In electron-muon scattering, we are interested in calculating the parameter M, which characterizes the scattering amplitude. Here, we present a derivation for M.

## Derivation

The scattering amplitude M can be expressed in terms of the Feynman diagrams. The lowest order Feynman diagram for electron-muon scattering involves the exchange of a virtual photon. Let’s denote this diagram as $M_0$.

## Kinematics

Before proceeding with the calculation, let’s define the kinematic variables involved. We have the initial electron momentum as $p_1$, the final electron momentum as $p_3$, the initial muon momentum as $p_2$, and the final muon momentum as $p_4$. The scattering angle between the initial and final electron momentum is denoted as $\theta$.

## Calculation

The scattering amplitude $M$ is given by the matrix element $\langle p_3, p_4 | M_0 | p_1, p_2 \rangle$. To calculate this, we need to evaluate the relevant Feynman integral. The expression for $M$ is:

$M$ = $\frac{-ie^2}{q^2} \bar{u}(p_3) \gamma^{\mu} u(p_1) \frac{-ig_{\mu\nu}}{q^2} \bar{u}(p_4) \gamma^{\nu} u(p_2)$

where $e$ is the elementary charge, $u(p)$ is the Dirac spinor for a particle with momentum $p$, and $q = p_1 – p_3$ is the momentum transfer.

We can simplify the expression by using the Dirac equation and the completeness relation for spinors:

$\bar{u}(p_i) \gamma^{\mu} u(p_j)$ = $\bar{u}(p_i) ( \gamma^0 p_j^{\mu} – \gamma^{\mu} p_j^0 ) u(p_j)$

Using this simplification, we obtain:

$M$ = $\frac{ie^2}{q^2} \left[ \bar{u}(p_3) \gamma^0 p_1^{\mu} u(p_1) – \bar{u}(p_3) \gamma^{\mu} p_1^0 u(p_1) – \bar{u}(p_4) \gamma^0 p_2^{\nu} u(p_2) + \bar{u}(p_4) \gamma^{\nu} p_2^0 u(p_2) \right] g_{\mu\nu}$

We can simplify the expression further by using the Dirac equation again:

$\bar{u}(p_i) \gamma^0 p_j^{\mu} u(p_j) = \bar{u}(p_i) \left[ \gamma^{\mu}, \gamma^0 \right] u(p_j) + \bar{u}(p_i) \left\{ \gamma^0, p_j^{\mu} \right\} u(p_j)$

The commutation and anticommutation relations of the Dirac matrices allow us to simplify this expression as follows:

$\bar{u}(p_i) \gamma^0 p_j^{\mu} u(p_j)$ = $\bar{u}(p_i) \left( 2p_j^{\mu} – \gamma^{\mu} \wp_{j} \right) u(p_j)$

where $\wp$ = $\gamma^{\mu} p_{\mu}$.

Using this simplification, we obtain:

$M$ = $\frac{2ie^2}{q^2} \left[ p_1^{\mu} \bar{u}(p_3) u(p_1) – p_1^0 \bar{u}(p_3) \gamma^{\mu} u(p_1) – p_2^{\nu} \bar{u}(p_4) u(p_2) + p_2^0 \bar{u}(p_4) \gamma^{\nu} u(p_2) \right] g_{\mu\nu}$

We can further simplify the expression by using the Dirac spinor normalization condition:

$\bar{u}(p_i) u(p_j)$ = $2m \delta_{ij}$

where $m$ is the mass of the particle. Using this relation, we get:

$M$ = $\frac{4ie^2}{q^2} \left[ p_1^{\mu} p_3^{\nu} – p_1^0 p_3^{\mu} g^{\mu\nu} – p_2^{\mu} p_4^{\nu} + p_2^0 p_4^{\mu} g^{\mu\nu} \right] g_{\mu\nu}$

Finally, we can simplify the expression by evaluating the metric tensor products:

$g^{\mu\nu} g_{\mu\nu}$ = 4

and

$g^{\mu\nu} g_{\mu\nu} p_1^0 p_3^{\mu} p_2^0 p_4^{\nu}$ = -4$(p_1 \cdot p_3)(p_2 \cdot p_4)$

where $(p_i \cdot p_j)$ = $p_i^{\mu} p_{j\mu}$ is the scalar product of two 4-vectors.

Finally, we arrive at the expression for M:

$M$ = $\frac{16ie^2}{q^2} \left[ (p_1 \cdot p_3)(p_2 \cdot p_4) – (p_1 \cdot p_2 )(p_3 \cdot p_4) \right]$

Hence, we calculated the parameter M for electron-muon scattering using the lowest order Feynman diagram involving the exchange of a virtual photon.