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

[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
\label{eqn:tensorComponents:20}
d\Bv = d\Bx \cdot \lr{ \spacegrad \otimes \Bv }

we found
\label{eqn:tensorComponents:40}
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2} d\Bx \cdot \lr{
+
} \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv }
+
},
\end{aligned}

and
\label{eqn:tensorComponents:60}
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2} d\Bx \cdot \lr{

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

}.
\end{aligned}

Let’s expand each of these in coordinates to verify that these are correct. For the symmetric component, that is
\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}

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.
\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}

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
\label{eqn:tensorComponents:120}
\BOmega = \inv{2} \lr{ \spacegrad \wedge \Bv } = \inv{2} \lr{ \partial_i v_j } \Be_i \wedge \Be_j,

which directly encodes the antisymmetric component of $$\spacegrad \otimes \Bv$$. We can also see that by fully expanding $$d\Bx \cdot \BOmega$$, which gives
\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}

as expected.

## PHY2403H Quantum Field Theory. Lecture 21, Part I: Dirac equation solutions, orthogonality conditions, direct products. Taught by Prof. Erich Poppitz

### DISCLAIMER: Rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.

## Overview.

See the PDF above for full notes for the first part of this particular lecture. We covered

• Normalization:
\begin{equation*}
u^{r \dagger} u^{s}
= 2 p_0 \delta^{r s}.
\end{equation*}
• Products of $$p \cdot \sigma, p \cdot \overline{\sigma}$$
\begin{equation*}
(p \cdot \sigma) (p \cdot \overline{\sigma})
=
(p \cdot \overline{\sigma}) (p \cdot \sigma)
= m^2.
\end{equation*}
• Adjoint orthogonality conditions for $$u$$
\begin{equation*}
\overline{u}^r(\Bp) u^{s}(\Bp) = 2 m \delta^{r s}.
\end{equation*}
• Solutions in the $$e^{i p \cdot x}$$ “direction”
\label{eqn:qftLecture21:99}
v^s(p)
=
\begin{bmatrix}
\sqrt{p \cdot \sigma} \eta^s \\
-\sqrt{p \cdot \overline{\sigma}} \eta^s \\
\end{bmatrix},

where $$\eta^1 = (1,0)^\T, \eta^2 = (0,1)^\T$$.
• $$v$$ normalization
\begin{equation*}
\begin{aligned}
\overline{v}^r(p) v^s(p) &= – 2 m \delta^{rs} \\
v^{r \dagger}(p) v^s(p) &= 2 p^0 \delta^{rs}.
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
\overline{u}^r(p) v^s(p) &= 0 \\
\overline{v}^r(p) u^s(p) &= 0.
\end{aligned}
\end{equation*}
• Dagger orthogonality conditions.
\begin{equation*}
\begin{aligned}
v^{r \dagger}(-\Bp) u^s(\Bp) &= 0 \\
u^{r\dagger}(\Bp) v^s(-\Bp) &= 0.
\end{aligned}
\end{equation*}
• Tensor product.

Given a pair of vectors
\begin{equation*}
x =
\begin{bmatrix}
x_1 \\
\vdots \\
x_n \\
\end{bmatrix},
y =
\begin{bmatrix}
y_1 \\
\vdots \\
y_n \\
\end{bmatrix},
\end{equation*}
the tensor product is the matrix of all elements $$x_i y_j$$

\begin{equation*}
x \otimes y^\T =
\begin{bmatrix}
x_1 \\
\vdots \\
x_n \\
\end{bmatrix}
\otimes
\begin{bmatrix}
y_1 \cdots y_n
\end{bmatrix}
=
\begin{bmatrix}
x_1 y_1 & x_1 y_2 & \cdots & x_1 y_n \\
x_2 y_1 & x_2 y_2 & \cdots & x_2 y_n \\
x_3 y_1 & \ddots & & \\
\vdots & & & \\
x_n y_1 & \cdots & & x_n y_n
\end{bmatrix}.
\end{equation*}

• Direct product relations.
\begin{equation*}
\begin{aligned}
\sum_{s = 1}^2 u^s(p) \otimes \overline{u}^s(p) &= \gamma \cdot p + m \\
\sum_{s = 1}^2 v^s(p) \otimes \overline{v}^s(p) &= \gamma \cdot p – m \\
\end{aligned}
\end{equation*}