= Solution
The spatial part of the <Lorentz force> is
$$
\frac d{dt}(\gamma m\mathbf v)
=q(\mathbf E+\mathbf v\times\mathbf B).
$$
Part (c), together with $\mathcal E=\gamma mc^2$, gives
$$
\frac{d\gamma}{dt}
=\frac q{mc^2}\mathbf E\mathbin{\cdot}\mathbf v.
$$
Writing $\mathbf a=d\mathbf v/dt$ and expanding the momentum derivative,
$$
m\gamma\mathbf a+m\mathbf v\frac{d\gamma}{dt}
=q(\mathbf E+\mathbf v\times\mathbf B).
$$
Substitution of $d\gamma/dt$ gives the <coordinate acceleration of a relativistic charged particle>:
$$
\boxed{
\frac{d\mathbf v}{dt}
=\frac q{m\gamma}
\left[
\mathbf E+\mathbf v\times\mathbf B
-\frac1{c^2}\mathbf v(\mathbf v\mathbin{\cdot}\mathbf E)
\right]
}.
$$
In the nonrelativistic limit $|\mathbf v|/c\to0$, one has $\gamma\to1$ and the final term is smaller by order $v^2/c^2$. Hence
$$
\boxed{
m\frac{d\mathbf v}{dt}
=q(\mathbf E+\mathbf v\times\mathbf B)
},
$$
which is precisely Newton's second law with the ordinary Lorentz force.
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