan
Because for classical electrodynamics the potentials are not the primitives. The electromagnetic fields are the primitives and the observables. To get the potentials from measurement you would have to work backwards from the knowledge of both the electric and magnetic fields. The potentials are handy in terms of their use in calculations and derivations of the behavior of the system's fields. But I do not think that it would be as useful to us to know the potentials if we already know the field behavior.