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Deflection angle of a particle integral

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the deflection angle of the particle if it is scattered by this central field.

    $$ U = \frac{α - β r - γ r^2}{r^2} $$


    2. Relevant equations
    Angle of deflection is given by:
    $$θ = ∫ \frac{Mdr}{r^2\sqrt{2m(E-U) - \frac{M^2}{r^2}}}$$


    3. The attempt at a solution

    $$θ = ∫ \frac{Mdr}{r^2\sqrt{2m(E-U) - \frac{M^2}{r^2}}}$$
    $$θ = ∫ \frac{Mdr}{r^2\sqrt{2m(E-\frac{α-βr-γr^2}{r^2}) - \frac{M^2}{r^2}}}$$
    $$θ = ∫ \frac{Mdr}{r\sqrt{2m(Er^2 - α - βr - γr^2) - M^2}}$$
    $$θ = ∫ \frac{Mdr}{r\sqrt{2m(E-γ)r^2 - 2mβr - (2mα - M^2)}}$$

    I am not sure how to solve this integral.
     
  2. jcsd
  3. Aug 4, 2013 #2

    HallsofIvy

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    Renaming the constants, it is
    [tex]\int \frac{dr}{r\sqrt{Ar^2+ Br+ C}}[/tex]
    Seeing that quadratic in the square root, my first thought would be to "complete the square" in the square root:
    [tex]Ar^2+ Br+ C= A(x^2+ (B/A)r )+ C= A(r^2+ (B/A)r+ B^2/4A^2)- B^2/4A+ C= A(r+B/2A)^2+ C- B^2/4A[/tex]
    Let u= r+ B/2A so that dr= du and r= u- B/2A.

    How you integrate that will depend upon whether [itex]C- B^2/4A[/itex] is positive or negative.
     
  4. Aug 4, 2013 #3
    How do you deal with the r outside the square root.

    When you sub in u you get

    $$ ∫\frac{du}{(u-\frac{B}{2A})\sqrt{Au^2 +C-\frac{B^2}{4A}}}$$

    The (u- B/2A) part throws me off how to solve this. I tried doing it by integration by parts but it just made it more complex.

    I am also assuming C - B^2/4A will be positive.
     
  5. Aug 4, 2013 #4
    Look up Euler's substitutions.
     
  6. Aug 4, 2013 #5

    haruspex

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    Check those signs.
     
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