( ) 2. To find the location of the steepest gradient, we need to solve the pair of equations ( ) θ = where n is integer.

Transcription

1 Homework 5 Solutions 1 For the function ( ( ( Φ, = ep, find the location(s at which its gradient is steepest (ie has the largest magnitude The algebra is easier if plane polar coordinates are used In terms of plane polar coordinates, we have and so The gradient in plane polar coordinates is Hence = ρcos θ, = ρsin θ, (, e ρ cos Φ ρθ = ρ Φ 1 Φ Φ = ρˆ + θˆ ρ ρ = cos θ sin θ ρ 1 ρ ep ρ ρep ρ cosθsinθθˆ ( ( ( ρˆ ( ρ ep( ρ ( 1 ρ cos θρˆ sin θθˆ = ( ( Φ = ρ ep ρ 1 ρ cos θ sin θ + To find the location of the steepest gradient, we need to solve the pair of equations { ρ e ( } ( ( ( Φ = ρ ρ sin θ + cos θ 1 ρ ρ ρ { } ρ = 8ρe 1 ρ sin θ + cos θ 1 ρ ρ cos θ 1 ρ = 0, Φ ρ ( = 16ρ e 1 1 ρ sin θ cos θ = 0 The second equation has solutions (a ρ = 0, (b ρ = ±, (c nπ θ = where n is integer Substituting these solutions into the first equation, we find the following sets of solutions (a ρ = 0 for an θ

2 (b (c ρ =, cos θ = 3 nπ θ = where n is integer 5 ± 17 or ρ = ρ = 1if n is even, 1 ρ = if n is odd These solutions can be obtained b the Mathematica commands: Φc[_,_]=(^-^*Ep[-^-^]; Φp[ρ_,θ_]=TransformedField[{"Cartesian"->"Polar"},Φc[,],{,}->{ρ,θ}] //Simplif; Grad[Φp[ρ,θ],{ρ,θ},"Polar"] //Simplif; Part[%,1]^+Part[%,]^; n[ρ_,θ_]=fullsimplif[%,assumptions->element[ρ,reals],assumptions- >Element[θ,Reals]]; e1[ρ_,θ_]=d[n[ρ,θ],ρ]; e[ρ_,θ_]=d[n[ρ,θ],θ]; Reduce[e1[ρ,θ]==0&&e[ρ,θ]==0,{ρ,θ}] Note the use of Part[%,1]^+Part[%,]^ instead of Norm[%]^ This is to avoid the appearance of the Abs function, which leads to unwield epressions on taking the partial derivatives Also Reduce is preferred over Solve to get the complete solution Using the solutions to evaluate Φ, we get (a Φ = 0 (b e Φ = 8 5± 17 1± 17 Φ = 5 ± 17 e and Φ = 0 if n is even, (c ( 1 Φ = e if n is odd 1 m + 1 The largest slope is case (c with n odd This corresponds to ρ = and θ = π, where m 1 1 is an integer In Cartesian co-ordinates the maimum slope occurs at the points ±, ± 1 m 1 e + ˆ The gradient at these points is ( Φ = 1 θ

3 A contour plot of Φ is shown below Verif b working in Cartesian coordinates that ( a b = b ( a a ( b Let a = a i+a j+a zk, with a similar epression for b Now

4 = + + z ( a b ( ab z ab z ( ab z ab z ( ab ab bz b z = bz + a b az z b bz + b + az bz a b b + b + a b a z z z z z z b b z b b bz b a a az z z = b a a b z z = b + b + bz ( ( 3 For an vector ( A r = A (a Verif this result using Cartesian coordinates (b Verif this result using spherical polar coordinates (a Let A = A i+a j+a zk, so that (b Let A= Arˆ+ A θˆ + A ˆ Then r θ z z = Ai+ A j+ Ak ( A r = A + A + A ( i+ j+ zk = A z 1 1 r θ r r rsinθ rˆ 1 rˆ = A ˆ rr + Aθ + A sinθ ( A r = A + A + A ( rrˆ Now rˆ = sinθcosi+ sinθsinj+ cos θk Hence

5 rˆ = cosθcosi + cosθsinj sin θk = θˆ rˆ = sinθsini + sinθcosj = sin θˆ We see that rˆ 1 rˆ A r = A ˆ ˆ ˆ ˆ rr+ Aθ + A = Arr+ Aθθ+ A= A sinθ ( B equating in Cartesian coordinates to in spherical polar coordinates, or otherwise, epress,, z in spherical polar coordinates We have ˆ 1 1 = i + j + k = rˆ + θ + ˆ z r r rsinθ Taking appropriate scalar products with the Cartesian unit vectors, we get ˆ = ir ˆ + iθ + i ˆ = sinθcos + cosθcos sin, r r rsinθ r r rsinθ ˆ = jr ˆ + jθ + j ˆ = sinθsin + cosθsin + cos, r r rsinθ r r rsinθ ˆ = k rˆ + k θ + k ˆ = cosθ sinθ z r r rsinθ r r 5 The quantum mechanical angular momentum operator is defined b (in certain units (a Show that in spherical polar coordinates L= i ( r 1 L= i eˆ ˆ θ e sinθ (b Resolving eˆθ and eˆ into Cartesian components, determine L, L and L z in terms of θ and and derivatives with respect to θ and (c From L = L + L + L show that z,

