Classical Wave Equations

Transcription

1 Classical Wave Equations Michael Fowle, Univesity of Viginia 5/16/07 Intoduction The aim of this section is to give a faily bief eview of waves in vaious shaped elastic media beginning with a taut sting, then going on to an elastic sheet, a dumhead, fist of ectangula shape then cicula, and finally consideing elastic waves on a spheical suface, like a balloon. The eason we look at this mateial hee is that these ae eal waves, hopefully not too difficult to think about, and yet mathematically they ae the solutions of the same wave equation the Schödinge wave function obeys in vaious contexts, so should be helpful in visualizing solutions to that equation, in paticula fo the hydogen atom. We begin with the stetched sting, then go on to the ectangula and cicula dumheads. We deive the wave equation fom F = ma fo a little bit of sting o sheet. The equation coesponds exactly to the Schödinge equation fo a fee paticle with the given bounday conditions. The most impotant section hee is the one on waves on a sphee. We find the fist few standing wave solutions. These waves coespond to Schödinge s wave function fo a fee paticle on the suface of a sphee. This is what we need to analyze to undestand the hydogen atom, because using sepaation of vaiables we split the electon s motion into adial motion and motion on the suface of a sphee. The potential only affects the adial motion, so the motion on the sphee is fee paticle motion, descibed by the same waves we find fo vibations of a balloon. (Thee is the genealization to complex nonstanding waves, paallel to the one-dimensional extension fom sinkx and coskx to e ikx and e -ikx, but this does not affect the stuctue of the equations.) Waves on a Sting Let s begin by eminding ouselves of the wave equation fo waves on a taut sting, stetched between x = 0 and x = L, tension T newtons, density ρ kg/mete. Assuming the sting s equilibium position is a staight hoizontal line (and, theefoe, ignoing gavity), and assuming it oscillates in a vetical plane, we use f(x,t) to denote its shape at instant t, so f(x,t) is the instantaneous upwad displacement of the sting at position x. We assume the amplitude of oscillation emains small enough that the sting tension can be taken constant thoughout. The wave equation is deived by applying F = ma to an infinitesimal length dx of sting (see the diagam below). We pictue ou little length of sting as bobbing up and down in simple hamonic motion, which we can veify by finding the net foce on it as follows. At the left hand end of the sting fagment, point x, say, the tension T is at a small angle df(x)/dx to the hoizontal, since the tension acts necessaily along the line of the sting.

2 Since it is pulling to the left, thee is a downwad foce component Tdf(x)/dx. At the ight hand end of the sting fagment thee is an upwad foce Tdf(x + dx)/dx. y T T 0 x x+dx x Putting f(x + dx) = f(x) + (df/dx)dx, and adding the almost canceling upwads and downwads foces togethe, we find a net foce T(d f/dx )dx on the bit of sting. The sting mass is ρ dx, so F = ma becomes f ( xt, ) f( xt, ) T dx= ρdx x t giving the standad wave equation f ( xt, ) 1 f( xt, ) = x c t with wave velocity given by c = T/ρ. (A moe detailed discussion is given in my Physics 15 Couse, plus an animation hee.) This equation can of couse be solved by sepaation of vaiables, f(x,t) = f(x)g(t), and the equation fo f(x) is identical to the time independent Schödinge equation fo a paticle confined to (0, L) by infinitely high walls at the two ends. This is why the eigenfunctions (states of definite enegy) fo a Schödinge paticle confined to (0, L) ae identical to the modes of vibation of a sting held between those points. (Howeve, it should be ealized that the time dependence of the sting wave equation and the Schödinge time-dependent equation ae quite diffeent, so a nonstationay state, one coesponding to a sum of waves of diffeent enegies, will develop diffeently in the two systems.) Waves on a Rectangula Dumhead Let us now move up to two dimensions, and conside the analogue to the taut sting poblem, which is waves in a taut hoizontal elastic sheet, like, say, a dumhead. Let us assume a ectangula dumhead to begin with. Then, paallel to the agument above, we would apply F = ma to a small squae of elastic with sides paallel to the x and y axes. The tension fom the est of the sheet tugs along all fou sides of the little squae, and we

