tensor product

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.

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

November 29, 2018 phy2403 , , , , ,

[Click here for a PDF of this notes with full details.]

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”
    \begin{equation}\label{eqn:qftLecture21:99}
    v^s(p)
    =
    \begin{bmatrix}
    \sqrt{p \cdot \sigma} \eta^s \\
    -\sqrt{p \cdot \overline{\sigma}} \eta^s \\
    \end{bmatrix},
    \end{equation}
    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*}
  • Dirac adjoint orthogonality conditions.
    \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},
    \qquad
    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*}