Editing Cross section (physics) (section)

== Examples ==

=== Elastic collision of two hard spheres ===
The following equations apply to two hard spheres that undergo a perfectly elastic collision.{{sfn | Taylor | 2005 | pp=564,574}} Let {{math|''R''}} and {{math|''r''}} denote the radii of the scattering center and scattered sphere, respectively. The differential cross section is
: <math>
\frac{d\sigma}{d\Omega} = \frac{R^2}{4},
</math>
and the total cross section is
: <math>
\sigma_\text{tot} = \pi \left(r + R\right)^2.
</math>
In other words, the total scattering cross section is equal to the area of the circle (with radius {{math|''r'' + ''R''}}) within which the center of mass of the incoming sphere has to arrive for it to be deflected.

=== Rutherford scattering ===
In [[Rutherford scattering]], an incident particle with charge {{math|''q''}} and energy {{math|''E''}} scatters off a fixed particle with charge {{math|''Q''}}. The differential cross section is
: <math>
\frac{d \sigma}{d \Omega} = \left(\frac{q \, Q}{16\pi\varepsilon_0 E \sin^2(\theta/2)} \right)^2
</math>
where <math>\varepsilon_0</math> is the [[vacuum permittivity]].{{sfn | Taylor | 2005 | p=576}} The total cross section is infinite unless a cutoff for small scattering angles <math>\theta</math> is applied.{{sfn | Griffiths | 2005 | p=409}} This is due to the long range of the <math>1/r</math> Coulomb potential.

=== Scattering from a 2D circular mirror ===

The following example deals with a beam of light scattering off a circle with radius {{math|''r''}} and a perfectly reflecting boundary. The beam consists of a uniform density of parallel rays, and the beam-circle interaction is modeled within the framework of [[geometric optics]]. Because the problem is genuinely two-dimensional, the cross section has unit of length (e.g., metre). Let {{math|''α''}} be the angle between the [[ray (optics)|light ray]] and the [[radius]] joining the reflection point of the ray with the center point of the mirror. Then the increase of the length element perpendicular to the beam is
: <math>\mathrm dx = r \cos \alpha \,\mathrm d \alpha.</math>
The reflection angle of this ray with respect to the incoming ray is {{math|2''α''}}, and the scattering angle is
: <math>\theta = \pi - 2 \alpha.</math>
The differential relationship between incident and reflected intensity {{math|''I''}} is
: <math>I \,\mathrm d \sigma = I \,\mathrm dx(x) = I r \cos \alpha \,\mathrm d \alpha = I \frac{r}{2} \sin \left(\frac{\theta}{2}\right) \,\mathrm d \theta = I \frac{\mathrm d \sigma}{\mathrm d \theta} \,\mathrm d \theta.</math>
The differential cross section is therefore ({{math|dΩ {{=}} d''θ''}})
: <math>\frac{\mathrm d \sigma}{\mathrm d \theta} = \frac{r}{2} \sin \left(\frac{\theta}{2}\right).</math>
Its maximum at {{math|''θ'' {{=}} π}} corresponds to backward scattering, and its minimum at {{math|''θ'' {{=}} 0}} corresponds to scattering from the edge of the circle directly forward. This expression confirms the intuitive expectations that the mirror circle acts like a diverging [[lens (optics)|lens]]. The total cross section is equal to the diameter of the circle:
: <math>\sigma = \int_0^{2 \pi} \frac{\mathrm d \sigma}{\mathrm d \theta} \,\mathrm d \theta = \int_0^{2 \pi} \frac{r}{2} \sin \left(\frac{\theta}{2}\right) \,\mathrm d \theta = 2 r.</math>

=== Scattering from a 3D spherical mirror ===
The result from the previous example can be used to solve the analogous problem in three dimensions, i.e., scattering from a perfectly reflecting sphere of radius {{math|''a''}}. 

The plane perpendicular to the incoming light beam can be parameterized by cylindrical coordinates {{math|''r''}} and {{math|''φ''}}. In any plane of the incoming and the reflected ray we can write (from the previous example):
: <math>\begin{align}
r &= a \sin \alpha,\\
\mathrm dr &= a \cos \alpha \,\mathrm d \alpha,
\end{align}</math>
while the impact area element is
: <math> \mathrm d \sigma = \mathrm d r(r) \times r \,\mathrm d \varphi = \frac{a^2}{2} \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right) \,\mathrm d \theta \,\mathrm d \varphi.</math>
In spherical coordinates,
: <math>\mathrm d\Omega = \sin \theta \,\mathrm d \theta \,\mathrm d \varphi.</math>
Together with the trigonometric identity
: <math>\sin \theta = 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right),</math>
we obtain
: <math>\frac{\mathrm d \sigma}{\mathrm d \Omega} = \frac{a^2}{4}.</math>
The total cross section is
: <math>\sigma = \oint_{4 \pi} \frac{\mathrm d \sigma}{\mathrm d \Omega} \,\mathrm d \Omega = \pi a^2.</math>