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As it is clear for everyone the deformation gradient $\mathbf{F}$ can be decomposed into,

$\mathbf{F}=\mathbf{Q} \cdot \mathbf{U}$

where $\mathbf{Q}$ is the rotation matrix.

Now if I consider $\mathbf{U}=\mathbf{1}$, I can write $\displaystyle Q_{ij}=\frac{\partial y_i}{\partial x_j}$.

According to the Schwarz integrability condition we have,

$\displaystyle \frac{\partial Q_{ij}}{\partial x_k}=\frac{\partial Q_{ik}}{\partial x_j}$.

Is it possible to show that, since $\mathbf{Q}$ is a rotation matrix, the only possibility consistent with the Schwatz integrability condition is $\mathbf{Q}=$constant?

The above has been taken from a book. But I don't know how it has deduced $\mathbf{Q}$ is constant.

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With tensor notation

You should use the property of rotational matrix $Q_{kl}Q_{jl}=Q_{li}Q_{lj}=\delta_{ij}$ to prove that tensor $T_{ijk}=Q_{li}Q_{lj,k}$ is antisymetric on first two indices: $$ 0 = \delta_{ij,k} = \partial_k(Q_{li}Q_{lj})=Q_{li}Q_{lj,k}+Q_{lj}Q_{li,k} = T_{ijk}+T_{jik}. $$ On the other hand, Schwarz condition implies $T_{ijk} = T_{ikj}$. Thus, you can show that it's impossible to have non-zero tensor of rank 3 with a symmetric pair of indices and an antisymmetric one: $$ T_{ijk} = T_{ikj} = -T_{kij} = -T_{kji} = T_{jki} = T_{jik} = -T_{ijk}. $$ Finally, you can show that if $T_{ijk}=0$, then $Q_{ij,k}$ should also be zero: $$ Q_{mi}T_{ijk}=Q_{mi}Q_{li}Q_{lj,k}=\delta_{ml}Q_{lj,k}=Q_{mj,k}. $$

Without tensor notation (almost)

Using property of rotation matrix: $QQ^T = Q^TQ=I$, we show that matrix $R_k = Q^T\partial Q/\partial x_k$ is antisymmetric: $$\frac{\partial}{\partial x_k}(Q^TQ) = Q^T\frac{\partial Q}{\partial x_k}+\frac{\partial Q^T}{\partial x_k}Q=Q^T\frac{\partial Q}{\partial x_k} + \left(Q^T\frac{\partial Q}{\partial x_k}\right)^T=0$$

Let's take consider one element $e$ of matrix $R_k$ in i-th row and j-th column. By definition of matrix multiplication we have: $$e= \sum_{\alpha=1}^{3} \frac{\partial y_\alpha}{\partial x_i}\frac{\partial^2 y_\alpha}{\partial x_j\partial x_k}.$$ Since matrix $R_k$ is antisymmetric, we can swap index $i$ with one of the indices $j$ or $k$ and multiply by $-1$: $$ \sum_{\alpha=1}^{3} \frac{\partial y_\alpha}{\partial x_i}\frac{\partial^2 y_\alpha}{\partial x_j\partial x_k} = -\sum_{\alpha=1}^{3} \frac{\partial y_\alpha}{\partial x_j}\frac{\partial^2 y_\alpha}{\partial x_i\partial x_k} = \sum_{\alpha=1}^{3} \frac{\partial y_\alpha}{\partial x_k}\frac{\partial^2 y_\alpha}{\partial x_i\partial x_j} = -\sum_{\alpha=1}^{3} \frac{\partial y_\alpha}{\partial x_i}\frac{\partial^2 y_\alpha}{\partial x_j\partial x_k}$$

That can be only possible when $e=0$. Since we had a full freedom in choosing $i$, $j$ and $k$, we conclude that all matrices $R_k=0$.

Finally, we see that: $$ QR_k = QQ^T\frac{\partial Q}{\partial x_k} = \frac{\partial Q}{\partial x_k} = 0. $$

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  • $\begingroup$ thanks for your response. But I mentioned only rank 2 matrices. For me this is a little bit confusing. $\endgroup$
    – KratosMath
    Commented Sep 14, 2018 at 16:39
  • $\begingroup$ I have added the explanation without tensor notation. Please, let me know what confuses you in it if it is not clear. $\endgroup$ Commented Sep 14, 2018 at 17:23

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