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