Deep Inelastic Scattering

[Nasa]

Homework Statement

Consider deep inelastic scattering, where a high energy electron collides with a proton. The initial momentum of the electron is $k^{\mu}$ , the final state momentum is $k'^{\mu}$ , and the scattering angle is $\theta$.
(1) Show that the transfer momentum $q(=k-k')$ is spatial in the limit where the electron mass is ignored.
(2) Find the scattering angle at which the electron mass cannot be ignored.

Relevant Equations

$$\frac{d^2\sigma}{d\Omega dE} $$

$$= \frac{\alpha^2}{4E^2}\frac{X}{\cos^2} \frac{\theta}{2}\sin^4$$

$$\frac{\theta}{2}(\frac{1}{\nu} F_2(x,Q)+\frac{2}{M}F_1(x,Q) \tan^2 \frac{\theta}{2})$$

where $Q^2=-q^2,\quad x=\frac{Q^2}{2p\cdot q}$

To make $q^0$ 0, we just need to boost it so that $|\boldsymbol{k}^{2}=|\boldsymbol{k}′|$ ~.

$$q^0=k−k′=\sqrt{m_e^2+\boldsymbol{k}^2}−\sqrt{m_e^2+\boldsymbol{k}′^2}$$

But this is possible regardless of the approximation $m_e\to0$. What does it mean when it says that $q^0$ becomes 0 by approximating the electron mass to 0?

Also this is possible for any scattering angle $\theta$, but since it says that the mass cannot be ignored depending on the scattering angle $\theta$, I assume that it is fixed to a Fixed Target frame and boosting is not taken into account.

 

[Orodruin]

Spacelike does not mean $q^0 = 0$. It means $q^2 < 0$.

 

[Nasa]

Spacelike does not mean $q^0 = 0$. It means $q^2 < 0$.

So "spatial" means spacelike ($q^2<0$). Now I have solved problem 1! Thank you.

 

[jedishrfu]

Hi Nasa,

I tried to fix your latex, which, I think, had an added close brace that caused other expressions to not render correctly.

Maybe you can look at your post and fix it the way you want

Also, notice the {X} I added because it seemed your \frac was missing a denominator.

Jedi