3 3 ealize that tension in a sheet of this kind must be defined in newtons pe mete, so the foce on one side of the little squae is given by multiplying this tension pe unit length by the length of the side. Following the sting analysis, we take the vetical displacement of the sheet at instant t to be given by f(x, y, t). We assume this displacement is quite small, so the tension itself doesn t vay, and that each bit of the sheet oscillates up and down (the sheet is not tugged to one side). Suppose the bottom left-hand cone (so to speak) of the squae is (x, y), the top ight-hand cone (x + dx, y + dy). Then the left and ight edges of the squae have lengths dy. Now, what is the total foce on the left edge? The foce is Tdy, in the local plane of the sheet, pependicula to the edge dy. Factoing in the slope of the sheet in the diection of the foce, the vetically downwad component of the foce must be Tdy f(x,y,t)/ x. By the same agument, the foce on the ight hand edge has to have an upwad component Tdy f(x+dx, y, t)/ x. Thus the net upwad foce on the little squae fom the sheet tension tugging on its left and ight sides is f ( x+ dx, y,) t f(, x y,) t f Tdy = Tdydx x x x The net vetical foce fom the sheet tension on the othe two sides is the same with x and y intechanged. The mass of the little squae of elastic sheet is ρ dxdy, and its upwad acceleation is f/ t. Thus F = ma becomes: f f f Tdydx Tdydx dxdy + = ρ x y t giving f f 1 f. + = x y c t with c = T/ρ. This equation can be solved by sepaation of vaiables, and the time independent pat is identical to the Schödinge time independent equation fo a fee paticle confined to a ectangula box. Waves on a Cicula Dumhead A simila agument gives the wave equation fo a cicula dumhead, this time in (, φ) coodinates (we use φ athe than θ hee because of its paallel ole in the spheical case, to be discussed shotly).

4 4 This time, instead of a tiny squae of elastic, we take the small aea ddφ bounded by the cicles of adius and + d and lines though the oigin at angles φ and φ + dφ. Now, the downwad foce fom the tension T in the sheet on the inwad cuved edge, which has length dφ, is Tdφ f(, φ, t)/. On putting this togethe with the upwad foce fom the othe cuved edge, it is impotant to ealize that the in Tdφ vaies as well as f/ on going fom to + d, so the sum of the two tems is Tdφ / ( f/ )d. To find the vetical elastic foces fom the staight sides, we need to find how the sheet slopes in the diection pependicula to those sides. The measue of length in that diection is not φ, but φ, so the slope is 1/. f/ φ, and the net upwad elastic foce contibution fom those sides (which have length d) is Tddφ / φ (1/. f/ φ). Witing F = ma fo this small aea of elastic sheet, of mass ρddφ, gives then Tdϕ d + Tddϕ = ρddϕ ϕ ϕ t f 1 f f which can be witten = ϕ c t f f f. This is the wave equation in pola coodinates. Sepaation of vaiables gives a adial equation called Bessel s equation, the solutions ae called Bessel functions. The coesponding electon standing waves have actually been obseved fo an electon captued in a cicula coal on a suface. Waves on a Spheical Balloon Finally, let us conside elastic waves on the suface of a sphee, such as an inflated spheical balloon. The natual coodinate system hee is spheical pola coodinates, with θ measuing latitude, but counting the noth pole as zeo, the south pole as π. The angle φ measues longitude fom some ageed oigin. We take a small elastic element bounded by longitude lines φ and φ + dφ and latitude θ and θ + dθ. Fo a sphee of adius, the sides of the element have lengths sinθ dφ, dθ etc. Beginning with one of the longitude sides, length dθ, tension T, the only slightly ticky point is figuing its deviation fom the local hoizontal, which is 1/sinθ.( f/ φ), since inceasing φ by dφ means moving an actual distance sinθ dφ on the suface, just analogous with the cicula case above. Hence, by the usual method, the actual vetical foce fom tension on the two longitude sides is Tdθ dφ. ( / φ)1/sinθ.( f/ φ). To find the foce on the latitude sides, taking the top one fist, the slope is given by 1/. f/ θ, so the foce is just Tsinθ dφ.1/. f/ θ. On putting this togethe with the opposite side, it is necessay to ecall that sinθ as well as f vaies with θ, so the sum is given by: Tdφdθ / θ sinθ.1/. f/ θ. We ae now eady to wite down F = ma once

