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3 Maxwell’s equations and material equations Contents 3.1 Macroscopic Maxwell’s equations 3.2 Material equations and LIH 3.3 The wave equation 3.4 Vector and scalar potentials Keywords: Maxwell equations, material equations, LIH, vector potential, scalar potential, wave equation and electromagnetic waves Ref: J. D. Jackson: Classical Electrodynamics; A. Sommerfeld: Electrodynamics. 3.1 Macroscopic Maxwell’s equations Any electromagnetic field in certain medium respects the following four equations in SI units, ∇ · D = ρ, ∇×E=− (3.1) ∂B , ∂t ∇ · B = 0, ∇×H=j+ (3.2) (3.3) ∂D , ∂t (3.4) 2 3 Maxwell’s equations and material equations where E and B are the electric and the magnetic fields respectively. The above equations are called macroscopic Maxwell’s equations. The first and the third equations are independently known as Gauss’ law for electrostatics and magnetostatics respectively. When we write them in Maxwell’s equation the status of these are raised from statics to dynamics. In other words the laws are valid even if the charges are moving. The second equation is nothing but the Faraday’s law whereas the fourth equation, the Ampere’s law modified by Maxwell with an additional term Ḋ, is called the Ampere-Maxwell law. It says that a by time varying magnetic field one can generate an electric field and it complements the Faraday’s law. These equations look slightly different in CGS and other units. The displacement vector D roughly tells how the medium is distorted when an electric field is applied or in other words it is a measure of response of the medium to the field. H is the analogue of D for the magnetic field. ρ represents the charge density and j is the current density i.e. current per unit area. These two densities are often termed as sources to the field. The ρ is measured in coulomb/m3 and j is expressed in ampere/m2 in SI units. The electric field is measured in newton/coulomb and the magnetic field SI unit is tesla. Permanent magnets are outside the scope of Maxwell’s electrodynamics which is essentially a classical description whereas the the permanent magnets necessarily need a quantum description. NPTEL Course: Wave Propagation in Continuous Media 3.2 Material equations and LIH 3 3.2 Material equations and LIH If one is supposed to find the electromagnetic field for a given set of ρ and j one has to solve the Maxwell’s equations. But the problem is that there are twelve unknowns since there are four vectors, E, D, B and H, each having three components and there are only eight equations(each vector equation is equivalent to three scalar equations). So to solve them we have to make some assumptions. These assumptions give us the material equations. Like Hooke’s law if the fields are not very strong one assumes D and H are linearly proportional to E and B. The E and B are analogues of the external or applied force (proportional to stress) and D and H are analogues of the strain in the medium. D = ǫE, (3.5) 1 B, µ (3.6) j = σE. (3.7) H= Such materials where these relations are valid are known as Linear-IsotropicHomogeneous or LIH medium. The proportionality constants ǫ and µ are known as permittivity and permeability of the medium respectively. Permittivity constant, ǫ0 , for the vacuum is approximately (1/36π) × 10−9 and the permeability, µ0 , for the vacuum is equal to 4π × 10−7 in SI units. The quantity, ǫr = ǫ/ǫ0 is called dielectric constant or the relative permittivity of NPTEL Course: Wave Propagation in Continuous Media 3 Maxwell’s equations and material equations 4 the medium. The dielectric constant usually depends on temperature of the medium. In case of alternating fields it also depends on the frequency of the applied field. Materials with permittivity greater than that of the vacuum, i.e. ǫ > ǫ0 , are called dielectrics whereas a material having permeability less than that of the vacuum, µ < µ0 , is known as a diamagnetic. Paramagnetic substances are those for which, µ > µ0 . Paramagnetic substances(like Magnesium, Manganese, Oxygen, Platinum) are attracted towards the high magnetic field whereas the diamagnetic materials (like Antimony, Bismuth, Carbon/Copper) are repelled by the magnetic fields. Problem 1: Find