- #1
etotheipi
Apparently the contravariant components of the 4-gradient ##\partial## are ##\partial^{\mu} = \left(\partial_t, -\nabla \right)## where ##\nabla## is the usual 3-gradient. We can use the metric to lower the index like ##\partial_{\mu} = \eta_{\mu \nu} \partial^{\nu}## and if the signature is ##(+,-,-,-)## the space signs get flipped, so we should get ##\partial_{\mu} = (\partial_t, \nabla)##.
Why then in this Wikipedia article do they use ##\partial_{\nu} = \left(\partial_t, -\nabla \right)## when they write$$\partial \cdot J = \partial_{\nu} J^{\nu} = \left(\partial_t, -\nabla \right) \cdot (\rho, \vec{j}) = 0$$It's the right way round, since when we use the Minkowski inner product we recover the continuity equation ##\partial_t \rho = \nabla \cdot \vec{j}##, but they put a negative sign inside ##\partial_{\nu}##. So is it actually the other way around, i.e. ##\partial^{\mu} = \left(\partial_t, \nabla \right)## and ##\partial_{\mu} = \left(\partial_t, -\nabla \right)##? Sorry if I missed something!
Why then in this Wikipedia article do they use ##\partial_{\nu} = \left(\partial_t, -\nabla \right)## when they write$$\partial \cdot J = \partial_{\nu} J^{\nu} = \left(\partial_t, -\nabla \right) \cdot (\rho, \vec{j}) = 0$$It's the right way round, since when we use the Minkowski inner product we recover the continuity equation ##\partial_t \rho = \nabla \cdot \vec{j}##, but they put a negative sign inside ##\partial_{\nu}##. So is it actually the other way around, i.e. ##\partial^{\mu} = \left(\partial_t, \nabla \right)## and ##\partial_{\mu} = \left(\partial_t, -\nabla \right)##? Sorry if I missed something!