6 1 1 sinθ L = sinθ sin θ = r + r r r (a In spherical polar coordinates, L ( ˆ ˆ ˆ i ir ˆ i ˆ = r = r + + = ˆ r r θ r rsinθ θ sinθ (b We have ( ( ( θˆ = iθi ˆ + jθ ˆ j+ kθk ˆ = cosθcosi+ cosθsinj sin θk, ( ( ( ˆ = ii ˆ + j ˆ j+ kk ˆ = sini+ cos j Hence 1 L= i( cosθcosi+ cosθsinj sinθk i( sini+ cosj sinθ = ii cotθcos + sin + ji cotθsin cos ki We see that L L L z = i cotθcos + sin = i cotθsin cos = i (c To avoid mistakes in working with operators, it is a good idea to appl them to a dumm function, f sa From the above, we find

7 f f L + L + L f = cotθcos + sin cotθcos + sin ( z f f f cotθsin cos cotθsin cos f f f = cot θcos cos cotθcos sin sincos cotθ f f f sin cot θsin sin cotθsin cos + θ + cossin cotθ cos θ f f f f f f f = cot θcos + cot θcossin cotθcossin cotθcos θ f cotθ f f sincoscotθ sincos sin θ f f f f cot θsin cot θsincos + cotθsincos cotθsin θ f cotθ f f + cossincotθ + cossin cos θ f Terms involving the cross derivative cancel out, and we get after some algebra 1 f 1 f + + = ( z L L L f sin θ sin θ sinθ Hence the operator is 1 1 L L L z sinθ + + = sinθ sin θ = r + r r r 6 Paraboloidal (also called 3D parabolic coordinates u, v, are related to Cartesian coordinates b 1 = uv cos, = uvsin, z = ( u v

8 Identif and describe the coordinate surfaces in the u, v, sstem Verif that each coordinate surface (eg u = constant intersects ever coordinate surface on which one of the other two coordinates (eg v is constant Show further that the sstem of coordinates is an orthogonal one and determine its scale factors Prove that the u-component of a is given b 1 a 1 v + v v uv ( u + v 1 Consider a surface of constant u Eliminating v and, this surface is, in Cartesian co-ordinates, uz= u + (1 ( This is a paraboloid of revolution, with z in the range (, u v is a paraboloid of revolution Similarl a surface of constant v z= v + ( +, ( with z in the range ( v, A surface of constant is a plane containing the z-ais with normal making an angle with the -ais Note, we can assume that u 0 and v 0 A surface of constant u will intersect a surface of constant v at z ( u v = on the circle + = u v Also each paraboloid intersects all the planes through the z-ais Hence, all coordinate surfaces intersect The figure shows the intersection of surfaces of constant u and v in the z plane 0 0 The orthogonalit of the coordinate sstem is apparent

9 The figure was obtained b using the Mathematic commands: ClearAll["Global`*"]; f[_,z_]=sqrt[^+z^]-z; g[_,z_]=sqrt[^+z^]+z; Show[ContourPlot[f[, z], {, -5, 5}, {z, -5, 5}, Contours -> {0, 0, 1,,, 6, 8}, ContourShading -> None, ContourStle -> Blue], ContourPlot[g[, z], {, -5, 5}, {z, -5, 5}, Contours -> {0, 0, 1,,, 6, 8}, ContourShading -> None, ContourStle -> Red]] To make the plot, epressions for u and v in terms of and z were obtained b solving equations (1 and ( with = 0 The position vector is 1 r = uv cosi+ uvsin j+ ( u v k Hence r = vcosi + vsinj + uk = eu u r = ucosi + usinj vk = ev v r = uvsini + uv cos j = e The matri of scalar products of pairs of the e s is u + v u + v 0, 0 0 uv and hence the co-ordinate sstem is orthogonal with scale factors h = h = u + v, h = uv u v Using the determinant epression for the curl in general curvilinear coordinates, we have 1 ( ( u va uva + v ( a = u + v u ( u + v uv v 1 v = ua uv u v + + uv u + v v 1 a 1 v = + u + v v v uv Paraboloid coordinates are useful in solving the Schrödinger equation for a hdrogen atom placed in a uniform electric field