# Deflection angle of a particle integral

1. Aug 4, 2013

### Lightf

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. Aug 4, 2013

### HallsofIvy

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

How you integrate that will depend upon whether $C- B^2/4A$ is positive or negative.

3. Aug 4, 2013

### Lightf

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.

4. Aug 4, 2013

### voko

Look up Euler's substitutions.

5. Aug 4, 2013

### haruspex

Check those signs.