5 5 moe, the mass of the element is ρ sinθ dθ dφ. Canceling out elements common to both sides of the equation, we find: 1 1 sinθ θ θ θ ϕ f f f sinθ + = sin c t. Again, this wave equation is solved by sepaation of vaiables. The time-independent solutions ae called the Legende functions. They ae the basis fo analyzing the vibations of any object with spheical symmety, fo example a planet stuck by an asteoid, o vibations in the sun geneated by lage sola flaes. Simple Solutions to the Spheical Wave Equation Recall that fo the two dimensional cicula case, afte sepaation of vaiables the angula dependence was all in the solution to f/ φ = λf, and the physical solutions must fit smoothly aound the cicle (no kinks, o it would not satisfy the wave equation at the kink), leading to solutions sinmφ and cosmφ (o e imφ ) with m an intege, and λ = m (this is why we took λ with a minus sign in the fist equation). Fo the spheical case, the equation containing all the angula dependence is 1 f 1 f sinθ + = λ f sinθ θ θ sin θ ϕ The standad appoach hee is, again, sepaation of vaiables. Taking the fist tem on the left hand side ove to the ight, and multiplying thoughout by sin θ isolates the φ tem: f 1 f = sin θ λf sinθ + ϕ sinθ θ θ Witing now ( θ, ϕ) = ( θ) ( ϕ ) f f f θ ϕ in the above equation, and dividing thoughout by f, we find as usual that the left hand side depends only on φ, the ight hand side only on θ, so both sides must be constants. Taking the constant as m, the φ solution is e ±imφ, and one can inset that in the θ equation to give fθ m f θ = λ f θ 1 sinθ sinθ θ θ sin θ

6 6 What about possible solutions that don t depend on φ? The equation would be the simple 1 f sinθ = λ f sinθ θ θ Obviously, f = constant is a solution (fo m = 0) with eigenvalue λ = 0. Ty f = cosθ. It is easy to check that this is a solution with λ =. Ty f = sinθ. This is not a solution. In fact, we should have ealized it cannot be a solution to the wave equation by visualizing the shape of the elastic sheet nea the noth pole. If f = sinθ, f = 0 at the pole, but ises linealy (fo small θ ) going away fom the pole. Thus the pole is at the bottom of a conical valley. But this conical valley amounts to a kink in the elastic sheet the slope of the sheet has a discontinuity if one moves along a line passing though the pole, so the shape of the sheet cannot satisfy the wave equation at that point. This is somewhat obscued by woking in spheical coodinated centeed thee, but locally the noth pole is no diffeent fom any othe point on the sphee, we could just switch to local (x,y) coodinates, and the cone configuation would clealy not satisfy the wave equation. Howeve, f = sinθ sinφ is a solution to the equation. It is a wothwhile execise to see how the φ tem gets id of the conical point at the noth pole by consideing the value of f as the noth pole is appoached fo vaious values of φ: φ = 0, π/, π, 3π/ say. The sheet is now smooth at the pole! We find f = sinθ cosφ, sinθ sinφ (and so sinθ e iφ ) ae solutions with λ =. It is staightfowad to veify that f = cos θ 1/3 is a solution with λ = 6. Finally, we mention that othe λ = 6 solutions ae sinθ cosθ sinφ and sin θ sinφ. We do not attempt to find the geneal case hee, but we have done enough to see the beginnings of the patten. We have found the seies of eigenvalues 0,, 6,. It tuns out that the complete seies is given by λ = l(l + 1), with l = 0, 1,,. This intege l is the analogue of the intege m in the wave on a cicle case. Recall that fo the wave on the cicle, if we chose eal wave functions (cosmφ, sinmφ, not e imφ ) then m gave the numbe of nodes the wave had (that is, m complete wavelengths fitted aound the cicle). It tuns out that on the sphee l gives the numbe of nodal lines (o cicles) on the suface. This assumes that we again choose the φ-component of the wave function to be eal, so that thee will be m nodal cicles passing though the two poles coesponding to the zeos of the cosmφ tem. We find that thee ae l m nodal latitude cicles coesponding to zeos of the function of θ.

