Ben Gurion University of the Negev www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter Physics 2 for Electrical Engineering Lecturers: Daniel Rohrlich , Ron Folman Teaching Assistants: Ben Yellin, Yoav Etzioni Grader: Gad Afek Week 11. Electromagnetic radiation – E • Divergence theorem • Maxwell’s equations in differential form • Current conservation • Electromagnetic waves • Poynting vector Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38. This set of four fundamental equations for E and B, E dA 0 q (Gauss’s law) B dA 0 d E dr B dt d B dr 0 I 0 0 E , dt are Maxwell’s equations in integral form. (Faraday’s law) (Ampère’s law as modified by Maxwell) In 1929 – 65 years after Maxwell’s prediction – M. R. Van Cauwenberghe was the first to measure “displacement current”, on a round parallel-plate capacitor of capacitance C = 100 pF and effective radius R = 40.0 cm. The potential across the capacitor, which alternated with frequency f = 50.0 Hz, reached V0 = 174 kV. We calculated that the maximum field B measureable at the edge of the capacitor was 2.73×10–8 T. You ask: What was so difficult about this experiment, that it took 65 years to do? After all, the two terms I and ε0 dΦE/dt in the Ampère-Maxwell law yield the same B, and it was not difficult to verify Ampère’s law! In 1929 – 65 years after Maxwell’s prediction – M. R. Van Cauwenberghe was the first to measure “displacement current”, on a round parallel-plate capacitor of capacitance C = 100 pF and effective radius R = 40.0 cm. The potential across the capacitor, which alternated with frequency f = 50.0 Hz, reached V0 = 174 kV. We calculated that the maximum field B measureable at the edge of the capacitor was 2.73×10–8 T. Two physicists who did the measurement in 1985 (thinking they were the first) answered this question in their paper: From the integral form of Maxwell’s equations, we anticipate electromagnetic radiation: Intuitively, we see that a changing electric flux will generate a transient magnetic field around it, which will generate a transient electric field, etc. etc. while the wave spreads through space. Electromagnetic radiation is a phenomenon that spans more than 18 orders of magnitude (in wavelength and frequency)! To calculate all the features of electromagnetic waves, we need Maxwell’s equations in their differential form. With the aid of Stokes’s theorem, we converted two of Maxwell’s equations to differential form: B E t E B 0 J 0 0 . t To convert the other two of Maxwell’s equations, we need the “divergence theorem”. E The divergence of a vector field E(x,y,z), written E, is the following: E Ex E y Ez x y z . Note that E is a scalar field. Note that the notation E is reasonable because E is (formally) the scalar product of the vectors , , x y z and E = (Ex, Ey, Ez), just as E is (formally) the vector product of the same two vectors. E Example 1: What is V ? E Example 1: What is V ? V V Answer: V x x y y 2 2 2 x 2 y 2 z 2 V z z V . This sum of second derivatives is called the “Laplacian” and written as “del-squared”: 2 2 2 2 . x 2 y 2 z 2 E Example 2: What is r ? E Example 2: What is r ? Answer: r x y z 3 . x y z E Example 3: E(r) is the electric field of a uniformly charged ball of radius R with total charge q. What is E(r) ? E Example 3: E(r) is the electric field of a uniformly charged ball of radius R with total charge q. What is E(r) ? Answer: As we know, the electric field E(r) of any electrostatic configuration is minus the gradient of the electric potential V(r), i.e. E(r) V (r). Hence E 2V , q 4 r , 0 V where (3R 2 r 2 )q , 8 0 R 3 E 2 q 8 0 R (3R x y z ) 2 3 r R . For r ≤ R we have r R 2 2 2 3q 4 0 R 3 . E Example 3: E(r) is the electric field of a uniformly charged ball of radius R with total charge q. What is E(r) ? Answer: As we know, the electric field E(r) of any electrostatic configuration is minus the gradient of the electric potential V(r), i.e. E(r) V (r). Hence E 2V , q , r R 4 r 0 V where (3R 2 r 2 )q . For r ≥ R we have , r R 8 0 R 3 2 1 x 1 x2 3 , so E 0. x 2 r x r 3 r3 r5 E Example 4: What is ( A) ? (A is any vector field.) E Example 4: What is ( A) ? (A is any vector field.) Answer: We write out A: A Az Ay , Ax Az , Ay Ax , z z x x y y so 2 2 2 A Az Ay Ax xy xz yz 2 2 2 Az Ay Ax yx zx zy 0. Divergence theorem The divergence theorem is Stokes’s theorem in 3D, but without curvature. We extend our notation: If V denotes any volume, then ∂V denotes the boundary of the volume (the surface bounding V ). ∂V V Divergence theorem To prove the divergence