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Isotropic Tensors

A tensor which has the special property that its components take the same value in all Cartesian coordinate systems is called an isotropic tensor. We have already encountered two such tensors: namely, the second-order identity tensor, $ \delta_{ij}$ , and the third-order permutation tensor, $ \epsilon_{ijk}$ . Of course, all scalars are isotropic. Moreover, as is easily demonstrated, there are no isotropic vectors (other than the null vector). It turns out that the most general isotropic Cartesian tensors of second-, third-, and fourth-order are $ \lambda\,\delta_{ij}$ , $ \mu\,\epsilon_{ijk}$ , and $ \alpha\,\delta_{ij}\,\delta_{kl} + \beta\,\delta_{ik}\,\delta_{jl} + \gamma\,\delta_{il}\,\delta_{jk}$ , respectively, where $ \lambda$ , $ \mu$ , $ \alpha $ , $ \beta $ , and $ \gamma$ are scalars. Let us prove these important results (Hodge 1961).

The most general second-order isotropic tensor, $ a_{ij}$ , is such that

$\displaystyle a_{ij}' = {\cal R}_{ip}\,{\cal R}_{jq}\,a_{pq} = a_{ij}$ (B.65)

for arbitrary rotations of the coordinate axes. It follows from Equation (B.24) that, to first order in the $ \delta\theta_i$ ,

$\displaystyle \delta \theta_m \left(\epsilon_{mis}\,a_{sj} + \epsilon_{mjs}\,a_{is}\right) = 0.$ (B.66)

However, the $ \delta\theta_i$ are arbitrary, so we can write

$\displaystyle \epsilon_{mis}\,a_{sj} + \epsilon_{mjs}\,a_{is}=0.$ (B.67)

Let us multiply by $ \epsilon_{mik}$ . With the aid of Equation (B.16), we obtain

$\displaystyle (\delta_{ii}\,\delta_{ks}-\delta_{is}\,\delta_{ki})\,a_{sj} + (\delta_{ij}\,\delta_{ks} -\delta_{is}\,\delta_{kj})\,a_{is} = 0,$ (B.68)

which reduces to

$\displaystyle 2\,a_{ij} + a_{ji} = a_{ss}\,\delta_{ij}.$ (B.69)

Interchanging the labels $ i$ and $ j$ , and then taking the difference between the two equations thus obtained, we deduce that

$\displaystyle a_{ij} = a_{ji}.$ (B.70)


$\displaystyle a_{ij} = \frac{a_{ss}}{3}\,\delta_{ij},$ (B.71)

which implies that

$\displaystyle a_{ij } =\lambda\,\delta_{ij}.$ (B.72)

For the case of an isotropic third-order tensor, Equation (B.67) generalizes to

$\displaystyle \epsilon_{mis}\,a_{sjk} + \epsilon_{mjs}\,a_{isk}+\epsilon_{mks}\,a_{ijs}=0.$ (B.73)

Multiplying by $ \epsilon_{mit}$ , $ \epsilon_{mjt}$ , and $ \epsilon_{mkt}$ , and then setting $ t=i$ , $ t=j$ , and $ t=k$ , respectively, we obtain

$\displaystyle 2\,a_{ijk} + a_{jik} + a_{kji}$ $\displaystyle =a_{ssk}\,\delta_{ij} + a_{sjs}\,\delta_{ik},$ (B.74)
$\displaystyle 2\,a_{ijk} + a_{jik} + a_{ikj}$ $\displaystyle = a_{ssk}\,\delta_{ij} + a_{iss}\,\delta_{jk},$ (B.75)
$\displaystyle 2\,a_{ijk} +a_{kji} +a_{ikj}$ $\displaystyle =a_{sjs}\,\delta_{ik} + a_{iss}\,\delta_{jk},$ (B.76)

respectively. However, multiplying the previous equations by $ \delta_{jk}$ , $ \delta_{ik}$ , and $ \delta_{ij}$ , and then setting $ i=i$ , $ j=i$ , and $ k=i$ , respectively, we obtain