7 7 Summay: Fist Few Standing Waves on the Balloon λ l m fom of solution (unnomalized) constant 1 0 cosθ 1 1 sinθ e iφ 1-1 sinθ e -iφ 6 0 cos θ 1/3 6 ±1 cosθ sinθ e ±iφ 6 ± sin θ e ±iφ The Schödinge Equation fo the Hydogen Atom: How Do We Sepaate the Vaiables? In thee dimensions, the Schödinge equation fo an electon in a potential can be witten: ψ ψ ψ + + V( x, y, z) ψ Eψ + = m x y z This is the obvious genealization of ou pevious two-dimensional discussion, and we will late be using the equation in the above fom to discuss electon wave functions in metals, whee the standad appoach is to wok with standing waves in a ectangula box. Recall that in ou oiginal deivation of the Schödinge equation, by analogy with the Maxwell wave equation fo light waves, we agued that the diffeential wave opeatos aose fom the enegy-momentum elationship fo the paticle, that is px + py + pz ψ ψ ψ ψ ψ + + = m m x y z m so that the time-independent Schödinge wave equation is nothing but the statement that E = K.E. + P.E. with the kinetic enegy expessed as the equivalent opeato.

8 8 To make futhe pogess in solving the equation, the only tick we know is sepaation of vaiables. Unfotunately, this won t wok with the equation as given above in (x, y, z) coodinates, because the potential enegy tem is a function of x, y and z in a nonsepaable fom. The solution is, howeve, faily obvious: the potential is a function of adial distance fom the oigin, independent of diection. Theefoe, we need to take as ou coodinates the adial distance and two paametes fixing diection, θ and φ. We should then be able to sepaate the vaiables, because the potential only affects adial motion. No potential tem will appea in the equations fo θ, φ motion, that will be fee paticle motion on the suface of a sphee. Momentum and Angula Momentum with Spheical Coodinates It is woth thinking about what ae the natual momentum components fo descibing motion in spheical pola coodinates (, θ, φ). The adial component of momentum, p, points along the adius, of couse. The θ-component p θ points along a line of longitude, away fom the noth pole if positive (emembe θ itself measues latitude, counting the noth pole as zeo). The φ-momentum component, p φ, points along a line of latitude. It will be impotant in undestanding the hydogen atom to connect these momentum components (p, p θ, p φ ) with the angula momentum components of the atom. Evidently, momentum in the -diection, which passes diectly though the cente of the atom, contibutes nothing to the angula momentum. Conside now a paticle fo which p = p θ = 0, only p φ being nonzeo. Classically, such a paticle is cicling the noth pole at constant latitude θ, say, so it is moving in space in a cicle o adius sinθ in a plane pependicula to the noth-south axis of the sphee. Theefoe, it has an angula momentum about that axis p ϕ sin θ = L z, say. (The standad tansfomation fom (x, y, z) coodinates to (, θ, φ) coodinates is to take the noth pole of the θ, φ sphee to be on the z-axis.) As we shall see in detail below, the wave equation descibing the φ motion is a simple one, with solutions of the fom e imφ with intege m, just as in the two-dimensional cicula well. This just means that the component of angula momentum along the z-axis is quantized, L z = mħ, with m an intege. Total Angula Momentum and Waves on a Balloon The total angula momentum is L= p, whee p is the component of the paticle s momentum pependicula to the adius, so p = p + ϕ pθ. Thus the squae of the total angula momentum is (apat fom a constant facto) the kinetic enegy of a paticle moving feely on the suface of a sphee. The equivalent

