# Why is scattering vector $\vec{q}$ called vector of 'momentum transfer'?

In the world of scattering the angle at which you detect the scattered radiation is known as $q$, where

$$\vec{q} = \frac{4\pi\eta}{\lambda}\sin(\theta/2)$$

I read in a lot of books that this is known as the scattering vector or "momentum transfer". What I am trying to understand is why it is called "momentum transfer"?

In the physical sense, say a beam of laser light hits a particle and creates dipole oscillations which re-radiate the scattered wave. What does the particular scattering vector have to do with the oscillations? Doesn't the particle just act as new source of light scattering uniformly in all directions? In this example ignore form-factor.

Microscopically, i.e. in the quantum theory the scattering with radiation is a collision of particles with photons such as $$e^- + \gamma \to e^- + \gamma$$ The momentum vectors of the particles above are $$\vec p_1+\vec p_2= \vec p_3 + \vec p_4$$ where the identity holds due to momentum conservation. But in general $\vec p_1\neq \vec p_3$ and $\vec p_2\neq \vec p_4$. But the equation above is equivalent to $$\vec p_3-\vec p_1= \vec p_2-\vec p_4\equiv \vec q$$ This momentum difference known as $\vec q$ was donated by the initial photon to the initial electron. So this $\vec q$ was subtracted from one particle and added to the other particle. That's why this momentum was "transferred".

In the center-of-mass inertial system, $\vec p_1+\vec p_2=\vec p_3+\vec p_4 = 0$ and the absolute values of all four vectors are the same by energy conservation. So only the angles change by $\pi-\theta$ in a plane and $\vec q=\vec p_3-\vec p_1$ is therefore a vector whose direction is given by $\theta/2$.

• Thank you. I am missing the QM approach as I only really have read the classical formulation of scattering so far. – Steve Hatcher Jul 8 '14 at 2:59

Here's a parallel answer to Luboš's but purely classical. Start by noting that the momentum vector of a plane wave with wavelength $\lambda$ is:

$$\vec{p} = \frac{2\pi}{\lambda}$$

In some elastic scattering experiment, e.g. X-ray or some other diffraction measurement, we have something like:

where $\vec{p}_{in}$ is the momentum of the incoming wave and $\vec{p}_{out}$ is the momentum of the scattered wave, and the scattering angle is $\theta$. The moduli of the momenta are the same because the scattering is elastic. $\vec{q}$ is the difference i.e.

$$\vec{p}_{in} + \vec{q} = \vec{p}_{out}$$

Since $|p_{in}| = |p_{out}| = 2\pi/\lambda$ the difference $q$ is:

\begin{align} q &= 2 p_{in} \sin\frac{\theta}{2} \\ &= \frac{4\pi}{\lambda} \sin\frac{\theta}{2} \end{align}

which is the same as the expression or $q$ that you give. So $q$ is called the momentum transfer because that's exactly what it is.

• Thanks, I actually understand this answer more than the other as I have only dealt with the theory classically so far. I actually am still a bit confused about something. Once the oscillations are occurring in the 'scatterer', is there not just a uniform electromagnetic field created that propagates outwards (think a dipole radiation pattern)? What exactly is changing with angle from this interpretation? – Steve Hatcher Jul 8 '14 at 3:02