PDEs in Spherical and Circular Coordinates

PDEs in Spherical and Circular Coordinates

PDEs in Spherical and Circular CoordinatesIntroduction to Partial Differential Equations

part of EM, Scalar and Vector Fields module (PHY2064)

Richard Sear

December 9, 2019

Richard Sear PDEs in Spherical and Circular Coordinates



This lecture

Laplacian in spherical & circular polar coordinates

Laplace’s PDE in electrostatics

Schrodinger’s (time independent) equation, for the hydrogenatom

Introduce spherical and circular polar coordinates, and apply themto two PDEs: Laplace’s and Schrodinger’s.



Spherical & cylindrical polar coordinates much easier thanCartesian coordinates for spheres & circles

We want to solve a PDE such as Laplace’s equation, the waveequation, Schrodinger’s equation etc, for a system that has eitherspherical or circular symmetry, e.g., a hydrogen atom.

For these systems we need to use a coordinate system suited to theproblem, for example spherical polar coordinates for the hydrogematom.



Laplacian in spherical polar coordinates

First thing we need to know is the Laplacian, ∇2, in spherical andcircular polar coordinates.In spherical polar coordinates, and for the function u(r , θ, φ),the Laplacian times u is

∇2u =1

r2∂

∂r

(r2∂u

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂u

∂θ

)+

1

r2 sin2 θ

∂2u

∂φ2

where r is the distance from the origin, θ is the angle between thevector r and the z axis, and φ is the angle between the projectionof r onto the xy plane and the x axis. These are the usualspherical polar coordinates, and are in the little booklet you getgiven in exams.



Laplacian in circular polar coordinates

In circular polar coordinates, and for the function u(r , θ), theLaplacian is

∇2u =1

r

∂

∂r

(r∂u

∂r

)+

1

r2∂2u

∂θ2

where r is the distance from the origin, and θ is the angle betweenr and the x axis. These are the usual circular polar coordinates.This is in the little booklet you get given in exams, but thenotation is different, there ρ is used instead of r and φ instead ofθ, but it is same expression just with different variable names.

I do not expect you to remember either of these expressions as theyare in the booklet. However, they are important, and the r parts ofthese expression are so common, it will help you to know them.



Laplacian in spherical polar coordinates: no θ or φdependence

If the function only depends of r , i.e., we want to solve for afunction u(r). Then the θ and φ derivatives are zero, and thespherical-polar-coordinates Laplacian simplifies to

∇2u =1

r2d

dr

(r2du

dr

)we can do the derivative of a product, and get

∇2u =d2u

dr2+

2

r

du

dr



Laplacian in circular polar coordinates: no θ dependence

In 2D the circular-polar-coordinates Laplacian simplifies to

∇2u =1

r

d

dr

(rdu

dr

)when u is only a function of r . We can again do the derivative of aproduct, and get

∇2u =d2u

dr2+

1

r

du

dr

Note that the 2D and 3D r parts of the Laplacian are the sameexcept that we have 1/r in front of the first derivative in 2D, and2/r in 3D.



Laplace’s equation for a system with spherical symmetry

As an example of Laplace’s equation in a spherical geometry, let usconsider a conducting sphere of radius R, that is at a potential VS .

VS

R

rφ(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.Richard Sear PDEs in Spherical and Circular Coordinates


Gauss’ Law and Laplace’s equation

The E field outside the sphere is given by Gauss’ Law (one ofMaxwell’s 4 equations):

∇.E(r) = ρ(r)/ε0

Now E = −∇φ, and as the charge density ρ = 0, we have

∇. [−∇φ(r)] = 0

∇2φ(r) = 0

Outside the sphere, where the charge density is zero, φ(r) obeysLaplace’s equation




In spherical polar coordinates

∇2φ(r) = 0 ord2φ(r)

dr2+

2

r

dφ(r)

dr= 0

Note that φ(r) is a function of r only, and hence this is an ODEnot an PDE.

This is a second-order Cauchy-Euler (CE) ODE.




