ELECTROMAGNETIC WAVES
|
|
- Eric Powell
- 5 years ago
- Views:
Transcription
1 Physics 4D ELECTROMAGNETIC WAVE Hans P. Paar 26 January 2006 i
2 Chapter 1 Vector Calculus 1.1 Introduction Vector calculus is a branch of mathematics that allows differentiation and integration of (scalar) functions and vector function in several variables at once. It is based upon multivariable calculus. A (scalar) function is a scalar whose value depends upon several variables. Examples are the temperature and pressure of the atmosphere, the density of an inhomogeneous solid and so on. A vector function (also called a vector field) is a vector whose components depend upon several variables. Examples of these are the electric and magnetic field, the velocity of a fluid, and so on. In the present case we are dealing with vectors in three-dimensional space so they have three components. The number of variables that functions and vector field can depend on is also three. In these notes we review the minimum amount of vector calculus needed to understand electromagnetic waves. We will often follow the notation of H.M. chey, Div, Grad, Curl, and All That (fourth edition). Wherever possible, a reference will be given to this book. 1.2 The Operator The fundamental operator we deal with in vector calculus is the operator. It is defined as = e x x + e y y + e z z (1.1) ii
3 see chey, p.44. The vectors e x, e y, e z are three mutually perpendicular unit vectors that form a right-handed triplet with e x along the positive x-axis and similarly e y along the positive y-axis and e z along the positive z-axis. We do not use chey s notation i, j, k for these. The partial differentiations are familiar from multivariable calculus. It is seen that the operator is a vector because it is the sum of three terms that are each a vector. The order of the factors in each term does not matter because the unit vectors are constant vectors, independent of position x, y, z so the partial differentiations will do nothing to them. This is all you need to know (besides some nomenclature) because everything in vector calculus we need follows from the definition (1.1) of the operator. 1.3 The Gradient of a calar Function The gradient of a (scalar) function f(x,y,z), defined as F, is evaluated as f = ( e x x + e y y + e ) z f(x, y, z) (1.2) z Working out the parentheses in (1.2) we get f = ( f e x x + e f y y + e f ) z z (1.3) see chey, p.121. chey muddles the gradient discussion a bit. It is seen that the gradient of a (scalar) function is a vector. 1.4 The Divergence of a Vector Function The divergence of a vector function v(x, y, z), defined as v, is evaluated as v = ( e x x + e y y + e ) z (vx e x + v y e y + v z e z ) (1.4) z Working out the parentheses in (1.4) we get nine terms. Using the fact that the three unit vectors e x, e y, e z form a triplet of mutually perpendicular unit vectors we get v = v x x + v y y + v z z (1.5) iii
4 see chey p. 41. We have used for example that e x e y = 0 and similar relations for the five other combinations of dot products of two different unit vectors. We also used for example that e x e x = 1 and similar relations for the other two combinations of dot products of a unit vector with itself. It is seen that the divergence of a vector function is a scalar. 1.5 The Curl of a Vector Function The curl of a vector function v(x, y, z), defined as v, is evaluated as v = ( e x x + e y y + e ) z (vx e x + v y e y + v z e z ) (1.6) z uch a cross product can be evaluated using a determinant e x e y e z v = x y z v x v y v z (1.7) to get v = ( v z y v y ) ex ( v z z x v x ) ey + ( v y z x v x ) ez (1.8) y see chey p It is seen that the curl of a vector function is a vector. 1.6 Examples The simplest example of a gradient is r where r = x 2 + y 2 + z 2 is the length of the vector r = xe x + ye y + ze z. We get r = ( e x x + e y y + e ) z (x 2 + y 2 + z 2 ) 1 2 (1.9) z Working out the parentheses in the first factor we get terms proportional to e x, e y, and e z. The first of these can be evaluated as e x x (x2 + y 2 + z 2 ) = e x 2 (x2 + y 2 + z 2 ) 1 x 2 2x = e x r (1.10) imilar expressions hold for the other two terms. This can be seen by an explicit calculation or by replacing x by y and replacing x by z respectively iv
5 in (1.10). It is a good idea to pick up these substitution tricks and save time. They also serve as a check if you did the explicit calculation after all. All three expressions have a factor 1/r in common which we take outside parentheses to get r = e xx + e y y + e z z r = r r We expect r to be a vector and it is. The units work out too. If f(x, y, z) = r n then the gradient of f is (1.11) f = ( e x x + e y y + e ) z (x 2 + y 2 + z 2 ) n 2 (1.12) z Working out the parentheses in the first factor we get terms proportional to e x, e y, and e z. The first of these can be evaluated as e x x (x2 + y 2 + z 2 ) n 2 = ex n 2 (x2 + y 2 + z 2 ) n 2 1 2x = e x nx r n 2 (1.13) imilar expressions hold for the other two terms. All three expressions have a factor nr n 2 in common which we take outside parentheses to get f = (e x x + e y y + e z z) nr n 2 = rnr n 2 (1.14) We expect f to be a vector and indeed it is. We could have guessed the answer by using f(r) = df r (1.15) dr which results in (1.14) as well. It is often true in vector calculus that when you guess a relation it is often correct but it bears checking. This feature is what makes vector calculus useful. The special case n = 1 corresponds to a 1/r potential such as we encounter in electrostatics and gravity. The force is proportional to the negative of the gradient of f. etting n = 1 in (1.14) we find that the force is proportional to r/r 3. The force is along r and has a magnitude proportional to 1/r 2 as expected. The proportionality constant can be positive or negative so the direction of the force can be in the direction of r or in the direction of r. Next consider a vector function v(x, y, z) = r/r n. The divergence of this vector function is v = ( e x x + e y y + e ) x z ( z r n e x + y r n e y + z r n e z) (1.16) v
6 Working out the parentheses we get nine terms. As in ec. 1.4, only three terms survive because e x, e y, e z are three mutually perpendicular vectors. One of the three terms is proportional to e x e x (which equals 1) and is of the form x ( x r n ) = rn xnr n 1 r x r 2n (1.17) This a good point to verify that the units work out (they do). The partial derivative of r with respect to x can be evaluated as r x = x (x2 + y 2 + z 2 ) 1 2 = x r compare (1.13), and (1.17) becomes ( x ) r n x 2 nr n 2 = x r n r 2n (1.18) = 1 n x2 r 2 r n (1.19) This is again a good point to verify that the units work out (they do). The other two terms from (1.16) are of the form 1 n y2 r 2 r n and 1 n z2 r 2 r n (1.20) This can be seen by an explicit calculation or by replacing x by y and replacing x by z respectively in (1.20), see the discussion below (1.10). To complete the calculation of v of (1.16) we must sum the term in (1.19) and the two terms in (1.20). They have a common denominator r n. The numerators neatly sum to 3 nr 2 /r 2 = 3 n so finally we get and the units work out. function is v = r we get r r n = 3 n r n (1.21) In the special case that n = 0 and the vector r = 3 (1.22) Next consider the curl of v(x, y, z) = r/r n. It can be evaluated as in (1.7) as e x e y e z v = x y z (1.23) x r n vi y r n z r n
