# e t e r Cylindrical and Spherical Coordinate Representation of grad, div, curl and 2

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1 Cylindical and Spheical Coodinate Repesentation of gad, div, cul and 2 Thus fa, we have descibed an abitay vecto in F as a linea combination of i, j and k, which ae unit vectos in the diection of inceasin, y and z espectively 0.1 F = F 1 i F 2 j F 3 k is called the Catesian epesentation of a vecto. The collection of unit vectos {i,j,k} along is called the Cateisan basis. We note fom definition of dot poduct the othonomality elations: 0.2 i j = 0 = j k = k i, i i = 1 = j j = k k Taking dot poduct of 0.1 with i, j and k espectively and using 0.2 that 0.3 F 1 = F i, F 2 = F j, F 3 = F k 1. Repesentation of vectos in cylindical basis y e t e t Figue 1. Unit vectos e and e θ at the tip of an abitay vecto in the diection of inceasing and θ We now seek an altenate epesentation of vecto in tems of unit vectos natual fo a cylindical coodinate system, θ, z. We ecall the elation between catesian and cylindical coodinates: 1.4 = 2 y 2, θ = actan y, z = z 1 = cosθ, y = sin θ, z = z

2 2 Note that the elations 1.4 implies that 1.5 = 2 y 2 = cosθ, = y = y 2 y = sin θ 2 y 2 y 2 = sin θ, y = 2 y 2 = 1 cosθ A unit vecto in the diection of inceasing will be defined e and a unit vecto in the diection of θ will be defined as e θ as shown in Fig 1. Since unit vecto e makes an angle of θ with espect to -ais, it follows that 1.6 e = cos θi sin θj = 2 y 2i y 2 y 2j. Also, since unit vecto e θ makes an angle of π/2 θ with espect to -ais, it follows that π π 1.7 e θ = cos 2 θ i sin 2 θ j = sin θi cos θj y = 2 y 2i 2 y 2j Fom 1.6 and 1.7, using dot poducts, we we have the following set of othogonomality elations fo {e,e θ,k}: 1.8 e e θ = 0 = e θ k = k e, e e = 1 = e θ e θ = k k Futhe, solving fo i and j by using 1.6 and 1.7, we note that y 1.9 i = cos θe sin θe θ = 2 y 2e 2 y 2e θ and 1.10 j = sin θe cosθe θ = Using 1.9 and 1.10, we obtain y 2 y 2e 2 y 2e θ 1.11 F = F 1 if 2 jf 3 k = [F 1 cosθ F 2 sin θ]e [ F 1 sin θ F 2 cos θ]e θ F 3 k F e F θ e θ F 3 k whee F, F θ, F 3 ae defined to be the cylindical coodinates of vecto F Note: Do not confuse subscipt and θ with deivatives with espect to and θ.

3 Based on 1.11 we have the following elation between cylindical and pola coodinates of an abitay vecto F: 1.12 F = F 1 cosθ F 2 sin θ = F 1 2 y 2 yf 2 2 y F θ = F 1 sin θ F 2 cos θ = yf 1 2 y 2 F 2 2 y 2 Going the othe way, using 1.12 and 1.13 and solving fo F 1 and F 2, we have y 1.14 F 1 = F cosθ F θ sin θ = 2 y 2F 2 y 2F θ 1.15 F 2 = F sin θ F θ cosθ = y 2 y 2F 2 y 2F θ By taking dot poduct of F given in 1.11 with espect to e and e θ, and using othogonality elations 1.8, it follows that 1.16 F = F e, F θ = F e θ, F 3 = F k This is a natual definition of the components F, F θ, F 3 of an abitay vecto F in cylindical coodinates. 2. Repesentation of in cylindical coodinates We note that in the Catesian basis 2.17 f = f i f y j f k Now, f is thought of as a function of, θ, z, then fom chain ule and using 1.4, f = f f f = f f y = f y f y f y = f 2 y f 2 y 2 y 2 = cosθ f sin θ f y 2 y f 2 2 y 2 = sin θ f cos θ f

4 4 Since 2.18 and 2.19 holds fo any f, we can abstact them into elations involving opeatos: 2.20 = cosθ sin θ 2.21 y = sin θ cosθ Using using 2.18 and 2.19, it follows fom 2.17 that 2.22 f = f [ f [cosθi sin θj] sin θ i cosθ ] j f k = f e 1 f e θ f k This is the last equality is the epesentation of f in cylindical coodinates. Since the epesentation 2.22 holds fo any f, it is customay to wite the identity moe abstactly involving the opeato: 2.23 = e e θ k Eample 1: If f = z calculate f in cylindical coodinates. 2 y 2 Solution: We note f, θ, z = cos θ z = 1 cosθ. So, fom fomula 2 1 sinθ 2.23, f = e cosθ e 2 θ k Repesentation of divegence and Laplacian in cylindical coodinates Suppose F = F 1 i F 2 j F 3 k in Catesian epesentation and F = F e F θ e θ F 3 k in cylindical basis epesentation F = F 1 F 2 y F 3 = F 1 F 1 = cos θ F 1 sin θ = F 2 y F 2 y F 3 F 1 sin θf 2 cosθ F 2 F 3