9 9 Schödinge equation fo such a paticle is the wave equation given in the last section fo waves on a balloon. (This can be established by the standad change of vaiables outine on the diffeential opeatos). Theefoe, the solutions we found fo elastic waves on a sphee actually descibe the angula momentum wave function of the hydogen atom. We conclude that the total angula momentum is quantized, L = l(l + 1)ħ, with l an intege. Angula Momentum and the Uncetainly Pinciple The conclusions of ou above waves on a sphee analysis of the angula momentum of a quantum mechanical paticle ae a little stange. We found that the component of angula momentum in the z-diection must be a whole numbe of ħ units, yet the squae of the total angula momentum L = l(l + 1)ħ is not a pefect squae! One might wonde if the component of angula momentum in the x-diection isn t also a whole numbe of ħ units as well, and if not, why not? The key is that in questions of this type we ae fogetting the essentially wavelike natue of the paticle s motion, o, equivalently, the uncetainty pinciple. Recall fist that the z- component of angula momentum, that is, the angula momentum about the z-axis, is the poduct of the paticle s momentum in the xy-plane and the distance of the line of that motion fom the oigin. Thee is no contadiction in specifying that momentum and that position simultaneously, because they ae in pependicula diections. Howeve, we cannot at the same time specify eithe of the othe components of the angula momentum, because that would involve measuing some component of momentum in a diection in which we have just specified a position measuement. We can measue the total angula momentum, that involves additionally only the component p θ of momentum pependicula to the p φ needed fo the z-component. Thus the uncetainty pinciple limits us to measuing at the same time only the total angula momentum and the component in one diection. Note also that if we knew the z- component of angula momentum to be mħ, and the total angula momentum wee L = l ħ with l = m, then we would also know that the x and y components of the angula momentum wee exactly zeo. Thus we would know all thee components, in contadiction to ou uncetainly pinciple aguments. This is the essential eason why the squae of the total angula momentum is geate than the maximum squae of any one component. It is as if thee wee a zeo point motion fuzzing out the diection. Anothe point elated to the uncetainty pinciple concens measuing just whee in its cicula (say) obit the electon is at any given moment. How well can that be pinned down? Thee is an obvious esemblance hee to measuing the position and momentum of a paticle at the same time, whee we know the fuzziness of the two measuements is elated by ΔpΔx ~ h. Naïvely, fo a cicula obit of adius in the xy-plane, p = L z and distance measued aound the cicle is θ, so ΔpΔx ~ h suggests ΔL z Δθ ~ h. That is to say, pecise knowledge of L z implies no knowledge of whee on the cicle the paticle is. This is not supising, because we have found that fo L z = mħ the wave has the fom e imφ, and so ψ, the elative pobability of finding the paticle, is the same anywhee in the cicle. On the othe hand, if we have a time-dependent wave function descibing a