7 The first term is proportional to e x and is [ ( z ) ( y )] ( n)z r e x = ex y r n z r n r n+1 y ( n)y r r n+1 z (1.24) We know from a calculation similar to the one leading to (1.10) that r x = x r (1.25) imilar relations hold for r/y and r/z. Again, all that is required is to replace x by y and x by z in (1.25). Thus (1.24) becomes [ ( z ) ( y )] ( n) e x = ex y r n z r n (yz zy) = 0 (1.26) rn+2 The other two terms proportional to e y and e z can be obtained in a similar fashion and are found to be zero as well. They are equal to ( n) e y r n+2 (zx xz) = 0 and e ( n) z (xy yx) = 0 (1.27) rn+2 To see this without doing an explicit calculation is somewhat harder than in the cases discussed earlier. Thus we find that r r n = 0 (1.28) This is true for all n so we have in particular that r = 0. vii
8 Chapter 2 Waves and their Properties 2.1 Plane Waves Waves were discussed in physics 4B. We review the discussion here, using a slightly different notation. In 4B the expression for a plane wave traveling in the +x-direction was written for example as D(x, t) = D M sin (kx ωt) (2.1) where the wave number k and the angular frequency ω are given by k = 2π λ and ω = 2π T with λ the wavelength and T the period. It was also shown there that (2.2) λ = vt (2.3) where v is the propagation velocity. The expression φ = kx ωt, the argument of the sine in (2.1), is called the phase of the wave. The expression (2.1) represents a plane wave (plane along the y and z direction) because (2.1) is independent of y and z and its amplitude for given values of x and t has the same value at all y and z. One could as well have used to cosine instead of the sine in (2.1). Instead of choosing one or the other, we employ a complex number notation using e iφ = cos φ + i sin φ (2.4) Using (2.4) we can write for a wave traveling in the +x direction D(x, t) = D M e i(kx ωt) (2.5) viii
9 where it is optional to take either the real or the imaginary part of (2.5), compare (2.1). We generalize the notation to represent a wave traveling in a direction along the vector k as follows u(r, t) = u 0 e i(k r ωt) (2.6) where u 0 is a constant and represents the amplitude of the wave. Here r is the position at which the amplitude of the wave is to be evaluated and φ = k r ωt is called the phase. The vector k is now called the wave vector. The wave is a plane wave traveling in the direction of k while the plane of the wave is perpendicular to k as can be seen as follows. We may choose the orientation of our coordinate system as we like without loss of generality. Therefore, let us choose a coordinate system with its z-axis along k so k = (0, 0, k) with k = k. The phase φ at a fixed time t 0 is given by φ = k r ωt 0 = kz ωt 0. The endpoints of all vectors r that satisfy the condition of constant phase φ = kz ωt 0 have a constant z = (φ + ωt 0 )/k with x and y unconstrained. This relation represents a plane perpendicular to the z-axis, parallel to the x, y plane. Thus the condition of constant phase φ forces the endpoints of all vectors r to lay in a plane that is perpendicular to the vector k. Keeping k parallel to the z-axis and keeping the phase φ constant (the observer rides with the traveling wave at a fixed position on the wave) we find that as z and t change, their change is such that φ = k(z + dz) ω(t + dt) = kz ωt. Canceling terms we find we find kdz ωdt = 0 or v = dz dt = ω k (2.7) where v is the propagation velocity of the traveling wave. To keep the phase φ constant, z and t must both increase (decreasing time is not possible of course) implying a wave traveling in the +z-direction. Had we written (2.6) with a plus sign as in u(r, t) = u 0 e i(k r+ωt) (2.8) we would have found that v would have had the opposite sign, corresponding to a wave traveling in the negative z-direction with our special choice of coordinate system. Of course it is the relative sign of k r and ωt that determines the direction of propagation, not their individual signs. If we keep t fixed (so we take a snapshot of the traveling wave) we can compare the amplitude of the wave at z and z + λ. These two amplitudes must be the same by the definition of the wavelength λ. Increasing z by an ix
10 amount λ at fixed t changes the phase but the phase change should be such that the amplitude of the wave is unchanged. Thus we must have Canceling terms we find k(z + λ) = kz + 2π (2.9) kλ = 2π or k = 2π λ = k (2.10) This is the first relation in (2.2) for k, the magnitude of the wave vector k. Instead of keeping t fixed, we can keep z fixed and compare the amplitude of the wave at t and t + T. These two amplitudes must be the same by the definition of the period T. Increasing t by an amount T at fixed z changes the phase but the phase change should be such that the amplitude of the wave is unchanged. Thus we must have Canceling terms we find ω(t + T ) = ωt + 2π (2.11) ωt = 2π or ω = 2π T (2.12) This is the second relation in (2.2). Using (2.10) and (2.12) in the second relation of (2.7) we find λ = vt, compare (2.3). Note that v is parallel to k. The waves considered in the previous section could be called scalar waves because the amplitude of the waves is a scalar. This is useful to distinguish those from vector waves where the amplitude of the wave is a vector. We write such a vector wave in general as v(r, t) = v 0 e i(k r ωt) (2.13) with v 0 a constant vector representing the vector amplitude of the vector wave. The meaning of the other quantities in (2.13) are the same as those for the scalar waves discussed above and the discussion following (2.6) applies here as well. The expressions (2.1), (2.5), (2.6), (2.8), and (2.13) satisfy wave equations that we discuss next. x
11 2.2 Wave Equations for calar Waves We start with a calculation of the gradient of a (scalar) wave (2.6) u(r, t) = u 0 e i(k r ωt) (2.14) = u 0 e i(k r ωt) (2.15) ( = u 0 ex x + e y y + e ) z e i(k xx+k yy+k zz ωt) (2.16) z where in (2.15) we exchanged the operator with the constant u 0. Working out the parentheses we get terms proportional to e x, e y and e z, compare (1.9). The first of these can be evaluated as u 0 e x x = u 0 e x (ik x ) e i(kxx+kyy+kzz ωt) (2.17) = e x (ik x ) u(r, t) (2.18) imilar expressions hold for the other two terms. This can be seen by an explicit calculation or by replacing x by y and replacing x by z respectively in (2.18). All three expressions have a factor iu(r, t) in common which we take outside parentheses to get u(r, t) = ( e x k x + e y k y + e z k z ) iu(r, t) (2.19) = ( ik) u(r, t) (2.20) We expect u(r, t) to be a vector and it is. Note that the result is exactly what you would expect intuitively: the operator leaves the exponential intact but the chain rule brings down a factor ik, something we are used to in single variable calculus. We can now evaluate 2 u(r, t) as 2 u(r, t) = u(r, t) (2.21) = (ik) u(r, t) (2.22) = (ik) u(r, t) (2.23) = (ik) 2 u(r, t) (2.24) = k 2 u(r, t) (2.25) = k 2 u(r, t) (2.26) xi
