High Energy Muon scattering - 4-momentum

1. Jan 15, 2015

Matt atkinson

1. The problem statement, all variables and given/known data
Explain what the term "four-momentum transfer $q$" is
Show that for a high energy muon scattering at an angle $\theta$, the value of $q^2$ is given approximately by;
$q^2=2E_iE_f(1-cos(\theta))$
where $E_i$ and $E_f$ are the initial and final values of the muon's energy. State when this approximation is justified.
2. Relevant equations
$\vec{p}=(p_1,p_2,p_3,iE)$
3. The attempt at a solution
The first part, four momentum transfer is just the momentum and energy lost/gained in a scattering process by the scattered particle?

And the second part, i know the approximation should be that $E>>m$ but i cant seem to work with 4-momentum.

What i tried to do was;
$q^2=(p_i-p_f)^2=|p_i|^2+|p_f|^2-2p_i\cdot p_f=|p_i|^2+|p_f|^2-E_iE_fcos(\theta)$
and i assume somehow i need to get $|p_i|^2+|p_f|^2=2E_iE_f$ but im not sure how.

Would it be because the $E>>m$ then the $p \approx E$ but then that would just mean $q^2=E_i^2+E_f^2-E_iE_fcos(\theta)$

2. Jan 15, 2015

PeroK

What about the energy component in the dot product?

3. Jan 15, 2015

Matt atkinson

So,
$2p_i \cdot p_f=2((p_{i_x},p_{i_y},p_{i_z},iE_i) \cdot (p_{f_x},p_{f_y},p_{f_z},iE_f))=-2E_iE_f+2((p_{i_x},p_{i_y},p_{i_z}) \cdot (p_{f_x},p_{f_y},p_{f_z})$
actually my momentum transfer would be this then;
$q^2=|p_i|^2+|p_f|^2+2E_iE_f-2((p_{i_x},p_{i_y},p_{i_z}) \cdot (p_{f_x},p_{f_y},p_{f_z})$
$q^2=|p_i|^2+|p_f|^2+2E_iE_f-2|{}^3p_i||{}^3p_f|cos(\theta)$
The subscript before just to symbolize that they are three vectors not the 4 vectors? not sure how to show that.

then using my assumption that $E \approx p$ for $E>>m$;

$q^2=|p_i|^2+|p_f|^2+2E_iE_f-2E_iE_fcos(\theta)$

Last edited: Jan 15, 2015
4. Jan 15, 2015

PeroK

You're not calculating the dot product of a 4-vector correctly. Try splitting the dot product into the energy component and the combined momentum components.

And remember that you're subtracting the two vectors and then taking the dot product of the result with itself.

5. Jan 16, 2015

Matt atkinson

I think i managed it, if you could just check my working!

$q^2=(p_i-p_f)\cdot(p_i-p_f)=(p_{i_x}-p_{f_x},p_{i_y}-p_{f_y},p_{i_z}-p_{f_z},i(E_i-E_f))\cdot(p_{i_x}-p_{f_x},p_{i_y}-p_{f_y},p_{i_z}-p_{f_z},i(E_i-E_f))$
$q^2=|p_i|^2+|p_f|^2-2p_i\cdot p_f-E_i^2-E_f^2+2E_iE_f$
Now using my condition that if, $E>>m$ then $p\approx E$;
$q^2=2E_iE_f-2p_i\cdot p_f$
$q^2=2E_iE_f-2E_iE_fcos(\theta)$

6. Jan 16, 2015

PeroK

Yes, you just needed to get the dot product sorted out.

7. Jan 16, 2015

Thank you!