Deriving integration identities for isotropic tensors

Question

For two spherical molecules, one at position $\mathbf{r}$ and the other at position $\mathbf{r}^{\prime}$, define $\mathbf{s}=\mathbf{r}^{\prime}-\mathbf{r}$. The pressure tensor of the system can be approximated by:

$$\mathbf{P}(\mathbf{r})=\rho(\mathbf{r})\,k_{\mathrm{B}}\,T\mathbf{\,I}-\tfrac{1}{2}\int\frac{\mathbf{ss}}{s}u^{\prime}(s)\,\rho(\mathbf{r})\sum_{i=0}^{\infty}\left[\frac{(\mathbf{s}\cdot\nabla)^{i}}{i!}\rho(\mathbf{r})\,g^{(2)}(s,\rho_{\mathrm{avg}})\right]\,d\mathbf{s}\,.\label{eq: P H series}$$

Davis [1] showed that the integrals are all of the form:

$$\mathbf{T}_{m}=\int\mathbf{s}\cdots\mathbf{s}\,f(s)\,d\mathbf{s}\,,$$

where $\mathbf{T}_{m}$ is an $m$th rank isotropic tensor, and $f(s)$ is an isotropic function of $\mathbf{s}$. If $m$ is odd, then the integral is equal to zero.

Davis then showed the following identities:

$$\mathbf{T}_{2}=\tfrac{1}{3}\left[\int s^{2}\,f(s)\,d\mathbf{s}\right]\mathbf{I}\,,$$

$$\mathbf{T}_{4}:\mathbf{ab}=\tfrac{1}{15}\int s^{4}\,f(s)\,d\mathbf{s}\left[\mathbf{ab}+\mathbf{ba}+(\mathbf{a}\cdot\mathbf{b})\mathbf{I}\right],$$

where $\mathbf{a}$ and $\mathbf{b}$ are arbitrary vectors.

How are these identities derived? I know the second-order isotropic tensor is the Kronecker delta $$, but I'm not sure where to start.

Ultimately, for my research I need to find a similar identity for $\mathbf{T}_{6}::\mathbf{abcd}$, where $\mathbf{c}$ and $\mathbf{d}$ are additional arbitrary vectors, which allows one to expand the pressure tensor to 4th order in density gradients.

Thank you for your time!

[1] Davis, H. Ted, Statistical mechanics of phases, interfaces and thin films, Center for Interfacial Engineering. New York, NY: VCH. xvi, 712 p. (1995). ZBL0884.00006

Answer 1

I think I figured this out with help from kipf . The forms of second, fourth, and sixth order isotropic tensors are given here. One can derive the $\mathbf{T}_{2}$ identity from the form of the second-order isotropic tensor

$$\mathbf{T}_{2}=\lambda\mathbf{I}\,,$$

where $\lambda$ is a scalar constant. The constant can be determined by taking the trace of both sides:

$$\int s^{2}\,f(s)\,d\mathbf{s}=\mathrm{Tr}(\lambda\mathbf{I})=3\lambda\,,$$ which gives

$$\lambda=\tfrac{1}{3}\int s^{2}\,f(s)\,d\mathbf{s}\,.$$

Deriving the $\mathbf{T}_{4}:\mathbf{ab}$ identity is similar. The most general form of a fourth-order isotropic tensor is

$$\mathbf{T}_{4}=\lambda\mathbf{\delta}_{ij}\mathbf{\delta}_{km}+\mu\mathbf{\delta}_{ik}\mathbf{\delta}_{jm}+\nu\mathbf{\delta}_{im}\mathbf{\delta}_{jk}\,,$$

where $\mu$ and $\nu$ are additional scalar constants. However, since $\mathbf{T}_{4}$ is symmetric in this case, the form must be invariant under permutation of indices, which means $\lambda=\mu=\nu$. The constant $\lambda$ can determined by taking the trace of both sides twice by setting $j=i$ and summing over $i$, and then setting $m=k$ and summing over $k$, which gives

$$\int s^{4}\,f(s)\,d\mathbf{s}=\lambda\left(\sum_{i=1}^{3}\sum_{k=1}^{3}1+2\mathbf{\delta}_{ik}\right)=\lambda\left(9+2*3\right)=15\lambda\,,$$

and therefore

$$\lambda=\tfrac{1}{15}\int s^{4}\,f(s)\,d\mathbf{s}\,.$$

The double dot product then follows as

$$\left(\mathbf{\delta}_{ij}\mathbf{\delta}_{km}+\mathbf{\delta}_{ik}\mathbf{\delta}_{jm}+\mathbf{\delta}_{im}\mathbf{\delta}_{jk}\right)a_{k}b_{m}=\mathbf{\delta}_{ij}a_{k}b_{k}+a_{i}b_{j}+b_{i}a_{j}=\mathbf{I}(\mathbf{a}\cdot\mathbf{b})+\mathbf{ab}+\mathbf{ba}\,.$$

Combining everything gives the $\mathbf{T}_{4}:\mathbf{ab}$ identity from Davis above.

