# Basic Summation Indices

If $j^\mu = ( j^0 , \vec{j} )$, why does

$\partial_\mu j^\mu = \partial_0 j^0 + \vec{\nabla} \cdot \vec{j}$

surely when you take a dot product of four vectors you get a subtraction as in
$a^\mu b_\mu = a^0 b_0 - \vec{a} \cdot \vec{b}$

Maybe I'm forgetting something

## Answers and Replies

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Suppose your metric is (+1,-1,-1,-1). Then

$$a^\mu b_\mu= a^0b_0+a^ib_i=a^0b^0-\sum_{i=1}^3a^ib^i$$

You flip the sign when you rise space-like indices. But with your continuity equation

$$\partial_\mu j^\mu=\partial_0 j^0+\sum_{i=1}^3\partial_i j^i$$

there is no reason to rise or lower the indices.

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