High Energy Muon scattering - 4-momentum

[Matt atkinson]

Homework Statement

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.

Homework Equations

$\vec{p}=(p_1,p_2,p_3,iE)$

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 can't 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 I am not sure how.

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

 

[PeroK]

What about the energy component in the dot product?

 

[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 $E>>m$;

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

 

[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.

 

[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 $p\approx E$;
$q^2=2E_iE_f-2p_i\cdot p_f$
$q^2=2E_iE_f-2E_iE_fcos(\theta)$

 

[PeroK]

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

 

[Matt atkinson]

Thank you!