Suppose we have $N$ particles whose coordinates are given by $\mathbf{r}_{i}$. These coordinates are confined to be within a three-dimensional unit cell defined by $$\mathcal{V}=\left[0,L_{x}\right]\times\left[0,L_{y}\right]\times\left[0,L_{z}\right]$$

(For simplicity, we may assume $L_{x}=L_{y}=L_{z}\equiv L$). We shall also assume periodic boundary conditions, i.e., for example, $f\left(x_i=0\right)=f\left(x_i=L_{x}\right)$ (and same for the other Cartesian components and particles). Consider the integral

$$ I = \int_{\Omega}dR\;f\left(R,t\right) $$

where $R$ represents the collection of coordinates $\mathbf{r}_1,\dots,\mathbf{r}_N$ and $dR=d\mathbf{r}_{1}\dots d\mathbf{r}_{N}$. The volume of integration $\Omega$ is essentially the hyper-volume $\mathcal{V}^{N}$ formed by the unit cell $\mathcal{\mathcal{V}}$.

Now, suppose we perform the change of coordinates $\bar{\mathbf{r}}_{i}=c\left(t\right)\mathbf{r}_{i}$, where $c\left(t\right)$ is some function of $t$. For example, $\bar{\mathbf{r}}_{i}=t^{\alpha}\mathbf{r}_{i}$, where $\alpha$ is some constant.

I would like to calculate the derivative of the transformed $I$ with respect to $t$, i.e.,

$$\frac{d}{dt}\int_{\bar{\Omega}\left(t\right)}d\bar{R}\;f\left(\bar{R},t\right) J(t)$$

where the transformed quantities are marked with a bar, and $J(t)$ is the Jacobian term that results from the change of coordinates. Note that after this substitution, the volume of integration depends on $t$, because $\bar{\mathcal{V}}=\left[0,c(t) L_{x}\right]\times\left[0,c(t) L_{y}\right]\times\left[0, c(t) L_{z}\right]$. According to Leibniz integral rule,

$$\frac{d}{dt}\int_{\bar{\Omega}\left(t\right)}d\bar{R}\;f\left(\bar{R},t\right)J(t)=\int_{\bar{\Omega}\left(t\right)}d\bar{R}\;\frac{\partial}{\partial t}\left[f\left(\bar{R},t\right)J(t)\right]+\underbrace{\int_{\partial\bar{\Omega}\left(t\right)}f\left(\bar{R},t\right)J(t)\;U\cdot d\mathbf{\Sigma}}_{\text{boundary term}}$$

where $\partial\bar{\Omega}\left(t\right)$ is the boundary of the integration hyper-volume, $U$ is the velocity of the boundary (rate of change with $t$) and $d\mathbf{\Sigma}$ is a surface element.

However, I have trouble evaluating $U\cdot d\mathbf{\Sigma}$ even for the simple cubic geometry provided here. For example, I am unsure how to calculate $U$, or how to properly account for the orientation of the surface element.


