Strain Tensor Based on Clifford Algebra

[Arash1950]

Let u=u_i e_i be the displacement field of a continuum body. Then the displacement gradient tensor H based on classical formulation is given by H=grad u = u_{i,j} e_i \otimes e_j, where \otimes represents tensor product.H is decomposed intro two parts. Namely, H=epsilon+Omega. The infinitesimal strain tensor is given by epsilon=(H+H^t)/2, where H^t=transose(H). In component form, we have epsilon_{ij}=(u_{i,j}+u_{j,i})/2. On the other hand, Omega=(H-H^t)/2 describes the rigid body rotation during deformation.

In Clifford analysis, the gradient operator (known as the Dirac operator) is defined as D=e_i \partial / \partial x_i. The displacement gradient MAY be written as H=Du=D.u+D^u (I think this is incorrect).

The first term is indeed D.u=div u that describes the dilatation (volume change).The second term is similar to curl(u) that describes rigid body rotation [it is related to Omega. Basically curl(u) is twice the axial vector of Omega].

I have also seen relations of the form epsilon=(Du+uD)/2, which is absolutely incorrect. indeed (Du+uD)/2=div u=trace(epsilon)

It seems that the gradient operator in Clifford analysis cannot produce the strain tensor in elasticity. Probably, a dyadic product of the form H=D \otimes u is needed. But as far as I know, this product is not standard in Clifford analysis.

I have not seen any good relation in papers and books. Moreover, ChatGPT was not able to provide useful information.

My questions:

• How to generate epsilon_{ij}=(u_{i,j}+u_{j,i})/2 based on the gradient operation in Clifford analysis?
• What is the general definition of the Gradient operator in Clifford analysis? It seems that the Dirac operator "D", as mentioned above, does not work properly.

 

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