The dirac equation;$$(i\gamma^\mu\partial_{\mu} - m)\psi=0 $$ is just; $$(i\gamma^{0}\partial_{0} - i\gamma^{i}\partial_{i} - m)\psi=0 $$ in a (+,---) metric right?
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The expression $A^{\mu}B_{\mu}$ simply means that $$A^{\mu}B_{\mu}=A^{0}B_{0}+A^{1}B_{1}+A^{2}B_{2}+A^{3}B_{3}$$ Using the Minkowski metric with signature $(+---)$ you write this as $$A^{\mu}B_{\mu}=A^{\mu}\eta_{\mu\nu}B^{\nu}=A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}$$ The metric simply tells you have how the components of a vector and its dual vector (covariant vector) are associated. With this specific metric and signature we have $B_0=B^0$ and $B_i=-B^i$, $i=1,2,3$. In Dirac equation we have $$\gamma^{\mu}\partial_{\mu}=\gamma^{0}\partial_{0}+\gamma^{1}\partial_{1}+\gamma^{2}\partial_{2}+\gamma^{3}\partial_{3}=\gamma^{0}\partial_{0}+\gamma^{i}\partial_{i}=\\ =\gamma^{\mu}\eta_{\mu\nu}\partial^{\nu}=\gamma^{0}\partial^{0}-\sum_{i=1}^{3}\gamma^{i}\partial^{i}$$ therefore $$(i\gamma^{0}\partial_{0} + i\gamma^{i}\partial_{i} - m)\psi=0 $$ $$%(i\gamma^{0}\partial^{0} - i\sum_{i=1}^{3}\gamma^{i}\partial^{i} - m)\psi=0 $$ or $$(i\gamma^{0}\partial^{0} - i\sum_{i=1}^{3}\gamma^{i}\partial^{i} - m)\psi=0 $$ $$\partial_0=\partial^0=\frac{\partial}{\partial t},\hspace{0.2cm}\partial_i=-\partial^i=\vec{\nabla}_i$$ |
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No, it's with a plus. You don't need to change the sign unless both indices are upper (or lower). |
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