Understanding the Continuity Equation in Special Relativity

[latentcorpse]

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

 

[arkajad]

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.