A vector field is conservative when it can be obtained from a scalar-valued potential function. If a differentiable scalar field $\phi$ satisfies

$$\mathbf F=\nabla\phi,$$

then $\phi$ is a scalar potential for $\mathbf F$.

This structure has a powerful consequence: line integrals of $\mathbf F$ depend only on their endpoints, not on the path taken between them.

The fundamental theorem for line integrals

Let a curve $C$ run from point $A$ to point $B$ and be parametrized by

$$\mathbf r(t)=(x(t),y(t),z(t)),\qquad a\le t\le b.$$

Along the curve, the potential becomes the single-variable function $\phi(x(t),y(t),z(t))$. Differentiating it gives

$$\frac{d}{dt}\phi(\mathbf r(t))

\frac{\partial\phi}{\partial x}x'(t) + \frac{\partial\phi}{\partial y}y'(t) + \frac{\partial\phi}{\partial z}z'(t).$$

In vector notation,

$$\frac{d}{dt}\phi(\mathbf r(t)) =\nabla\phi(\mathbf r(t))\cdot\mathbf r'(t) =\mathbf F(\mathbf r(t))\cdot\mathbf r'(t).$$

Integrating from $a$ to $b$ therefore gives

$$\boxed{ \int_C\mathbf F\cdot d\mathbf r =\phi(B)-\phi(A)}. $$

This is the fundamental theorem for line integrals. Instead of accumulating the field along every point of the path, a conservative field lets us compare the potential at the two endpoints.

Path independence and closed loops

If two curves $C_1$ and $C_2$ connect the same points $A$ and $B$, then

$$\int_{C_1}\mathbf F\cdot d\mathbf r =\phi(B)-\phi(A) =\int_{C_2}\mathbf F\cdot d\mathbf r.$$

The integral is therefore path-independent.

For a closed curve, the starting and ending point are the same, so

$$\oint_C\mathbf F\cdot d\mathbf r=0.$$

Conversely, on a connected region where line integrals are path-independent, a potential can be constructed by choosing a reference point $A$ and defining

$$\phi(P)=\int_A^P\mathbf F\cdot d\mathbf r.$$

Path independence makes this definition unambiguous.

Finding a potential from a field

Consider

$$\mathbf F(x,y,z)=(2xy+3,;x^2+2z,;2y).$$

We seek $\phi$ such that

$$\nabla\phi=\mathbf F.$$

From

$$\frac{\partial\phi}{\partial x}=2xy+3,$$

integrating with respect to $x$ gives

$$\phi=x^2y+3x+g(y,z),$$

where $g$ may still depend on $y$ and $z$.

Now compare the $y$-component:

$$\frac{\partial\phi}{\partial y}=x^2+\frac{\partial g}{\partial y} =x^2+2z.$$

Thus

$$\frac{\partial g}{\partial y}=2z,$$

so

$$g(y,z)=2yz+h(z).$$

Finally,

$$\frac{\partial\phi}{\partial z}=2y+h'(z)=2y,$$

which gives $h'(z)=0$. Therefore one potential is

$$\boxed{\phi(x,y,z)=x^2y+3x+2yz}.$$

Any additive constant would give the same vector field because the gradient of a constant is zero.

We can now evaluate a line integral without choosing or parametrizing a path. From $A=(0,0,0)$ to $B=(1,2,3)$,

$$\int_A^B\mathbf F\cdot d\mathbf r =\phi(1,2,3)-\phi(0,0,0) =17.$$

Sign conventions in applications

Mathematically, writing $\mathbf F=\nabla\phi$ is convenient. In physics a conservative force field is often written

$$\mathbf F=-\nabla U,$$

so the force points toward decreasing potential energy. Electric potential uses the same minus-sign convention for the electrostatic field. This is a sign convention, not a different mathematical structure.

Conservative fields connect a local description, the gradient at each point, with a global description, path-independent accumulation between endpoints.