PHYS 110B - HW #6 for (2), Fall 2005, Solutions by David Pace Equations Referenced As ”Eq

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PHYS 110B - HW #6 For (2), Fall 2005, Solutions by David Pace Equations referenced as ”Eq. #” are from Grifﬁths −µ J~ = 2A~ − ∇~ L Problem statements are paraphrased o (11)

∂2A~ ∂V [1.] Problem 10.1 from Grifﬁths = ∇2A~ − µ − ∇~ ∇~ · A~ + µ o o ∂t2 o o ∂t (12) Show that the following, ∂ ρ 2 ~ ~ 2 ~ ! ∇ V + (∇ · A) = − (1) 2 ∂ A ∂t o −µ J~ = ∇ A~ − µ − o o o ∂t2 ! ∂2A~ 2 ~ ∇ A − µoo 2 − ∂V ∂t ∇~ ∇~ · A~ + µ (13) o o ∂t X

∂V ∇~ ∇~ · A~ + µoo = −µoJ~ (2) where in the last line I added parentheses to make this ∂t expression match the original. where (1) is Eq. 10.4 and (2) is Eq. 10.5, can be written as, [2.] Problem 10.4 from Grifﬁths 2 ∂L ρ V + = − (3) ∂t o

Let V = 0 and A~ = Ao sin(kx − ωt)y ˆ. Solve for the 2 ~ ~ ~ fields in this case and show that they satisfy Maxwell’s A − ∇L = −µoJ (4) vacuum equations. State the relation between ω and k where, that is required. 2 2 2 ∂ ≡ ∇ − µoo (5) ∂t2 Solution ∂V L ≡ ∇~ · A~ + µoo (6) ∂t The fields are given by, Solution ∂A~ E~ = −∇~ V − Eq. 10.3 (14) It will suffice to write out the new forms of these equa- ∂t tions and then verify that they are equivalent to the original forms. B~ = ∇~ × A~ Eq. 10.2 (15)

For (1), Since the potential is zero, it’s gradient is also zero. We ρ 2 ∂L need only be concerned with the vector potential in this − = V + (7) o ∂t problem. Solving for the electric ﬁeld,

∂2V ∂ ∂V ∂A~ = ∇2V − µ + ∇~ · A~ + µ E~ = − (16) o o ∂t2 ∂t o o ∂t ∂t (8) ∂ = − A sin(kx − ωt)y ˆ (17) ∂t o ∂2V ∂ ∂2V = ∇2V − µ + (∇~ · A~) + µ o o ∂t2 ∂t o o ∂t2 = −Ao(cos(kx − ωt))(−ω)y ˆ (18) (9)

= Aoω cos(kx − ωt)y ˆ (19) ρ 2 ∂ ~ ~ − = ∇ V + (∇ · A) X (10) o ∂t Determining the magnetic ﬁeld is simpliﬁed by noting 2 only one term in the curl expression is non-zero. taneously, ~ ∂Ay ~ ~ ∂B B~ = zˆ (20) ∇ × E = − (31) ∂x ∂t ∂ = A sin(kx − ωt)z ˆ (21) ∂Ey ∂ ∂x o zˆ = − A k cos(kx − ωt)z ˆ ∂x ∂t o (32) = Ao(cos(kx − ωt))k zˆ (22)

∂ = Aok cos(kx − ωt)z ˆ (23) A ω cos(kx − ωt)z ˆ = −A k(− sin(kx − ωt))(−ω)z ˆ ∂x o o (33) At this point it is worth noting that the electric and magnetic fields are perpendicular to each other. The next part of the problem is to show that these fields satisfy Aoω(− sin(kx − ωt))k zˆ = −Aokω sin(kx − ωt)z ˆ the vacuum Maxwell’s equations, which would require (34) that they fit the form of electromagnetic waves. Such waves also feature perpendicular fields, so we move on −A kω sin(kx − ωt)z ˆ = −A kω sin(kx − ωt)z ˆ to the next part of the problem with confidence that so o o (35) far we have the correct answer. Verifying the final equation will allow us to show a con- 2 Maxwell’s equations in vacuum: dition placed between ω and k. Recall that µoo = 1/c . ∂E~ ∂B~ µ = ∇~ × B~ (36) ∇~ · E~ = 0 ∇~ × E~ = − o o ∂t ∂t (24) 1 ∂ ∂Bz ∂E~ Aoω cos(kx − ωt)y ˆ = − yˆ (37) ∇~ · B~ = 0 ∇~ × B~ = µ c2 ∂t ∂x o o ∂t A ω ∂ where (24) is given in Griffiths as Eq. 9.40. o (− sin(kx − ωt))(−ω)y ˆ = − A k cos(kx − ωt)y ˆ c2 ∂x o (38) Verify these equations one at a time. Since the electric field of (19) has only a y component, its divergence is, 2 Aoω 2 sin(kx − ωt)y ˆ = −Aok(− sin(kx − ωt))k yˆ ∂Ey c ∇~ · E~ = (25) (39) ∂y

