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Argument for transversality of EM wave from absence of monopole radiation: Consider a hypothetical monopole electromagnetic wave. Then, a radially pulsating spherical charge distribution would emit such spherical outgoing waves. As the fundamental equations of physics are symmetrical under time reversal (t → −t), an incoming imploding spherical monopole wave would make an otherwise static spherical charge distribution pulsate radi- ally. Each charge pulsates radially because of a radial force acting on it. The form of the Lorentz force law then implies there were a radial E field acting on the charges. But a radial E field is perpendicular to the imploding spherical wave front and therefore in the same direc- tion of its motion, so that the E field would have a longitudinal component. This qualitative argument implies that a monopole electromagnetic wave is necessarily longitudinal. We do know, however, that outside a spherical charge distribution the electromagnetic field is static, as the monopole piece of the electromagnetic field is non-radiative. Students are exposed to this argument also in Newtonian gravity: because of the inverse square force law (common to both Newton’s and Coulomb’s laws) the field strength of any spherical charge (or mass) distribution is the same as if all the charge (or mass) were concentrated at the center. Specifically, the electric field outside a radially pulsating spherical charge distribution is static. Therefore, the longitudinal component of the E field cannot exist, as such a longitudinal component would have to be radiated by a pulsating spherical charge distribution. Show that E does not have a longitudinal component: Consider a plane electromagnetic wave traveling in vacuum in the direction of ˆx. (Our coordinates may be rotated to this orientation.) Because of the planar symmetry the most general electric field strength is E = E(t, x). In what follows we omit the explicit time dependence of fields. We first show there can be no component Ex: construct a gaussian surface enclosing a volume V as in Fig.
- According to Gauss’s law, (^) ∮ ∂V E · dA = q ! 0
where q is the total charge inside the gaussian surface, and! 0 is the permittivity of vacuum. Here, dA is a surface element normal to the (orientable) surface ∂V , defined conventionally such that it is positive when pointing outward. Calculate the integral on the LHS: take E = Ex(x)ˆx + Ey(x)ˆy + Ez (x)ˆz. Then ∮ ∂V E · dA = [Ex(x + dx) − Ex(x)] dy dz = 0 as there are no charges. The components Ey and Ez identically do not contribute to the LHS integral, as these components are not functions of y or z, respectively, and as they are in the plane of the faces of the surface whose normals are in the direction of ˆx. Notice that there
is no flux of Ex through the faces of the gaussian surfaces in the ˆy or ˆz directions because of the orthogonality of Ex to the surface’s normals. This immediately implies ∂Ex/ ∂x = 0, or Ex does not obey a non-trivial wave equation. Therefore, the most general electric field is E = Ey(x)ˆy + Ez (x)ˆz. Without loss of generality, we may rotate our coordinate system so that E = Ey(x)ˆy. Our demonstration depends on the lack of charges. Presence of charges requires a non- vanishing gradient of Ex. Indeed, Gauss’s law would not be violated for longitudinal waves without assuming vacuum, as there are charges and inhomogeneities in matter. Therefore, this demonstration does not rule out longitudinal waves in matter or inhomogeneous media. Fig. 1.— A gaussian surface constructed in a shape of a rectangular box whose faces parallel the axes. The ˆx components of either E or B have to be independent of x. The arrows indicate a case in which they are not. Then, the surface integral must be nonzero, and proportional to their x derivatives. In the case of E this implies free charges inside the box, i.e., a continuous charge distribution, which contradicts the vacuum assumption. In the case of B that distribution would be of magnetic monopoles. Show that E may have a transverse component: Consider next an electric field E = Ey(x)ˆy, as in Fig. 2. Applying Gauss’s law does not yield a restriction on Ey, because the LHS of the integral vanishes identically, as Ey = Ey(x): ∮ ∂V E · dA = [Ey(x, y + dy) − Ey(x, y)] dx dz = 0. This step may of course be combined with the preceding one by taking E = Ex(x)ˆx +
Fig. 2.— A transverse component to either E or B leads to no contradiction: as the fields are functions of only the spatial coordinate x (in addition to the time t), the surface integral of the fields vanish along all faces, either because of the field being parallel to the face (faces parallel to the x − y and y − z planes) or because the field exiting the surface completely balances the field entering (faces parallel to the x − z plane). It is therefore important that the fields are only functions of x, otherwise the cancellation would not be exact. Fig. 3.— An Ampere loop constructed so that the ˆz component of the B field is perpendicular to the entire loop, so that it does not contribute to the line integral around it. When applied to the Ampere–Maxwell law, an electric field in the direction of ˆy has zero flux through the loop. The line integral around the loop is proportional to the x derivative of the ˆy component of B. When applied to the Faraday law, the x derivative of the ˆy component of E is proportional to the time derivative of the ˆz component of B. The arrow indicates a ˆy component to either E or B.