Deriving the $\mathbf{T}_{6}::\mathbf{abcd}$ identity is more tedious because the form of a symmetric sixth-order isotropic tensor has 15 linearly independent terms:

$$\mathbf{T}_{6} = \lambda\left(\mathbf{\delta}_{ij}\mathbf{\delta}_{km}\mathbf{\delta}_{pq}+\mathbf{\delta}_{ij}\mathbf{\delta}_{kp}\mathbf{\delta}_{mq}+\mathbf{\delta}_{ij}\mathbf{\delta}_{kq}\mathbf{\delta}_{mp}+\mathbf{\delta}_{ik}\mathbf{\delta}_{jm}\mathbf{\delta}_{pq}+\mathbf{\delta}_{ik}\mathbf{\delta}_{jp}\mathbf{\delta}_{mq}\right.$$ $$+\mathbf{\delta}_{ik}\mathbf{\delta}_{jq}\mathbf{\delta}_{mp}+\mathbf{\delta}_{im}\mathbf{\delta}_{jk}\mathbf{\delta}_{pq}+\mathbf{\delta}_{im}\mathbf{\delta}_{jp}\mathbf{\delta}_{kq}+\mathbf{\delta}_{im}\mathbf{\delta}_{jq}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ip}\mathbf{\delta}_{jk}\mathbf{\delta}_{mq}$$ $$\left.+\mathbf{\delta}_{ip}\mathbf{\delta}_{jm}\mathbf{\delta}_{kq}+\mathbf{\delta}_{ip}\mathbf{\delta}_{jq}\mathbf{\delta}_{km}+\mathbf{\delta}_{iq}\mathbf{\delta}_{jk}\mathbf{\delta}_{mp}+\mathbf{\delta}_{iq}\mathbf{\delta}_{jm}\mathbf{\delta}_{kp}+\mathbf{\delta}_{iq}\mathbf{\delta}_{jp}\mathbf{\delta}_{km}\right)\,.$$

The constant $\lambda$ can determined by taking the trace of both sides three times by setting $j=i$ and summing over $i$, and then setting $m=k$ and summing over $k$, and setting $q=p$ and summing over $p$ which gives

$$\int s^{6}\,f(s)\,d\mathbf{s} = \lambda\sum_{i=1}^{3}\sum_{k=1}^{3}\sum_{p=1}^{3}\left(\mathbf{\delta}_{ii}\mathbf{\delta}_{kk}\mathbf{\delta}_{pp}+\mathbf{\delta}_{ii}\mathbf{\delta}_{kp}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ii}\mathbf{\delta}_{kp}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ik}\mathbf{\delta}_{ik}\mathbf{\delta}_{pp}+\mathbf{\delta}_{ik}\mathbf{\delta}_{ip}\mathbf{\delta}_{kp}\right.$$ $$+\mathbf{\delta}_{ik}\mathbf{\delta}_{ip}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ik}\mathbf{\delta}_{ik}\mathbf{\delta}_{pp}+\mathbf{\delta}_{ik}\mathbf{\delta}_{ip}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ik}\mathbf{\delta}_{ip}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ip}\mathbf{\delta}_{ik}\mathbf{\delta}_{kp}$$ $$\left.+\mathbf{\delta}_{ip}\mathbf{\delta}_{ik}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ip}\mathbf{\delta}_{ip}\mathbf{\delta}_{kk}+\mathbf{\delta}_{ip}\mathbf{\delta}_{ik}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ip}\mathbf{\delta}_{ik}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ip}\mathbf{\delta}_{ip}\mathbf{\delta}_{kk}\right)\,.$$

Simplifying gives

$$\int s^{6}\,f(s)\,d\mathbf{s}=\lambda\sum_{i=1}^{3}\sum_{k=1}^{3}\sum_{p=1}^{3}\left(1+2\mathbf{\delta}_{kp}+2\mathbf{\delta}_{ik}+2\mathbf{\delta}_{ip}+8\mathbf{\delta}_{ik}\mathbf{\delta}_{ip}\mathbf{\delta}_{kp}\right)=\lambda\left(27+6*9+8*3\right)=105\lambda\,.$$