∂ 2 = Aoω cos(kx − ωt) (26) Aoω 2 ∂y sin(kx − ωt)y ˆ = Aok sin(kx − ωt)y ˆ c2 (40) = 0 X (27) where there is only satisfied if k2 = ω2/c2. The divergence of the magnetic field follows the same pattern, We do know, however, that electromagnetic waves in vacuum travel at the speed of light and that this veloc- ∂Bz ∇~ · B~ = (28) ity is related to the angular frequency and wave vector ∂z as ω/k = c. This is the condition imposed on ω and k. ∂ = A k cos(kx − ωt) (29) ∂z o [3.] Problem 10.5 from Griffiths

= 0 (30) Use the gauge, qt The following writes out the curl of the electric ﬁeld and λ = − (41) the negative time derivative of the magnetic ﬁeld simul- 4πor 3 to transform the potentials, with a linearly increasing current. The major change here is that the problem involves the two dimensional ~ qt form of the vector potential. After making this adjust- V (~r, t) = 0 A(~r, t) = − 2 rˆ (42) 4πor ment, the rest of the process matches what has been done previously.

The two dimensional form of the vector potential is Solution given by Eq. 5.64, Z µo K~ The scalar potential is transformed according to, A~(~r, t) = da0 (51) 4π R ∂λ V 0 = V − (43) ~ ∂t where K is a surface current density. To solve for the electric and magnetic ﬁelds we must evaluate this vector potential expression at the retarded time, ∂ −qt = 0 − (44) Z µo Kotr ∂t 4πor A~(~r, t) = da0 xˆ (52) 4π R q = (45) Z 4π r µoKo t − R/c o = da0 xˆ (53) 4π R The vector potential is transformed according to,

0 Z A = A + ∇~ λ (46) µoKo t 1 = xˆ − da0 (54) 4π R c −qt ∂λ where the second integral simplifies to the area of cur- = 2 rˆ + rˆ (47) 4πor ∂r rent seen at any given time. Notice that I extracted the xˆ term from the integral. At this point I am still treat- da0 −qt −qt 1 ing the in Cartesian coordinates and the vector does = rˆ + (−1) rˆ (48) not change the integral. I will eventually write this inte- 4π r2 4π r2 o o gral in a different coordinate frame while keeping the xˆ term. We now need to determine what the current looks −qt qt like as time advances. I am only going to consider the = rˆ + rˆ 2 2 (49) z > 0 4πor 4πor region extending through . At the end of the solution I will describe the fields in the region z < 0. = 0 (50) As in the wire example problems, we need to determine These potentials represent a stationary charge. We see how much of the current is seen by the observer at any from this example that gauge choices allow us to rep- instant of time. An observer situated above the current resent electromagnetic systems in equivalent, though carrying plane by a distance z would see a circular area hopefully simpler, ways. Notice that the originally pro- of the plane, as shown below. vided potentials return the same electric and magnetic field values as these potentials. Such fields are more eas- ily calculated from the expressions derived in this problem.

[4.] Suppose at t = 0 a uniform surface current K~ (t) = Kot xˆ is established everywhere in the xy plane. Find the electric and magnetic ﬁelds everywhere in space for t > 0.

