Compton Shift: Why are Both Incident & Scattered Wavelengths Present?

[gnome]

Example 2.9 in Serway's Modern Physics, 2nd ed., asks
Why are x-ray photons used in the Compton experiment, rather than visible light photons? To answer this question, we shall first calculate the Compton shift for scattering at 90^o from graphite for the following cases: (1) very high energy γrays from cobalt, λ = 0.0106A; (2) x-rays from molybdenum, λ = 0.712A; and (3) green light from a mercury lamp, λ = 5461A. (A = angstrom units)

They calculate the wavelength shift in each case using the Compton shift formula, then calculate the fractional change in wavelength in each case, showing that:
for x-rays from cobalt, Δλ/λ₀ = 2.29
for x-rays from molybdenum, Δλ/λ₀ = .0341
for visible light from mercury, Δλ/λ₀ = 4.45 x 10^-6

and conclude that, "Because both incident and scattered wavelengths are simultaneously present in the beam, they can be easily resolved only if Δλ/λ₀ is a few percent or if λ₀ ≤ 1A."

FINALLY my question: the computations are straightforward and the conclusion about resolution makes perfect sense, but...
Why are both incident and scattered wavelengths present simultaneously in the beam? This beam is at a 90^o angle to the original direction of the beam from the x-ray source. So didn't all the photons in this beam get there by scattering? How do photons of the original, incident wavelength get there?

Same question, another way:
on pg. 81 of the same text, there are graphs of the distribution of intensity vs. wavelength for x-rays scattered at 0, 45, 90 and 135 degrees. All except the 0-degree graph show two peaks in the distribution: at λ₀ and at λ'. Why not just λ'?

 

[swansont]

FINALLY my question: the computations are straightforward and the conclusion about resolution makes perfect sense, but...
Why are both incident and scattered wavelengths present simultaneously in the beam? This beam is at a 90^o angle to the original direction of the beam from the x-ray source. So didn't all the photons in this beam get there by scattering? How do photons of the original, incident wavelength get there?

Same question, another way:
on pg. 81 of the same text, there are graphs of the distribution of intensity vs. wavelength for x-rays scattered at 0, 45, 90 and 135 degrees. All except the 0-degree graph show two peaks in the distribution: at ?₀ and at ?'. Why not just ?'?

There is scattering present that doesn't shift the wavelength. Compton scattering happens off of free or loosely bound electrons. Photons can also scatter off of inner-shell electrons without ionizing, and behave like they are scattering off of the whole atom, so there will be a minimal change in energy (use the mass of an atom in the Compton formula to see this).

Also - another reason why mercury light at 5461A would be bad is that its energy may be too low to Compton scatter in the first place. It may cause the photoelectric effect or just excitation if it interacts with a bound electron. You could only see the Compton effect for free electrons (and you wouldn't see it anyway for the reason given in the book)

 

[gnome]

So the photons that remain at the incident wavelength have in effect "collided" with entire atoms?

But if those photons have changed direction (as they obviously have), conservation of momentum requires that they must have imparted some momentum to whatever they collided with. And if that is true, then they must have transferred some kinetic energy to those objects. And if they have given up some energy, their wavelengths must have changed. So I've just gone around in a circle, showing that for the photons whose wavelengths haven't changed, their wavelengths must have changed.

Also, that leads me to this question: photons have no charge, so what force is acting between the photon and the atom, or the electron for that matter?

Maybe this is the same question in different words: when an electromagnetic wave exerts "radiation pressure" on something, what actual force is involved?

 

[swansont]

So the photons that remain at the incident wavelength have in effect "collided" with entire atoms?

But if those photons have changed direction (as they obviously have), conservation of momentum requires that they must have imparted some momentum to whatever they collided with. And if that is true, then they must have transferred some kinetic energy to those objects. And if they have given up some energy, their wavelengths must have changed. So I've just gone around in a circle, showing that for the photons whose wavelengths haven't changed, their wavelengths must have changed.