As this is a Cauchy ODE, its solutions are power laws. Thus we trythe power law solution

φ(r) = rβ

and find that the two solutions are β = 0 and β = −1. See theCauchy-Euler notes. The general solution is

φ(r) = C + D1

r

where C and D are the constants.NB General solution of a 2nd-order ODE always needs twounknown constants — whose values are set by the two BCs.



BCs for Laplace’s equation for a system with sphericalsymmetry

We have two BCS:

φ(r = R) = VS (surface of sphere)

φ(r →∞) = 0.

What is the particular solution?

[ poll everywhere ]



Schrodinger equation for the ground state of the hydrogenatom

As a second example of a PDE where we need to use sphericalpolar coordinates, we will determine the ground state wavefunctionand energy of the hydrogen atom. The (time-independent)Schrodinger equation for the hydrogen atom is

− ~2

2m∇2ψ − e2

4πε0rψ = Eψ

where ψ is the wavefunction and E the energy. m is the mass ofthe electron. The first and second terms on the left-hand side arethe kinetic energy term and the Coulomb attraction between theproton and the electron, respectively.




Now, to simplify the bunches of constants above it is best tomultiply both sides of this equation by 4πε0/e

2. Then we have

−1

2

h2ε0πme2

∇2ψ − 1

rψ = E

(4πε0/e

2)ψ

where we used ~ = h/2π. Now, we define the Bohr radius as

a =h2ε0πme2

= 0.53A

which is a length. We also define an energy U in units of 4πε0/e2,

by setting U = E (4πε0/e2).




Then we get the neater equation

−1

2a∇2ψ − 1

rψ = Uψ

Now, this equation yields all the wavefunctions of the hydrogenatom and many of them depend on all three of r , θ and φ.

However, some depend only on r , these are called the swavefunctions.




If we restrict ourselves to studying the s wavefunctions we canreplace the Laplacian by just the r -derivative terms - the θ and φderivatives are zero and so these terms are zero. Then we have

−1

2a

(d2ψ

dr2+

2

r

dψ

dr

)− 1

rψ = Uψ

−1

2ad2ψ

dr2− a

r

dψ

dr− 1

rψ = Uψ

This is the equation that yields the s wavefunctions and theircorresponding energies.




To obtain a simple solution we will try substituting in anexponential into this equation, i.e., we will try

ψ(r) = exp (−r/d)

where d is the decay length for the exponential, whose value wedon’t know at the moment.

Here we know that in the 1s state, the electron is bound to theproton, which means that ψ → 0 as r →∞, and an exponentialdecay is then a sensible guess, which turns out to be correct.




Withψ(r) = exp (−r/d)

as the trial function the derivatives are

dψ

dr= −exp(−r/d)

d= −ψ

dand

d2ψ

dr2=

exp(−r/d)

d2=

ψ

d2

Substituting into Schrodinger’s equation, we get

−1

2a× ψ

d2− a

r×(−ψd

)− 1

rψ = Uψ

or, if we cancel the ψ’s and rearrange a bit

− a

2d2+

1

r

( a

d− 1

)= U



The LHS must equal the RHS at all values of r

So, we have

− a

2d2+

1

r

( a

d− 1

)= U +

0

rat all values of r

Solution must work at all values of r so LHS=RHS at all values ofr . So the constant terms must be equal (to make LHS=RHS asr →∞)

− a

2d2= 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

2d2= − 1

2a




So the 1s state of the H atom has an energy

E =1

(4πε0/e2)U =

1

(4πε0/e2)×(− 1

2a

)= − me4

8ε20h2

soE = −13.6eV = −2.2× 10−18J

and a (not normalised) wavefunction

ψ(r) = exp(−r/a) = exp(−r/0.53)

for r in A’s.




This lecture

Laplacian in spherical & circular polar coordinates

Laplace’s PDE in electrostatics

Schrodinger’s (time independent) equation, for the hydrogenatom


PDEs in Spherical and Circular Coordinates

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