5 5 Using 1.14 and 1.15, we obtain 3.25 F = cosθ [F cosθ F θ sin θ] sin θ [F cosθ F θ sin θ] sin θ [F sin θ F θ cosθ] cosθ [F sin θ F θ cos θ] F 3 = F 1 F 1 F θ F 3 == 1 F 1 F θ F 3 Combining 2.22 and 3.25 f f = f = e 1 f e θ f k = 1 f 1 1 f 2 f 2 = 2 f 1 f f 2 2 f 2 2 Eample 2: If F = y i jzk, calculate F using cylindical 2 y 2 2 y 2 coodinates. Solution: Note that in pola coodinates F = cos θ i sin θ j zk = 1 e zk. So, F = 1, F θ = 0 and F 3 = z. Using fomula 3.25, we have fo 0, F = 1 [F ] F 3 = 0 1 Note this computation is not valid when = 0 since the denominato is zeo. Eample 3: Show that ln 2 y 2 is a solution of of Laplace s equation 2 f = 0, ecept at = 0. Solution: Note that f = ln 2 y 2 = ln 2 = 2 ln. So, using fomula 3.26 fo 2 f, we have fo 0, 2 f = 2 2 ln 1 2 ln = = = 0 2 The calculation is not valid fo = 0 because of division by 0. Recall that 4.27 F = 4. Cul in cylindical coodinates F3 y F 2 F1 i F 3 F2 j F 1 k y

6 6 Using 2.20 and 2.21, it follows 4.28 F = sin θ F 3 cosθ F1 F 3 F 2 cosθf 3 sin θ i F 3 j cosθ F 2 sin θ F 2 sin θf 1 cosθ F 1 k Using 1.14 and 1.15, it follows that 4.29 F = sin θ F 3 cosθ F 3 {sin θf cosθf θ } i {cos θf sin θf θ } cosθ F 3 sin θ F 3 j cosθ {sin θf cosθf θ } sin θ {sin θf cosθf θ } sin θ {cosθf sin θf θ } cosθ {cos θf sin θf θ } Using 1.9 and 1.10, we obtain 4.30 F = sin θ F 3 cosθ F 3 {sin θf cosθf θ } [e cos θ e θ sin θ] {cos θf sin θf θ } cosθ F 3 sin θ F 3 j [e sin θ e θ cosθ] cosθ {sin θf cosθf θ } sin θ {sin θf cosθf θ } sin θ {cosθf sin θf θ } cosθ On simplifying above, 4.31 F = e [ 1 F 3 F ] θ {cos θf sin θf θ } [ F e θ F ] 3 k Fθ 1 F θ 1 k k F

7 5. Spheical Coodinate Repesentation In spheical coodinates, an onomal set of unit vectos {e ρ,e φ,e φ } defined in the diections of inceasing ρ, θ and φ espectively. Recall the tansfomation between catesian and spheical coodinates: 5.32 = ρ sin φ cosθ, y = ρ sin φ sin θ, z = ρ cosφ and 5.33 ρ = 2 y 2 z 2, θ = actan y, = φ = accos z 2 y 2 z 2 We note that position vecto of an abitay point 5.34 = ρ sin φ cosθi ρ sin φ sin θj ρ cosφk We note that 5.35 ρ φ Theefoe, it follows that = sin φ cosθ i sin φ sin θ j cosφ k, = ρ sin φ sin θ i ρ sin φ cosθ j = ρ cosφcosθ i ρ cosφsin θ j ρ sin φ k 5.36 e ρ = 1 = sin φ cosθ i sin φ sinθ j cos φk ρ ρ e φ = 1 = cos φ cosθ i cosφsin θ j sin φ k φ φ e θ = 1 = sin θ i cosθ j Using 5.36, epessions fo {i,j,k} may be found in tems of {e ρ,e φ,e θ } 5.37 i = e ρ sin φ cosθ e φ cosφcosθ e θ sin θ j = e ρ sin φ sinθ e φ cosφsin θ e θ cosθ k = e ρ cosφ e φ sin φ Thus, fo an abitay vecto F, F = F 1 i F 2 j F 3 j = F 1 sin φ cosθ F 2 sin φ sin θ F 3 cosφe ρ F 1 cosφcosθ F 2 cosφsin θ F 3 sin φe φ F 1 sin θ F 2 cosθe θ e ρ F ρ e φ F φ e θ F θ 7

