A Virial theorem and translational invariance

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The virial theorem, expressed as the relationship between total kinetic energy and forces in a system, faces challenges when applied to periodic boundary conditions due to the behavior of position vectors under translation. Specifically, translating a system by one period alters the position vectors, leading to inconsistencies in the calculated average kinetic energy. This raises questions about the validity of the theorem in systems with periodic potentials, where the periodicity of the potential does not align with the expected behavior of derivatives. The discussion references Tuckerman's "Statistical Mechanics" to highlight that certain terms in the virial theorem are valid only for bound systems, not for those exhibiting translational invariance. The resolution of these contradictions may lie in understanding the distinction between free and bound vectors in the context of transformations.
gjk
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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?
 
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gjk said:
It is often claimed that this expression is not valid for systems with periodic boundary conditions
Can you give a specific reference that makes this claim?
 
PeterDonis said:
Can you give a specific reference that makes this claim?
p. 465 (Section 12.6.3) in Tuckerman's "Statistical Mechanics". There he talks about the path-integral generalization of the virial theorem, but the idea is pretty much the same. You have terms of the form ##x_k (\partial U / \partial x_k)## which are only valid for bound (not translationally-invariant) systems. I didn't want to delve into path-integral formalism because I believe the question is more general and has to do with vectors and general properties of transformations.
 
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