PDEs in Spherical and Circular Coordinates

Transcription

1 Introduction to Partial Differential Equations part of EM, Scalar and Vector Fields module (PHY2064)

2 This lecture Laplacian in spherical & circular polar coordinates Laplace s PDE in electrostatics Schrödinger s (time independent) equation, for the hydrogen atom So we will introduce spherical and circular polar coordinates and apply them to two PDEs: Laplace s and Schrödinger s.

3 Spherical & cylindrical polar coordinates much easier than Cartesian coordinates for spheres & circles Often, we wish to solve a PDE such as Laplace s equation, the wave equation, Schrödinger s equation etc, for a system that has either spherical or circular symmetry, e.g., a hydrogen atom, the temperature distribution inside a sphere, the waves on the circular membrane on the top of a drum, etc. Although in principle we could obtain solutions in these cases in Cartesian coordinates (i.e. x, y and z), in practice it is much much easier to obtain solutions using a coordinate system suited to the problem: either spherical or circular polar coordinates.

4 Laplacian in spherical polar coordinates In order to do this, the first thing we need to know is what the Laplacian, 2, is in spherical and circular polar coordinates. In spherical polar coordinates, and for the function u(r, θ, φ), the Laplacian times u is 2 u = 1 r 2 r ( r 2 u r ) + 1 r 2 sin θ ( sin θ u ) + θ θ 1 2 u r 2 sin 2 θ φ 2 where r is the distance from the origin, θ is the angle between the vector r and the z axis, and φ is the angle between the projection of r onto the xy plane and the x axis. These are the usual spherical polar coordinates, and are in the little booklet you get given in exams.

5 Laplacian in circular polar coordinates In circular polar coordinates, and for the function u(r, θ), the Laplacian is 2 u = 1 ( r u ) u r r r r 2 θ 2 where r is the distance from the origin, and θ is the angle between r and the x axis. These are the usual circular polar coordinates. This in the little booklet you get given in exams, but the notation is different, there ρ is used instead of r and φ instead of θ, but it is same expression just with different variable names. I do not expect you to remember either of these expressions as they are in the booklet. However, they are important, and the r parts of these expression are so common, it will help you to know them.

6 Laplacian in spherical polar coordinates: no θ or φ dependence Often, for spherically symmetric systems the function only depends of r, i.e., we want to solve for a function u(r). Then the θ and φ derivatives are zero, and the spherical-polar-coordinates Laplacian simplifies to 2 u = 1 ( d r 2 r 2 du ) dr dr we can do the derivative of a product, and get 2 u = d2 u dr du r dr

7 Laplacian in circular polar coordinates: no θ dependence In 2D the circular-polar-coordinates Laplacian simplifies to 2 u = 1 ( d r du ) r dr dr when u is only a function of r. We can again do the derivative of a product, and get 2 u = d2 u dr du r dr Note that the 2D and 3D r parts of the Laplacian are the same except that we have 1/r in front of the first derivative in 2D, and 2/r in 3D.

8 Laplace s equation for a system with spherical symmetry As an example of Laplace s equation in a spherical geometry, let us consider a conducting sphere of radius R, that is at a potential V S. φ(r) r V S R The sphere is in a large volume with no charges, and we assume that the potential at infinity is 0 V. So, the two BCs for the volume of space outside the sphere are at the surface of the sphere and at r. We centre the coordinates on the sphere, and then the problem has spherical symmetry.

9 Gauss Law and Laplace s equation Gauss Law (one of Maxwell s four equations) is.e(r) = ρ(r)/ɛ 0 Now E = φ, by definition, and if the charge density ρ = 0, we have. [ φ(r)] = 2 φ(r) = 0 or 2 φ(r) = 0 Outside the sphere, where the charge density is zero, φ(r) obeys Laplace s equation

10 Laplace s equation for a system with spherical symmetry In spherical polar coordinates 2 φ(r) = 0 or d 2 φ(r) dr r dφ(r) dr = 0 Note that φ(r) is a function of r only, and hence this is an ODE not an PDE. When the system is spherically symmetric although it has three Cartesian coordinates, x, y and z, once we move over into spherical coordinates it is only a function of one variable. The PDE simplifies into an ODE which makes life much easier. This is a second-order Cauchy-Euler (CE) ODE.

11 Laplace s equation for a system with spherical symmetry As this is a Cauchy ODE, its solutions are power laws. d 2 φ(r) dr r Thus we try the power law solution dφ(r) dr = 0 φ(r) = r β We need the first and second derivatives, which are dφ(r) dr = βr β 1 and Substituting these in we get d 2 φ(r) dr 2 = β(β 1)r β 2 β(β 1)r β r βr β 1 = 0 or β(β 1)r β 2 + 2βr β 2 = 0

12 Laplace s equation for a system with spherical symmetry Cancelling the r β 2 we get a simple equation for β β 2 β + 2β = 0 or β(β + 1) = 0 The solutions of this equation are β = 0 and β = 1, i.e., a constant and r 1. We multiply each of these solutions by a constant and add them together to obtain the general solution: φ(r) = C + D 1 r where C and D are the constants. The values of the two constants are set by the two BCs, i.e., to determine the value of the potential everywhere outside the sphere we need to not only solve the PDE but also apply two BCs. Two BCs because it is a second order differential equation.

