Klein–Gordon equation with time dependent boundary conditions.

[Spinnor]

Klein–Gordon equation with time dependent boundary conditions.

Suppose we look for solutions to the Klein–Gordon equation with the following time dependent boundary conditions,

psi(r,theta,phi,t) = 0 zero at infinity

psi(on surface of small ball, B_1,t) = C*exp[i*omega*t]

psi(on surface of small ball, B_2,t) = C*exp[-i*omega*t]

where the small balls B_1 and B_2 have radius delta, do not overlap, and are a large distance D apart. C is a complex number and omega is some angular frequency. Assume omega > m the mass in the Klein–Gordon equation.

Will the field energy decrease as D gets smaller? Is this hard to show?

Thank you for any help!

 

[Spinnor]

Change things a bit. Let there be one ball below. Changes follow,

Suppose we look for solutions to the Klein–Gordon equation with the following time dependent boundary conditions,

psi(r,theta,phi,t) = 0 zero at infinity

psi(on surface of small ball, B_1,t) = C*exp[i*omega*t]

where the small ball B_1 has radius delta. C is a complex number and omega is some angular frequency. Assume omega > m the mass in the Klein–Gordon equation.

This makes an interesting "surface" in spacetime? Spacetime X phasespace?

It "twists" in the radial direction about the boundary? Phase has some r dependence (for any omega but for omega >> m things get very twisted)?

Thanks for any thoughts!

 

[genneth]

You should find the spectral decomposition of the (3D) Laplacian on the space described. For a single sphere, you should get the usual spherical harmonics times r^-n. I'm not sure what it is for two spheres.

 

[Spinnor]

Lets make it more interesting and add the electromagnetic field. We will still "drive" the Klein Gordon field on the surface of some small ball but now we will include the electromagnetic field. Assume the driving force starts up very quickly at t = 0 so that on the surface of a small ball the Klein Gordon field is C*exp[i*omega*t]. After a long time has gone by what rough fraction of the energy will be in the massless electromagnetic field compared with the massive Klein Gordon field and what is its general form.

5 dollars for the first answer that sounds right by me. Any takers %^)

Thanks for any help!

Change things a bit. Let there be one ball below. Changes follow,

Suppose we look for solutions to the Klein–Gordon equation with the following time dependent boundary conditions,

psi(r,theta,phi,t) = 0 zero at infinity

psi(on surface of small ball, B_1,t) = C*exp[i*omega*t]

where the small ball B_1 has radius delta. C is a complex number and omega is some angular frequency. Assume omega > m the mass in the Klein–Gordon equation.

This makes an interesting "surface" in spacetime? Spacetime X phasespace?

It "twists" in the radial direction about the boundary? Phase has some r dependence (for any omega but for omega >> m things get very twisted)?

Thanks for any thoughts!

 

[Spinnor]

Change things a bit. Let there be one ball below. Changes follow,

Suppose we look for solutions to the Klein–Gordon equation with the following time dependent boundary conditions,

psi(r,theta,phi,t) = 0 zero at infinity

psi(on surface of small ball, B_1,t) = C*exp[i*omega*t]

where the small ball B_1 has radius delta. C is a complex number and omega is some angular frequency. Assume omega > m the mass in the Klein–Gordon equation.

This makes an interesting "surface" in spacetime? Spacetime X phasespace?

It "twists" in the radial direction about the boundary? Phase has some r dependence (for any omega but for omega >> m things get very twisted)?

Thanks for any thoughts!

Assume the solution to the above problem has the form

phi(r,t) = psi(r)*E(t), and with the boundary condition,

psi(R)*E(t) = psi(R)*C*exp[i*omega*t]

We want to solve,

[Box^2 + m^2]phi(r,t) = 0, or
[Box^2 + m^2]psi(r)*E(t) = 0, or
[m^2 - omega^2]psi(r)*E(t) = E(t)[Laplacian*psi(r)], or
[m^2 - omega^2]psi(r) = Laplacian*psi(r)

Assume psi(r) = r^-1*exp(-M*r) with help from Wolfram Alpha,

http://www.wolframalpha.com/ we find out

Laplacian*psi(r) = M^2*psi(r) so

[m^2 - omega^2] = M^2

as omega > m , M = +or- i*[omega^2 - m^2]^.5

I think we throw away the negative root?

so phi(r,t) = r^-1*exp(-M*r)*C*exp[i*omega*t]

It appears this function only "twists" in the radial direction for omega > m , bad guess on my part. With this function now we might be able to come up with an answer to my last question above where we now want to "turn on" the electromagnetic field.

Thanks for any help!

 

[Spinnor]

http://www.wolframalpha.com/[/url] we find out

Laplacian*psi(r) = M^2*psi(r) so

[m^2 - omega^2] = M^2

as omega > m , M = +or- i*[omega^2 - m^2]^.5

I think we throw away the negative root?

so phi(r,t) = r^-1*exp(-M*r)*C*exp[i*omega*t]...

Depending on whether the energy in the field is increasing or decreasing we will need both signs for M?

 

[Spinnor]

Lets make it more interesting and add the electromagnetic field. We will still "drive" the Klein Gordon field on the surface of some small ball but now we will include the electromagnetic field. Assume the driving force starts up very quickly at t = 0 so that on the surface of a small ball the Klein Gordon field is C*exp[i*omega*t]. After a long time has gone by what rough fraction of the energy will be in the massless electromagnetic field compared with the massive Klein Gordon field and what is its general form.

5 dollars for the first answer that sounds right by me. Any takers %^)

Thanks for any help!

Change things a bit. Let there be one ball below. Changes follow,

Given

J_mu=i(phi* partial_u phi - phi partial_u phi*)

and

phi(r,t) = r^-1*exp(-M*r)*C*exp[i*omega*t]

thenedit mistake?

J_r goes as M/r^2
J_0 goes as omega/r^2

With this and

Box^2 A_u = J_u

We should be able to find A_r and A_0 ?

 

[Spinnor]

thenedit mistake?

J_r goes as M/r^2
J_0 goes as omega/r^2

With this and

Box^2 A_u = J_u

We should be able to find A_r and A_0 ?

So A_r and A_0 go as ln(r)?

 

[Deric Boyle]

Spinnor you are asking doubt or teaching us :-) ? by the way Nice problem to ask.

 

[Spinnor]

Spinnor you are asking doubt or teaching us :-) ? by the way Nice problem to ask.

I'm glad I did not waste your time %^) thank you. I now think I can tackle the original question in this thread assuming I have not made any major errors?

Thanks for any help!

 

[Deric Boyle]

Are you joking Spinnor! Its an ace level problem.Your approach seems to be correct to me but I'm not an expert.Therefore you must consult sir A.Neumaier or Sir tom.stoer.
Thank you!