12 where we used (2.20) in (2.21) and (2.23). The 2 is a scalar. It multiplies the scalar u(r, t) so the result 2 u(r, t) should be a scalar and (2.26) shows that it is a scalar. It is much easier to differentiate the scalar wave (2.6) with respect to time. We get u(r, t) = u 0e i(k r ωt) (2.27) = u 0 (2.28) = u 0 ( iω) e i(k r ωt) (2.29) = ( iω)u 0 e i(k r ωt) (2.30) = ( iω) u(r, t) (2.31) We can now evaluate the second derivative with respect to time of the scalar wave (2.6) as 2 u(r, t) = u(r, t) (2.32) 2 = ( iω)u(r, t) (2.33) = ( iω) u(r, t) (2.34) = ( iω) 2 u(r, t) (2.35) = ω 2 u(r, t) (2.36) where we used (2.31) in (2.32) and (2.34). Using (2.26) and (2.36) we see that the scalar wave u(r, t) satisfies 2 u(r, t) k2 2 ) u(r, t) ω 2 2 = ( 2 k2 ω 2 u(r, t) = 0 (2.37) or with (2.7) 2 u(r, t) 1 v 2 2 u(r, t) 2 = ( 2 1v 2 ) u(r, t) = 0 (2.38) The expressions (2.37) and (2.38) are called wave equations because their solutions describe (traveling) waves. These wave equations are pretty complicated expressions: they are second order partial differential equations. xii
13 They are also linear and homogeneous. Linear because all u(r, t) and its derivatives appear to the first power, homogeneous because all terms involve u(r, t). uch linear homogeneous differential equations have an extremely important property: if we have two linearly independent solutions u 1 (r, t) and u 2 (r, t) (never mind where we got them) then the linear superposition u(r, t) = a 1 u 1 (r, t) + a 2 u 2 (r, t) is also a solution. The proof is easy and you should try to prove this property. Verify that the proof fails when the equation is non-linear or inhomogeneous (or both). Because a second order differential equation has at most two linearly independent solutions, this linear superposition of two linearly independent solutions is also the most general solution of the wave equation. The two linearly independent solutions of (2.37) and (2.38) are (2.6) and (2.8) that is traveling waves in the +r and r directions respectively. Under certain circumstances their linear superposition can form standing waves as discussed in Physics 4B. 2.3 Wave Equations for Vector Waves We follow the development of the previous section for scalar waves. calculate the divergence of the vector wave (2.13) We v(r, t) = v 0 e i(k r ωt) (2.39) = v 0 e i(k r ωt) (2.40) = v 0 (ik)e i(k r ωt) (2.41) = i k v(r, t) (2.42) where we exchanged the operator with the constant vector v 0 and used (2.20) in (2.40). We can now evaluate 2 v(r, t) as 2 v(r, t) = v(r, t) (2.43) = [ (ik)] v(r, t) (2.44) = (ik) v(r, t) (2.45) = (ik) 2 v(r, t) (2.46) = k 2 v(r, t) (2.47) = k 2 v(r, t) (2.48) xiii
14 where we used (2.42) in (2.43) and (2.45). The 2 is a scalar. It multiplies the vector v(r, t) so the result 2 v(br, t) should be a vector and (2.48) shows that it is a vector. This result can be compared with (2.26). We now differentiate the vector wave (2.13) with respect to time. We get v(r, t) = v 0e i(k r ωt) (2.49) = v 0 (2.50) = v 0 ( iω) e i(k r ωt) (2.51) = ( iω)v 0 e i(k r ωt) (2.52) = ( iω) v(r, t) (2.53) We can now evaluate the second derivative with respect to time of the vector wave (2.13) as 2 v(r, t) = v(r, t) (2.54) 2 = ( iω)v(r, t) (2.55) = ( iω) v(r, t) (2.56) = ( iω) 2 v(r, t) (2.57) = ω 2 v(r, t) (2.58) This result can be compared with (2.36). Using (2.48) and (2.58) we see that the vector wave v(r, t) satisfies 2 v(r, t) k2 2 ) v(r, t) ω 2 2 = ( 2 k2 ω 2 v(r, t) = 0 (2.59) or with (2.7) 2 v(r, t) 1 v 2 2 v(r, t) 2 = ( 2 1v 2 ) v(r, t) = 0 (2.60) The expressions (2.59) and (2.60) are called vector wave equations because their solutions describe (traveling) vector waves. The same comments apply here that were made below (2.38). xiv
15 Chapter 3 Gauss and tokes Theorems 3.1 Gauss Theorem Consider a volume V that is enclosed by a surface. Unit vectors ˆn are normal to the surface and are directed outward. Gauss Theorem states v dv = v ˆn d (3.1) V where v(r) is a vector function, also called a vector field. ometimes the notation d = ˆnd is used, see for example Giancoli s formulation of Maxwell s equations. The vector v may also depend upon other variables such as time but those are irrelevant for Gauss Theorem. Gauss Theorem is also called the Divergence Theorem. As an example we consider the case where the surface is a sphere with radius R and the vector field is the radius vector r. For the left hand side of (3.1) we must evaluate v = r. We find r = 3 according to (1.22). Thus the integrand of the volume integral has the constant value 3 and can be taken outside the volume integral. The integral over the volume is 4πR 3 /3 so the left hand side of (3.1) is 4πR 3. For the right hand side of (3.1) we use that ˆn = r/r so that v ˆn = r r/r = r. Thus the integrand of the surface integral has the constant value R and can be taken outside the surface integral. The integral over is 4πR 2 so the right hand side of (3.1) is 4πR 3. The two sides of (3.1) are seen to be equal. xv
16 3.2 tokes Theorem Consider a surface that is bounded by a line s. The direction for the line integral is defined by the direction of unit vectors ˆτ tangent to the curve. Unit vectors ˆn are normal to the surface in a direction relative to the direction of the line integral defined by the right-hand rule. This definition is different from the one for Gauss Theorem because here there is no definition possible for inside or outside the surface. tokes Theorem states v ˆn d = v ˆτ ds (3.2) where v(r) is a vector function as above. ometimes the notation ds = ˆτ ds is used, see for example Giancoli s formulation of Maxwell s equations.. The vector v may also depend upon other variables such as time but those are irrelevant for tokes Theorem. tokes Theorem is also called the Curl Theorem. As an example we consider the case where the line s is a circle of radius R. The vector r has its endpoint laying on the circle so r = x e x + y e y with x 2 + y 2 = R 2. Here e x and e y are the unit vectors along the positive x and y axes respectively. The vector s = y e x x e y is tangential to the circle as can be seen by evaluating s r, which is zero so s and r are perpendicular. For the left hand side of (3.2) we must evaluate v = s. We find e x e y e z s = x y z y x 0 [ = e x y 0 z ( x)] [ e y x 0 z y] [ + e z x ( x) y y] = 2e z (3.3) At the point (x, y) = (0, R) the vector s = Re y. We choose ˆτ parallel to s. Using the right-hand rule we find that the normal on the plane of the circle is given by ˆn = e z. Thus s ˆn = 2e z ( e z ) = +2 and the integrand of the surface integral has the constant value 2 and can be taken outside the surface integral. The integral over the surface is πr 2 so the left hand side of (3.2) is 2πR 2. For the right hand side of (3.2) we use that s and ˆτ are parallel and that the length of the vector s = R so that s ˆτ = R. Thus the integrand of the line integral has the constant value R and can be taken outside the line integral. The integral along s is 2πR so the right hand side of (3.2) is 2πR 2. The two sides of (3.2) are seen to be equal. s xvi