theorem, we assume that we can reduce any volume V to infinitesimal cubes: Double pyramid reduced to cubes Heart made of cubes Divergence theorem The divergence theorem tells us that E dA over any closed V surface ∂V is equal to the sum nE dA over all the cubes n in the volume V that has ∂V as its boundary. Divergence theorem To see why the divergence theorem is true, all we have to do is look at two neighboring cubes and at E dA for each one: Divergence theorem To see why the divergence theorem is true, all we have to do is look at two neighboring cubes and at E dA for each one: The contributions of their common side cancel! Divergence theorem When we put all the cubes together, the only sides that contribute (that are not cancelled by facing sides) are the faces along ∂V ! The contributions of their common side cancel! Divergence theorem When we put all the cubes together, the only sides that contribute (that are not cancelled by facing sides) are the faces along ∂V ! The divergence theorem thus tells us E dA V nE dA, n i.e. that the integral over ∂V equals the sum of the contribution of each cube. We will now compute the contribution of an infinitesimal cube. In the limit of infinitely many cubes, the sum over cubes will become an integral over the volume V. Divergence theorem We compute the flux E dA for an infinitesimal cube. The cube is centered at a point r = (x,y,z) with the area elements dA pointing along these axes. There are six contributions to the total flux: Ey(x,y+Δy/2,z)ΔxΔz y z Ez(x,y,z+Δz/2)ΔxΔy x Ex(x+Δx/2,y,z)ΔyΔz –Ex(x–Δx/2,y,z)ΔyΔz –Ey(x,y–Δy/2,z)ΔxΔz –Ez(x,y,z–Δz/2)ΔxΔy Divergence theorem We compute the flux E dA for an infinitesimal cube. The cube is centered at a point r = (x,y,z) with the area elements dA pointing along these axes. There are six contributions to the total flux, and they add up to E x ( x x/ 2, y, z ) E x ( x x/ 2, y, z ) E dA xyz x E y ( x, y y/ 2, z ) E y ( x, y y/ 2, z ) y xyz E z ( x, y, z z/ 2) E z ( x, y, z z/ 2) xyz z Divergence theorem We compute the flux E dA for an infinitesimal cube. The cube is centered at a point r = (x,y,z) with the area elements dA pointing along these axes. There are six contributions to the total flux, and they add up to E x E y E z E dA xyz E xyz . y z x The sum of E dA over all infinitesimal cubes then goes over to a volume integral, and we obtain… Divergence theorem …the divergence theorem: E dA ( E) dxdydz , V V i.e. the flux of E through the boundary ∂V of the volume V equals the integral of E over the volume. Maxwell’s equations in differential form Gauss’s law says E dA 0 0 (r)dxdydz , q 1 V V where ρ(r) is charge density. But the divergence theorem says E dA ( E) dxdydz V so V (r ) E dxdydz dxdydz . V V 0 . This is true for any V, hence E 0 , Maxwell’s equations in differential form Gauss’s law for B says B dA 0 . V But the divergence theorem says B dA ( B) dxdydz , V therefore V B dxdydz 0 . This is true for any V, hence V B 0 . Maxwell’s equations in differential form Using Stokes’s theorem and the divergence theorem, we have converted all four Maxwell equations to differential form: E 0 B 0 B E t E B 0 J 0 0 . t Current conservation What happens when we apply the div operator to Ampère’s law, with and without Maxwell’s modification? E yields Applying to B 0 J 0 0 t E B 0 J 0 0 . t But we proved ( A) 0 for any A, and Maxwell’s equations say E = ρ/ε0, so we conclude 0 J t What does this equation mean? . Current conservation The equation 0 J is called the “continuity equation”. t What does it mean? Let’s integrate both sides of this equation over the volume V of dq , an infinitesimal box. We obtain 0 J dxdydz dt V where q is the amount of charge in the box. Next, using the divergence theorem, we can rewrite this equation as dq J dA . dt V Current conservation dq The left side of the equation J dA is the net flux of dt V current out of the box. The right side is the rate of change of the total charge in the box. Jy(x,y+Δy/2,z)ΔxΔz y z Jz(x,y,z+Δz/2)ΔxΔy x Jx(x+Δx/2,y,z)ΔyΔz –Jx(x–Δx/2,y,z)ΔyΔz –Jy(x,y–Δy/2,z)ΔxΔz –Jz(x,y,z–Δz/2)ΔxΔy Current conservation dq The left side of the equation J dA is the net flux of dt V current out of the box. The right side is the rate of change of the total charge in the box. Ampere’s law, without Maxwell’s modification, implies that the net flux of current out of any box vanishes: J d A 0. V No charge can build up anywhere. That is certainly not true for a circuit with a capacitor inside. Current conservation dq The left side of the equation J dA is