What is the property of a counterpart of conservative vector field in Minkowski space?

Question

As we know, a conservative vector field is defined by a vector field ${\displaystyle \vec {v} :U\to \mathbb {R} ^{n}}$, where ${\displaystyle \vec {v} =\nabla \varphi}$. It is also an irrotational vector field ($\nabla \times \vec v = 0$). The line integral of such field is path independent. The gravitational field, electric field are all such kind of fields.

And, what if we have a vector field in Minkowski space with similar definition: $j^\mu = \partial^\mu \varphi$, then what will be properties for this 4-vector field $\vec j$, is there some examples for such field?

Answer 1

The analogue is a closed differential form. This works not only over Minkowski space, but over any curved manifold of any dimension. In fact, the metric is not neccessary to establish this result.

Given a closed diffferential form $\alpha$ and which means it's differential vanishes:

$d\alpha = 0$

Then the Poincare lemma says locally it is exact. That is given any point $p$ of the manifold, there is an open $U$ containing this point such there is a differential form $\beta$ on $U$ such that:

$d\beta = \alpha_U$

Personally, I don't like the name 'forms' in physics as the terminology comes from mathematics whereas the term 'field' is traditional in physics. Since forms are built from the cotangent bundle, the term 'cofield' suggests itself. Thus, I would say a closed cofield is the correct analogue and so a conservative cofield is simply a closed cofield. Then the Poincare lemma says that locally, a conservative cofield has a copotential (which is again a cofield, but of a degree lower).

Moreover, the analogue of the property you listed above, ie path independency, also holds in this context where itvis proven via Stokes theorem. To be more precise:

First, let $\alpha$ be a conservative cofield. Let $M$ be a closed manifold, hence there is another manifold $N$ whose boundary is $M$. In symbols, $\partial N = M$. Then:

$\int_M \alpha = \int_{\partial N} \alpha = \int_N d\alpha = \int_N 0 = 0$

Now notice that a circle on the surface of a ball partitions that surface into two pieces that is diffeomorphic to disks. The analogous property holds for higher dimensional spheres. Whilst in ordinary language spheres and balls are the same, for mathematicians spheres are what they call the surface of balls.

More precisely, let $S$ be a sphere of dimension the degree of $\alpha$. Then every sphere $B$ of dimension one lower of $S$, partitions $S$ into two balls with boundary $B$ which we write as $S^+$ and $S^-$.

Now let $\alpha$ be a conservative cofield. Then:

$\int_S \alpha = 0$

by the above. And by the decomposition of $S$, we also have:

$\int_S = \int_{S^+ \sqcup S^-} \alpha = (\int_{S^+} \alpha) + (\int_{S^-} \alpha)$      Thus:

$\int_{S^+} \alpha= - \int_{S^-} \alpha = \int_{-S^-} \alpha$

Now, in 1d the circle $S:=S^1$ is partitioned into two intervals $S^{1+}$ and $S^{1-}$ by the zero sphere, $S^0$ which is simply two points. And hence the final formula above is directly analogous to path independence in this special case.