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How to derive Dirac $\gamma$ matrices in spherical polar coordinates?

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    $\begingroup$ Sure, see here, or XV.7 here , the usual way. But why? $\endgroup$ Commented Jan 31, 2021 at 20:08
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    $\begingroup$ In english. $\endgroup$ Commented Jan 31, 2021 at 20:14
  • $\begingroup$ Thank you sir.Actually I have seen the arxiv paper but i couldnt reach the expression of matrices in this ..I think I am missing something...tetrad basis are 1,1,$\frac{1}{r}$,$\frac{1}{rsin\theta}$.If we multiply the cartesian gamma matrices with these basis then how the term like exp(i$\phi$) will come? $\endgroup$
    – Novice
    Commented Feb 2, 2021 at 14:04
  • $\begingroup$ Actually I don't have so much knowledge about this..honestly i have studied dirac equation in cartesian coordinates . I am looking into these things for the first time. $\endgroup$
    – Novice
    Commented Feb 2, 2021 at 14:10

1 Answer 1

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In flat space, you only need to rotate bases in the spacelike part, and leave $\gamma^0$ alone; it stays the same.

That is, you just express the spacelike dot product $\vec \gamma \cdot \nabla$ in spherical coordinates, $$ \vec \gamma= \hat{\mathbf r} \gamma_ r + \hat{\boldsymbol\theta} \gamma_\theta + \hat{\boldsymbol\varphi} \gamma _\varphi, \\ \nabla = \hat{\mathbf r} \partial_ r + \hat{\boldsymbol\theta} {1 \over r}{\partial_\theta} + \hat{\boldsymbol\varphi}{1 \over r\sin\theta}{ \partial _\varphi}, $$ where $$ \begin{align} \hat{\mathbf r} &= \sin \theta \cos \varphi \,\hat{\mathbf x} + \sin \theta \sin \varphi \,\hat{\mathbf y} + \cos \theta \,\hat{\mathbf z}, \\ \hat{\boldsymbol\theta} &= \cos \theta \cos \varphi \,\hat{\mathbf x} + \cos \theta \sin \varphi \,\hat{\mathbf y} - \sin \theta \,\hat{\mathbf z}, \\ \hat{\boldsymbol\varphi} &= - \sin \varphi \,\hat{\mathbf x} + \cos \varphi \,\hat{\mathbf y} \end{align} $$ an (orthogonal!) rotation , $$ \begin{pmatrix} \hat{\mathbf r} \\ \hat{\boldsymbol\theta} \\ \hat{\boldsymbol\varphi} \end{pmatrix}= R \begin{pmatrix} \hat{\mathbf x} \\ \hat{\mathbf y}\\ \hat{\mathbf z}\end{pmatrix} \\ R =\begin{pmatrix} \sin\theta\cos\varphi&\sin\theta\sin\varphi& \cos\theta\\ \cos\theta\cos\varphi&\cos\theta\sin\varphi&-\sin\theta\\ -\sin\varphi&\cos\varphi &0 \end{pmatrix},\\ $$ hence $$ \begin{pmatrix} \gamma_ r \\ \gamma_ \theta \\ \gamma_\varphi \end{pmatrix}= R \begin{pmatrix} \gamma_ 1 \\ \gamma_2\\ \gamma_3 \end{pmatrix}, $$ the very matrices (5) and (6) of Dzhunushaliev cited, (XV-16 of Vaudon), dependent on $\varphi$, alright.

As you might expect for all directional gradients in spherical coords, $$ \vec \gamma \cdot \nabla= \gamma_ r \partial_ r + \gamma_\theta {1 \over r}{\partial_\theta} + \gamma _\varphi {1 \over r\sin\theta}{ \partial _\varphi}~~, $$ which probably addresses your tetrad puzzlement, in practice.

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