$\displaystyle 2\,a_{iss} + a_{sis} + a_{ssi}$ $\displaystyle =a_{ssi}+a_{sis},$ (B.77)
$\displaystyle 2\,a_{sis} + a_{iss} + a_{ssi}$ $\displaystyle = a_{ssi}+a_{iss},$ (B.78)
$\displaystyle 2\,a_{ssi} +a_{iss} +a_{sis}$ $\displaystyle =a_{sis} + a_{iss},$ (B.79)

respectively, which implies that

$\displaystyle a_{iss}=a_{sis}= a_{ssi} = 0.$ (B.80)

Hence, we deduce that

$\displaystyle 2\,a_{ijk} + a_{jik} + a_{kji}$ $\displaystyle = 0,$ (B.81)
$\displaystyle 2\,a_{ijk} + a_{jik} + a_{ikj}$ $\displaystyle = 0,$ (B.82)
$\displaystyle 2\,a_{ijk} +a_{kji} +a_{ikj}$ $\displaystyle =0.$ (B.83)

The solution to the previous equation must satisfy

$\displaystyle a_{ikj} = a_{jik}=a_{kji} = -a_{ijk}.$ (B.84)

This implies, from Equation (B.8), that

$\displaystyle a_{ijk} = \mu\,\epsilon_{ijk}.$ (B.85)

For the case of an isotropic fourth-order tensor, Equation (B.73) generalizes to

$\displaystyle \epsilon_{mis}\,a_{sjkl} + \epsilon_{mjs}\,a_{iskl}+\epsilon_{mks}\,a_{ijsl} +\epsilon_{mls}\,a_{ijks}=0.$ (B.86)

Multiplying the previous by $ \epsilon_{mit}$ , $ \epsilon_{mjt}$ , $ \epsilon_{mkt}$ , $ \epsilon_{mlt}$ , and then setting $ t=i$ , $ t=j$ , $ t=k$ , and $ t=l$ , respectively, we obtain

$\displaystyle 2\,a_{ijkl} + a_{jikl} + a_{kjil}+a_{ljki}$ $\displaystyle =a_{sskl}\,\delta_{ij} + a_{sjsl}\,\delta_{ik}+ a_{sjks}\,\delta_{il},$ (B.87)
$\displaystyle 2\,a_{ijkl} + a_{jikl} + a_{ikjl} +a_{iljk}$ $\displaystyle = a_{sskl}\,\delta_{ij} +a_{isks}\,\delta_{jl}+ a_{issl}\,\delta_{jk},$ (B.88)
$\displaystyle 2\,a_{ijkl} +a_{kjil} +a_{ikjl}+a_{ijlk}$ $\displaystyle =a_{ijss}\,\delta_{kl}+a_{sjsl}\,\delta_{ik} + a_{issl}\,\delta_{jk},$ (B.89)
$\displaystyle 2\,a_{ijkl}+a_{ljki} +a_{ilkj}+a_{ijlk}$ $\displaystyle =a_{ijss}\,\delta_{kl}+a_{isks}\,\delta_{jl}+a_{sjks}\,\delta_{il},$ (B.90)

respectively. Now, if $ a_{ijkl}$ is an isotropic fourth-order tensor then $ a_{sskl}$ is clearly an isotropic second-order tensor, which means that is a multiple of $ \delta_{kl}$ . This, and similar arguments, allows us to deduce that

$\displaystyle a_{sskl}$ $\displaystyle =\lambda\,\delta_{kl},$ (B.91)
$\displaystyle a_{sjsl}$ $\displaystyle =\mu\,\delta_{jl},$ (B.92)
$\displaystyle a_{sjks}$ $\displaystyle = \nu\,\delta_{jk}.$ (B.93)

Let us assume, for the moment, that

$\displaystyle a_{ijss}$ $\displaystyle =a_{ssij},$ (B.94)
$\displaystyle a_{isks}$ $\displaystyle =a_{sisk},$ (B.95)
$\displaystyle a_{issl}$ $\displaystyle = a_{sils}.$ (B.96)

