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No, the situation you’ve described in the body of your question is not periodic. If there exists a $T\in \mathbb{R}$ such that the motion of a particle is described by

$$\mathbf{r}(t + T) = \mathbf{r}(t) $$

and

$$ \mathbf{p}(t + T) = \mathbf{p}(t)$$

for all $t$ in the domain of $\mathbf{r}$, then your motion may be considered to be periodic.

More generally, one can consider phase space $\{ (\mathbf{r, p}) \}$ of your system. Periodic motion corresponds to an orbit in phase space parametrized by time.

So while the position will come back on itself, the momentum never will. Hence the trajectory in phase space is not an orbit.

No, the situation you’ve described in the body of your question is not periodic. If there exists a $T\in \mathbb{R}$ such that the motion of a particle is described by

$$\mathbf{r}(t + T) = \mathbf{r}(t) $$

and

$$ \mathbf{p}(t + T) = \mathbf{p}(t)$$

for all $t$ in the domain of $\mathbf{r}$, then your motion may be considered to be periodic.

More generally, one can consider phase space $\{ (\mathbf{r, p}) \}$ of your system. Periodic motion corresponds to an orbit in phase space parametrized by time.

So while the position will come back on itself, the momentum never will. Hence the trajectory in phase space is not an orbital.

Not quite. If there exists a $T\in \mathbb{R}$ such that the motion of a particle is described by

$$\mathbf{r}(t + T) = \mathbf{r}(t) $$

and

$$ \mathbf{p}(t + T) = \mathbf{p}(t)$$

for all $t$ in the domain of $\mathbf{r}$, then your motion may be considered to be periodic.

More generally, one can consider phase space $\{ (\mathbf{r, p}) \}$ of your system. Periodic motion corresponds to an orbit in phase space parametrized by time.

No, the situation you’ve described in the body of your question is not periodic. If there exists a $T\in \mathbb{R}$ such that the motion of a particle is described by

$$\mathbf{r}(t + T) = \mathbf{r}(t) $$

and

$$ \mathbf{p}(t + T) = \mathbf{p}(t)$$

for all $t$ in the domain of $\mathbf{r}$, then your motion may be considered to be periodic.

More generally, one can consider phase space $\{ (\mathbf{r, p}) \}$ of your system. Periodic motion corresponds to an orbit in phase space parametrized by time.

So while the position will come back on itself, the momentum never will. Hence the trajectory in phase space is not an orbit.

Yes. If there exists a $T\in \mathbb{R}$ such that the motion of a particle is described by

$$\mathbf{r}(t + T) = \mathbf{r}(t) $$

and

$$ \mathbf{p}(t + T) = \mathbf{p}(t)$$

for all $t$ in the domain of $\mathbf{r}$, then your motion may be considered to be periodic.

More generally, one can consider phase space $\{ (\mathbf{r, p}) \}$ of your system. Periodic motion corresponds to an orbit in phase space parametrized by time.

Not quite. If there exists a $T\in \mathbb{R}$ such that the motion of a particle is described by

$$\mathbf{r}(t + T) = \mathbf{r}(t) $$

and

$$ \mathbf{p}(t + T) = \mathbf{p}(t)$$

for all $t$ in the domain of $\mathbf{r}$, then your motion may be considered to be periodic.

More generally, one can consider phase space $\{ (\mathbf{r, p}) \}$ of your system. Periodic motion corresponds to an orbit in phase space parametrized by time.