 # on the Dirac equation, expands the gamma^nu d_nu term as: gamma^nu d_nu=gamma^0 d/dt + gamma * nabla where gamma=(gamma^1, gamma^2, gamma^3), but to my knolwdge gamma^nu d_nu=gamma^nu mu_(nu v) d^v=gamma^0 d/dt + gamma * nabla Paxton Hoffman 2022-07-22 Answered
on the Dirac equation, expands the ${\gamma }^{\mu }{\mathrm{\partial }}_{\mu }$ term as:
${\gamma }^{\mu }{\mathrm{\partial }}_{\mu }={\gamma }^{0}\frac{\mathrm{\partial }}{\mathrm{\partial }t}+\stackrel{\to }{\gamma }\cdot \stackrel{\to }{\mathrm{\nabla }}$
where $\stackrel{\to }{\gamma }=\left({\gamma }^{1},{\gamma }^{2},{\gamma }^{3}\right)$, but to my knowledge,
${\gamma }^{\mu }{\mathrm{\partial }}_{\mu }={\gamma }^{\mu }{\eta }_{\mu \nu }{\mathrm{\partial }}^{\nu }={\gamma }^{0}\frac{\mathrm{\partial }}{\mathrm{\partial }t}-\stackrel{\to }{\gamma }\cdot \stackrel{\to }{\mathrm{\nabla }}$
using the convention ${\eta }_{\mu \nu }=\mathrm{diag}\left(+,-,-,-\right)$.
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Yes. You are missing the fact that he is using the convention
$\mathrm{\nabla }=\left({\mathrm{\partial }}_{1},{\mathrm{\partial }}_{2},{\mathrm{\partial }}_{3}\right)$
as opposed to
$\mathrm{\nabla }=\left({\mathrm{\partial }}^{1},{\mathrm{\partial }}^{2},{\mathrm{\partial }}^{3}\right)$
The first convention is by far the most common in my experience.