 Solution

ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2021/section-b/5/b/solution

The variable initial energy

E_{0}

replaces

m_{π}

in the kinematic expression before differentiation. The charged-lepton mass remains fixed:

p (E_{0}) = \frac{E _{0}^{2} - m _{ℓ}^{2}}{2 E _{0}} = \frac{E _{0}}{2} - \frac{m _{ℓ}^{2}}{2 E _{0}} .

Thus the required Jacobian determinant in this one-dimensional change of variables is

\frac{d p}{d E _{0}}_{E_{0} = m_{π}} = \frac{1}{2} + \frac{m _{ℓ}^{2}}{2 m _{π}^{2}} = \frac{m _{π}^{2} + m _{ℓ}^{2}}{2 m _{π}^{2}} .

Equivalently,

E_{0} = p^{2} + m_{ℓ}^{2} + p

gives

d E_{0} / d p = 1 + p / E_{ℓ}

, so

d p / d E_{0} = E_{ℓ} / (E_{ℓ} + p) = E_{ℓ} / m_{π}

, the same result. It is dimensionless in natural units.

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