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I was performing some calculations with Geometric Algebra today and I've found myself stuck with a simple operation that I don't know how to answer.

Considering that you have the usual vector derivative $$\nabla = \gamma^\mu \partial_\mu = \gamma_0(\partial_0 + \vec{\nabla}) =(\partial_0 - \vec{\nabla})\gamma_0,\tag{1}$$ and performing the scalar multiplication by $\gamma_0$. One one side one would have

$$ \gamma_0 \cdot \nabla = \partial_0\tag{2}$$

but on the other hand, if one performs a time-split before the multiplication

$$\gamma_0 \cdot \nabla = \gamma_0 \cdot \gamma_0(\partial_0 + \vec{\nabla}) = \partial_0 + \vec{\nabla}\tag{3}$$

Both results don't seem compatible to be, so I would like to ask. What I am missing?

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  • $\begingroup$ @NDewolf It comes from the anticommutativity of the $\gamma_\mu$. Because $\vec{\nabla} = \gamma_0 \wedge \gamma_i \partial_i$ $\endgroup$ Jan 25 at 17:03

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I believe that there is no incompatibility here, everything is to be expected. What is happening is that you are comparing two cases of operator precedences and finding they are not equivalent, which is fine.

In the first case, you are considering the direct application of the inner product of $\gamma_0$ and $\nabla$, $$ \gamma_0\cdot\nabla=\gamma_0\cdot\gamma_\mu\partial_\mu=\partial_0.\tag{1} $$ However, in the second case, you are inserting $\gamma_0^2=1$ in between the two terms, then multiplying this to the left and to the right: \begin{align} \gamma_0\cdot\nabla &= \gamma_0\cdot\left(\gamma_0^2\right)\nabla \\ &= \left(\gamma_0\cdot\gamma_0\right)\left(\gamma_0\nabla\right) \tag{2}\\ &= \gamma_0\nabla \end{align} And this is different from Equation (1) here because you've changed the order of products between the first and second lines. Instead of dotting the frame with the gradient (Eq 1 here), you've dotted frame with itself and multiplied that to the spacetime split gradient (Eq 2).

So it seems to me that all that you have discovered is that, $$ a\cdot bc\equiv\left(a\cdot b\right)c\neq a\cdot\left(bc\right) $$ which is fine, since product orders matter.

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  • $\begingroup$ I supposed that it was some kind of order multiplication mistake, but I could not spot it. Thank you for your answer. $\endgroup$ Jan 27 at 9:16

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