S.E in cylinder - Bessel's equation

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In summary, the problem asks for the lowest energy a particle of mass m can have in a right circular cylinder of radius R and height H. The wave function satisfies the Schrödinger equation, and the condition that the wave function go to zero over the surface of the cylinder gives the solutions of the three differential equations. The first two equations give the solution to the laplace operator in cylindrical coordinates, and the third equation gives the bessels equation for a particle at a particular point in space. The solutions of the first two equations are functions of z, and the solutions of the third equation are bessels functions of the first kind. The wave function is given by \Psi(r,\phi,
  • #1
JohanL
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Im trying to solve the following problem:

A particle of mass m is contained in a right circular cylinder(pillbox) of radius R and height H. The particle is described by a wave function satisfying the Schrödinger wave equation

[tex]
-\frac{\hbar^2}{2m}\nabla^2\Psi(r,\phi,z) = E\Psi(r,\phi,z)
[/tex]

and the condition that the wave function go to zero over the surface of the pillbox. Find the lowest(zero point) permitted energy.

Solution:

First i express the laplace operator in cylindrical coordinates and use separation of variables. This gives three differential equations.

[tex]
\frac {d^2Z(z)} {dz^2} = \lambda^2z
[/tex]

[tex]
\frac {d^2\Phi(\phi)} {d\phi^2} = -m^2\Phi
[/tex]

[tex]
r\frac {d} {dr}(r\frac {dR(r)} {dr}) + (\mu^2r^2-m^2)R(r) = 0
[/tex]

with

[tex]
k^2 + \lambda^2 = \mu^2
[/tex]

and [tex]\lambda, \mu, m [/tex] are the separation constants

The periodicity and the boundary conditions Z(0)=0 and Z(H)=0 gives the solutions of the first two D.E are

[tex]
Z(z) = C_1sinh(\pi lz/H)
[/tex]
[tex]
\Phi(\phi) = C_2exp(im\phi)
[/tex]

where l is an integer and m can take on the positive and negative integers and zero.

The third differential equation is bessels equation and because R(0) should be finite the solutions are bessels functions of the first kind.

The wave function is then...i think.

[tex]
\Psi(r,\phi,z) = const.*J_m(\sqrt{\frac{2mE}{\hbar^2}}r)exp(im\phi)sinh(\pi lz/H)
[/tex]

Usually you then use the other boundary condition R(r=R)=0 to get the energies. If the wave function above didnt include the sinh-term (the z dependence) i know that the energies are given by the condition:


[tex]
J_m(\sqrt{\frac{2mE}{\hbar^2}}r) = 0
[/tex]

which gives

[tex]
E(m,n) = \frac{\hbar^2}{2mR^2}r_m_n
[/tex]

where r_mn is the n:th root of the m:th Bessel function.

But with this problem's wave function i don't know how to calculate the energies.
Any ideas?

The answer is:

[tex]
E(m,n) = \frac{\hbar^2}{2m}[(\frac{r_m_n}{R})^2+(\frac{l\pi}{H})^2]
[/tex]
 
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  • #2
[tex]\frac {d^2Z(z)} {dz^2} = \lambda^2z[/tex]

Perhaps [itex] \lambda [/itex] needs to be imaginary to satisfy your boundary conditions??
 
  • #3
OlderDan's point is that sinh(x) is a one to one function and is not 0 at two different points. If λ is imaginary (equivalently: if &lamda;2 is negative), you get sine and cosine solutions which are periodic.
 
  • #4
Thank you.
ok...lets see if i understand.

we have

[tex]\frac {d^2Z(z)} {dz^2} = \lambda^2z[/tex]

My book says that if [tex] \lambda > 0[/tex] the general solution is

[tex]C_1cosh (\lambda z) + C_2sinh (\lambda z)[/tex]

But in the problem we have Z=0 at two different points

"OlderDan's point is that sinh(x) is a one to one function and is not 0 at two different points."

so [tex]\lambda[/tex] must be imaginary (or am i something wrong when i eliminate C1cosh because of the boundary condition Z(0)=0. Either way the main problem remains.)

"If λ is imaginary you get sine and cosine solutions which are periodic."

and the general solution is

[tex]C_1cos (\lambda z) + C_2sin (\lambda z)[/tex]



Then you get

[tex]Z(z) = C_1sin(\pi lz/H)[/tex]

[tex]\Psi(r,\phi,z) = R(r) \Phi(\phi)Z(z)[/tex]

[tex]\Psi(r,\phi,z) = const.*J_m(\sqrt{\frac{2mE}{\hbar^2}}r)exp(im\phi) sin(\pi lz/H)[/tex]

sin instead of sinh in the wave function.
where

[tex]R(r) = C*J_m(\sqrt{\frac{2mE}{\hbar^2}}r)[/tex]

One problem solved...i hope.
But i still don't know how to continue.

Now you should use the other boundary condition
R(r=R)=0 to get the energies. Is this correct:

[tex]const.*J_m(\sqrt{\frac{2mE}{\hbar^2}}R)exp(im\phi) sin(\pi lz/H) = 0[/tex]

and then you solve for E and get the energies?

I don't really know how to do this tho.
 
  • #5
Assuming you now have the correct eigenfunction, if you put it back into the original equation

[tex]-\frac{\hbar^2}{2m}\nabla^2\Psi(r,\phi,z) = E\Psi(r,\phi,z) [/tex]

and operate on the left hand side you should get a constant times the original function. The const. in your function divides out. You just need to associate the constant that comes from doing the operation on the left hand side with the E on the right hand side.
 
  • #6
I solved it. Thank you!
 

Related to S.E in cylinder - Bessel's equation

1. What is "S.E in cylinder - Bessel's equation"?

"S.E in cylinder - Bessel's equation" is a mathematical equation used to describe the behavior of waves in a cylindrical environment. It is named after the mathematician Friedrich Bessel who first studied these types of equations.

2. What does the "S.E" in "S.E in cylinder - Bessel's equation" stand for?

The "S.E" stands for "second order differential equation". This refers to the type of equation that Bessel's equation is, which includes second order derivatives of a function.

3. How is "S.E in cylinder - Bessel's equation" used in science?

Bessel's equation is used in a variety of scientific fields, such as physics, engineering, and mathematics. It can be used to describe the behavior of waves in cylindrical systems, such as heat transfer, fluid dynamics, and electromagnetic waves.

4. What are the applications of "S.E in cylinder - Bessel's equation"?

One of the main applications of Bessel's equation is in solving boundary value problems in cylindrical systems. It is also used in analyzing the behavior of waves in cylindrical systems and can be applied to various real-world problems in different scientific fields.

5. Is "S.E in cylinder - Bessel's equation" a difficult concept to understand?

The concept of Bessel's equation can be challenging to grasp for those without a strong mathematical background. However, with proper study and practice, it can be understood and applied in various scientific contexts.

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