17 Chapter 4 Differential Form of Maxwell s Equations Maxwell s equations are ɛ 0 E ˆn d = ρ dv (4.1) B ˆn d = 0 (4.2) E τ ds = B ˆn d (4.3) s B τ ds = µ 0 j ˆn d + µ 0 ɛ 0 E ˆn d (4.4) s We will not discuss the Maxwell equations but instead refer to Giancoli. We use Gauss Theorem to get E ˆn d = E dv (4.5) B ˆn d = B dv (4.6) and tokes Theorem to get E τ ds = s B τ ds = s xvii E ˆn d (4.7) B ˆn d (4.8) (4.9)
18 Combining (4.1) with (4.5), (4.2) with (4.6), (4.3) with (4.7), and (4.5) with (4.8) we get ɛ 0 E dv = ρ dv (4.10) B dv = 0 (4.11) E ˆn d = B ˆn d (4.12) B ˆn d = µ 0 j ˆn d + µ 0 ɛ 0 E ˆn d (4.13) The relations (4.10) through (4.13) must hold for arbitrary volumes, surfaces, and lines so in each equation, the integrands must be equal. Thus we get E = ρ ɛ 0 (4.14) B = 0 (4.15) E = B (4.16) E B = µ 0 j + µ 0 ɛ 0 (4.17) In vacuum ρ = j = 0 and the Maxwell equations become E = 0 (4.18) B = 0 (4.19) E = B (4.20) E B = µ 0 ɛ 0 (4.21) There is a certain symmetry in these equations. Of particular note is the minus sign in (4.20), a manifestation of Lenz law. If this minus sign was not there, or more accurately if (4.20) and (4.21) did not have opposite signs, electromagnetic waves would not exist! We will show next the existence of traveling electromagnetic waves. xviii
19 Chapter 5 Electromagnetic Waves 5.1 Wave Equations for E and B The relations (4.20) and (4.21) are two coupled equations because each involves E and B. We can separate E and B, at the cost of obtaining higher derivatives, by taking the curl of either equation then substitute the other equation. Let s start with taking the curl of both sides of (4.21) to get ( B) = µ 0 ɛ 0 E (5.1) = µ 0 ɛ 0 ( E) (5.2) = µ 0 ɛ 0 ( B ) (5.3) = µ 0 ɛ 0 2 B 2 (5.4) where we have used (4.20) in (5.2). We now use the general relation for a vector triple product a (b c) = (a c) b (a b) c (5.5) Using this relation to work out the triple product on the left hand side of (5.1), we must take care with the order of the operators. The operator works only on vectors to its right. Therefore v(r) is not the same as v(r). The two operators in (5.1) need to be kept to the left of B xix
20 when using (5.5). Thus we get ( B) = ( B) 2 B (5.6) = 2 B (5.7) where we have used (4.19) in (5.6). ubstitution of (5.7) in (5.1) gives 2 B µ 0 ɛ 0 2 B 2 = 0 (5.8) We have succeeded in decoupling E from B and found that B satisfies a vector wave equation, compare (2.59) and (2.60). The solutions of this wave equation are traveling waves of the form (2.13) with v(r, t) = B(r, t) and with either a plus or a minus sign in the exponent, see the discussion near (2.8). Comparing (5.8) with (2.60) we see immediately that the propagation velocity v of the vector wave B(r, t) is given by v = 1 ɛ0 µ 0 (5.9) With 1/(4πɛ 0 ) = and µ 0 /(4π) = 10 7 we find that v = m/s, the speed of light. This is a strong indication that light is nothing else but an electromagnetic wave phenomenon. We use the letter c for the velocity of light so (5.9) becomes c = 1 ɛ0 µ 0 (5.10) Instead of starting with taking the curl of both sides of (4.17) we could have started with (4.16). We find ( E) = ( B ) (5.11) = E ( B) (5.12) = ( E µ0 ɛ 0 ) (5.13) = µ 0 ɛ 0 2 E 2 (5.14) xx
21 where we have used (4.21) in (5.12). As in (5.7) we have that ( E) = ( E) 2 E (5.15) = 2 E (5.16) where we have used (4.18) in (5.15). ubstitution of (5.16) in (5.15) gives 2 E µ 0 ɛ 0 2 E 2 = 0 (5.17) We find that in decoupling E and B that E satisfies the same vector wave equation as B, see (5.8) and thus the solution of (5.17) are traveling waves of the form (2.13) with a propagation velocity given by (5.9) and (5.10). 5.2 Plane Wave olutions We try solutions of the form (2.13) except that we factor the vector amplitude v 0 in (2.13) into a unit vector ɛ and a scalar that represents the amplitude of the vector wave. Because E and B may point in different directions and have different amplitudes we introduce two units vectors ɛ 1 and ɛ 2 and two amplitudes E m and B m. Thus we write E(r, t) = ɛ 1 E m e i(k r ωt) (5.18) B(r, t) = ɛ 2 B m e i(k r ωt) (5.19) We now want to find expressions for ɛ 1 and ɛ 2 and for E m and B m. We obtain these by substitution of the vector waves (5.18) and (5.19) in Maxwell s equations (4.18) through (4.21). ubstitution of (5.18) and (5.19) in (4.18) and (4.19) respectively we find E = [ɛ 1 E m e i(k r ωt)] = ik E = 0 (5.20) E = [ɛ 2 B m e i(k r ωt)] = ik B = 0 (5.21) where we have used (2.42). We find therefore that k is perpendicular to ɛ 1 and perpendicular to ɛ 2 and thus thta k is perpendicular to E and perpendicular to B. Because k point in the direction of propagation of the xxi
22 traveling vector wave, we find that E and B are perpendicular to the direction of propagation. We call such waves transverse (plane) waves and electromagnetic waves are thus transverse waves. Next we want to find the relative orientation of ɛ 1 and ɛ 2 and thus of E and B. To simplify the calculation we use the freedom we have to choose our coordinate system. Without loss of generality we choose the z-axis parallel to k so k = (0, 0, k) where k = k. The dot product k r = kz. We can still choose the orientation of for example the x-axis without affecting the orientation of the z-axis. We choose the x-axis parallel to ɛ 1 so ɛ 1 = (1, 0, 0) and therefore E = (E m e i(kz ωt), 0, 0) (5.22) We now use (4.20). We need E which can be evaluated as E = e x e y e z x y z E m e i(k z ωt) 0 0 = e y E m (ik)e i(k z ωt) (5.23) We also need the derivative of B with respect to time. This can be evaluated as B = ɛ 2B m ( iω)e i(k z ωt) = iωɛ 2 B m e i(k z ωt) (5.24) compare (2.53). ubstitution of (5.23) and (5.24) in (4.20) gives e y E m (ik)e i(k z ωt) = iωɛ 2 B m e i(k z ωt) (5.25) Canceling common factors in (5.25) we get or e y E m k = ωɛ 2 B m (5.26) ɛ 2 = e y ke m ωb m (5.27) We find that ɛ 2 is parallel to the y-axis. We already have that ɛ 1 is parallel to the x-axis so we find that the vectors (ɛ 1, ɛ 2, k) form a right handed triplet xxii
23 and therefore also (E, B, k) form a right handed triplet. We now require that ɛ 2 = 1. With (5.27) we find that or using (2.7) with v = c we find B m = E m c ke m = ωb m (5.28) = ɛ 0 µ 0 E m (5.29) The electric and magnetic fields are in phase, that at each position they oscillate in phase. The amplitude of the magnetic field is much smaller than the amplitude of the electric field. However, the stored energies in each are equal, see Giancoli. xxiii
DIVERGENCE AND CURL THEOREMS
This document is stored in Documents/4C/Gausstokes.tex. with LaTex. Compile it November 29, 2014 Hans P. Paar DIVERGENCE AND CURL THEOREM 1 Introduction We discuss the theorems of Gauss and tokes also