1 Answer 1


This question could be asked just as well for any integral of the form $${\cal I}:=\int_{\bar{\Omega}(t)}d\bar{R}\;G(\bar{R},t).$$ Since there are $N$ particles, the box $\bar{\Omega}(t)$ will be a product of $3N$ intervals: $$ \bar{\Omega}(t)=([0,c(t) L_{x}]\times[0,c(t) L_{y}]\times[0, c(t) L_{z}])^N. $$ If you increase $t$ by an infinitesimal amount $dt$, $\cal I$ will change in two ways: first because the integrand changes, giving an increment of $$ dt \int_{\bar{\Omega}(t)}d\bar{R}\;\frac{\partial }{\partial t}G(\bar{R},t), $$ and second because the volume of integration changes. For example, if you take the $y$ coordinate of the $i$th particle, its interval of integration will change from $[0, c(t)L_y]$ to $[0, c(t)L_y + (d c/d t) L_y dt]$, adding (assuming that $d c/d t>0$) the interval $[c(t)L_y, c(t)L_y + (d c/d t) L_y dt]$. Taking the product of this with the $3N-1$ other intervals of those whose product equals $\bar\Omega(t)$ then gives a small extra slice of volume that you must integrate over. You can approximate this by a surface integral over the portion of $\partial \bar\Omega(t)$ where $\bar y_i=c(t)L_y$, giving an increment of $$\frac{d c}{d t} L_y dt \int_{\partial\bar\Omega(t),\ \bar y_i=c(t)L_y} d{\bar R'_{y,i}} G(\bar R, t)$$ where you are now integrating only over the $3N-1$ coordinates $\bar R'_{y,i}$ which exclude $\bar y_i$. Since the box $\bar\Omega(t)$ has $3N$ moving boundaries, you will get $3N$ increments of this sort. Adding them up will give \begin{eqnarray*} d{\cal I}&=&dt \int_{\bar{\Omega}(t)}d\bar{R}\;\frac{\partial }{\partial t}G(\bar{R},t)\\ &+&\sum_{1\le i\le N}\frac{d c}{d t} L_x dt \int_{\partial\bar\Omega(t),\ \bar x_i=c(t)L_x} d{\bar R'_{x,i}} G(\bar R, t)\\ &+&\sum_{1\le i\le N}\frac{d c}{d t} L_y dt \int_{\partial\bar\Omega(t),\ \bar y_i=c(t)L_y} d{\bar R'_{y,i}} G(\bar R, t)\\ &+&\sum_{1\le i\le N}\frac{d c}{d t} L_z dt \int_{\partial\bar\Omega(t),\ \bar z_i=c(t)L_z} d{\bar R'_{z,i}} G(\bar R, t). \end{eqnarray*} You can express the $3N$ boundary terms as a surface integral $$dt \int_{\partial\bar{\Omega}\left(t\right)}G\left(\bar{R},t\right) U\cdot d\mathbf{\Sigma}$$ if you let $d\mathbf{\Sigma}$ have the same magnitude as $d{\bar R'_{x,i}}$, $d{\bar R'_{y,i}}$ or $d{\bar R'_{z,i}}$ (so, it measures $3N-1$-dimensional hypervolume in the usual way), and let it point outwards from the cube (for example, towards the negative $x$ axis for ${\bf\bar r}_i$ on the hyperface $\bar x_i=0$, and towards the positive $z$ axis for ${\bf\bar r}_j$ on the hyperface $\bar z_j=c(t) L_z.$) $U$ can be taken to be zero on the $3N$ hyperfaces where $\bar x_i=0$, $\bar y_i=0$, or $\bar z_i=0$, since they are not moving, and you can take it to have the same direction as $d\mathbf{\Sigma}$ and magnitude $L_x (dc/dt)$, $L_y (dc/dt)$, and $L_z (dc/dt)$ on the $3N$ hyperfaces where $\bar x_i=c(t)L_x$, $\bar y_i=c(t)L_y$, and $\bar z_i=c(t)L_z$, respectively. The choice of $U$ is somewhat arbitrary since you could add to it any vector perpendicular to $d\mathbf{\Sigma}$, corresponding to a sidewise translation of the cube hyperface, without changing the result. (If $dc/dt<0$, the box will be shrinking instead of expanding, so $U$ should point in the opposite direction to $d\mathbf{\Sigma}$.)

  • $\begingroup$ Thank you. Do periodic boundary conditions allow for any simplification in the evaluation of the surface integral corresponding to the boundary term? $\endgroup$
    – user186483
    Sep 16, 2023 at 20:54
  • $\begingroup$ Seeing that there were periodic boundary conditions made me wonder if there was something that hadn't been explained about the problem that I was missing. But based on what's written here, no, they don't. $\endgroup$ Sep 17, 2023 at 2:18

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