the dimensions of ǫ and µ in SI units. The third equation, (3.7), above is known as Ohm’s law, where the current density is assumed to be proportional to the electric field. Problem 2: Show the familiar form of the Ohm’s law (i.e. V=IR, where V is the voltage across the sample, I is the current through the sample and R is the resistance of the sample) by taking a sample as a straight wire of some length with some cross-section. If we take the divergence of the equation (3.4), we have ∇ · (∇ × H) = ∇ · j + ∂ρ = 0, ∂t (3.8) where we have used the equation (3.1) and the fact that the divergence of a curl vanishes. This is the continuity equation and it shows the charge NPTEL Course: Wave Propagation in Continuous Media 3.3 The wave equation 5 conservation. With the materials equations now all the equations can be written in terms E and B and can be solved with the constraint (3.3). 3.3 The wave equation In the subsequent chapters we shall be mostly interested in the electromagnetic waves. We shall restrict ourselves to the source free regions, i.e. ρ = 0 and j = 0, unless otherwise specifically mentioned. With D = ǫE and H = B/µ we have the source free Maxwell’s equations: ∇ · E = 0, ∇×E=− (3.9) ∂B , ∂t ∇ · B = 0, ∇ × B = µǫ (3.10) (3.11) ∂E . ∂t (3.12) The interesting part of the electromagnetic field is that even for vanishing sources it has non trivial solutions, which are known as electromagnetic waves. Now if we take the curl of the equation (3.10) we have, ∇ × (∇ × E) = ∇(∇ · E) − ∇2 E = − ∂(∇ × B) . ∂t (3.13) Now we use equations (3.9) and (3.12) to obtain the wave equation for the electric field, ∂ 2E 1 ∂ 2E 2 µǫ 2 − ∇ E = 2 2 − ∇2 E = E = 0, ∂t c ∂t (3.14) NPTEL Course: Wave Propagation in Continuous Media 3 Maxwell’s equations and material equations 6 √ √ where c = 1/ µǫ (for vacuum c = 1/ µ0 ǫ0 ≡ 3 × 108 m/s) is the speed of the electromagnetic wave in the medium and the operator is known as d’ Alembertian. Problem 3: Obtain the wave equation for the magnetic field, i.e. B = 0. In the next chapter we shall find the solutions of these wave equations which will be nothing but the electromagnetic waves. 3.4 Vector and scalar potentials Potentials for the fields are defined in such a manner that their space gradients determine the electric and magnetic fields uniquely and they are also consistent with the Maxwell’s equations. The form of the equation (3.3) suggests that one can always define a vector potential, A, such that curl of which will produce the magnetic field B, B = ∇ × A. (3.15) Note that divergence of B will be vanishing by the vector identity mentioned above. Substituting the vector potential in the equation (3.2), we have with a little rearrangement ∇ × (E + ∂A ) = 0. ∂t NPTEL Course: Wave Propagation in Continuous Media (3.16) 3.4 Vector and scalar potentials 7 Now one can use the vector identity ‘curl of gradient vanishing’ to define a scalar potential, φ, such that E+ ∂A = −∇φ, ∂t (3.17) so that the electric field is given by, E=− ∂A − ∇φ. ∂t (3.18) Problem 4: Eliminate E and B using (3.15) and (3.18) from Maxwell’s equations and derive the equations for scalar and vector potentials. Answers 4: ρ ∂(∇ · A) =− , and ∂t ǫ 1 ∂ 2A 1 ∂φ = µj. − ∇2 A + ∇ ∇ · A + 2 2 2 c ∂t c ∂t ∇2 φ + (3.19) (3.20) Problem 5: Show that if we change A by A + ∇Λ and φ by φ − ∂Λ ∂t (Λ being a scalar function of space-time) the E and B remain unchanged hence the scalar and vector potentials are not unique for a given electromagnetic field. The above freedom in the choice of scalar and vector potentials is known as the gauge degree of freedom. Usually one fixes this freedom either by choosing ∇ · A = 0 (Coulomb, radiation or transverse gauge) or by setting ∇·A+ 1 ∂φ c2 ∂t = 0 (Lorentz gauge). Using the latter condition the potential equations become inhomogeneous wave equations i.e. equations with source terms, φ = ρ/ǫ and A = µj. NPTEL Course: Wave Propagation in Continuous Media 8 3 Maxwell’s equations and material equations Problem 6: Show that Lorentz condition does not fix the gauge completely: the potentials are not determined uniquely. Two different sets of potentials, both satisfying Lorentz condition can produce the same electromagnetic field. NPTEL Course: Wave Propagation in Continuous Media