the net flux of dt V current out of the box. The right side is the rate of change of the total charge in the box. Ampere’s law with Maxwell’s modification simply states that dq . Anywhere that the net charge is conserved: J dA dt V flux of current out of a box is not zero, charge must build up in that box, or leave that box. Electromagnetic waves Let’s learn more tricks with the curl and div operators, and with Maxwell’s equations. Example 1: What is ( E) ? Electromagnetic waves Let’s learn more tricks with the curl and div operators, and with Maxwell’s equations. Example 1: What is ( E) ? Answer: We have E x E z E z E y ( E) x , ( E) y and y z z x E y E x ( E) z , from which we can compute the x y x-component of ( E). It is ( E)x E y E x E x E z . y x y z z x Electromagnetic waves We have ( E)x Ez E y y z E y E x E x E z y x y z z x E x E y E z 2 2 2 Ex 2 2 2 x x y z x y z E 2 E x , x where we added and subtracted ∂2Ex/∂x2 on the right side. Electromagnetic waves Since ( E)x E 2 E x , it is easy to guess x how the other components will turn out; we obtain ( E) E 2E . Both terms on the right side are vectors, as they must be since the left side is a vector. Electromagnetic waves Now let’s consider Maxwell’s equations in the vacuum (no charge ρ or current density J): E 0 B 0 B E t E B 0 0 . t Electromagnetic waves B Let’s apply to the Maxwell equation E and t E . insert the vacuum Maxwell equation B 0 0 t B 2E We get ( E) B 0 0 , 2 t t t and since ( E) E 2 E 2 E in the vacuum, we arrive at a wave equation: 2 E 0 0 2E t 2 0. Electromagnetic waves Let’s try a plane wave solution to this wave equation: E(r,t) = E cos (k·r – ωt + δ) , where E is a constant vector that fixes the amplitude and the polarization of the wave, k = 2π/λ is the wave number of the wave, and ω is its angular frequency. Substituting this plane wave into E 0 0 2 2E 0, we obtain a solution if and t only if – k2 + μ0ε0 ω2 = 0, hence ω/k = 1 / 0 0 c. 2 Electromagnetic waves We found that E(r,t) = E cos (k·r – ωt + δ) is a solution if and only if ω/k = 1 / 0 0 c. We can determine the speed of this wave by checking at what speed a crest of the wave moves. At a crest of the wave, the phase k·r – ωt + δ is unchanging over time. For the component of r in the k-direction, we have d dr dr ω 0 kr – ωt k ω, hence dt dt dt k 1 0 0 c . Around 1864, Maxwell calculated the speed of these waves. Imagine how he felt when he obtained the speed of light! Electromagnetic waves The speed of electromagnetic waves is v = ω/k = 1 / 0 0 c. Today’s data: μ0 = 4π × 10–7 T·m/A, ε0 = 8.854187817 × 10–12 F/m, c = 2.99792458 × 108 m/s. “We can scarcely avoid the conclusion that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena.” (J. C. Maxwell) Electromagnetic waves Heinrich Hertz was the first (in 1886) to verify Maxwell’s prediction of electromagnetic waves travelling at the speed of light. Receiver Spark Gap Transmitter Receiver Spark Gap Transmitter Electromagnetic waves Summary: Maxwell’s equations predict that electromagnetic waves (including light waves) travel at speed c in vacuum. The accepted value of c today is 299,792,458 m/s. B 0 E t Faraday’s law B 0 E t Vector identity ( B) ( E) E t Gauss’s law in a vacuum: 2 •E = 0 Ampère’s law, as modified by Maxwell 2 E 1 E 2 2 E 0 0 2 2 E t t c t Electromagnetic waves What else can we find out about electromagnetic waves? Applying the Gauss law 0 E to our wave solution E(r,t) = E cos (k·r – ωt + δ), we obtain 0 E E x E y E z cos(k r t ) y z x k E sin(k r t ) . Since sin (k·r – ωt + δ) does not vanish, it follows that that k·E = 0: the electric field of an electromagnetic wave in a vacuum is transverse, i.e. perpendicular to the wave vector k. Electromagnetic waves What else can we find out about electromagnetic waves? B Applying Faraday’s law E to our wave solution t E(r,t) = E cos (k·r – ωt + δ), we obtain B k E sin(k r t ) E ; t if we try B(r,t) = B cos (k·r – ωt + δ), we obtain a solution provided k × E = ωB! Thus B is perpendicular to E and to k (i.e. B is transverse) and propagates in phase with E. Since we have seen that ω = ck, we have E = cB. Electromagnetic waves B is perpendicular to E and to k (i.e. B is transverse) and propagates in phase with E. E k B Electromagnetic waves Example: Prove using one of Maxwell’s equations that B is perpendicular to k (i.e. B is transverse). E k B Electromagnetic waves Example: Prove using one of Maxwell’s equations that B is perpendicular to k (i.e. B is transverse). E k B Answer: From 0 B we derive 0 B cos(k r t ) k B sin(k r t ) . Since the sine function does not vanish identically, it follows that k·B = 0, i.e. B is transverse. Poynting vector EB The Poynting vector S, defined as S , has units W/m2. 