EM Radiation oscillating charged mass

[roeb]

Homework Statement

A particle of mass m and charge q is attached to a spring with force constant k, hanging from the ceiling. Its equilibrium position is a distance h above the floor. It is pulled down a distance d below equilibrium and released, at time t = 0;

Under the usual assumptions (d << lambda << h) calculate the intensity of the radiation hitting the floor as a function of the distance R from the point directly below q.

Homework Equations

The Attempt at a Solution

I see that this is a harmonic oscillator that could be described by x(t) = d cos(wt)
and a = -w*w*dcos(wt)

I would like to use Larmor's formula: P = u_0 q^2 a^2 / ( 6 pi m c) but I believe that I may need to revert back the to Poynting vector because we are only trying to find intensity [W/area] so: S = u_0 q^2 a^2 / ( 32 pi m c).

I think I'm having a hard time determining how to draw the picture to visualize this situation. How do I handle finding the intensity on the floor? I see that the distance from the mass to the floor is (R^2 + (h+delta)^2)^1/2.

Does anyone have any tips on how to get started with this? Do I need to rewrite the electric and magnetic fields and recalculate the Poynting vector from scratch (re-derive the electric dipole equations essentially?)

 

[Spinnor]

Draw the charge at a height h above a plane. Draw a ring of radius R and thickness dR in said plane centered on the charge. Draw the angles theta and theta + d(theta) that the ring makes.

From http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node146.html

We find the differential power, dP, into a solid angle d(omega) goes as:

dP/d(omega) is proportional to a^2*sin^2(theta) eq. 18.35

d(omega) = sin(theta)*d(theta)*d(phi) See solid angle:

http://en.wikipedia.org/wiki/Solid_angle

Convince your self that the power that goes through the ring is proportional to:

a^2*sin^3(theta)*d(theta)

There is a geometric relationship between dR and d(theta) that you should be able to find from your drawing.

You know dP/d(omega) but you want dP/dA where dA is the area of the ring.

But dP/dA = dP/d(omega)*d(omega)/dA

All you need is the relationship between d(omega) and dA.

Corrections welcome, hope this helps!

 

[roeb]

Hello Spinnor,

Thanks for your excellent response, I've almost figured it out.

From the picture I see that: dR² = 2(h² + R²) + dR² + 2RdR - 2 sqrt(h² +R²) sqrt( h² + (R+dR)²) cos( d(theta) )

I decided that this was a mess but we could probably simplify it to where (R+dR)² ~= R² so I get: dR = (h² + R²) d(theta).

Then using the solid angle d(omega) = sin(theta) d(theta) d(phi) I re-arrange and plug it in.

d(omega) = sin(theta) dR d(phi) / sqrt(h² + R²)

Now there is where I've struggled. I can't seem to translate that into a useful d(omega)/dA expression. Is my approximation for dR correct? Or am I missing something else?

 

[Spinnor]

I think we want dA = R*d(phi)*dR

d(omega)/dA =[sin(theta)*d(theta)*d(phi)]/ R*d(phi)*dR

dR is related to d(theta)

[h^2+R^2]*d(theta)/h = dR , so

d(omega)/dA = [sin(theta)*d(theta)]/R*dR

=[h*sin(theta)*d(theta)]/R*[h^2+R^2]*d(theta)

=[h*sin(theta)]/(R*[h^2+R^2])

So it looks like dP/dA goes as

a^2*sin^2(theta)*d(omega)/dA

We want to get sin(theta) in terms of R. Let straight down be theta = 0, then R/[h^2+R^2]^.5 = sin(theta)

This was pretty sloppy. Better next time, hope it helps.

 

[paranormal]

I would approach this problem by first understanding the physical situation described. Based on the information provided, it seems that a charged particle attached to a spring is oscillating and releasing electromagnetic radiation. The intensity of this radiation is being measured at a certain distance from the point directly below the particle.

To calculate the intensity at a distance R from the point below the particle, we can use the formula for the Poynting vector, which represents the energy flux density of electromagnetic radiation. This vector is given by S = (1/μ0)E x B, where E is the electric field and B is the magnetic field.

In this case, we can use the equations for the electric and magnetic fields of a dipole to calculate the Poynting vector at a distance R. We can then use the Poynting vector to calculate the intensity of the radiation at that distance.

It is important to note that the assumptions given in the problem (d << λ << h) refer to the distance d being much smaller than the wavelength of the radiation and the distance h being much smaller than the distance between the particle and the floor. These assumptions allow us to simplify the equations and make calculations easier.

In summary, to calculate the intensity of the radiation hitting the floor at a distance R from the point directly below the particle, we can use the Poynting vector formula and the equations for the electric and magnetic fields of a dipole. We can also use the given assumptions to simplify the calculations.