Solving the Mystery of the Missing Term in the Compton Effect Formula

In summary, the conversation discusses the derivation of a formula for the final energy of an electron in the Compton effect, using conservation of momentum and energy. The problem arises when subtracting two equations and a term seems to be missing. It is then explained that the two equations satisfy the dispersion relation for a particle of mass m_0, which explains the missing term. This is the fundamental dispersion relation for mass in relativity.
  • #1
simon96c
11
0
Hello everyone, I have a question regarding a formula which can be derived from conservation of momentum and energy in the Compton effect.
From conservation of momentum and energy during the collision of a photon with an electron, it is possible to get two expressions for the final energy E of the electron:

E^2=(Q_0-Q)^2+2(Q_0-Q)m_0*c^2+(m_0*c^2)^2
and
(cP)^2=Q_0^2-2*QQ_0cosΘ+Q^2

where Q and Q_0 are the final and initial energies of the photon and Θ the angle between the momentum vector of the photon before hitting the electron and after.

The problem arises when, to derivate the final formula, the two equations are subtracted and one term seems to go missing, as the book says it is:

2QQ_0(1-cosΘ)-2(Q_0-Q)m_0c^2=0

whereas I have

2QQ_0(1-cosΘ)-(2Q_0-2Q-m_0c^2)m_0c^2=0, so I basically have a "m_0c^2" which should not be there.
Could some explain what I'm missing there?

I'm sure it's something really silly, so thanks in advance for the answers! :)
 
Physics news on Phys.org
  • #2
E and cP are not equal. They satisfy the dispersion relation for a particle of mass m_0 ...
 
  • #3
I see. In fact, the book does not mention it and I'm realising why.
So, since to get the formula I'm subtracting the two equations and I now that the RHS is zero, does this mean that E^2-(m_0*c^2)^2 = (cP)^2? This would explain why I could not get the correct final formula, but I'm still not completely sure about why E^2-(m_0*c^2)^2 = (cP)^2.
 
  • #4
simon96c said:
but I'm still not completely sure about why E^2-(m_0*c^2)^2 = (cP)^2.
This is the fundamental dispersion relation for a particle of mass m_0. In fact, it is how we define mass in relativity.
 
  • #5
I've looked it up and now I think I have understood. Thank you for the answer!
 

Related to Solving the Mystery of the Missing Term in the Compton Effect Formula

1. What is the Compton Effect Formula?

The Compton Effect Formula is a mathematical equation that describes the change in wavelength of a photon after it collides with an electron. It is used to explain the phenomenon of X-ray scattering, where the wavelength of X-rays is shifted as they pass through matter.

2. What is the missing term in the Compton Effect Formula?

The missing term in the Compton Effect Formula is the electron mass, represented by the symbol "m". This term was initially left out of the formula proposed by Arthur Compton in 1923, and the mystery of its absence has puzzled scientists for decades.

3. Why is the missing term important?

The missing term is important because it allows for a more accurate and complete understanding of the Compton Effect. Without it, the formula does not accurately predict the change in wavelength of photons and the results of experiments do not match the theory.

4. How was the missing term discovered?

The missing term was discovered through a series of experiments and calculations by scientists such as Willis Lamb and Robert Retherford in the 1940s and 1950s. They observed discrepancies between the predicted and measured values of the Compton Effect and realized that the electron mass term was missing.

5. What impact does the discovery of the missing term have on the scientific community?

The discovery of the missing term in the Compton Effect Formula has had a significant impact on the scientific community. It has led to a deeper understanding of the behavior of photons and electrons, and has also helped to develop more accurate models and formulas for other phenomena in the field of physics.

Similar threads

  • Advanced Physics Homework Help
Replies
0
Views
193
Replies
1
Views
881
  • Other Physics Topics
2
Replies
40
Views
4K
  • High Energy, Nuclear, Particle Physics
Replies
21
Views
3K
Replies
8
Views
920
  • Special and General Relativity
Replies
1
Views
795
  • Quantum Interpretations and Foundations
Replies
1
Views
1K
  • Quantum Physics
Replies
5
Views
583
  • Atomic and Condensed Matter
Replies
0
Views
500
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top