0 Poynting vector EB The Poynting vector S, defined as S , has units W/m2. 0 What is the energy density in an electromagnetic wave? We have formulas for the energy densities uE and uB in constant electric and magnetic fields E and B, respectively; the total electromagnetic energy density in constant E and B is uE + uB = ε0E2/2 + B2/2μ0. In a wave, the energy density is half as much, because it is the time average of the oscillating fields. Now using E = cB and c2 = 1/ε0μ0, we can write the energy density in an electromagnetic wave as EB/2μ0c. Poynting vector EB The Poynting vector S, defined as S , has units W/m2. 0 What is the current of energy density in an electromagnetic wave? It is the energy density times the speed of light, c, namely EB/2μ0. The direction of the energy flow is the same as the direction of S, which is parallel to k. E B Thus the time-averaged Poynting vector Sav= , also 20 called the intensity, represents the current of energy density in an electromagnetic wave. Poynting vector Let electromagnetic radiation fall upon a totally absorbing surface of area A. If we multiply the time-averaged Poynting vector S av E B by the area A, we obtain the rate of energy 2 0 absorption by the surface. It can be shown that SavA/c is the force on the surface. Pressure is force per unit area, thus (SavA/c)/A = Sav/c is the electromagnetic radiation pressure on the surface. If the surface totally reflects the radiation instead of absorbing it, the pressure is doubled, i.e. to 2Sav/c, since the change in momentum per unit surface is doubled. Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13: The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is polarized along the y-axis and its wavelength is λ = 3.18 m. The amplitude of the wave is E = 288 V/m. (a) What is the frequency f of the wave? (b) What are the magnitude and the direction of the magnetic part of the wave? (c) What are k and ω? (d) What is the intensity? (e) If the wave falls upon a perfectly absorbing sheet of area 1.85 m2, at what rate is momentum delivered to the sheet and what is the radiation pressure? Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13: The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is polarized along the y-axis and its wavelength is λ = 3.18 m. The amplitude of the wave is E = 288 V/m. (a) What is the frequency f of the wave? (b) What are the magnitude and the direction of the magnetic part of the wave? (c) What are k and ω? (d) What is the intensity? (e) If the wave falls upon a perfectly absorbing sheet of area 1.85 m2, at what rate is momentum delivered to the sheet and what is the radiation pressure? Answer: (a) f = c/λ = (3.00 × 108 m/s)/(3.18 m) = 94.3 MHz. (b) B = E/c = (288 V/m)/(3.00 × 108 m/s) = 9.60 × 10–7 T and B points along the z-axis. (c) k = 2π/λ = 1.98/m and ω = ck = (3.00 × 108 m/s) (1.98/m) = 5.93 × 108 Hz. Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13: The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is polarized along the y-axis and its wavelength is λ = 3.18 m. The amplitude of the wave is E = 288 V/m. (a) What is the frequency f of the wave? (b) What are the magnitude and the direction of the magnetic part of the wave? (c) What are k and ω? (d) What is the intensity? (e) If the wave falls upon a perfectly absorbing sheet of area 1.85 m2, at what rate is momentum delivered to the sheet and what is the radiation pressure? Answer: (d) The time-averaged Poynting vector magnitude is Sav = EB/2μ0 = (288 V/m)(9.60 × 10–7 T)/2(4π × 10–7 T·m/A) = 110 W/m2, and this is the intensity. Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13: The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is polarized along the y-axis and its wavelength is λ = 3.18 m. The amplitude of the wave is E = 288 V/m. (a) What is the frequency f of the wave? (b) What are the magnitude and the direction of the magnetic part of the wave? (c) What are k and ω? (d) What is the intensity? (e) If the wave falls upon a perfectly absorbing sheet of area 1.85 m2, at what rate is momentum delivered to the sheet and what is the radiation pressure? Answer: (e) The radiation pressure is EB/2μ0c = B2/2μ0 = (9.60 × 10–7 T)2 / 2(4π × 10–7 T·m/A) = 3.67 × 10–7 N/m2, and the rate of transfer of momentum is the pressure times the area, namely (3.67 × 10–7 N/m2) (1.85 m2) = 6.78 × 10–7 N.