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Using Conservation of 4 momentum

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data



    Hi Physics forum members,

    The following is questions on how to use 4 momentum and momentum 4 vector conservation. From what I've learned of it, it is essentially the same as classical momentum conservation. However, I'm still not quite sure what you can, and should, be able to do with it given initial conditions.

    The question in particular is if a particle like an electron, (moving relativistically so it's energy can be approximated as momentum), scatters of a proton initially at rest, say you're given the initial energy of the electron, from which obviously you could find it's initial momentum.

    Keeping it simple by having the proton initially moving only along the x-axis, the electron and proton scatter at angles 0e and 0p relative to the axis. If you only know the initially electron energy and momentum, along the x axis, before it scatters and the angle 0e of the electron after, is it possible to use 4 momentum conservation to find the scattering angle of the proton, Op and in addition the final momenta of the electron and proton after scattering? Or is more info needed?

    Keeping it simple by having the proton initially moving only along the x-axis, the electron and proton scatter at angles 0e and 0p relative to the axis. If you only know the initially electron energy and momentum, along the x axis, before it scatters and the angle 0e of the electron after, is it possible to use 4 momentum conservation to find the scattering angle of the proton, Op and in addition the final momenta of the electron and proton after scattering? Or is more info needed?

    2. Relevant equations
    I started simply using 4 momentum vectors, (E/c, px, py, pz) For initial and final momentum. Then, for the electron, I had thought the 4 momentum was [Pe, Pe, 0,0] with momentum Pe in the x axis and, for the electron moving relativistically, E estimated to be Pe, the same as momentum, (the electron's mass is neglegible). The 4 momentum of the proton at rest is [Mp, 0, 0, 0], i.e. simply the mass of the proton-and yes, this is in c=1 units.


    The 4 momentum of the electron after collision is then [((Me^2) + (Pef^2))^.5, Pef * Cos(0e), Pef * Sin(0e), 0] and for the proton is [((Mp^2) + (Ppf^2))^.5, Ppf * Cos(0p), Ppf * Sin(0p), 0]. Pef is the final momentum of the electron, Ppf is the final momentum of the proton, Oe and Op are scattering angles of electron and proton , (Me and Mp are electron and proton mass).


    3. The attempt at a solution

    I had thought that by equating the each of the 4 components for the total momentum before and after the collision you could find 0p, scattering angle of proton, along with the final momenta Pef and Ppf of electron and proton after collision, knowing only Pe, the initial electron momentum and 0e, the scattering angle of the electron. But it hasn't worked very well, and I was wondering if there's some subtle aspect to 4 momentum conservation I'm missing or something I'm overlooking.
     
  2. jcsd
  3. Jun 2, 2009 #2

    Hao

    User Avatar

    The utility of 4 vectors is in the fact that you can apply the rules of vector manipulation such as dot products.

    'p' will refer to a 4-vector.

    Together with some useful identities, such as [tex]p \bullet p = m^2 c^2[/tex] (prove this!), it can be a very powerful tool <--- This is the subtle point you are missing.

    For example, given some initial conditions for a two particle collision, suppose we want to find the final outcome of the collision for one particle (3), and we aren't too concerned about the other particle (4). Also suppose we are given the rest mass of particle (4).

    An algebraic approach would be to equate [tex]p_{1} + p_{2} = p_{3} + p_{4}[/tex] and solve for rest mass of (3), and such. Not fun.

    However, if we recognize that these are vectors, we can 'eliminate' the irrelevant [tex]p_{4}[/tex].

    [tex]p_{1} + p_{2} - p_{3}= p_{4}[/tex]
    Taking dot products,
    [tex]|p_{1} + p_{2} - p_{3}|^2= |p_{4}|^2[/tex]
    [tex]p_1\bullet p_1 + p_2\bullet p_2+ p_3\bullet p_3 + 2 p_1\bullet p_2 - 2 p_1\bullet p_3 - 2 p_2\bullet p_3= p_4\bullet p_4[/tex]
    Using our identify:
    [tex]m_1^2 c^2 + m_2^2 c^2 + m_3^2 c^2 + 2 p_1\bullet p_2 - 2 p_1\bullet p_3 - 2 p_2\bullet p_3= m_4^2 c^2[/tex]

    All of a sudden, we notice that all quantities associated with particle 4 (except its rest mass) have disappeared!