8 8 It follows that the catesian and pola coodinates of an abitay vecto F ae elated though 5.38 F ρ F φ F θ = F 1 sin φ cosθ F 2 sin φ sin θ F 3 cos φ = F 1 cos φ cosθ F 2 cosφsin θ F 3 sin φ = F 1 sin θ F 2 cosθ Futhe since F = F ρ e ρ F φ e φ F θ e θ, it follows fom elation 5.36 that 5.39 F 1 F 2 F 3 = F ρ sin φ cosθ F φ cos φ cosθ F θ sin θ = F ρ sin φ sinθ F φ cos φ sinθ F θ cos θ = F ρ cosφ F φ sin φ Routine calculations using chain ule shows that 5.32 and 5.33 imply ρ = ρ = sin φ cosθ, ρ y = y ρ = sin φ sinθ, ρ = z ρ = cosφ = y 2 y = sin θ 2 ρ sin φ, y = 2 y = cosθ 2 ρ sin φ, = 0 φ = z ρ ρ 2 sin φ = cosφcosθ ρ φ z ρ = y ρ 2 sin φ y = cosφsin θ ρ φ = 1 ρ sin φ z ρ ρ 2 sin φ = sin φ ρ 5.1. Epession fo gadient in Spheical coodinate. We note that 5.40 f = f i f y i f k { f ρ = ρ f φ φ f } { f i ρ ρ y f φ φ y f { f ρ ρ f φ φ f } y j } k Using 5.37, given elationship between deivatives of {ρ, φ, θ} and {, y, z} and some tedious but outine algeba, 5.40 implies 5.41 f = f ρ e ρ 1 f ρ φ e φ 1 f ρ sin φ e θ

9 5.2. Divegence and Laplacian 2 in Spheical Coodinates. Recall if F = F 1 i F 2 j F 3 k, then using 5.39 and chainule 5.42 F = F 1 F 2 y F 3 [ ρ = ρ φ φ ] {F ρ sin φ cosθ F φ cosφcosθ F θ sin θ} [ ρ = y ρ φ y φ ] {F ρ sin φ sin θ F φ cosφsin θ F θ cosθ} y [ ρ ρ φ φ ] {F ρ cos φ F φ sin φ} Again using known deivatives of {ρ, φ, θ} with espect to {, y, z} as aleady woked out and some outine but tedious algebea, 5.42 implies that in tems of spheical coodinate epesentation 5.43 F = F ρ ρ 2 ρ F ρ 1 F φ ρ φ cot φ ρ F φ 1 F θ ρ sin φ Combining the epessions of gadients and divegence, it follows that f = 2 f ρ 2 f 2 ρ ρ 1 2 f ρ 2 φ cot φ 2 ρ 2 f φ 1 ρ 2 sin 2 φ 2 f 2 Eample 4: Using spheical coodinates calculate the divegence of the vecto field F = 3 Solution: Note fom the epesentation of e ρ that 9 So, = ρ sin φ cosθi ρ sin φ sin θj ρ cosφk = ρe ρ F = 1 ρ 2e ρ So, the spheical coodinate components F ρ = ρ 2, F φ = 0 = F θ. Theefoe, it follows fom 5.43 that in this eample F = ρ 2 ρ 2ρ 2 ρ = 2 ρ 3 2 ρ 3 = 0 fo ρ 0. 1 Eample 5: Show that is a solution to the Laplace s equation 2 f = 0, ecept at the oigin. 2 y 2 z2 Solution: Note fo ρ 0, fom 5.44, it follows that 2 f = 21 ρ = 2 ρ 1 ρ 2 2 ρ 1 ρ ρ = 2 ρ 2 3 ρ = 0 3

10 Cul in Spheical Coodinates. Recall if F = F 1 if 2 jf 3 k, then 5.45 F = i ρ = i y ρ F 1 j ρ φ k F3 y F 2 F1 j F 3 F2 k F 1 y F 3 ρ φ F 3 y φ F 3 y ρ F 1 φ F 1 ρ F 3 ρ F 2 ρ φ F 2 φ F 2 ρ F 1 y F 2 ρ φ F 2 φ ρ φ ρ φ y F 3 φ F 1 φ y F 2 F 3 F 1 Using 5.37 and 5.39 and the epessions fo patial deivaties of {ρ, φ, θ} in tems of {, y, z}, it follows afte some tedious though outine algeba that 5.45 implies in spheical coodinates 1 F θ 5.46 F = e ρ ρ φ cot φ ρ F θ 1 1 F ρ e φ ρ sin φ F θ ρ F θ ρ ρ sin φ F φ Fφ e θ ρ F φ ρ F ρ φ Eample: Show that a cental foce F which is diected along the line of the position vecto and only dependent on the adial distance must be consevative. Solution: Note that fom given condition F = F ρ e ρ, with othe components F φ = 0 = F θ. Futhe F ρ only depends on ρ. Plugging into 5.46, it is clea that F = 0, hence fom Theoem in tet, F is consevative.

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