Solution

This problem is very similar to the examples done in lec- ture and discussion. Those examples were for a wire 4

It takes time for the ﬁelds produced by the sheet current Bringing it all together, to reach this observer, so this person will see more of the current as time progresses. Even though the current density is provided in Cartesian coordinates, it is eas- ier to describe the area of current observed using cylin- µ K 1 drical coordinates. The following image illustrates this A~(~r, t) = o o 2πt(ct − z) − π(c2t2 − z2) xˆ geometry. 4π c (61) 2 µoKo z = 2ct2 − 2zt − ct2 + xˆ (62) 4 c

µoKo = c2t2 − 2czt + z2 xˆ (63) 4c

µoKo = (ct − z)2 xˆ (64) 4c

This only spatial dependence in this result is z. The cylindrical z coordinate is the same as the Cartesian z, so where z is the location of the observer, s is the radius of I do not have to worry about further adjustments. This the current area observed at any time t, and R ≡ R is answer is fully Cartesian. the distance from the observer to any current element. Due to the finite propagation speed of electromagnetic fields, there is an upper limit on R at any time. This is given by Rmax = ct and allows us to write the ra- One other point to mention here is the physical meaning dius of the circular current sheet at a generic time as of this result. It appears as though the expression in (64) p 2 2 claims that at z = ∞ the vector potential is infinite. This smax = (ct) − z . Our solution method still matches well with that applied to the wire problem. is not the case, however, because there is a limit to the applicability of this result. When we are at any position there is a minimum amount of time that must pass be- The second integral in (54) reduces to 1/c multiplied by fore we realize there is a current sheet. The current situ- the area of current because the integrand is a constant. 2 2 2 2 ated directly below (or above) us will be the first current This area is πsmax = π(c t − z ). The first integral sim- element that we observe. Such an observation is made plifies remarkably using the identity (or solving it on after a time t = z/c has passed. This means that our ex- your own), pression for the vector potential, and later for the elec- Z x p tric and magnetic fields, is only valid for t ≥ z/c. When √ dx = a2 + x2 (55) z = ∞ t = ∞ a2 + x2 , we must have at least , which gives a vector potential of zero (though to be more technically cor- So the first integral is, rect I would now have to discuss the solution for times Z t Z smax Z 2π t greater than infinity, which don’t matter for us). da0 = √ s ds dφ (56) 2 2 R 0 0 s + z

Z smax s Our solution for the vector potential is, = 2πt √ ds (57) 2 2 0 s + z

√ p c2t2−z2 2 2 µoKo = 2πt s + z (58) A~(~r, t) = (ct − z)2 xˆ ct ≥ z (65) 0 4c

p p = 2πt c2t2 − z2 + z2 − 0 + z2 (59) where the solution is zero for all other times. On with = 2πt(ct − z) (60) the electric and magnetic ﬁeld calculations. Applying 5

(14) and (15), given by Eq. 5.64 (which also displays the 2D form),

Z ∂ µoKo µo I~ E~ (~r, t) = − (ct − z)2 xˆ (66) A~(~r, t) = dl0 (73) ∂t 4c 4π R

A~ µoKo The correct value of is determined by evaluating this = − (2)(ct − z)(c)x ˆ (67) 4c expression at the retarded time. We make such an eval- uation by making the replacement t → tr. Z µoKo µo ktr = − (ct − z)x ˆ (68) A~(~r, t) = d~l0 (74) 2 4π R

µoKo 2 Z R B~ (~r, t) = ∇~ × (ct − z) xˆ (69) µok t − c 0 4c = d~l (75) 4π R

∂ µ K o o 2 Z = (ct − z) yˆ (70) µok t 1 ∂z 4c = − d~l0 (76) 4π R c

µoKo = (2)(ct − z)(−1)y ˆ (71) Consider the second integral above, 4c Z 1 I 1 d~l0 = d~l0 (77) µoKo c c = − (ct − z)y ˆ (72) 2c = 0 (78) and once again these expressions are zero when the time inequality (ct ≥ z) is not satisfied. In addition, this is where (77) states that this integral is taken around a only the solution for the region of positive z. From sym- closed loop. Since the integrand is constant, the integral metry (the right-hand rule) we know that these fields are may be performed around the entire loop in one pass in the opposite directions in the region below the sheet. (for the first integral we will have break this loop into all of its separate components). Vector integrations performed around closed loops return a value of zero. Note This problem is suggesting the principle of radiation. that this means, Notice from the electric and magnetic field expressions |B~ | = |E~ |/c E~ × B~ → zˆ I that and that . In the region ~0 above the current sheet it appears as though EM waves dl = 0 (79) are propagating outward. This is exactly the case, which makes this problem a good example of how expressing Returning to the non-zero integral, fields in terms of their retarded potentials describes the Z consequences of radiation and why this material imme- µok t 0 A~(~r, t) = d~l (80) diately precedes the Griffiths’ chapter on radiation. 4π R