More informationThis document is stored in Documents/4C/nablaoperator.tex. Compile it with LaTex. VECTOR CALCULUS
This document is stored in Documents/4C/nablaoperator.tex. Compile it with LaTex. November 12, 2014 Hans P. Paar VECTOR CALCULUS 1 Introduction Vector calculus is a branch of mathematics that allows differentiation
More informationThis document is stored in Documents/4C/nablaoperator.tex. Compile it with LaTex. VECTOR CALCULUS
This document is stored in Documents/4C/nablaoperator.tex. Compile it with LaTex. November 12, 2014 Hans P. Paar VECTOR CALCULUS 1 Introduction sec:intro Vector calculus is a branch of mathematics that
More informationContents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9
MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Contents Multiple Integrals 3 2 Vector Fields 9 3 Line and Surface Integrals 5 4 The Classical Integral Theorems 9 MATH 32B-2 (8W)
More informationElectromagnetic Theory (Hecht Ch. 3)
Phys 531 Lecture 2 30 August 2005 Electromagnetic Theory (Hecht Ch. 3) Last time, talked about waves in general wave equation: 2 ψ(r, t) = 1 v 2 2 ψ t 2 ψ = amplitude of disturbance of medium For light,
More informationSolutions for the Practice Final - Math 23B, 2016
olutions for the Practice Final - Math B, 6 a. True. The area of a surface is given by the expression d, and since we have a parametrization φ x, y x, y, f x, y with φ, this expands as d T x T y da xy
More informationThe Divergence Theorem Stokes Theorem Applications of Vector Calculus. Calculus. Vector Calculus (III)
Calculus Vector Calculus (III) Outline 1 The Divergence Theorem 2 Stokes Theorem 3 Applications of Vector Calculus The Divergence Theorem (I) Recall that at the end of section 12.5, we had rewritten Green
More informationMultiple Integrals and Vector Calculus: Synopsis
Multiple Integrals and Vector Calculus: Synopsis Hilary Term 28: 14 lectures. Steve Rawlings. 1. Vectors - recap of basic principles. Things which are (and are not) vectors. Differentiation and integration
More informationElectro Magnetic Field Dr. Harishankar Ramachandran Department of Electrical Engineering Indian Institute of Technology Madras
Electro Magnetic Field Dr. Harishankar Ramachandran Department of Electrical Engineering Indian Institute of Technology Madras Lecture - 7 Gauss s Law Good morning. Today, I want to discuss two or three
More informationElectromagnetic Waves Retarded potentials 2. Energy and the Poynting vector 3. Wave equations for E and B 4. Plane EM waves in free space
Electromagnetic Waves 1 1. Retarded potentials 2. Energy and the Poynting vector 3. Wave equations for E and B 4. Plane EM waves in free space 1 Retarded Potentials For volume charge & current = 1 4πε
More informationMITOCW 8. Electromagnetic Waves in a Vacuum
MITOCW 8. Electromagnetic Waves in a Vacuum The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources
More informationA Brief Revision of Vector Calculus and Maxwell s Equations
A Brief Revision of Vector Calculus and Maxwell s Equations Debapratim Ghosh Electronic Systems Group Department of Electrical Engineering Indian Institute of Technology Bombay e-mail: dghosh@ee.iitb.ac.in
More information2.20 Fall 2018 Math Review
2.20 Fall 2018 Math Review September 10, 2018 These notes are to help you through the math used in this class. This is just a refresher, so if you never learned one of these topics you should look more
More information+ f f n x n. + (x)
Math 255 - Vector Calculus II Notes 14.5 Divergence, (Grad) and Curl For a vector field in R n, that is F = f 1, f 2,..., f n, where f i is a function of x 1, x 2,..., x n, the divergence is div(f) = f
More informationMath 11 Fall 2016 Final Practice Problem Solutions
Math 11 Fall 216 Final Practice Problem olutions Here are some problems on the material we covered since the second midterm. This collection of problems is not intended to mimic the final in length, content,
More informationClass 30: Outline. Hour 1: Traveling & Standing Waves. Hour 2: Electromagnetic (EM) Waves P30-
Class 30: Outline Hour 1: Traveling & Standing Waves Hour : Electromagnetic (EM) Waves P30-1 Last Time: Traveling Waves P30- Amplitude (y 0 ) Traveling Sine Wave Now consider f(x) = y = y 0 sin(kx): π
More informationMAT389 Fall 2016, Problem Set 4
MAT389 Fall 2016, Problem Set 4 Harmonic conjugates 4.1 Check that each of the functions u(x, y) below is harmonic at every (x, y) R 2, and find the unique harmonic conjugate, v(x, y), satisfying v(0,
More informationfree space (vacuum) permittivity [ F/m]
Electrostatic Fields Electrostatic fields are static (time-invariant) electric fields produced by static (stationary) charge distributions. The mathematical definition of the electrostatic field is derived
More informationTutorial 3 - Solutions Electromagnetic Waves
Tutorial 3 - Solutions Electromagnetic Waves You can find formulas you require for vector calculus at the end of this tutorial. 1. Find the Divergence and Curl of the following functions - (a) k r ˆr f
More informationKeble College - Hilary 2015 CP3&4: Mathematical methods I&II Tutorial 4 - Vector calculus and multiple integrals II
Keble ollege - Hilary 2015 P3&4: Mathematical methods I&II Tutorial 4 - Vector calculus and multiple integrals II Tomi Johnson 1 Prepare full solutions to the problems with a self assessment of your progress
More informationMultiple Integrals and Vector Calculus (Oxford Physics) Synopsis and Problem Sets; Hilary 2015
Multiple Integrals and Vector Calculus (Oxford Physics) Ramin Golestanian Synopsis and Problem Sets; Hilary 215 The outline of the material, which will be covered in 14 lectures, is as follows: 1. Introduction
More informationS12.1 SOLUTIONS TO PROBLEMS 12 (ODD NUMBERS)
OLUTION TO PROBLEM 2 (ODD NUMBER) 2. The electric field is E = φ = 2xi + 2y j and at (2, ) E = 4i + 2j. Thus E = 2 5 and its direction is 2i + j. At ( 3, 2), φ = 6i + 4 j. Thus the direction of most rapid
More informationMath 234 Exam 3 Review Sheet
Math 234 Exam 3 Review Sheet Jim Brunner LIST OF TOPIS TO KNOW Vector Fields lairaut s Theorem & onservative Vector Fields url Divergence Area & Volume Integrals Using oordinate Transforms hanging the
More informationGauss s Law & Potential
Gauss s Law & Potential Lecture 7: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Flux of an Electric Field : In this lecture we introduce Gauss s law which happens to
More informationMath 11 Fall 2007 Practice Problem Solutions
Math 11 Fall 27 Practice Problem olutions Here are some problems on the material we covered since the second midterm. This collection of problems is not intended to mimic the final in length, content,
More informationJim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt
Jim Lambers MAT 28 ummer emester 212-1 Practice Final Exam olution 1. Evaluate the line integral xy dx + e y dy + xz dz, where is given by r(t) t 4, t 2, t, t 1. olution From r (t) 4t, 2t, t 2, we obtain
More informationPRACTICE PROBLEMS. Please let me know if you find any mistakes in the text so that i can fix them. 1. Mixed partial derivatives.
