1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photon-Photon scattering

  1. Sep 2, 2011 #1

    wukunlin

    User Avatar
    Gold Member

    This is an assignment I'm doing that is due next month but I would like to make an early start because I find most of the materials in this assignment unfamiliar.

    The whole questions is composed of 5 parts, most of them I have a rough idea of but there are some bits and pieces I don't really know about since its been a year since my last quantum mechanics paper and 2 years since I last had a lecture in relativistic effects.

    The problem is to do with two 5mW laser beams points at each other as shown in the attached picture as the red and green arrows, representing laser beams of 650nm and 532nm respectively. The orange bits are where the photons interact.

    So the first part of the problem.

    1. The problem statement, all variables and given/known data
    Calculate the velocity (magnitude and direction) of the CM (I assume centre of mass) System.

    2. Relevant equations
    [tex]E=h\nu[/tex]
    and
    [tex]m=\frac{E}{c^2}[/tex]
    for relativistic mass

    3. The attempt at a solution
    I think to work out the direction I can just treat the photon like two point particals in classical mechanics, which will be similar to the black arrow I have drawn but leaning towards the particle with more relativistic mass.
    The magnitude is something I'm not sure about. In classical mechanics, knowing the directions means I can just use geometry to work out the magnitude, but in this case I believe I need to take relativistic effects into account since photons travel at the speed of light.
    I looked at this: http://www.numericana.com/answer/relativity.htm#photons
    but I'm not sure if that is relavent.

    I have also looked at this: http://en.wikipedia.org/wiki/Mass–energy_equivalence#Massless_particles
    That doesn't seem to tell me anyway about the velocity of the CoM though,
    unless, the CoM also move at the speed of light?
     

    Attached Files:

  2. jcsd
  3. Sep 3, 2011 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi wukunlin! :smile:
    that's correct :smile:

    the simplest way to find the CM frame (and yes, that does mean centre of mass) is to note that it's the frame in which the photons have the same frequency (and opposite directions) :wink:
     
  4. Sep 3, 2011 #3

    wukunlin

    User Avatar
    Gold Member

    ah, that makes sense. Looks like I was overthinking a bit. thanks tim

    What puzzled me earlier was that I had the same misconception of OP in this thread, I was thinking that I might run into problems with the possibility of their seperation distance changing by >c in the CM frame.
     
  5. Sep 6, 2011 #4

    wukunlin

    User Avatar
    Gold Member

    arrrgghh, not quite working

    so the red laser shoots from right to left, I say along the x axis, the green laser shoots from left to right with 20 degree angle pointing upwards, I did the following:

    let the red and green photon's effective mass be:
    [tex]
    m_r = \frac{h}{c \times 650nm}
    m_g = \frac{h}{c \times 532nm}
    [/tex]
    their instantaneous positions:
    [tex]
    \vec r_r = \left[ \begin{array}{c} x_r(t) \\ y_r(t) \end{array} \right] = \left[ \begin{array}{c} -ct \\ 0 \end{array} \right]
    [/tex]
    [tex]
    \vec r_g = \left[ \begin{array}{c} x_g(t) \\ y_g(t) \end{array} \right] = \left[ \begin{array}{c} ct \: sin 20^O \\ ct \: cos 20^O \end{array} \right]
    [/tex]

    and for the centre of mass:

    [tex]
    \vec r_{cm} = \frac{m_r \vec r_r + m_g \vec r_g}{m_r+m_g} = \frac{ct}{m_r+m_g}\left[ \begin{array}{c} m_g \: cos 20^O - m_r \\ m_g \: sin 20^O \end{array} \right]
    [/tex]
    [tex]
    \vec v_{cm} = \frac{d \vec r_{cm}}{dt} =
    \vec r_{cm} = \frac{m_r \vec r_r + m_g \vec r_g}{m_r+m_g} = \frac{c}{m_r+m_g}\left[ \begin{array}{c} m_g \: cos 20^O - m_r \\ m_g \: sin 20^O \end{array} \right]
    [/tex]
    the velocity I worked out has magnitude of around 0.2c, direction is around 70.5 degrees above the positive x-axis

    I plotted the trajectory of the photons and the CoM, they look okay, at all points the CoM, green photon and red photon are colinear and lean towards the green photon

    I then tested to see if both photons appear to have the same wavelength, using the doppler shift formula:
    [tex]
    \frac{\lambda_o}{\lambda_s}=\frac{\sqrt{1+v/c}}{\sqrt{1-v/c}}
    [/tex]
    for the green photon, because the green photon is 'chasing' the observer so the wavelengh appears longer. the speed v in this case is 0.2*cos(50.5) ie the projection of the CoM velocity on the velocity of the photon.
    What I did for the red photon is a similar process.

    my problem is the green photon becomes 604nm after the doppler shift, the red photon becomes 608nm after the doppler shift. all calculations are done in matlab without any rounding, have I done something wrong? :uhh:
     

    Attached Files:

    • a21.jpg
      a21.jpg
      File size:
      7.6 KB
      Views:
      80
  6. Sep 6, 2011 #5

    wukunlin

    User Avatar
    Gold Member

    nevermind
    I didn't know there was such thing as transverse doppler effect, taking that into account and turns out my answer is correct
     
  7. Sep 7, 2011 #6

    wukunlin

    User Avatar
    Gold Member

    oh, another questions.

    The next two parts of the assignment involves calculating the collision wavelength [tex]\lambda _{CM}[/tex] and the collision energy [tex]\sqrt{s}[/tex] in CM system

    since the net momentum in CM system is zero, wouldn't the collision energy just be the collision frequency multiplied by the Planck's constant?
     
    Last edited: Sep 7, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Photon-Photon scattering
Loading...