Z ~0 µokt dl [5.] Problem 10.10 from Grifﬁths A~(~r, t) = (81) 4π R

A piece of wire bent into a loop is shown in Griffiths’ fig- where the time dependence is extracted from the inte- ure 10.5. This wire carries a current that varies linearly grand because the integral is taken over space. in time as I(t) = kt. Calculate the retarded vector potential at the center. Find the electric field at this point. Why Since this is a vector integration we must consider each does this neutral wire produce an electric field? Why is it portion of the loop separately. The separation is applied not possible to solve for the magnetic field at this point because the vector (shown by the arrows in the loop) is using your expression for the vector potential? not continuous. 0 0 ! µ kt Z dl~ dx0 dx0 dl~ Solution A~(~r, t) = o a + l xˆ + r xˆ + b 4π a x x b The one dimensional form of the vector potential is (82) 6 where the integrand represents the circular region of ra- lengths of current that generate the observed vector po- dius a, the linear region to the left of the origin, the lin- tential at the center. ear region to the right of the origin, and the circular re- b gion of radius respectively. When integrating over the We may now return to the expression for the vector po- circular regions, the distance to the point of interest is tential, constant (as shown in the denominator of these terms). Z 0 0 µokt dx dx A~(~r, t) = l xˆ + r xˆ (87) I have used a Cartesian coordinate system even though 4π x x it appears that a spherical or cylindrical coordinate system is better suited. The reason for this is that the circu- Z −a Z b ! µokt dx dx lar paths simplify to Cartesian results. For these vector = xˆ + xˆ (88) integrals along the circular paths, notice that the verti- 4π −b x a x cal components cancel. This is illustrated in the figure below, which represents the circular loop of radius a. where there is an interesting application of the R 1/x terms. The correct expression for this integral is,  Z 1  ln(x) x > 0 dx = (89) |x|  − ln(x) x < 0

where this integral is usually written in terms of the absolute value of x and this realization in the x < 0 region is not encountered. Consider the derivative of 1/|x| if you wish to verify this.

Continuing on,

Z −b Z b ! ~ µokt dx dx The black arrows are tangent to the loop and represent A(~r, t) = − xˆ + xˆ (90) 4π −a x a x the actual vector component d~l0 from the integrations. As we integrate along this path, the vertical components (denoted y and colored blue in the figure) cancel each µokt −b = − (− ln(x)) + ln(x)b xˆ (91) other out because they are directed oppositely on the left 4π −a a and right halves of the loop. The horizontal components (denoted by x and colored red in the figure) are always µokt b b directed to the right and add constructively. For the cir- = ln + ln xˆ (92) cular loop of radius b, the vertical components also can- 4π a a cel, though the horizontal components add in the −xˆ direction. µokt b = ln xˆ (93) 2π a Neglecting the vertical (y) contributions, the integrations of the circular regions are, Now we may use (14) to solve for the electric field. There is no net charge on the wire loop, so the scalar potential Z ~ 0 ~ 0 ! Z a Z −b dla dlb dx dx is zero, which leaves, + = xˆ + xˆ (83) a b −a a b b ∂A~ E~ = − (94) ∂t 1 1 = x|a xˆ + x|−b xˆ (84) a −a b b ∂ µokt b = − ln xˆ (95) ∂t 2π a 1 1 = (2a)x ˆ + (−2b)x ˆ (85) a b µok b = − ln xˆ (96) = 0 (86) 2π a The contributions to the vector potential from the cir- where this shows that a neutral wire can indeed pro- cular regions cancel out entirely. It is the two colinear duce an electric field. From previous work this quarter 7 we know that a changing current in the wire leads to a because applying the derivative to the rest of the Taylor changing magnetic field around it. This changing mag- expansion in (97) would generate a second order term. netic field induces an electric field through Faraday’s ~ ~ ~ law, ∇ × E = −∂B/∂t. Making these replacements in (99),