PRACTICE PROBLEMS Please let me know if you find any mistakes in the text so that i can fix them. 1.1. Let Show that f is C 1 and yet How is that possible? 1. Mixed partial derivatives f(x, y) = {xy x
More informationMATH 308 COURSE SUMMARY
MATH 308 COURSE SUMMARY Approximately a third of the exam cover the material from the first two midterms, that is, chapter 6 and the first six sections of chapter 7. The rest of the exam will cover the
More informationSolutions to Sample Questions for Final Exam
olutions to ample Questions for Final Exam Find the points on the surface xy z 3 that are closest to the origin. We use the method of Lagrange Multipliers, with f(x, y, z) x + y + z for the square of the
More informationFiber Optics. Equivalently θ < θ max = cos 1 (n 0 /n 1 ). This is geometrical optics. Needs λ a. Two kinds of fibers:
Waves can be guided not only by conductors, but by dielectrics. Fiber optics cable of silica has nr varying with radius. Simplest: core radius a with n = n 1, surrounded radius b with n = n 0 < n 1. Total
More informationDifferential Operators and the Divergence Theorem
1 of 6 1/15/2007 6:31 PM Differential Operators and the Divergence Theorem One of the most important and useful mathematical constructs is the "del operator", usually denoted by the symbol Ñ (which is
More informationPhysics GRE: Electromagnetism. G. J. Loges 1. University of Rochester Dept. of Physics & Astronomy. xkcd.com/567/
Physics GRE: Electromagnetism G. J. Loges University of Rochester Dept. of Physics & stronomy xkcd.com/567/ c Gregory Loges, 206 Contents Electrostatics 2 Magnetostatics 2 3 Method of Images 3 4 Lorentz
More informationElectromagnetic Waves
May 7, 2008 1 1 J.D.Jackson, Classical Electrodynamics, 2nd Edition, Section 7 Maxwell Equations In a region of space where there are no free sources (ρ = 0, J = 0), Maxwell s equations reduce to a simple
More informationRadio Propagation Channels Exercise 2 with solutions. Polarization / Wave Vector
/8 Polarization / Wave Vector Assume the following three magnetic fields of homogeneous, plane waves H (t) H A cos (ωt kz) e x H A sin (ωt kz) e y () H 2 (t) H A cos (ωt kz) e x + H A sin (ωt kz) e y (2)
More informationIntegral Theorems. September 14, We begin by recalling the Fundamental Theorem of Calculus, that the integral is the inverse of the derivative,
Integral Theorems eptember 14, 215 1 Integral of the gradient We begin by recalling the Fundamental Theorem of Calculus, that the integral is the inverse of the derivative, F (b F (a f (x provided f (x
More informationTime-varying electromagnetic field
Chapter 5 Time-varying electromagnetic field In this chapter we take up the time-varying electromagnetic field. By this study, we will be able to understand in a systematic way that actually the electric
More informationReview of Vector Analysis in Cartesian Coordinates
Review of Vector Analysis in Cartesian Coordinates 1 Scalar: A quantity that has magnitude, but no direction. Examples are mass, temperature, pressure, time, distance, and real numbers. Scalars are usually
More informationMath Review for Exam 3
1. ompute oln: (8x + 36xy)ds = Math 235 - Review for Exam 3 (8x + 36xy)ds, where c(t) = (t, t 2, t 3 ) on the interval t 1. 1 (8t + 36t 3 ) 1 + 4t 2 + 9t 4 dt = 2 3 (1 + 4t2 + 9t 4 ) 3 2 1 = 2 3 ((14)
More informationPHYS 408, Optics. Problem Set 1 - Spring Posted: Fri, January 8, 2015 Due: Thu, January 21, 2015.
PHYS 408, Optics Problem Set 1 - Spring 2016 Posted: Fri, January 8, 2015 Due: Thu, January 21, 2015. 1. An electric field in vacuum has the wave equation, Let us consider the solution, 2 E 1 c 2 2 E =
More informationMaxwell s Equations and Electromagnetic Waves W13D2
Maxwell s Equations and Electromagnetic Waves W13D2 1 Announcements Week 13 Prepset due online Friday 8:30 am Sunday Tutoring 1-5 pm in 26-152 PS 10 due Week 14 Friday at 9 pm in boxes outside 26-152 2
More informationMR. YATES. Vocabulary. Quadratic Cubic Monomial Binomial Trinomial Term Leading Term Leading Coefficient
ALGEBRA II WITH TRIGONOMETRY COURSE OUTLINE SPRING 2009. MR. YATES Vocabulary Unit 1: Polynomials Scientific Notation Exponent Base Polynomial Degree (of a polynomial) Constant Linear Quadratic Cubic Monomial
More informationENGI 4430 Gauss & Stokes Theorems; Potentials Page 10.01
ENGI 443 Gauss & tokes heorems; Potentials Page.. Gauss Divergence heorem Let be a piecewise-smooth closed surface enclosing a volume in vector field. hen the net flux of F out of is F d F d, N 3 and let
More informationMathematics (Course B) Lent Term 2005 Examples Sheet 2
N12d Natural Sciences, Part IA Dr M. G. Worster Mathematics (Course B) Lent Term 2005 Examples Sheet 2 Please communicate any errors in this sheet to Dr Worster at M.G.Worster@damtp.cam.ac.uk. Note that
More informationMATH 332: Vector Analysis Summer 2005 Homework
MATH 332, (Vector Analysis), Summer 2005: Homework 1 Instructor: Ivan Avramidi MATH 332: Vector Analysis Summer 2005 Homework Set 1. (Scalar Product, Equation of a Plane, Vector Product) Sections: 1.9,
More informationLecture 10. (2) Functions of two variables. Partial derivatives. Dan Nichols February 27, 2018
Lecture 10 Partial derivatives Dan Nichols nichols@math.umass.edu MATH 233, Spring 2018 University of Massachusetts February 27, 2018 Last time: functions of two variables f(x, y) x and y are the independent
More informationMa 1c Practical - Solutions to Homework Set 7
Ma 1c Practical - olutions to omework et 7 All exercises are from the Vector Calculus text, Marsden and Tromba (Fifth Edition) Exercise 7.4.. Find the area of the portion of the unit sphere that is cut
More informationr r 1 r r 1 2 = q 1 p = qd and it points from the negative charge to the positive charge.
