gjk
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- TL;DR
- Apparent paradox when translating vectors.
According to the virial theorem,
$$\left\langle T\right\rangle =-{\frac {1}{2}}\,\sum _{k=1}^{N}{\bigl \langle }\mathbf {F} _{k}\cdot \mathbf {r} _{k}{\bigr \rangle }$$
where ##N## is the number of particles in the system and ##T## is the total kinetic energy. It is often claimed that this expression is not valid for systems with periodic boundary conditions due to the ##\mathbf{r}_{k}## terms in the sum. And it makes sense, because if the system is periodic and we translate it by one period ##\mathbf{L}## then ##\mathbf{r}_k \to \mathbf{r}_k + \mathbf{L}##, so ##\left\langle T\right\rangle## before the shift is not equal to ##\left\langle T\right\rangle## after the translation.
On the other hand, we can write ##\mathbf{r}_{k}=\mathbf{r}_{k}-\mathbf{0}##, but then the same translation gives
$$
\mathbf{r}_{k}=\mathbf{r}_{k}-\mathbf{0}\to\left(\mathbf{r}_{k}+\mathbf{L}\right)-\left(\mathbf{0}+\mathbf{L}\right)=\mathbf{r}_{k}-\mathbf{0}=\mathbf{r}_{k}
$$
which contradicts the previous statement. Perhaps this silly "paradox" has something to do with the distinction between free and bound vectors?
A similar problem arises if we consider some periodic potential ##V(\mathbf{r})=V(\mathbf{r}+\mathbf{L})##. Assume we perform the change of coordinates ##\mathbf{r}=a\mathbf{r}^{\prime}## where ##a \in \mathbb{R}## is nonzero. Since ##V## is periodic, ##\partial_{a} V## should be periodic as well. However, using the chain rule, we get
$$
\frac{\partial}{\partial a}V\left(a\mathbf{r}^{\prime}\right)=\frac{\partial V\left(a\mathbf{r}^{\prime}\right)}{\partial\left(a\mathbf{r}^{\prime}\right)}\cdot\frac{\partial\left(a\mathbf{r}^{\prime}\right)}{\partial a}=\frac{\partial V\left(a\mathbf{r}^{\prime}\right)}{\partial\left(a\mathbf{r}^{\prime}\right)}\cdot\mathbf{r}^{\prime}
$$
and the RHS of the last expression is clearly not periodic. How this apparent contradiction can be resolved?
$$\left\langle T\right\rangle =-{\frac {1}{2}}\,\sum _{k=1}^{N}{\bigl \langle }\mathbf {F} _{k}\cdot \mathbf {r} _{k}{\bigr \rangle }$$
where ##N## is the number of particles in the system and ##T## is the total kinetic energy. It is often claimed that this expression is not valid for systems with periodic boundary conditions due to the ##\mathbf{r}_{k}## terms in the sum. And it makes sense, because if the system is periodic and we translate it by one period ##\mathbf{L}## then ##\mathbf{r}_k \to \mathbf{r}_k + \mathbf{L}##, so ##\left\langle T\right\rangle## before the shift is not equal to ##\left\langle T\right\rangle## after the translation.
On the other hand, we can write ##\mathbf{r}_{k}=\mathbf{r}_{k}-\mathbf{0}##, but then the same translation gives
$$
\mathbf{r}_{k}=\mathbf{r}_{k}-\mathbf{0}\to\left(\mathbf{r}_{k}+\mathbf{L}\right)-\left(\mathbf{0}+\mathbf{L}\right)=\mathbf{r}_{k}-\mathbf{0}=\mathbf{r}_{k}
$$
which contradicts the previous statement. Perhaps this silly "paradox" has something to do with the distinction between free and bound vectors?
A similar problem arises if we consider some periodic potential ##V(\mathbf{r})=V(\mathbf{r}+\mathbf{L})##. Assume we perform the change of coordinates ##\mathbf{r}=a\mathbf{r}^{\prime}## where ##a \in \mathbb{R}## is nonzero. Since ##V## is periodic, ##\partial_{a} V## should be periodic as well. However, using the chain rule, we get
$$
\frac{\partial}{\partial a}V\left(a\mathbf{r}^{\prime}\right)=\frac{\partial V\left(a\mathbf{r}^{\prime}\right)}{\partial\left(a\mathbf{r}^{\prime}\right)}\cdot\frac{\partial\left(a\mathbf{r}^{\prime}\right)}{\partial a}=\frac{\partial V\left(a\mathbf{r}^{\prime}\right)}{\partial\left(a\mathbf{r}^{\prime}\right)}\cdot\mathbf{r}^{\prime}
$$
and the RHS of the last expression is clearly not periodic. How this apparent contradiction can be resolved?