^^ [[Schutz - 2 - Differentiable Manifolds and Tensors|2. Differentiable Manifolds and Tensors]] <<[[Schutz - 2.10 - Examples of Fiber Bundles|2.10 Examples of Fiber Bundles]] | [[Schutz - 2.13 - Exponentiation of the Operator ddlambda|2.12 Exponentiation of the Operator d/d\lambda]]>> # Vector Fields and Integral Curves Every [[Schutz - 2.5 - Curves#^5c64cb|curve]] has a [[Schutz - 2.7 - Vectors and Vector Fields#^ab2ebc|vector]] at every point, defining a [[Schutz - 2.7 - Vectors and Vector Fields#^a9b237|vector field]], which can be thought of as as cross-section on the [[Schutz - 2.9 - Fiber Bundles#^7a66b2|tangent bundle]]. The question now arises, given a vector field, is it possible to start at a point $P$ and find a curve whose tangent vector is always the vector field at whatever point the curve passes through? This turns out to be true for $C^1$ vector fields. These curves are known as *integral curves* of the vector field. Let the components of the vector field be $V^i(P)$, in some coordinate system $\{x^i\}$, we have $V^i(P) = v^i(x^j)$. We can write the components to the tangent vector to a curve with parameter $\lambda$ as: $ \frac{d x^i}{d\lambda} = v^i(x^j) $ which is a first order ODE with a unique solution which always exists in the neighborhood of the initial point $P$ by the existence and uniqueness theorems. The paths of different integral curves can never cross except possibly at points where $V^i = 0$ due to the uniqueness of solutions. By solving the equation above with initial conditions for every point $P \in M$, we see that the integral curves "fill" $M$. Such a manifold filling set of curves is called a *congruence*. Integral curves can be thought of as "joining the arrows" of a vector field. Integral curves are invariant sets, i.e. they map into themselves under time evolution. Take a point along the curve, it will map into the next point under time evolution and so on, giving the whole integral curve. The unions and intersections of invariant sets are invariant sets.