Special Relativity: Photon emission by a moving atom

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SUMMARY

The discussion centers on the photon emission by a moving atom, specifically an atom in an excited state with energy Q_0 traveling towards a scintillation counter at speed v. The goal is to demonstrate that the emitted photon energy Q can be expressed as Q = Q_0(1 + Q_0/(2mc^2)). Key equations utilized include conservation of energy and momentum, with the kinetic energy of the atom represented as K = (\gamma - 1)m_0c^2. The initial energy in the lab frame is established as (Q_0 + mc^2)\gamma, while the initial momentum is (Q_0/c^2 + m)\gamma v.

PREREQUISITES
  • Understanding of special relativity concepts, particularly Lorentz transformations.
  • Familiarity with conservation laws in physics, specifically energy and momentum conservation.
  • Knowledge of relativistic energy equations, including E^2 = (cp)^2 + m_0^2c^4.
  • Basic proficiency in algebraic manipulation and solving equations.
NEXT STEPS
  • Study the derivation of Lorentz transformations in special relativity.
  • Explore detailed examples of conservation of energy and momentum in relativistic contexts.
  • Learn about the implications of photon emission in quantum mechanics and relativity.
  • Investigate the role of rest mass and relativistic mass in energy calculations.
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Students of physics, particularly those studying special relativity, as well as educators and researchers interested in the dynamics of photon emission and atomic behavior in relativistic scenarios.

Helgi
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Homework Statement


The problem involves an atom (Said to be in an excited state of energy Q_0) traveling towards a scintillation counter with speed v. The atom then emits a photon of energy Q and stops completely. The rest mass of the atom is m. I'm supposed to show that
[tex]Q = Q_0(1+\frac{Q_0}{2mc^2})[/tex]

Homework Equations


It's mostly just conservation of energy and momentum stuff.
The kinetic energy of the atom is
[tex]K = (\gamma -1)m_0 c^2[/tex]
I have a feeling this formula
[tex]E^2 = (cp)^2+m_0^2c^4[/tex]
has to be used but I don't see where.

The Attempt at a Solution


What I figured I should do was to use conservation of energy and momentum. So I set up 2 equations

[tex]Q_0 + (\gamma-1)m_0c^2 = m_0 c^2+Q[/tex] Which is the energy conservation
[tex]\gamma\cdot m_0v = \frac{Q}{C}[/tex] and the conservation of momentum

Which I'm pretty sure is set up alright but when I try to solve for Q I don't get anything that is simplifies to what it's supposed to be. The [tex]\gamma[/tex] and v usually get in the way and they're not supposed to be in the answer.

Now what I tried was to just simplify the first formula, and I ended up with
[tex]2Q_0 = 2m_c^2 + Q[/tex]
this looks pretty reasonable but it's missing the second power on the [tex]Q_0[/tex].
 
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Helgi said:
The problem involves an atom (Said to be in an excited state of energy Q_0) traveling towards a scintillation counter with speed v. The atom then emits a photon of energy Q and stops completely. The rest mass of the atom is m. I'm supposed to show that
[tex]Q = Q_0(1+\frac{Q_0}{2mc^2})[/tex]

The key point to note here that the excited state of energy Q0 is in the rest frame of the atom, not the lab frame as you have considered. This energy Q0 makes the rest energy of the atom (Q0+mc^2). So, the initial energy of the system in labframe is [tex](Q0+mc^2)\gamma[/tex], and the initial momentum is [tex](Q0/c^2+m)\gamma v[/tex].

The rest is as you have done. It's a two liner. Put c=1 to keep it neat, and put it back in the end to make it dimensionally correct.
 

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