10 10 paticle obiting the nucleus, so that the pobability of finding the paticle at a paticula place vaies with time, the paticle cannot be in a definite angula momentum state. This is just the same as saying that a paticle descibed by a wave packet cannot have a definite momentum. The Schödinge Equation in (, θ, φ) Coodinates It is woth witing fist the enegy equation fo a classical paticle in the Coulomb potential: 1 1 m p p p e d + + E 4 = θ ϕi πε This makes it possible to see, tem by tem, what the vaious pats of the Schödinge equation signify. In spheical pola coodinates, Schödinge s equation is: ψ 1 ψ 1 e ( ψ ) + sinθ ψ E + = ψ m sinθ θ θ sin θ ϕ 4πε0 Sepaating the Vaiables: the Messy Details We look fo sepaable solutions of the fom ψ (, θ, ϕ) = R() Θ( θ) Φ ( ϕ) We now follow the standad technique. That is to say, we substitute RΘΦ fo ψ in each tem in the above equation. We then obseve that the diffeential opeatos only actually opeate on one of the factos in any given pat of the expession, so we put the othe two factos to the left of these opeatos. We then divide the entie equation by RΘΦ, to get Θ 1 1 Φ 1 ( R) + sinθ + sin sin m R Θ θ θ θ θ Φ ϕ 4πε0 e = E Sepaating Out and Solving the Φ(φ) Equation The above equation can be eaanged to give: Θ 1 1 Φ m 1 e ( R) + sin E θ + = + R Θ sinθ θ θ sin θ Φ ϕ 4πε0 Futhe eaangement leads to:

11 11 1 Φ m 1 e Θ = sin θ E ( R) sinθ + Φ ϕ 4πε0 R Θsinθ θ θ At this point, we have achieved the sepaation of vaiables! The left hand side of this equation is a function only of φ, the ight hand side is a function only of and θ. The only way this can make sense is if both sides of the equation ae in fact constant (and of couse equal to each othe). Taking the left hand side to be equal to a constant we denote fo late convenience by -m, Φ ( ϕ) ϕ = m Φ We wite the constant m because we know that as a facto in a wave function Φ(φ) must be single valued as φ inceases though π, so an intege numbe of oscillations must fit aound the cicle, meaning Φ is sinmφ, cosmφ o e imφ with m an intege. These ae the solutions of the above equation. Of couse, this is vey simila to the paticle in the cicle in two dimensions, m signifies units of angula momentum about the z-axis. Sepaating Out the Θ(θ) Equation Backing up now to the equation in the fom ( ϕ) Θ 1 1 Φ m 1 e ( R) + sin E θ + = + R Θ sinθ θ θ sin θ Φ ϕ 4πε0 we can eplace the 1 Φ Φ Φ tem by m, and move the tem ove to the ight, to give 1 1 Θ m m 1 e 1 1 sin θ E ( R) = + Θ sinθ θ θ sin θ 4πε0 R We have again managed to sepaate the vaiables the left hand side is a function only of θ, the ight hand side a function of. Theefoe both must be equal to the same constant, which we set equal to λ. This gives the Θ(θ) equation: 1 Θ( θ ) m sin θ Θ ( θ) = λθ( θ ) sinθ θ θ sin θ

12 1 This is exactly the wave equation we discussed above fo the elastic sphee, and the allowed eigenvalues λ ae l(l+1), whee l = 0, 1,,.. with l m. The R() Equation Replacing the θ, φ opeato with the value found just above in the oiginal Schödinge equation gives the equation fo the adial wave function: 1 ll ( + 1) 1 e ( R()) R() R() ER() = m 4πε0 The fist tem in this adial equation is the usual adial kinetic enegy tem, equivalent to p /m in the classical pictue. The thid tem is the Coulomb potential enegy. The second tem is an effective potential epesenting the centifugal foce. This is claified by econsideing the enegy equation fo the classical case, The angula momentum squaed d 1 1 m p + p + p e E 4 = θ ϕ πε i ( θ ϕ) ( ) L = p + p = l l+ 1. Thus fo fixed angula momentum, we can wite the above classical equation as 0 1 ll ( + 1) + 1 e = E. p m 4πε0 The paallel to the adial Schödinge equation is then clea. We must find the solutions of the adial Schödinge equation that decay fo lage. These will be the bound states of the hydogen atom. In natual units, measuing lengths in units of the fist Boh adius, and enegies in Rydbeg units 4 ρ 4πε 0 me = a0 = ρ, E = E ε ε R =. me (4πε ) Finally, taking u() = R(), the adial equation becomes d u( ρ) l( l + 1) + u( ρ) u( ρ) = εu( ρ). dρ ρ ρ 0