Given:
Euler's second law for the conservation of angular momentum
![{\displaystyle {\text{(1)}}\qquad \int _{\partial B}e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~dS+\int _{B}\rho ~e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\left(\int _{B}\rho ~e_{ijk}~x_{j}~v_{k}~dV\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66cb75a042ac1250439c8bb26222a25dfd2322b9)
The divergence theorem
![{\displaystyle {\text{(2)}}\qquad \int _{\partial B}n_{i}~\sigma _{ij}~dS=\int _{B}{\frac {\partial \sigma _{ij}}{\partial x_{i}}}~dV}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d03bfd4a463acf7c5384d3cc741998f195b9db58)
The equilibrium equation (Cauchy's first law)
![{\displaystyle {\text{(3)}}\qquad {\frac {\partial \sigma _{ij}}{\partial x_{i}}}+\rho ~b_{j}={\frac {d}{dt}}\left(\rho ~v_{j}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fc51e2169c82c23e51e7cb660e1559b9b260d40)
Show:
![{\displaystyle {\text{(4)}}\qquad \sigma _{ij}=\sigma _{ji}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/003f69b577703751abe6e10dc9c52a0ae97bacaf)
Let us first look at the first term of equation~(1) and apply the
divergence theorem (2).
Thus,
![{\displaystyle {\begin{aligned}\int _{\partial B}e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~dS&=\int _{\partial B}n_{l}~(e_{ijk}~x_{j}~\sigma _{lk})~dS\\&=\int _{B}{\frac {\partial {(e_{ijk}~x_{j}~\sigma _{lk})}}{\partial x_{l}}}~dV\\&=\int _{B}\left(e_{ijk}~{\frac {\partial x_{j}}{\partial x_{l}}}~\sigma _{lk}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ijk}~\delta _{jl}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c7e8de84abea77becca2adf75bdb3c43190ce57)
Plugging this back into equation~(1) gives
![{\displaystyle \int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV+\int _{B}\rho ~e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\left(\int _{B}\rho ~e_{ijk}~x_{j}~v_{k}~dV\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb78b60b6fe5e4fbba1eb16c0af30943c4c42ac4)
Therefore, bringing all terms to the left hand side,
![{\displaystyle {\text{(5)}}\qquad \int _{B}\left[e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}\left({\frac {\partial \sigma _{lk}}{\partial x_{l}}}+\rho ~b_{k}-{\frac {d}{dt}}\left(\rho ~v_{k}\right)\right)\right]~dV=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68d12638060b3f14207830faf6d3d0fbd6a00d40)
Using the equilibrium equations~(3), equation~(5) reduces
to
![{\displaystyle {\text{(6)}}\qquad \int _{B}e_{ilk}~\sigma _{lk}~dV=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec9635ab792de5e64d7073060df872cc3a12b914)
Since this holds for any
, we have
![{\displaystyle {\text{(7)}}\qquad e_{ilk}~\sigma _{lk}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d66149d744567f6daa58058af42068d18a70ac3)
If you work this expression out, you will see that
.
Hence, the stress tensor is symmetric.