MP204, Important Equations page 1 Below is a list of important equations that we meet in our study of Electromagnetism in the MP204 module. For your exam, you are expected to understand all of these, and
More informationCoordinate systems and vectors in three spatial dimensions
PHYS2796 Introduction to Modern Physics (Spring 2015) Notes on Mathematics Prerequisites Jim Napolitano, Department of Physics, Temple University January 7, 2015 This is a brief summary of material on
More informationThe Divergence Theorem
Math 1a The Divergence Theorem 1. Parameterize the boundary of each of the following with positive orientation. (a) The solid x + 4y + 9z 36. (b) The solid x + y z 9. (c) The solid consisting of all points
More informationMATH Calculus IV Spring 2014 Three Versions of the Divergence Theorem
MATH 2443 008 Calculus IV pring 2014 Three Versions of the Divergence Theorem In this note we will establish versions of the Divergence Theorem which enable us to give it formulations of div, grad, and
More informationMaxwell s equations for electrostatics
Maxwell s equations for electrostatics October 6, 5 The differential form of Gauss s law Starting from the integral form of Gauss s law, we treat the charge as a continuous distribution, ρ x. Then, letting
More informationDivergence Theorem Fundamental Theorem, Four Ways. 3D Fundamental Theorem. Divergence Theorem
Divergence Theorem 17.3 11 December 213 Fundamental Theorem, Four Ways. b F (x) dx = F (b) F (a) a [a, b] F (x) on boundary of If C path from P to Q, ( φ) ds = φ(q) φ(p) C φ on boundary of C Green s Theorem:
More informationElectromagnetism. Christopher R Prior. ASTeC Intense Beams Group Rutherford Appleton Laboratory
lectromagnetism Christopher R Prior Fellow and Tutor in Mathematics Trinity College, Oxford ASTeC Intense Beams Group Rutherford Appleton Laboratory Contents Review of Maxwell s equations and Lorentz Force
More informationLine, surface and volume integrals
www.thestudycampus.com Line, surface and volume integrals In the previous chapter we encountered continuously varying scalar and vector fields and discussed the action of various differential operators
More informationFourier Approach to Wave Propagation
Phys 531 Lecture 15 13 October 005 Fourier Approach to Wave Propagation Last time, reviewed Fourier transform Write any function of space/time = sum of harmonic functions e i(k r ωt) Actual waves: harmonic
More informationFinal Exam May 4, 2016
1 Math 425 / AMCS 525 Dr. DeTurck Final Exam May 4, 2016 You may use your book and notes on this exam. Show your work in the exam book. Work only the problems that correspond to the section that you prepared.
More informationMATHS 267 Answers to Stokes Practice Dr. Jones
MATH 267 Answers to tokes Practice Dr. Jones 1. Calculate the flux F d where is the hemisphere x2 + y 2 + z 2 1, z > and F (xz + e y2, yz, z 2 + 1). Note: the surface is open (doesn t include any of the
More informationUNIT 1. INTRODUCTION
UNIT 1. INTRODUCTION Objective: The aim of this chapter is to gain knowledge on Basics of electromagnetic fields Scalar and vector quantities, vector calculus Various co-ordinate systems namely Cartesian,
More informationRadiation Integrals and Auxiliary Potential Functions
Radiation Integrals and Auxiliary Potential Functions Ranga Rodrigo June 23, 2010 Lecture notes are fully based on Balanis [?]. Some diagrams and text are directly from the books. Contents 1 The Vector
More information2. Waves and the Wave Equation
2. Waves and the Wave Equation What is a wave? Forward vs. backward propagating waves The one-dimensional wave equation Phase velocity Reminders about complex numbers The complex amplitude of a wave What
More information09 The Wave Equation in 3 Dimensions
Utah State University DigitalCommons@USU Foundations of Wave Phenomena Physics, Department of --2004 09 The Wave Equation in 3 Dimensions Charles G. Torre Department of Physics, Utah State University,
More informationScalar and vector fields
Scalar and vector fields What is a field in mathematics? Roughly speaking a field defines how a scalar-valued or vectorvalued quantity varies through space. We usually work with scalar and vector fields.
More informationMathematical Concepts & Notation
Mathematical Concepts & Notation Appendix A: Notation x, δx: a small change in x t : the partial derivative with respect to t holding the other variables fixed d : the time derivative of a quantity that
More informationReinterpreting the Solutions of Maxwell s Equations in Vacuum
1 Reinterpreting the Solutions of Maxwell s Equations in Vacuum V.A.I. Menon, M.Rajendran, V.P.N.Nampoori International School of Photonics, Cochin Univ. of Science and tech., Kochi 680022, India. The
More information- HH Why Can Light Propagate in Vacuum? Hsiu-Hau Lin (Apr 1, 2014)
- HH0124 - Why Can Light Propagate in Vacuum? Hsiu-Hau Lin hsiuhau@phys.nthu.edu.tw (Apr 1, 2014) Sunlight is vital for most creatures in this warm and humid planet. From interference experiments, scientists
More information[variable] = units (or dimension) of variable.
Dimensional Analysis Zoe Wyatt wyatt.zoe@gmail.com with help from Emanuel Malek Understanding units usually makes physics much easier to understand. It also gives a good method of checking if an answer
More informationRigid body simulation. Once we consider an object with spatial extent, particle system simulation is no longer sufficient
Rigid body dynamics Rigid body simulation Once we consider an object with spatial extent, particle system simulation is no longer sufficient Rigid body simulation Unconstrained system no contact Constrained
More informationSOLUTIONS TO THE FINAL EXAM. December 14, 2010, 9:00am-12:00 (3 hours)
SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN December 14, 2010, 9:00am-12:00 (3 hours) 1) For each of (a)-(e) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please
More informationTypical anisotropies introduced by geometry (not everything is spherically symmetric) temperature gradients magnetic fields electrical fields
Lecture 6: Polarimetry 1 Outline 1 Polarized Light in the Universe 2 Fundamentals of Polarized Light 3 Descriptions of Polarized Light Polarized Light in the Universe Polarization indicates anisotropy
More informationBasics of Radiation Fields
Basics of Radiation Fields Initial questions: How could you estimate the distance to a radio source in our galaxy if you don t have a parallax? We are now going to shift gears a bit. In order to understand
More informationMath 5BI: Problem Set 9 Integral Theorems of Vector Calculus
Math 5BI: Problem et 9 Integral Theorems of Vector Calculus June 2, 2010 A. ivergence and Curl The gradient operator = i + y j + z k operates not only on scalar-valued functions f, yielding the gradient
More information18.02 Multivariable Calculus Fall 2007
MIT OpenourseWare http://ocw.mit.edu 18.02 Multivariable alculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.02 Lecture 21. Test for
More informationFluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 15 Conservation Equations in Fluid Flow Part III Good afternoon. I welcome you all
More informationToday in Physics 218: back to electromagnetic waves
Today in Physics 18: back to electromagnetic waves Impedance Plane electromagnetic waves Energy and momentum in plane electromagnetic waves Radiation pressure Artist s conception of a solar sail: a spacecraft
More informationPeriodic functions: simple harmonic oscillator
Periodic functions: simple harmonic oscillator Recall the simple harmonic oscillator (e.g. mass-spring system) d 2 y dt 2 + ω2 0y = 0 Solution can be written in various ways: y(t) = Ae iω 0t y(t) = A cos
More information1. 4 2y 1 2 = x = x 1 2 x + 1 = x x + 1 = x = 6. w = 2. 5 x
.... VII x + x + = x x x 8 x x = x + a = a + x x = x + x x Solve the absolute value equations.. z = 8. x + 7 =. x =. x =. y = 7 + y VIII Solve the exponential equations.. 0 x = 000. 0 x+ = 00. x+ = 8.
