differential rate of strain

Almost an academic author: Appendix B: Representation of Dyadics via Geometric Algebra

May 29, 2022 math and physics play , , ,

I corresponded for a bit with the author of a paper on variations and inconsistencies with dyadic notation that is used in some fluid flow and other problems.  In the end I ended up contributing an appendix (Appendix B: Representation of Dyadics via Geometric Algebra) to their paper, which has been submitted to an academic journal (I forget which one.)

I’ve written thousands of pages of independent musings as blog posts, and even a book (self published, and also available for free in pdf form), but this is the first time anything that I’ve written has ended up in an academic journal.

You can find a preprint here on arxiv

Verifying the GA form for the symmetric and antisymmetric components of the different rate of strain.

March 8, 2022 math and physics play , , , , , , , ,

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

We found geometric algebra representations for the symmetric and antisymmetric components for a gradient-vector direct product. In particular, given
\begin{equation}\label{eqn:tensorComponents:20}
d\Bv = d\Bx \cdot \lr{ \spacegrad \otimes \Bv }
\end{equation}
we found
\begin{equation}\label{eqn:tensorComponents:40}
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2} d\Bx \cdot \lr{
\spacegrad \otimes \Bv
+
\lr{\spacegrad \otimes \Bv }^\dagger
} \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv }
+
\gpgradeone{ \spacegrad d\Bx \Bv }
},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:tensorComponents:60}
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2} d\Bx \cdot \lr{
\spacegrad \otimes \Bv

\lr{\spacegrad \otimes \Bv }^\dagger
} \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv }

\gpgradeone{ d\Bx \Bv \spacegrad }
}.
\end{aligned}
\end{equation}

Let’s expand each of these in coordinates to verify that these are correct. For the symmetric component, that is
\begin{equation}\label{eqn:tensorComponents:80}
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2}
\lr{
dx_i \partial_j v_j \Be_i
+
\partial_j dx_i v_k \gpgradeone{ \Be_j \Be_i \Be_k }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i
+
\partial_j v_k \lr{ \delta_{ji} \Be_k + \lr{ \Be_j \wedge \Be_i } \cdot \Be_k }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i
+
\partial_j v_k \lr{ \delta_{ji} \Be_k + \delta_{ik} \Be_j – \delta_{jk} \Be_i }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i
+
\partial_i v_k \Be_k
+
\partial_j v_i \Be_j

\partial_j v_j \Be_i
} \\
&=
\inv{2} dx_i
\lr{
\partial_i v_k \Be_k
+
\partial_j v_i \Be_j
} \\
&=
dx_i \inv{2} \lr{ \partial_i v_j + \partial_j v_i } \Be_j.
\end{aligned}
\end{equation}
Sure enough, we that the product contains the matrix element of the symmetric component of \( \spacegrad \otimes \Bv \).

Now let’s verify that our GA antisymmetric tensor product representation works out.
\begin{equation}\label{eqn:tensorComponents:100}
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2}
\lr{
dx_i \partial_j v_j \Be_i

dx_i \partial_k v_j \gpgradeone{ \Be_i \Be_j \Be_k }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i

\partial_k v_j
\lr{ \delta_{ij} \Be_k + \delta_{jk} \Be_i – \delta_{ik} \Be_j }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i

\partial_k v_i \Be_k

\partial_k v_k \Be_i
+
\partial_i v_j \Be_j
} \\
&=
\inv{2} dx_i
\lr{
\partial_i v_j \Be_j

\partial_k v_i \Be_k
} \\
&=
dx_i
\inv{2}
\lr{
\partial_i v_j

\partial_j v_i
}
\Be_j.
\end{aligned}
\end{equation}
As expected, we that this product contains the matrix element of the antisymmetric component of \( \spacegrad \otimes \Bv \).

We also found previously that \( \BOmega \) is just a curl, namely
\begin{equation}\label{eqn:tensorComponents:120}
\BOmega = \inv{2} \lr{ \spacegrad \wedge \Bv } = \inv{2} \lr{ \partial_i v_j } \Be_i \wedge \Be_j,
\end{equation}
which directly encodes the antisymmetric component of \( \spacegrad \otimes \Bv \). We can also see that by fully expanding \( d\Bx \cdot \BOmega \), which gives
\begin{equation}\label{eqn:tensorComponents:140}
\begin{aligned}
d\Bx \cdot \BOmega
&=
dx_i \inv{2} \lr{ \partial_j v_k }
\Be_i \cdot \lr{ \Be_j \wedge \Be_k } \\
&=
dx_i \inv{2} \lr{ \partial_j v_k }
\lr{
\delta_{ij} \Be_k

\delta_{ik} \Be_j
} \\
&=
dx_i \inv{2}
\lr{
\lr{ \partial_i v_k } \Be_k

\lr{ \partial_j v_i }
\Be_j
} \\
&=
dx_i \inv{2}
\lr{
\partial_i v_j – \partial_j v_i
}
\Be_j,
\end{aligned}
\end{equation}
as expected.