Momentum of photons in CM frame.

In summary, two photons with frequencies \nu_1 and \nu_2 are in the laboratory system with an angle of \theta between their propagation directions. The expressions for total energy and momentum in the laboratory system are given by the four momentum equation. The frequency of the photons in the center of mass system can be found by doing a Lorentz boost and setting the spatial total momentum to zero. The x-axis can be aligned to point along ##\vec{p}_1 + \vec{p}_2## to simplify the calculation.
  • #1
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Homework Statement



Two photons in the laboratory system have frequencies [itex]\nu_1[/itex] and [itex]\nu_2[/itex]. The angle between the propagation directions is [itex]\theta[/itex].

a) Write down the expressions for the total energy and momentum of the photons in the laboratory system.

b) Find the photons’ frequency in the center of mass system.

c) Is it allways possible to find a center of mass system for the photons?


Homework Equations


Four momentum
[tex] \underline P = (E/c, \vec p)[/tex]

Lorentz transformation
[tex] P'^\mu = \Lambda^\mu_{\ \ \nu} x^\nu [/tex]

Duality equations
[tex] p = h/\lambda \ \ \ E = h\nu [/tex].

The Attempt at a Solution


a)
I've taken the x-axis in the laboratory system to be parallel with photon 2 such that

[tex] \vec p_1 = \frac{h\nu_1}c(\cos \theta, \sin \theta, 0) \ \ \vec p_1 = \frac{h\nu_2}c(1,0,0)[/tex]

The total four momentum would be

[tex] \underline P = (E_{tot}/c, \vec p_{tot})[/tex]

where
[tex] E_{tot} = \frac{h(\nu_1 + \nu_2}c[/tex].

b) To find the momentum in the CM system I would have to do a lorentz boost into that system which is defined so that the spatial total momentum is zero. I do however have problems in finding the correct [itex]\beta = v/c[/itex] for this transformation.
I know the velocity of this frame would have to satisfy
[tex] \vec v = \frac{(\vec p_1 + \vec p_2)c^2}E[/tex]

but that frame would not have it's x-axis parallel with my original one, so I do not
know how to find the correct transformation. Have I chosen my axis badly in this problem
or is there a clever way out?

Thanks for any help.
 
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  • #2
Try aligning the x-axis to point along ##\vec{p}_1 + \vec{p}_2##. You'll have to find the angle each photon makes with the axis, but that should be straightforward.
 

1. What is the concept of momentum in the context of photons in the center-of-mass frame?

The momentum of a photon in the center-of-mass (CM) frame refers to the amount of motion or inertia it possesses. In other words, it is a measure of the photon's resistance to changes in its velocity. In the CM frame, the total momentum of all the particles involved should be zero, so the momentum of the photon plays a crucial role in balancing out the momenta of other particles.

2. How is the momentum of a photon calculated in the CM frame?

The momentum of a photon can be calculated using the equation p = hν/c, where h is Planck's constant, ν is the frequency of the photon, and c is the speed of light. In the CM frame, the frequency of the photon will be different than its frequency in the lab frame, so the momentum will also differ.

3. Can the momentum of a photon in the CM frame be negative?

No, according to the laws of physics, the momentum of a photon cannot be negative. This is because the direction of a photon's momentum is always the same as its direction of motion, and a photon always moves in a forward direction. Therefore, its momentum can only have positive values.

4. How does the momentum of a photon in the CM frame affect its energy?

The momentum of a photon is directly proportional to its energy. This means that as the momentum of a photon in the CM frame increases, so does its energy. Conversely, as the momentum decreases, so does the energy. This relationship is described by the famous equation E = pc, where E is the energy, p is the momentum, and c is the speed of light.

5. Is the momentum of a photon in the CM frame affected by its wavelength?

Yes, the momentum of a photon is inversely proportional to its wavelength. This means that as the wavelength increases, the momentum decreases, and vice versa. This relationship can be seen in the equation p = h/λ, where h is Planck's constant and λ is the wavelength of the photon.

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