# Radiation from an infinite current sheet

The Feynman LECTURES ON PHYSICS (NEW MILLENNIUM EDITION) by FEYNMAN•LEIGHTON•SANDS
VOLUME II discusses radiation from an infinite sheet of switched-on constant current in section "18-4 A travelling field" on page 18-15. The solution shows a constant E field and constant B field at a given point x after the wave front has passed x. The constant E field as been disputed.

A now closed PF discussion found at "https://www.physicsforums.com/threads/infinite-current-sheet-current-suddenly-turned-on.419209/" makes claims that there is no steady E field and thus Feynman is in error.

This same scenario is given as a problem in "Griffiths, David J. (2007), Introduction to Electrodynamics,. 3rd Edition; Pearson Education - Chapter 11, problem 24" and the solution is given at
"physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2011.24.pdf"
in Example 1 on page 4 which for the case a steady current. The result in this article agrees with Feynman's result of a constant E field.

Is Feynman right or wrong? I believe that Feynman is correct (although there isn't much discussion about the resulting E field). Can anyone comment on which answer is correct?

TSny
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• scoomer
Thank you TSny! Previously, I misread the responses. I now see they all agree that Feynman is correct, I should have been more careful.

vanhees71
Gold Member
Well, the calculation looks formally correct.

However, one must make sure that this very unphysical and singular source is at least formally fulfilling the continuity equation, which is an integrability condition for the Maxwell equations. If I understand right, what's calculated, it's a surface current in the ##yz## plane, i.e., the correct current density is
$$\vec{j}(t,\vec{r})=K(t) \delta(x),$$
where ##(x,y,z)## are the usual Cartesian coordinates for the position vector ##\vec{r}##
Since thus
$$\vec{\nabla} \cdot \vec{j}=0,$$
the tacid assumption that the charge density
$$\rho=0$$
is consistent with the continuity equation.

Supposed there's no mistake in the lengthy calculation, everything looks fine (formally), since they simply use the retarded Green's function for the d'Alembertian for the four-potential in the Lorenz gauge. It's of course a highly idealized academic solution of the Maxwell equations, never realizable fully in nature.

Some more caveats concerning the possible choices for ##K## are given in the above linked pdf too!

• QuantumQuest
Thank you vanhees71. I appreciate your response.

Staff Emeritus
The solution shows a constant E field and constant B field at a given point x after the wave front has passed x.

Why do you call this "radiation"?

vanhees71
Gold Member
It's radiative in the transient state. In the long-time limit, of course you get the static field for the stationary situation.

TSny
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It is very interesting that the long-time limit includes a uniform electric field as well as the magnetic field. The Poynting vector is nonzero at all points in these fields, which indicates a flux of field energy away from the current sheet. This energy flux must be associated with the work required to maintain the current in the sheet. That is, an external agent must input energy to the system to maintain the steady current, and this input energy is balanced by the flux of field energy away from the sheet.

tech99
Gold Member
It is very interesting that the long-time limit includes a uniform electric field as well as the magnetic field. The Poynting vector is nonzero at all points in these fields, which indicates a flux of field energy away from the current sheet. This energy flux must be associated with the work required to maintain the current in the sheet. That is, an external agent must input energy to the system to maintain the steady current, and this input energy is balanced by the flux of field energy away from the sheet.
Does the energy all come back again if the current stops? This is surely the test for radiation or not.

TSny
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Gold Member
Does the energy all come back again if the current stops? This is surely the test for radiation or not.
No, I don't think so. Feynman discusses suddenly stopping the current at some time. The energy that has been radiated up to that time just keeps propagating outward from the sheet. http://www.feynmanlectures.caltech.edu/II_18.html (section 4)