    The only quantities that appear in our equation are either relevant to particle 3, or are already known (particles 1 & 2).


    Another example would be completely inelastic collisions. We form [tex]p_1 + p_2 = p_3[/tex] and we want to find the rest mass of [tex]p_3[/tex].

    The standard procedure is to Plug & Chug from conversation of mass-energy and conservation of relativistic momentum, and hope for the best, but with 4-vectors, we can take a shortcut.

    Taking dot products,
    [tex]|p_1 + p_2|^2 = |p_3|^2[/tex]
    [tex]p_1\bullet p_1 + p_2\bullet p_2 + 2 p_1\bullet p_2 = p_3\bullet p_3[/tex]
    [tex]m_1^2 c^2 + m_2^2 c^2 + 2 p_1\bullet p_2 = m_3^2 c^2[/tex]

    !!!?? And there is our answer.

    These are just two example of how 4-vectors are applied. Recognizing what the shortcuts are will come with experience.

    Technically, the above was just an application of [tex]E^2 + p^2 c^2 = m^2 c^4[/tex], but the 4-vector notation makes the mathematics much cleaner.
     
    Last edited: Jun 2, 2009
  4. Jun 2, 2009 #3

    jtbell

    User Avatar

    Staff: Mentor

    I'm a bit suspicious of this approximation, especially since you don't use it for the outgoing electron. Note that if you use this approximation for both the incoming and outgoing electron, what you have is almost identical with Compton scattering (photon "colliding" with an initially stationary electron), which is discussed in many textbooks and Web pages.

    I think this should be quite do-able without the approximation, probably just a bit more algebra.

    OK. By the way, your equations would be much easier to read if you use our LaTeX feature. See this thread for an introduction and links to tutorials. Click on any of my equations below to get a pop-up that shows the code for it.

    [tex] [\sqrt{M_e^2 + P_{ef}^2}, P_{ef} \cos \theta_e, P_{ef} \sin \theta_e, 0] [/tex]

    Likewise OK.

    [tex] [\sqrt{M_p^2 + P_{pf}^2}, P_{pf} \cos \theta_p, P_{pf} \sin \theta_p, 0] [/tex]


    Yes, that's the straightforward way to do it. You get three equations, for conservation of energy, conservation of x-momentum and conservation of y-momentum. After you eliminate the energies using [itex]E = \sqrt{M^2 + P^2}[/itex], you have three unknowns: [itex]P_{ef}[/itex], [itex]P_{pf}[/itex], and [itex]\theta_p[/itex] (you're assuming all the other quantities are "given"). Solving three equations in three unknowns is a somewhat messy application of algebra and a trig identity or two. Let's check your equations to make sure your starting point is OK. Here's the x-momentum conservation equation:

    [tex]P_e + 0 = P_{ef} \cos \theta_e + P_{pf} \cos \theta_p[/tex]

    Can you show me the y-momentum and energy conservation equations?
     
  5. Jun 2, 2009 #4
    Thanks for the help regarding the LaTex feature.

    Just to clarify, I was using the approximation E = Pe for the relativistic electron before AND after the scattering, in both situations presuming the electron's mass, which is needless to say extremely tiny, is neglegible when the electron is moving relativistically.

    Since you wanted to see that i can write them, the equations are

    [tex]
    P_{pf} \sin \theta_p = P_{ef} \sin \theta_e
    [/tex]

    for y momentum and

    [tex]
    sqrt{M_e^2 + P_{ei}^2} + M_p = sqrt{M_e^2 + P_{ef}^2} + sqrt{M_p^2 + P_{pf}^2}
    [/tex]

    For energy.

    I would imagine from there it's just a matter of grinding it out unless there's tricks to solving these types of equations I most likely forgot about.
     
  6. Jun 2, 2009 #5
    For the energy, it should read

    [tex]

    \sqrt{M_e^2 + P_{ei}^2} + M_p = \sqrt{M_e^2 + P_{ef}^2} + \sqrt{M_p^2 + P_{pf}^2}

    [/tex]
     
  7. Jun 2, 2009 #6

    jtbell

    User Avatar

    Staff: Mentor

    It looks like your starting point is OK, then! The algebra is a bit beastly because you have to get rid of those square roots in the energy equation.

    Here's a tip: the trig identity [itex]\sin^2 \theta + cos^2 \theta = 1[/itex] comes in handy for eliminating one of the angles.
     
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