Show that the E and B fields satisfy the wave equation with speed c: Without the pre- ceding parts of the argument this part is merely a demonstration of consistency of the trans- verse wave equation with the Maxwell equations, not their unavoidability. We now use the loop in Fig. 4 to evaluate the Ampere–Maxwell law. On the LHS, ∮ ∂A B · dS = [−Bz (x + dx) + Bz (x)] dz = − ∂Bz ∂x dx dz. The flux of the E field is simply ΦE = Ey(x) dx dz, so that the RHS is evaluated to equal ! 0 μ 0 dΦE dt = −! 0 μ 0 ∂Ey ∂t dx dz , so that the Ampere–Maxwell law yields ∂Bz ∂x = −! 0 μ 0 ∂Ey ∂t
Fig. 4.— An Ampere loop constructed so that the ˆz component of the B field is in the plane of the loop. The line integral around the loop is proportional to the x derivative of the ˆz component of B, which by Faraday’s law is proportional to the time derivative f the electric field’s flux through it. The arrow indicates the ˆz component of the B field. The E field (which is in the direction of ˆy) is not shown. We next consider again the Ampere loop in Fig. 3, but this time apply it to the Faraday law, (^) ∮ ∂A E · dS = − dΦB dt
Fig. 5.— A current sheet in the y − z plane, carrying current density j. The direction of the B field is as indicated. The Ampere loop is in the x − z plane. together, one finds that B = 12 μ 0 j. Notably, B is independent of the distance from the current sheet. Consider next a current sheet that is abruptly turned on at time t = 0. That is, j = −j Θ(t) ˆy, where Θ(t) is the Heaviside step function. Before the current is turned on, the B field is zero everywhere. Following the abrupt turning on of the current, a non–zero B field is starting to fill up space, the wavefront propagating at the speed c. Therefore, we have a propagating wavefront, so that in front of the wavefront the B field vanishes, and behind it is uniform as shown above. To find the accompanying E field we use the Faraday law for the Ampere loop in Fig. 6. Only the part of the Amp`ere loop behind the wavefront has non-vanishing B–field flux. As the wavefront is propagating with speed c, the area of the loop where there is a B field is increasing, so that there is a time–changing B–field flux through the loop. The E field between the wavefront and the current sheet is parallel to the current sheet, and counter-parallel to the current. That is, E = E yˆ. (See Fig. 6.) Using the Faraday law, (^) ∮ ∂A E · dS = − dΦB dt
one finds that the RHS equals
∂A E^ ·^ dS^ =^ −E^ ∆y, and the flux of the^ B^ field through the loop at the time t > t 0 is ΦB = B c(t − t 0 ) ∆y, so that the RHS of the Faraday law equals − dΦB / dt = cB ∆y. Setting the two hand sides equal to zero, we find E ∆z = cB ∆z, or E = cB.
Although our argument was made only for the configuration of a current sheet that is abruptly turned on, the result that E = cB for an electromagnetic wave is general. The general result, however, is beyond the scope of the introductory course. An important consequence is that in an electromagnetic wave the E and B fields are in phase. That is, as at any given time E = cB, when E is maximal so is B, when E vanishes B is zero too, etc. The result that E = cB appears in many introductory texts. However, in such books an extra assumption is typically made, namely that the fields are harmonic and are in phase. Students of the introductory course, who normally are not familiar with Fourier theory, often find it unconvincing to base an important general physical result on the mathematical properties of sinusoidal waves, in addition to having to memorize yet another factoid, namely that the fields are in phase. At this point one is in a position to make the following discussion. We found that E = cB. this means that the ratio E/B for an electromagnetic wave equals c, which depends in magnitude on the system of units. E.g., in SI units this ratio is 3 × 108 m s−^1 , and in cgs units is it 3 × 1010 cm s−^1. It is therefore natural to use units that put the E and B fields on equal footing, that is use units in which distance is measured in light-seconds. In such units the speed of light is c = 1 (light-)second/second, or simply c = 1. That is, in these units E = B. This choice of units becomes further motivated physically when one considers the energy density E in the electromagnetic field (not necessarily that of an electromagnetic wave), E = 12! 0 (E^2 + c^2 B^2 ). For an electromagnetic wave E = cB, so that the energy densities stored in the E and B fields are equal. It is therefore suggestive to make the two fields symmetrical by using units in which c = 1. It is often instructive to remind students that SI or cgs units were developed because of their convenience in describing everyday phenomena involving humans, and while all unit systems are in principle equivalent, these are not necessarily the units in which physical phenomena take their simplest or most natural form. We therefore showed that electromagnetic waves are transverse, and that the electric and magnetic fields are perpendicular to each other. Also the directions are so that E × B is in the direction of propagation. We have also shown they have the same speed c, that they are in phase, and that the magnitudes of the E and B is related by E = cB. This demonstration makes use of only the integral Maxwell equations, and is appropriate for the level of most calculus–based physics courses, and includes only arguments already available for the perspective students. We believe its main strength is that it avoids giving the students factoids without deep understanding, and instead empowers the student to gain deeper insight.