The quadruple dot produce is evaluated as

$$\mathbf{T}_{6}::\mathbf{abcd} = \lambda\left(\mathbf{\delta}_{ij}\mathbf{\delta}_{km}\mathbf{\delta}_{pq}+\mathbf{\delta}_{ij}\mathbf{\delta}_{kp}\mathbf{\delta}_{mq}+\mathbf{\delta}_{ij}\mathbf{\delta}_{kq}\mathbf{\delta}_{mp}+\mathbf{\delta}_{ik}\mathbf{\delta}_{jm}\mathbf{\delta}_{pq}+\mathbf{\delta}_{ik}\mathbf{\delta}_{jp}\mathbf{\delta}_{mq}\right.$$ $$+\mathbf{\delta}_{ik}\mathbf{\delta}_{jq}\mathbf{\delta}_{mp}+\mathbf{\delta}_{im}\mathbf{\delta}_{jk}\mathbf{\delta}_{pq}+\mathbf{\delta}_{im}\mathbf{\delta}_{jp}\mathbf{\delta}_{kq}+\mathbf{\delta}_{im}\mathbf{\delta}_{jq}\mathbf{\delta}_{kp}+\mathbf{\delta}_{ip}\mathbf{\delta}_{jk}\mathbf{\delta}_{mq}$$ $$\left.+\mathbf{\delta}_{ip}\mathbf{\delta}_{jm}\mathbf{\delta}_{kq}+\mathbf{\delta}_{ip}\mathbf{\delta}_{jq}\mathbf{\delta}_{km}+\mathbf{\delta}_{iq}\mathbf{\delta}_{jk}\mathbf{\delta}_{mp}+\mathbf{\delta}_{iq}\mathbf{\delta}_{jm}\mathbf{\delta}_{kp}+\mathbf{\delta}_{iq}\mathbf{\delta}_{jp}\mathbf{\delta}_{km}\right)a_{k}b_{m}c_{p}d_{q}\,.$$

Simplifying gives

$$\mathbf{T}_{6}::\mathbf{abcd} = \lambda\left(\mathbf{\delta}_{ij}a_{k}b_{k}c_{p}d_{p}+\mathbf{\delta}_{ij}a_{k}b_{m}c_{k}d_{m}+\mathbf{\delta}_{ij}a_{k}b_{m}c_{m}d_{k}+a_{i}b_{j}c_{p}d_{p}+a_{i}b_{m}c_{j}d_{m}\right.$$ $$+a_{i}b_{m}c_{m}d_{j}+a_{j}b_{i}c_{p}d_{p}+a_{k}b_{i}c_{j}d_{k}+a_{k}b_{i}c_{k}d_{j}+a_{j}b_{m}c_{i}d_{m}$$ $$\left.+a_{k}b_{j}c_{i}d_{k}+a_{k}b_{k}c_{i}d_{j}+a_{j}b_{m}c_{m}d_{i}+a_{k}b_{j}c_{k}d_{i}+a_{k}b_{k}c_{j}d_{i}\right)\,.$$

Putting everything together shows

$$\mathbf{T}_{6}::\mathbf{abcd} =$$ $$\tfrac{1}{105}\int s^{6}\,f(s)\,d\mathbf{s}\left[\mathbf{I}(\mathbf{a}\cdot\mathbf{b})(\mathbf{c}\cdot\mathbf{d})+\mathbf{I}(\mathbf{a}\cdot\mathbf{c})(\mathbf{b}\cdot\mathbf{d})+\mathbf{I}(\mathbf{a}\cdot\mathbf{d})(\mathbf{b}\cdot\mathbf{c})+\mathbf{ab}(\mathbf{c}\cdot\mathbf{d})+\mathbf{ac}(\mathbf{b}\cdot\mathbf{d})\right.$$ $$+\mathbf{ad}(\mathbf{b}\cdot\mathbf{c})+\mathbf{ba}(\mathbf{c}\cdot\mathbf{d})+\mathbf{bc}(\mathbf{a}\cdot\mathbf{d})+\mathbf{bd}(\mathbf{a}\cdot\mathbf{c})+\mathbf{ca}(\mathbf{b}\cdot\mathbf{d})$$ $$\left.+\mathbf{cb}(\mathbf{a}\cdot\mathbf{d})+\mathbf{cd}(\mathbf{a}\cdot\mathbf{b})+\mathbf{da}(\mathbf{b}\cdot\mathbf{c})+\mathbf{db}(\mathbf{a}\cdot\mathbf{c})+\mathbf{dc}(\mathbf{a}\cdot\mathbf{b})\right]\,.$$