Compton scattering when E mc^2

In summary, we are considering Compton scattering of a photon by a moving electron. The photon has a wavelength λ and is moving in the positive x direction, while the electron is moving in the negative x direction with a total energy E (including rest energy mc^2). After the head-on collision, both the photon and electron are moving in the negative x direction. By manipulating the Compton equation and using conservation of energy, we can derive an expression for the wavelength λ' of the scattered photon. However, this formula assumes that the electron is initially at rest and may not apply to our situation. Alternatively, we can set up conservation of momentum and energy for the collision and solve for λ' using messy algebra. In the end, if
  • #1
Upallnight
1
0

Homework Statement


Consider Compton scattering of a photon by a moving electron. Before the collision the photon has a wavelength λ and is moving in the positive x direction. The electron is moving in the negative x direction with a total energy E ( including rest energy mc^2). The photon and electron collide head on. After the collision both are moving in the negative x direction.
Derive an expression for the wavelength λ' of the scattered photon. Show that if E>>mc^2, where m is the rest mass of the electron, your result reduces to
λ'=(hc\λ)(1+m^2c^4/4hcE)


Homework Equations



λ'-λ=h/mc(1-cosθ)

The Attempt at a Solution


I've tried many things to manipulate the Compton equation and conservation of energy but I can't seem to get anywhere.
 
Physics news on Phys.org
  • #2
Hello, Upallnight.
Upallnight said:
Show that if E>>mc^2, where m is the rest mass of the electron, your result reduces to
λ'=(hc\λ)(1+m^2c^4/4hcE)

That doesn't look right dimensionally. Is there a typo here?

Relevant equations

λ'-λ=h/mc(1-cosθ)

This Compton formula assumes that the electron is initially at rest. So, it doesn't apply to your situation (unless you want to first Lorentz transform to the rest frame of the electron, apply the Compton formula, and then transform back to the original frame.)

Or, you could just start from scratch and set up conservation of momentum and energy for the collision and do the messy algebra.
 

1. What is Compton scattering when E=mc^2?

Compton scattering is a phenomenon in which a photon (particle of light) collides with an electron, causing the photon to lose energy and change direction. This occurs due to the conversion of the photon's energy into the electron's mass, as described by Einstein's famous equation E=mc^2.

2. How does Compton scattering relate to Einstein's theory of relativity?

Compton scattering provides evidence for Einstein's theory of relativity by demonstrating the conversion of energy into mass, as described by the equation E=mc^2. This phenomenon supports the idea that mass and energy are interchangeable, and that the speed of light is a fundamental constant.

3. What is the significance of Compton scattering in physics?

Compton scattering is significant because it allows scientists to study the properties of photons and electrons, as well as their interactions. It also provides evidence for the concepts of mass-energy equivalence and the constancy of the speed of light, which are important principles in physics.

4. How is Compton scattering observed in experiments?

In experiments, Compton scattering is observed by directing a beam of photons at a target material, such as a metal. The scattered photons are then detected and their energies are measured. By analyzing the change in energy and direction of the scattered photons, scientists can determine the properties of the electrons in the target material.

5. Can Compton scattering be applied in real-world applications?

Yes, Compton scattering has various applications in fields such as medical imaging, astronomy, and materials science. For example, in medical imaging, Compton scattering can be used to create images of the inside of the body by detecting the scattered photons after they pass through the body. This technique is commonly used in X-ray imaging.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
3K
  • Advanced Physics Homework Help
Replies
0
Views
168
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
Replies
8
Views
1K
Replies
1
Views
2K
Replies
1
Views
800
  • Advanced Physics Homework Help
2
Replies
58
Views
5K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
445
Back
Top