Exterior derivative of a 2-form in Minkowski space

Question

I'm working on a problem in Minkowski space that involves the exterior product of a two-form. $$d(A\ dx\wedge dt)$$ Were this a problem in simple 3D space, the obvious answer would be to take the derivative of $A$ with respect to the third coordinate in the space. However, I am unsure of how to do this for a 4D space like the one I'm using.

Answer 1

The fact that you're living in Minkowski space is irrelevant, as the exterior derivative is not an operation defined from the metric tensor. If your coordinates are $(x,y,z,t)$ and $A$ depends on those coordinates, then $${\rm d}(A\,{\rm d}x \wedge {\rm d}t) = \partial_yA \,{\rm d}y\wedge{\rm d}x \wedge {\rm d}t + \partial_zA\,{\rm d}z \wedge \,{\rm d}x \wedge {\rm d}t,$$since ${\rm d}x \wedge {\rm d}x = {\rm d}t \wedge {\rm d}t = 0$.