Also, that leads me to this question: photons have no charge, so what force is acting between the photon and the atom, or the electron for that matter?

Maybe this is the same question in different words: when an electromagnetic wave exerts "radiation pressure" on something, what actual force is involved?

Since the atom is massive compared to an electron, the wavelength shift is small, so it's not unchanged. It's just that the change is so small one can typically ignore it. Use the Compton equation and see - the mass is going to be 4 or 5 orders of magnitude higher, so the change in wavelength will be proportionally smaller.

Photons are electromagnetic radiation - they interact with the electric field of the electron or atom. They exert a pressure because they can exchange momentum (and energy), and changing momentum over time is a force. (F=dp/dt). Usually radiation pressure is put in terms of many photon interactions over a period of time.

 

[gnome]

Thanks swansont. I just ran some numbers for a 7.12 nm photon colliding with a stationary proton. If I'm doing this right, if the photon scatters at 90^o, it's change in wavelength is 1.32 x 10^-15m. After the collision the photon's speed is still c (but at 90^o to it's original path). The proton ends up moving at 45^o to the photon's original path, at 78.6 m/s.


Regarding the radiation pressure:
>> They exert a pressure because they can exchange momentum (and energy), and >> changing momentum over time is a force. (F=dp/dt)
This is exactly what I was thinking. But, does this force have no name?

 

[swansont]

Regarding the radiation pressure:
>> They exert a pressure because they can exchange momentum (and energy), and >> changing momentum over time is a force. (F=dp/dt)
This is exactly what I was thinking. But, does this force have no name?

Radiation pressure seems to be descriptive enough. When done in the context of laser cooling (usually 6 intersecting beams) it's called optical molasses.

 

[gnome]

Thanks again. I don't think I ever came across the term "laser cooling" before.

Found a real cute description of it, though:
http://www.colorado.edu/physics/2000/bec/lascool1.html

There's a lot of interesting introductory-level material on many other modern physics topics there as well.

 

[gnome]

>> Radiation pressure seems to be descriptive enough. When done in the context of laser >> cooling (usually 6 intersecting beams) it's called optical molasses.

I read that there are four fundamental forces: gravity, electro-magnetic force, and strong and weak nuclear forces.

But the "electro-magnetic force" seems to be 3 different forces: coulomb force between charged particles, magnetic force, and now this "radiation pressure" force. In what sense are these a single force?

 

[turin]

But the "electro-magnetic force" seems to be 3 different forces: coulomb force between charged particles, magnetic force, and now this "radiation pressure" force. In what sense are these a single force?

The interaction is electromagnetic. The force that the interaction exerts on matter can manifest in various ways. The electromagnetic interaction is characterized by a 2nd rank tensor (Faraday tensor), which leads to the Lorentz force (that is the electric and magnetic forces together in one equation). Conservation of momentum gives radiation pressure.

 

[gnome]

Well, I have no idea what a 2nd rank tensor is, but thanks anyway. Maybe eventually...

 

[turin]

Well, I have no idea what a 2nd rank tensor is, but thanks anyway. Maybe eventually...

Try to imagine a 6-D vector.

The Faraday tensor is kind of like a 6-D vector. You could also think of it as a matrix, but that isn't "safe" for a few reasons. The 6-D vector isn't really 6-D, it's sort of an "unfolded" 4-D vector. The 4-D are the space-time dimensions. We "see" 3 of them as the 3-D space, and the other one is the time dimension.

The components of the E-field are in the 3-D space directions. The components of the B-field are in the other 3 directions in the 6-D vector.

When you transform from a frame in which the 6-D vector has only E-field components to a frame in which the 6-D vector has some B-field components, this is kind of like a rotation of the 6-D space so that the 6-D vector now projects onto the B-field axes.

Again, it is really a rotation in 4-D space, but I am trying to avoid the full-blown tensor discussion. It's not that it's so complicated, but I think you probably wouldn't find it all that intuitive.