Thus, we get

$\displaystyle 2\,a_{ijkl} + a_{jikl} + a_{kjil}+a_{ljki}$ $\displaystyle =\lambda\,\delta_{ij}\,\delta_{kl} + \mu\,\delta_{ik}\,\delta_{jl} +\nu\,\delta_{il}\,\delta_{jk},$ (B.97)
$\displaystyle 2\,a_{ijkl} + a_{jikl} +a_{ikjl} +a_{ilkj}$ $\displaystyle =\lambda\,\delta_{ij}\,\delta_{kl} + \mu\,\delta_{ik}\,\delta_{jl} +\nu\,\delta_{il}\,\delta_{jk},$ (B.98)
$\displaystyle 2\,a_{ijkl} +a_{kjil} +a_{ikjl}+a_{ijlk}$ $\displaystyle =\lambda\,\delta_{ij}\,\delta_{kl} + \mu\,\delta_{ik}\,\delta_{jl} +\nu\,\delta_{il}\,\delta_{jk},$ (B.99)
$\displaystyle 2\,a_{ijkl}+a_{ljki} +a_{ilkj}+a_{ijlk}$ $\displaystyle =\lambda\,\delta_{ij}\,\delta_{kl} + \mu\,\delta_{ik}\,\delta_{jl} +\nu\,\delta_{il}\,\delta_{jk}.$ (B.100)

Relations of the form

$\displaystyle a_{ijkl} = a_{jilk} = a_{klij}=a_{lkji}$ (B.101)

can be obtained by subtracting the sum of one pair of Equations (B.97)-(B.100) from the sum of the other pair. These relations justify Equations (B.94)-(B.96). Equations (B.97) and (B.101) can be combined to give

$\displaystyle 2\,a_{ijkl}+(a_{ijlk}+a_{ikjl}+a_{ilkj})$ $\displaystyle =\lambda\,\delta_{ij}\,\delta_{kl} + \mu\,\delta_{ik}\,\delta_{jl} +\nu\,\delta_{il}\,\delta_{jk},$ (B.102)
$\displaystyle 2\,a_{iklj}+(a_{ikjl}+a_{ilkj}+a_{ijlk})$ $\displaystyle =\lambda\,\delta_{ik}\,\delta_{jl} + \mu\,\delta_{il}\,\delta_{jk} +\nu\,\delta_{ij}\,\delta_{kl},$ (B.103)
$\displaystyle 2\,a_{iljk}+(a_{ilkj}+a_{ijlk}+a_{ikjl})$ $\displaystyle =\lambda\,\delta_{il}\,\delta_{jk} + \mu\,\delta_{ij}\,\delta_{kl} +\nu\,\delta_{ik}\,\delta_{jl}.$ (B.104)

The latter two equations are obtained from the first via cyclic permutation of $ j$ , $ k$ , and $ l$ , with $ i$ remaining unchanged. Summing Equations (B.102)-(B.104), we get

$\displaystyle 2\,(a_{ijkl}+a_{iklj}+a_{iljk}) + 3\,(a_{ijlk}+a_{ikjl} +a_{ilkj}...
...,(\delta_{ij}\,\delta_{kl} +\delta_{ik}\,\delta_{jl}+\delta_{il}\,\delta_{jk}).$ (B.105)

It follows from symmetry that

$\displaystyle a_{ijkl}+a_{iklj}+a_{iljk}= a_{ijlk}+a_{ikjl}+a_{ilkj} = \frac{1}...
...,(\delta_{ij}\,\delta_{kl} +\delta_{ik}\,\delta_{jl}+\delta_{il}\,\delta_{jk}).$ (B.106)

This can be seen by swapping the indices $ k$ and $ l$ in the previous expression. Finally, substitution into Equation (B.102) yields

$\displaystyle a_{ijkl} = \alpha\,\delta_{ij}\,\delta_{kl} + \beta\,\delta_{ik}\,\delta_{jl} +\gamma\,\delta_{il}\,\delta_{jk},$ (B.107)


$\displaystyle \alpha$ $\displaystyle = (4\,\lambda-\mu-\nu)/10,$ (B.108)
$\displaystyle \beta$ $\displaystyle = (4\,\mu-\nu-\lambda)/10,$ (B.109)
$\displaystyle \gamma$ $\displaystyle = (4\,\nu-\lambda-\mu)/10.$ (B.110)

next up previous
Next: Exercises Up: Cartesian Tensors Previous: Tensor Fields
Richard Fitzpatrick 2016-03-31