13 BCs for Laplace s equation for a system with spherical symmetry We have two BCS: φ(r = R) = V S (surface of sphere) and φ(r ) = 0. Applying the BC at r φ(r ) = C + D = C = 0 so C = 0, and the function is now φ(r) = D 1 r if the potential at the surface of a sphere of radius R is V S, then φ(r = R) = D 1 R = V S D = V S R and the potential is φ(r) = V S R r

14 Coulomb s Law derived from Maxwell s equation By solving Laplace s equation and applying the BCs appropriate to a charged sphere we obtained the potential φ(r) = V S R r This is just Coulomb s Law. Coulomb s Law was discoved before Maxwell proposed his equation, but Coulomb s Law is not independent of Maxwell s equations, it can (as we have just showed) be derived from one of them.

15 Schrödinger equation for the ground state of the hydrogen atom As a second example of a PDE where we need to use spherical polar coordinates, we will determine the ground state wavefunction and energy of the hydrogen atom. The (time-independent) Schrödinger equation for the hydrogen atom is 2 2m 2 ψ e2 4πɛ 0 r ψ = Eψ where ψ is the wavefunction and E the energy. m is the mass of the electron. The first and second terms on the left-hand side are the kinetic energy term and the Coulomb attraction between the proton and the electron, respectively.

16 Schrödinger equation for the ground state of the hydrogen atom Now, to simplify the bunches of constants above it is best to multiply both sides of this equation by 4πɛ 0 /e 2. Then we have 1 h 2 ɛ 0 2 πme 2 2 ψ 1 r ψ = E ( 4πɛ 0 /e 2) ψ where we used = h/2π. Now, we define the Bohr radius as a = h2 ɛ 0 πme 2 = 0.53Å which is a length. We also define an energy U in units of 4πɛ 0 /e 2, by setting U = E(4πɛ 0 /e 2 ).

17 Schrödinger equation for the ground state of the hydrogen atom Then we get the neater equation 1 2 a 2 ψ 1 r ψ = Uψ Now, this equation yields all the wavefunctions of the hydrogen atom and many of them depend on all three of r, θ and φ. However, some depend only on r, these are called the s wavefunctions.

18 Schrödinger equation for the ground state of the hydrogen atom If we restrict ourselves to studying the s wavefunctions we can replace the Laplacian by just the r-derivative terms - the θ and φ derivatives are zero and so these terms are zero. Then we have 1 2 a ( d 2 ψ dr r dψ dr ) 1 r ψ = Uψ 1 ψ 2 ad2 dr 2 a dψ r dr 1 r ψ = Uψ This is the equation that yields the s wavefunctions and their corresponding energies.

19 Schrödinger equation for the ground state of the hydrogen atom To obtain a simple solution we will try substituting in an exponential into this equation, i.e., we will try ψ(r) = exp ( r/d) where d is the decay length for the exponential, whose value we don t know at the moment. I don t expect you to know that exponential is the right functional form to guess. Here we know that in the 1s state, the electron is bound to the proton, which means that ψ 0 as r, and an exponential decay is then a sensible guess, which turns out to be correct. Note that guessing the solutions is a perfectly respectable way to solve differential equations, so long as you then substitute it into the PDE, check whether it is a solution, and disgard it if it does not work of course.

20 Schrödinger equation for the ground state of the hydrogen atom With ψ(r) = exp ( r/d) as the trial function the derivatives are dψ dr = exp( r/d) = ψ d d and d 2 ψ dr 2 Substituting into Schrödinger s equation, we get 1 2 a ψ d 2 a ( r ψ ) 1 d r ψ = Uψ or, if we cancel the ψ s and rearrange a bit a 2d ( a ) r d 1 = U = exp( r/d) d 2 = ψ d 2

21 The LHS must equal the RHS at all values of r So, we have a 2d ( a ) r d 1 = U + 0 r at all values of r For LHS=RHS at all values of r, the constant terms must be equal a 2d 2 = U and the coefficients of 1/r must be equal a d 1 = 0 so d = a using d = a in equation for U gives U = a 2d 2 = 1 2a

22 Schrödinger equation for the ground state of the hydrogen atom So the 1s state of the H atom has an energy so E = 1 (4πɛ 0 /e 2 ) U = 1 (4πɛ 0 /e 2 ) ( 1 2a E = 13.6eV = J and a (not normalised) wavefunction for r in Å s. ψ(r) = exp( r/a) = exp( r/0.53) ) = me4 8ɛ 2 0 h2