# Are “Streamlines” Anything Like Level Curves?

Streamlines are found in complex space, and represent constant values that are tangent to the outgoing velocity vector.

Level curves are found in real (perhaps complex, as well?) space, and represent constant values of a multivariate function.

Is there any connection between them in general? In particular cases?

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The streamlines of a gradient flow are orthogonal with the potential's level curves... – anon Aug 1 '11 at 18:43

Streamlines can be also be defined on any arbitrary differentiable manifold (in particular, in real and complex spaces of arbitrary dimension).

Definition Given a (continuous) vector field $v$, a streamline is a (differentiable) curve $\gamma(s)$ such that the vector $\dot{\gamma}(s) = v\circ \gamma(s)$.

In other words, the notion of streamlines are identical to the notion of an integral curve. (Same thing, different names.) (Okay, one caveat, if you deal with fluid mechanics, the streamlines are the integral curves for the flow vector field at a fixed instant in time. You can constrast that with pathlines, which allow the vector fields to also change in time.)

In the case of working on the real and complex spaces $\mathbb{R}^n$ and $\mathbb{C}^n$, a vector field can be thought of as a function taking values in the space itself: $v:\mathbb{R}^n \to \mathbb{R}^n$ or $\mathbb{C}^n \to \mathbb{C}^n$. So this is why for a complex valued function defined on the complex plane, you can define a streamline, when the object is a function and not a vector field (because there is a one-to-one correspondance between vector-valued functions and vector fields).

Level curves can also be defined on any differentiable manifold.

Definition Given a differentiable real-valued function $f$ such that its critical points (points where the differential $df = 0$) are isolated, the level sets of $f$ are given by $f^{-1}(c)$ for some constant value $c$.

We can extend the definition to complex valued functions and functions that are vector valued. The important thing is:

The dimension of the level sets is equal to the dimension of the space/manifold minus the dimension of the codomain of $f$, whenever we are away from the critical points of $f$.

So if you have a real valued function on a $k$ dimension real space, the level surfaces are $k-1$ dimensional. If you have a vector-valued function on a real space, the level sets are $0$ dimensional (i.e. they are isolated points).

If you are working over a Riemannian manifold, there is a canonical way of identifying covectors (things that are like differentials of functions) and vectors. Which is how we can define the gradient vector $\nabla f$ of a function $f$: it is the vector field associated to the differential $df$.

Then in this situation, we have (like anon mentioned in his comment) that the streamlines of $\nabla f$ are orthogonal to the level sets of $f$: let $v$ be a vector. By definition $v\cdot \nabla f = df(v)$ is the partial derivative of $f$ in the direction of $v$. So the vanishing of the partial derivative (which is saying that $f$ is constant in the direction of $v$, i.e. that $v$ is tangent to the level surface of $f$) is equivalent to $v$ being orthogonal to $\nabla f$.

One connection you can make is the following. For convenience we'll work over $\mathbb{R}^k$. But the conclusion can be modified for suitable differentiable manifolds.

Claim Let $v$ be any nonvanishing smooth vector field (i.e. $v:\mathbb{R}^k\to\mathbb{R}^k$), we can find a vector valued function $w:\mathbb{R}^k\to\mathbb{R}^{k-1}$ such that the streamlines of $v$ coincide with the level curves of $w$. Conversely, if $w$ is a smooth vector valued function $\mathbb{R}^k\to\mathbb{R}^{k-1}$ with no critical points (in the sense that $dw$ is not surjective), we can find a vector field $v$ such that the level curves of $w$ coincides with the streamlines of $v$.

I will omit the proof here for space. But remark that the condition that $v$ is nonvanishing or $w$ is non-critcal is necessary. For example, consider the case in $\mathbb{R}^2$, where $v$ is the radial vector field. Then all streamlines get infinitely close to the origin. If there were a suitable $w$ (which would take values in $\mathbb{R}^1$; hence scalar valued), continuity of $w$ at the origin requires that $w$ takes the same value on al the streamlines, and hence $w$ is the constant function, but then $dw = 0$ everywhere and you don't have "level curves". For the other case, consider the function on $\mathbb{R}^3$ defined by $w(x,y,z) = (x^2 + y^2, (x^2+y^2)z)$. The level curves of $w$ are the set of all circles in planes parallel to the $xy$-plane, and the $z$ axis. One cannot find a continuous vector field such that along the $z$ axis it is in the $z$ direction, but away from the $z$ axis it is orthogonal to the $z$ direction.

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Consider an analytic function $f(z)$ with real and imaginary parts $u(x,y)$ and $v(x,y)$ considered as functions of the real and imaginary parts $x$ and $y$ of $z$. The level curves $v(x,y) = \text{constant}$ are streamlines of the vector field $\nabla u$, and the level curves $u(x,y) = \text{constant}$ are streamlines of the vector field $\nabla v$.

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