Solución

$$S[{{\boldsymbol{x}}}(t)]= \int dt \left( \frac{1}{2}m\dot{{\bolds... ...bol{A}}}({{\boldsymbol{x}}},t) -q\phi({{\boldsymbol{x}}},t) \right) = \int dt L$$ (3.2)

a)

$${{\boldsymbol{p}}}=\frac{\partial L}{\partial {{\boldsymbol{x}}}}... ...dot{{\boldsymbol{x}}} + \frac{q}{c}{{\boldsymbol{A}}}({{\boldsymbol{x}}},t) \,.$$ (3.3)

Usamos las ecuaciones de Euler-Lagrange

$$\frac{d{{\boldsymbol{p}}}}{dt}=\frac{\partial L}{\partial {{\boldsymbol{x}}}} \,,$$ (3.4)

$$m\ddot x_i+ \frac{q}{c}(\partial_tA_i+\dot x_j\partial_jA_i) = \frac{q}{c}\dot x_j\partial_iA_j-q\partial_i\phi$$ (3.5)

de donde

$$m\ddot x_i=F_i= -q\partial_i\phi -\frac{q}{c} \partial_tA_i +\frac{q}{c}\dot x_j(\partial_iA_j-\partial_jA_i) \,.$$ (3.6)

Teniendo en cuenta que

$$E_i=-\partial_i\phi-\frac{1}{c} \partial_tA_i \,,\qquad B_i=\epsilon_{ijk}\partial_jA_k$$ (3.7)

así como

$$(\dot{{\boldsymbol{x}}}\times {{\boldsymbol{B}}})_i = \epsilon_{ijk}\dot x_jB_k= \epsilon_{ijk}\dot x_j\epsilon_{k\ell ... ... (\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})\dot x_j\partial_\ell A_m$$

$$= \dot x_j(\partial_i A_j-\partial_j A_i) \,,$$ (3.8)

se obtiene la fuerza de Lorentz

$${{\boldsymbol{F}}}=q {{\boldsymbol{E}}}+\frac{q}{c}\dot{{\boldsymbol{x}}}\times {{\boldsymbol{B}}}$$ (3.9)

b) Para obtener el hamiltoniano

$${{\boldsymbol{p}}}=m\dot{{\boldsymbol{x}}}+\frac{q}{c}{{\boldsymb... ...{\boldsymbol{x}}}=\frac{1}{m}({{\boldsymbol{p}}}-\frac{q}{c}{{\boldsymbol{A}}})$$ (3.10)

$$H = \dot{{\boldsymbol{x}}}{{\boldsymbol{p}}}-L$$

$$= \frac{1}{m}({{\boldsymbol{p}}}-\frac{q}{c}{{\boldsymbol{A}}}){{\b... ...1}{m}({{\boldsymbol{p}}}-\frac{q}{c}{{\boldsymbol{A}}}){{\boldsymbol{A}}}+q\phi$$

$$= \frac{1}{2m} ({{\boldsymbol{p}}}-\frac{q}{c}{{\boldsymbol{A}}})^2 +q\phi \,.$$ (3.11)