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Collision of two photons using four-momentum

  1. Nov 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Two photon of energy ##E_1 ## and ## E_2## collide with their trajectory at an angle $\theta$ with respect to each other.
    a) Total four-momentum before collision?
    b) square length of 4-momentum in lab frame (LB)and in center of momentum frame (CM)?
    c) 4-momentum of two photon system in CM?
    d) After photons annihilate they form two identical particle with mass m. what can we say about the spatial components of their 4-momentum in CM?
    e) requirement of this process is that ##\sqrt{E_1E_2}\geq E_{min}##. find ##E_{min}## in terms of rest energy of particles and angle ##\theta##.

    2. Relevant equations
    -Conservation of Energy
    -Conservation of Momentum
    -For a photon square of 4-momentum is zero (light like particle)




    3. The attempt at a solution

    Attempt at Answer:(assuming without loss of generality that particle 1 is moving in + x direction and the whole motion is in x-y plane)
    a)
    - The initial momentum for each particle is:
    $$p^{\mu}_1=(E_1, E_1,0,0) \\ p^{\mu}_2=(E_2,-E_2cos\theta,E_2sin\theta,0) \\$$
    hence the total initial momentum is: $$p^{\mu}_T=(E_1+E_2,E_1-E_2cos\theta,E_2sin\theta,0)$$
    b)
    - squared length of 4-momentum in lab frame (LB): $$(p_{T\mu}p^{\mu}_T)_{LB}=(E_1+E_2)^2- (E_1-E_2cos\theta)^2-(E_2sin\theta)^2=2E_1E_2(1+cos\theta)$$
    - square 4-momentum in COM = square length of 4-momentum in LB (** is it right??!!):
    $$(p_{T\mu}p^{\mu}_T)_{LB}=(p_{T\mu}p^{\mu}_T)_{COM}=2E_1E_2(1+cos\theta)$$
    c)
    Now to find the momentum in COM I use the above fact that square of the total four momentum is invariant under frame transformation (again is it right!!) and combining this with the fact that the square of 4-momentum of photon is zero we get:
    $$p^{\mu}_{1(COM)}=(E',E'_x,E'_y,0)\\p^{\mu}_{2(COM)}=(E',-E'_x,-E'_y,0)$$
    -So the total 4-momentum in COM is:
    $$p^{\mu}_{T(COM)}=(2E',0,0,0)$$
    squaring it and equating to square of total 4-momentum in COM, gives: $$E'^2=\frac{E_1E_2}{2}(1+cos\theta)=E_1E_2cos^2{\frac{\theta}{2}}$$
    As I have the magnitude and I know that $$E'^2_x+E'^2_y=E_1E_2cos^2{\frac{\theta}{2}}$$
    its evident that the two components will have the following form:$$E'_x=\sqrt{E_1E_2}\cos{\frac{\theta}{2}}\cos{\theta'}\\E'_y=\sqrt{E_1E_2} \cos{\frac{\theta}{2}} \sin{\theta'}$$
    but here I face a problem. I still don't know $$\theta'$$
    Any help or hint is appreciated. Thank you so much for your time.


     
  2. jcsd
  3. Nov 25, 2015 #2

    mfb

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    I don't think θ was meant in the way you used it. Two photons moving in the same direction should correspond to θ=0, not θ=pi.
    Right.

    You can calculate θ' based on the 4-vector of the two-photon system in your lab system.
     
  4. Nov 26, 2015 #3
    Can we use the Lorentz boost ti find it? I mean looking for a transformation that makes the spatial components of the total four momentum vanish?
     
  5. Nov 27, 2015 #4

    vela

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    You don't need to know or find ##\theta'## to solve the problem. You already wrote down the four-momentum of the two-photon system in the COM frame — (2E', 0, 0, 0) — and you used the answer for part (d) in your calculations — the three-momentum of the particles are equal in magnitude and opposite in direction.
     
  6. Nov 27, 2015 #5
    Then how do we do we solve the last part (part d), I thought i was gonna need it for determining the minimum energy.
     
  7. Nov 27, 2015 #6
    Thank you so much I have found it. it will be as following:
    the four momentum of system after the collision and creation of two identical particle will be:
    $$p^{\mu}_T=(2 \gamma mc,0,0,0)$$
    now using $$\gamma=1$$ and using the invariance of the square of the total momentum in a reaction we get to the following for minimum energy:
    $$\frac{4E_1E_2}{c^2}\geq4m^2c^2 $$ hence $$\sqrt{E_1E_2}\geq\frac{mc^2}{cos\frac{\theta}{2}}$$

    have I missed anything?
     
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