Editing Scalar potential (section)

==Integrability conditions==
If {{math|'''F'''}} is a [[conservative vector field]] (also called ''irrotational'', ''[[Curl (mathematics)|curl]]-free'', or ''potential''), and its components have [[continuous function|continuous]] [[partial derivative]]s, the '''potential''' of {{math|'''F'''}} with respect to a reference point {{math|'''r'''{{sub|0}}}} is defined in terms of the [[line integral]]:
<math display="block">V(\mathbf r) = -\int_C \mathbf{F}(\mathbf{r})\cdot\,d\mathbf{r} = -\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt,</math>
where {{mvar|C}} is a [[Parametrization (geometry)|parametrized]] path from {{math|'''r'''{{sub|0}}}} to {{math|'''r'''}},
<math display="block">\mathbf{r}(t), a\leq t\leq b, \mathbf{r}(a)=\mathbf{r_0}, \mathbf{r}(b)=\mathbf{r}.</math>

The fact that the line integral depends on the path {{mvar|C}} only through its terminal points {{math|'''r'''{{sub|0}}}} and {{math|'''r'''}} is, in essence, the '''path independence property''' of a conservative vector field. The [[Gradient theorem|fundamental theorem of line integrals]] implies that if {{mvar|V}} is defined in this way, then {{math|1='''F''' = –∇''V''}}, so that {{mvar|V}} is a scalar potential of the conservative vector field {{math|'''F'''}}. Scalar potential is not determined by the vector field alone: indeed, the gradient of a function is unaffected if a constant is added to it. If {{mvar|V}} is defined in terms of the line integral, the ambiguity of {{mvar|V}} reflects the freedom in the choice of the reference point {{math|'''r'''{{sub|0}}}}.