I am trying to calculate the dispersion relation for a fermion in a gravitational field. So far, I have computed the equation of motion, but I am stuck trying to figure out a determinant I just can't get.... I am having trouble calculating the final determinant.

My derivation is below:

Equation of motion:

\begin{equation} \left(i\gamma^a\partial_a - m -\frac{1}{2} \gamma^a\gamma^5B_a\right)\Psi = 0 \end{equation}

If we take $\Psi = u(\vec{p_a})e^{-ip_ax^a}$ as our ansatz we get the result \begin{align} \left(i\gamma^a(-ip_a) -m -\frac 12\gamma^a\gamma^5B_a \right)u(\vec{p_a}) &= 0 \\ \left(\gamma^ap_a -m -\frac 12\gamma^a\gamma^5B_a \right)u(\vec{p_a}) &= 0 \end{align} We multiply this by expression by $\left(\gamma^ap_a +m -\frac 12\gamma^a\gamma^5B_a \right)$ and expand to get \begin{equation} \bigg[\gamma^b\gamma^ap_ap_b -m^2 + m(\gamma^ap_a - \gamma^bp_b) - \frac{m}{2}(\gamma^b\gamma^5B_b-\gamma^a\gamma^5B_a) - \gamma^b\gamma^a\gamma^5B_aP_b ~~ - \notag \end{equation} \begin{equation} -~~ \gamma^a\gamma^5\gamma^bB_bP_a + \frac 14 \gamma^b\gamma^5\gamma^a\gamma^5B_aB_b\bigg]u(\vec{p}_a) = 0 \end{equation} \begin{equation} \bigg[p^ap_a -m^2 + \frac 12[\gamma^a,\gamma^b]\gamma^5B_aP_b - \frac 14B^aB_a\bigg]u(\vec{p}_a) = 0 \end{equation} This is a matrix times spinor equal to zero, meaning the determinant of the matrix must be zero (since the spinor being zero is of no interest to us). Therefore:

$$ \text{det}\bigg[(p^ap_a -m^2 - \frac 14B^aB_a){\mathcal{\large\mathbb{1}}}+ \frac 12[\gamma^a,\gamma^b]\gamma^5B_aP_b\bigg] = 0 $$

Anyone have any tips on calculating such a monster? Or even a different way of finding the relation?

  • 2
    $\begingroup$ I think you may choose a frame where the spatial part of $p$ is zero (so only $p^0 $ is non-zero) and use the [Weyl Basis](en.wikipedia.org/wiki/Gamma_matrices#Weyl_basis). The interesting matrices $[\gamma^0, \gamma^i]\gamma^5$ are block-diagonal, so it is easy to find the determinant (Each block is a matrix $\lambda ~ \mathbb{Id} + \vec \mu . \vec \sigma$, with determinant $\lambda^2-\vec \mu^2$). Then, you have to express the final result in a Lorentz-invariant way, using $p^2, B^2, (p.B)^2$ $\endgroup$ – Trimok Dec 5 '13 at 21:02

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