As I understand it, according to the CPT theorem an antiparticle can be thought of as having either the opposite charge (C) or the opposite spacetime parity (PT). For example in Feynman diagrams a positron going forwards in spacetime is equivalent to an electron going backwards in spacetime.

Let us assume that we have a particle of mass $m$ and charge $q$ moving in the x-direction under the influence of an electric field $\vec{E}$ directed along the x-axis.

The velocity $\vec{v}=d\vec{x}/dt$ and acceleration $\vec{a}=d^2\vec{x}/dt^2$ of the particle are given by

\begin{eqnarray}

\frac{d\vec{x}}{dt} &=& \frac{\vec{p}}{m},\tag{1}\\

\frac{d^2\vec{x}}{dt^2} &=& \frac{q\vec{E}}{m}.\tag{2}

\end{eqnarray}

Now consider the particle's antiparticle. Instead of assuming it has the opposite charge we can assume that it has the opposite spacetime parity. This means that it has negative momentum and negative mass implying that it is traveling backwards in both space and time.

When we apply the following transformations to Eqn.$(1)$ and Eqn.$(2)$ they remain the same:

\begin{eqnarray}

\vec{p} &\rightarrow& -\vec{p},\\

m &\rightarrow& -m,\\

d\vec{x} &\rightarrow& -d\vec{x},\\

dt &\rightarrow& -dt.

\end{eqnarray}

Thus the antiparticle has the same velocity $\vec{v}$ and acceleration $\vec{a}$ as the particle.

Is this correct?