" ˙ ˙ # Our work here does not allow us to calculate the mag- µ Z J~(t) + (t − t)J~(t) J~(t) B~ (~r, t) = o r + × Rˆ dτ 0 netic field produced by this wire. The reason is that we 4π R2 cR have only solved for one numerical value of the vector potential, while the magnetic field is determined by the (101) spatial dependence of the vector potential, B~ = ∇~ × A~. ~ We need to know the functional form of A in order to  ˙  Z J~(t) + (t − R − t)J~(t) ~˙ solve for B~ . Notice that the problem has us solve for the µo c J(t) 0 =  +  × Rˆ dτ vector potential at the ”origin” of the loop so that we 4π R2 cR may take advantage of symmetry. To solve for the vector (102) potential generally would have been a much more dif- ficult problem (one added difficulty is that the integrals for the circular loops would not have constant denom- Z " ˙ ˙ # µo J~(t) J~(t) J~(t) inators and therefore require lengthy algebra to solve). = − + × Rˆ dτ 0 (103) 4π R2 cR cR

Z 0 µo J~(r~ , t) × Rˆ [6.] Problem 10.12 from Griffiths = dτ 0 (104) 4π R2 X Let the current density change slowly enough that it may be approximated in an expansion by keeping terms only through the first derivative. [7.] Problem 10.21 from Griffiths ˙ J~(tr) = J~(t) + (tr − t)J~(t) (97) Begin with a plastic ring of radius a and glue charges The spatial dependence does not need to be explicitly to it. The charge density may now be described as written here because it will not be relevant to the prob- λ = λo| sin(θ/2)|. Spin this loop about its axis with an- lem. Show that the Biot-Savart law holds with the cur- gular velocity ω. Solve for the exact scalar and vector rent density evaluated at the non-retarded time, i.e., potentials at the center of the ring. Z ~ ~0 ˆ ~ µo J(r , t) × R 0 B(~r, t) = 2 dτ (98) 4π R Griffiths’ Hint: the vector potential is, Griffiths’ Note: This problem illustrates that for slowly µ λ ωa h ωa ωa i changing currents the magnetostatic approximation is A~ = o o sin ωt − xˆ − cos ωt − yˆ quite good. The errors in such an approximation include 3π c c (105) neglecting retardation and dropping the time derivative factor in equation 10.31 from the text, but these mistakes cancel each other out. Solution

Solution The charge density changes in time due to the spinning of the ring. This is accounted for in the description of the charge density by letting θ → φ − ωt. Consider φ Considering retardation effects, the magnetic field is to be the cylindrical coordinate and represent the initial given by Eq. 10.31, position of an individual charge. At a later time t, this specific charge will be at a new angular position given Z " 0 ˙ 0 # µo J~(r~ , tr) J~(r~ , tr) B~ (~r, t) = + × Rˆ dτ 0 (99) by φ − ωt. The charge density is, 4π R2 cR φ − ωt ρ(~r, t) = λo sin (106) Since we are told to keep only the first derivative for the 2 current density we have, This problem does not allow us to neglect the scalar po- ˙ ˙ J~(tr) = J~(t) (100) tential because the ring is not charge neutral. The exact 8 scalar potential is given by, Granted, after the charges are initially placed on the ring it will take a time t = a/c for an observer at the center Z 0 1 ρ(r~ , tr) V (~r, t) = dτ 0 Eq. 10.19 (107) to notice the scalar potential. After this time has passed, 4πo R however, the scalar potential will be observed to be constant. where expressing the charge density in terms of the retarded time requires only exchanging t with tr. The vector potential depends on the current, which in This problem features charge distributed along only the turn depends on the amount of charge flowing past a φ coordinate, so the integral is taken in one dimen- certain point. This means that the location of the charges sion. In the cylindrical coordinate system the differential affects the solution and we must carefully incorporate length along the φ direction is dl = s dφ, furthermore, in their time dependence. this problem the radius of the ring is fixed so R = s = a.