More informationChapter 1. Vector Algebra and Vector Space
1. Vector Algebra 1.1. Scalars and vectors Chapter 1. Vector Algebra and Vector Space The simplest kind of physical quantity is one that can be completely specified by its magnitude, a single number, together
More informationG G. G. x = u cos v, y = f(u), z = u sin v. H. x = u + v, y = v, z = u v. 1 + g 2 x + g 2 y du dv
1. Matching. Fill in the appropriate letter. 1. ds for a surface z = g(x, y) A. r u r v du dv 2. ds for a surface r(u, v) B. r u r v du dv 3. ds for any surface C. G x G z, G y G z, 1 4. Unit normal N
More informationLecture 7. Please note. Additional tutorial. Please note that there is no lecture on Tuesday, 15 November 2011.
Lecture 7 3 Ordinary differential equations (ODEs) (continued) 6 Linear equations of second order 7 Systems of differential equations Please note Please note that there is no lecture on Tuesday, 15 November
More informationin Electromagnetics Numerical Method Introduction to Electromagnetics I Lecturer: Charusluk Viphavakit, PhD
2141418 Numerical Method in Electromagnetics Introduction to Electromagnetics I Lecturer: Charusluk Viphavakit, PhD ISE, Chulalongkorn University, 2 nd /2018 Email: charusluk.v@chula.ac.th Website: Light
More informationAPPM 2350 Final Exam points Monday December 17, 7:30am 10am, 2018
APPM 2 Final Exam 28 points Monday December 7, 7:am am, 28 ON THE FONT OF YOU BLUEBOOK write: () your name, (2) your student ID number, () lecture section/time (4) your instructor s name, and () a grading
More informationDivergence Theorem December 2013
Divergence Theorem 17.3 11 December 2013 Fundamental Theorem, Four Ways. b F (x) dx = F (b) F (a) a [a, b] F (x) on boundary of If C path from P to Q, ( φ) ds = φ(q) φ(p) C φ on boundary of C Green s Theorem:
More informationEE2007: Engineering Mathematics II Vector Calculus
EE2007: Engineering Mathematics II Vector Calculus Ling KV School of EEE, NTU ekvling@ntu.edu.sg Rm: S2-B2b-22 Ver 1.1: Ling KV, October 22, 2006 Ver 1.0: Ling KV, Jul 2005 EE2007/Ling KV/Aug 2006 EE2007:
More informationSecond Order ODEs. Second Order ODEs. In general second order ODEs contain terms involving y, dy But here only consider equations of the form
Second Order ODEs Second Order ODEs In general second order ODEs contain terms involving y, dy But here only consider equations of the form A d2 y dx 2 + B dy dx + Cy = 0 dx, d2 y dx 2 and F(x). where
More informationMathematical Tripos, Part IB : Electromagnetism
Mathematical Tripos, Part IB : Electromagnetism Proof of the result G = m B Refer to Sec. 3.7, Force and couples, and supply the proof that the couple exerted by a uniform magnetic field B on a plane current
More informationMATH 228: Calculus III (FALL 2016) Sample Problems for FINAL EXAM SOLUTIONS
MATH 228: Calculus III (FALL 216) Sample Problems for FINAL EXAM SOLUTIONS MATH 228 Page 2 Problem 1. (2pts) Evaluate the line integral C xy dx + (x + y) dy along the parabola y x2 from ( 1, 1) to (2,
More informationChapter 5. The Differential Forms of the Fundamental Laws
Chapter 5 The Differential Forms of the Fundamental Laws 1 5.1 Introduction Two primary methods in deriving the differential forms of fundamental laws: Gauss s Theorem: Allows area integrals of the equations
More informationChapter 6: Vector Analysis
Chapter 6: Vector Analysis We use derivatives and various products of vectors in all areas of physics. For example, Newton s 2nd law is F = m d2 r. In electricity dt 2 and magnetism, we need surface and
More informationMUDRA PHYSICAL SCIENCES
MUDRA PHYSICAL SCIENCES VOLUME- PART B & C MODEL QUESTION BANK FOR THE TOPICS:. Electromagnetic Theory UNIT-I UNIT-II 7 4. Quantum Physics & Application UNIT-I 8 UNIT-II 97 (MCQs) Part B & C Vol- . Electromagnetic
More informationAPPENDIX 2.1 LINE AND SURFACE INTEGRALS
2 APPENDIX 2. LINE AND URFACE INTEGRAL Consider a path connecting points (a) and (b) as shown in Fig. A.2.. Assume that a vector field A(r) exists in the space in which the path is situated. Then the line
More informationx + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the
1.(8pts) Find F ds where F = x + ye z + ze y, y + xe z + ze x, z and where T is the T surface in the pictures. (The two pictures are two views of the same surface.) The boundary of T is the unit circle
More informationJim Lambers MAT 280 Fall Semester Practice Final Exam Solution
Jim Lambers MAT 8 Fall emester 6-7 Practice Final Exam olution. Use Lagrange multipliers to find the point on the circle x + 4 closest to the point (, 5). olution We have f(x, ) (x ) + ( 5), the square
More informationWaves, the Wave Equation, and Phase Velocity. We ll start with optics. The one-dimensional wave equation. What is a wave? Optional optics texts: f(x)
We ll start with optics Optional optics texts: Waves, the Wave Equation, and Phase Velocity What is a wave? f(x) f(x-) f(x-) f(x-3) Eugene Hecht, Optics, 4th ed. J.F. James, A Student's Guide to Fourier
More information8.03 Lecture 12. Systems we have learned: Wave equation: (1) String with constant tension and mass per unit length ρ L T v p = ρ L
8.03 Lecture 1 Systems we have learned: Wave equation: ψ = ψ v p x There are three different kinds of systems discussed in the lecture: (1) String with constant tension and mass per unit length ρ L T v
More informationWaves in Linear Optical Media
1/53 Waves in Linear Optical Media Sergey A. Ponomarenko Dalhousie University c 2009 S. A. Ponomarenko Outline Plane waves in free space. Polarization. Plane waves in linear lossy media. Dispersion relations
More informationChapter 1. Maxwell s Equations
Chapter 1 Maxwell s Equations 11 Review of familiar basics Throughout this course on electrodynamics, we shall use cgs units, in which the electric field E and the magnetic field B, the two aspects of
More informationVector Analysis. Electromagnetic Theory PHYS 401. Fall 2017
Vector Analysis Electromagnetic Theory PHYS 401 Fall 2017 1 Vector Analysis Vector analysis is a mathematical formalism with which EM concepts are most conveniently expressed and best comprehended. Many
More informationFLUX OF VECTOR FIELD INTRODUCTION
Chapter 3 GAUSS LAW ntroduction Flux of vector field Solid angle Gauss s Law Symmetry Spherical symmetry Cylindrical symmetry Plane symmetry Superposition of symmetric geometries Motion of point charges
More informationMath 212-Lecture 8. The chain rule with one independent variable
Math 212-Lecture 8 137: The multivariable chain rule The chain rule with one independent variable w = f(x, y) If the particle is moving along a curve x = x(t), y = y(t), then the values that the particle
More informationCHAPTER 6 VECTOR CALCULUS. We ve spent a lot of time so far just looking at all the different ways you can graph
CHAPTER 6 VECTOR CALCULUS We ve spent a lot of time so far just looking at all the different ways you can graph things and describe things in three dimensions, and it certainly seems like there is a lot
More information