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Homework Help: Special relativity - collision of a photon and an electron

  1. Feb 14, 2008 #1
    Just wanted to say that this is my first post here, and I thank in advance anyone who spends time on this. Hopefully I will be able to return the favor someday.

    I thought I knew how to do this, but apparently I was wrong... I need to analyze the collision of a photon at an electron.

    1. The problem statement, all variables and given/known data

    A photon hits an electron in it's rest system. After the collision, the photon's direction is perpendicular to it's prior direction, having half the starting energy (first it was going x+, now it is going y+).
    After the collision, the electron is traveling at some speed v at an angle -a to the x axis (i.e. x+ y-)

    1. write the relevant preservation equations.
    2. Find the hitting photon's energy
    ... (if I'll understand these I'll do fine with the rest)

    The "knowns" are C - the speed of light and E0 - the electron's rest energy.

    2. Relevant equations
    3. The attempt at a solution

    I have to say that I'm new to using the Lorentz transformation and it's consequences so I likely made a few mistakes, but..:

    I had the conservation of energy as:
    (E1 is the photon's starting energy)
    E_0 + E_1 = E_0 * \omega + (E1 / 2);

    I wrote the conservation of momentum as follows:
    (E_1 / C) = (E_0 / C) * \beta * cos a (X axis)
    0 = (E_1 / 2C) - (E_0 / C) * \beta * sin a (Y axis)

    Then (after hours of thought and tackling this) I realized that E=pc is only true for particles traveling very close to (or at) the speed of light, and revised it to:
    P_1 = P_e * cos a (X axis)
    0 = P_2 - P_e sin a (Y axis)
    and used:
    (E_1 / 2) = P_2 * C
    P_2 = ([P_e]^2 + [m_e]^2 * C^2)^(1/2)

    However, it got somewhat messy and then I got an impossible answer, so I suspect something is wrong here as well...

    I would really appreciate if someone explains the general way of solving this kind of problems (or rather how to treat the momentum conservation, I guess) and/or what is wrong about my way.

    Thanks in advance,
  2. jcsd
  3. Feb 14, 2008 #2

    Shooting Star

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    Let p1 and p2 denote the magnitude of the momenta of the photon before and after, and g=gamma(v), m0 the rest mass of the electron. So, m=m0g and E0=m0c^2.

    From energy consvn,

    p1c + m0c^2 = p2c + mgc^2.

    From momentum consvn toward the initial direction pf photon, that is, the x-axis,

    p1 = mv(cos a) = m0gv(cos a).

    From momentum consvn toward the final direction pf photon, that is, the y-axis,

    p2 - m0gv(sin a) = 0.

    Solve for p1 and E1=p1*c.
    Last edited: Feb 14, 2008
  4. Feb 15, 2008 #3
    Thanks for the reply!

    These are my exact equations (from the second time around), so I know I did something right, but on the other hand, I still don't know what I did wrong because I didn't get a solution.

    I will double check all my calculations though, since your post confirms that the initial equations were correct...

    Thanks again - Hopefully I will reply in a short time, saying I've done it...
    Last edited: Feb 15, 2008
  5. Feb 16, 2008 #4
    Alright, I have reached a solution, and it seems somewhat too simple, esp. after all the time I put into this. Can someone please confirm this is correct?

    Here is what I did:
    P_1 = 2P_2 = The photon's momentum before the collision.
    P_2 = The Photon's momentum after the collision.
    E_1 = P_1*C = The energy of the photon prior to the collision.
    m_0 = The electron's rest mass.
    E_0 = m_0*C^2 = The electrons rest energy.
    v = the electron's velocity after the collision.
    g = (1-[v^2/C^2])^(-1/2)
    a = The angle between the X axis and the velocity of the electron after the collision.

    Conservation of momentum:
    Along X:
    P_1 = 2P_2 = m_0*g*v*cos a (i)

    Along Y:
    0 = P_2 - m_0*g*v*sin a =>
    P_2 = m_0*g*g*sin a (ii)

    Conservation of Energy:
    P_1*C + m_0*C^2 = P_2*C + m_0*g*C^2 =>
    2P_2*C + m_0*C^2 = P_2*C + m_0*g*C^2 =>
    2P_2 + m_0*C = P_2 + m_0*g*C =>
    m_0*C = m_0*g*C - P_2 (iii)

    4(P_2)^2 + (P_2)^2 = (m_0*g*v)^2 =>
    P_2 = [5^(-1/2)]*m_0*g*v (iv)

    (iv) -> (iii):
    m_0*C = m_0*g*C - [5^(-1/2)]*m_0*g*v =>
    C*5^(1/2) = g(C*5^(1/2) - v) =>
    5C^2 = [5C^2 - 2*(5^{1/2})*C*v + v^2]/[1 - (v^2)/(C^2)] =>
    5(C^2 - v^2) = 5C^2 - 2*[5^(1/2)]*c*v + v^2 =>
    6v^2 = 2*[5^(1/2)]*C*v =>
    v = sqrt(5/9)*C; (v)
    g = 1/sqrt[1-5/9]=1/(2/3) = 1.5

    (v) -> (iv):
    P_2 = sqrt(1/9)*m_0*1.5*C =>
    E_1 = 2P_2*C = 2*1.5*(1/3)*m_0*C^2 = m_0*C^2 = E_0;

    So the final answer is that the photon's energy prior to the collision is equal to the electron's rest energy.

    Is that correct?
    Last edited: Feb 16, 2008
  6. Feb 17, 2008 #5
    Alright, now for the really hard part - how do I find the photon's energy (prior to the collision) in the center of mass system?
  7. Feb 19, 2008 #6

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    That is correct. If you had done Compton scattering, that can be derived in a very short time. Good work.

    In relativity, a “centre of momentum frame” or CM frame is better defined than the centre of mass. It’s an IFR in which the total momentum is zero. The “centre of mass” is not a unique point but frame dependant.

    If the speed of the CM frame is ‘u’, wrt the lab frame, then u = [(total momentum in lab frame/total energy in lab frame)]c^2 = c^2*p1/(E1 + m0c^2). From there, you can find the energy of the photon in a frame travelling with speed u to the right.
  8. Feb 20, 2008 #7
    Thanks, got it :)
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