φ−ωt 2π λ sin r 1 Z o 2 The simplified form of the vector potential in this ring V (~r, t) = Rdφ (108) 4πo 0 R geometry is given by Eq. 5.64, Z 2π 1 φ − ωtr = λo sin dφ (109) 4πo 0 2 Z µo I~ A~ = dl0 (117) Use the following substitution to solve the integral; con- 4π R sidering this at any fixed value of tr, Z 2π ˆ µo I φ θ = φ − ωtr dθ = dφ (110) = R dφ (118) 4π 0 R Z 2π 1 φ − ωtr V (~r, t) = λo sin dφ (111) 4πo 0 2 Recall that for current in a wire I = λv, where v is the Z 2π velocity of the charge flow. For a spinning ring this is 1 θ v = ωR = ωa = λo sin dθ (112) given by . Therefore, 4πo 0 2 where the limits are unchanged because at any one φ−ωt moment of time we want to consider the entire ring. Z 2π λoωa sin µo 2 Through these limits the value of sin(θ/2) is always A~ = a dφ φˆ (119) greater than zero, so the absolute value signs are no 4π 0 a longer an extra difficulty. Z 2π µoλoωa θ ˆ 2π = sin dθ φ (120) λo θ 4π 0 2 V (~r, t) = −2 cos (113) 4πo 2 0

λo = (114) taking advantage of all the steps used in the calculation πo of the scalar potential. We could have solved this problem without any concern for the time dependence. The scalar potential depends only on how far away the charges are from the observa- At this point we reach the difﬁculty in taking integrals tion point. Since these charges are moving as the ring over an angular vector. The φˆ vector changes direc- spins and we want the potential they create at the cen- tion through the integration and is dealt with here by ter of that ring, the distance between each charge and switching to Cartesian coordinates. Also, because of the the observation point is constant. We can use the charge change of variables used, φ = θ + ωtr. distribution at time t = 0 and treat it as though this is ﬁxed since even as the charges move they will always make the same contribution to the potential at the cen- φˆ = − sin φ xˆ + cos φ yˆ ter. (121)

φ Z 2π λo sin 1 2 = − sin(θ + ωtr)x ˆ + cos(θ + ωtr)y ˆ (122) V (~r, t) = a dφ (115) 4πo 0 a

λo = (116) Now write this out and solve for the integrals separately, πo 9

Z 2π µoλoωa θ A~ = sin [− sin(θ + ωtr)x ˆ + cos(θ + ωtr)y ˆ] dθ (123) 4π 0 2 Z 2π θ First Integral = − sin sin(θ + ωtr)x ˆ dθ (124) 0 2 1 Z 2π θ 3θ = − cos + ωtr − cos + ωtr dθ xˆ (125) 2 0 2 2 1 θ 2 3θ 2π = − 2 sin + ωtr − sin + ωtr xˆ (126) 2 2 3 2 0 1 1 = − sin(π + ωt ) − sin(ωt ) − sin(3π + ωt ) + sin(ωt ) xˆ (127) r r 3 r 3 r 2 = − −2 sin(ωt ) + sin(ωt ) xˆ (128) r 3 r 4 = sin(ωt )x ˆ (129) 3 r Z 2π θ Second Integral = sin cos(θ + ωtr)y ˆ] dθ (130) 0 2 1 Z 2π θ 3θ = − sin + ωtr + sin + ωtr dθ yˆ (131) 2 0 2 2 1 θ 2 3θ 2π = 2 cos + ωtr − cos + ωtr yˆ (132) 2 2 3 2 0 1 1 = cos(π + ωt ) − cos(ωt ) − cos(3π + ωt ) + cos(ωt ) yˆ (133) r r 3 r 3 r 2 = −2 cos(ωt ) + cos(ωt ) yˆ (134) r 3 r 4 = − cos(ωt )y ˆ (135) 3 r

The ﬁrst integral made use of the following trigonomet- Insert the integral results into (123), ric identities, 1 sin(u) sin(v) = [cos(u − v) − cos(u + v)] (136) 2 cos(u) = cos(−u) (137) µoλoωa 4 4 A~ = sin(ωt )x ˆ − cos(ωt )y ˆ (142) sin(u + v) = sin(u) cos(v) + cos(u) sin(v) (138) 4π 3 r 3 r

The second integral made use of these trigonometric µoλoωa = [sin(ωt )x ˆ − cos(ωt )y ˆ] (143) identities, 3π r r 1 sin(u) cos(v) = [sin(u + v) + sin(u − v)] (139) 2

sin(u) = − sin(−u) (140) Finally, expressing the retarded time in terms of the non- retarded time will put (143) in the form of Grifﬁths’ cos(u + v) = cos(u) cos(v) − sin(u) sin(v) (141) given solution, (105).