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What is the physical significance of curl $$\nabla\times\boldsymbol{V}~?$$ I mean I read 'curl V represents the rotation of the vector $V$. My question what is it about the term $\nabla\times\boldsymbol{V}$ that it represents the rotation of the vector?

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    $\begingroup$ Related: physics.stackexchange.com/q/202923/2451 and links therein. $\endgroup$ – Qmechanic Oct 10 '15 at 16:21
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    $\begingroup$ Now people are voting to close this because it is a duplicate of another question that has been closed as being too broad - insane. $\endgroup$ – Rob Jeffries Oct 10 '15 at 17:18
  • $\begingroup$ I think that this question is superior than the proposed duplicate and is not unclear. $\endgroup$ – Kyle Kanos Oct 10 '15 at 17:48
  • $\begingroup$ I think the current version of the question is slightly unclear, but that it is easily repaired. $\endgroup$ – dmckee Oct 10 '15 at 19:14
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Curl can be equated with the closed line integral in the limit that the encircled area $\Delta S$ goes to zero. However, we would have to do this in three components because curl is a vector. $$ (\nabla \times \vec{v})_x = \lim_{\Delta S \rightarrow 0} \frac{1}{\Delta S} \oint \vec{v}\cdot d\vec{l} $$ in the $yz$ plane and so on.

But what does it mean? Well it is easy to show that $$ \nabla \times \vec{v} = 2 \vec{ \omega} $$ As follows: $$ (\nabla \times \vec{v})_x = \partial_y v_z - \partial_z v_y = \partial_y (\vec{\omega} \times \vec{r})_z - \partial_z (\vec{\omega} \times \vec{r})_y $$ $$ (\nabla \times \vec{v})_x = \partial_y (\omega_x y - \omega_y x) - \partial_z ( \omega_z x - \omega_x z) = 2 \omega_x$$

and ditto for the other components $$ (\nabla \times \vec{v})_y = \partial_z v_x - \partial_x v_z = 2\omega_y$$ $$ (\nabla \times \vec{v})_z = \partial_x v_y- \partial_y v_x = 2\omega_z$$ i.e. the curl of a velocity field equals twice the angular velocity at that point. In other words it is angular velocity within a fluid flow that creates curl! You can imagine constructing a ``curl meter'' out of a little (infinitesimally small) paddle wheel which could be inserted into the fluid flow. If the paddle wheel turns then there is curl.

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  • $\begingroup$ 'Curl meter'; an exotic device! Sounds like Purcell's explanation:p $\endgroup$ – user36790 Oct 10 '15 at 18:45
  • $\begingroup$ @user36790 It's mine. From my lecture notes. Though the paddle wheel is a commonly used aid to understanding. $\endgroup$ – Rob Jeffries Oct 10 '15 at 18:47
  • $\begingroup$ Don't get bothered, sir. Sorry, if it in anyway disturbs you. What I wanted to tell you is that the same 'curl-meter' had been mentioned by Edward Purcell in his Berkeley Classic: Electricity & Magnetism; that's it. I am in no way against you. Sorry again:( $\endgroup$ – user36790 Oct 10 '15 at 18:50
  • $\begingroup$ Yes! This integral relation and the corresponding $\nabla \cdot$ relation are practically the most important things to know about vector calculus. $\endgroup$ – user12029 Oct 10 '15 at 21:42
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You've got a bunch of good answers now, let me add two things that I did not see:

The $\delta$-function curl: things can look twisty without being so.

Consider the field $$\vec V(x, y, z) = \frac{1}{x^2 + y^2}\begin{bmatrix}-y\\x\\0\end{bmatrix} = \hat\theta / r.$$Strangely, $\nabla\times\vec V=0$ at every point except the line $(x,y,z)=(0, 0, z)$ where it is not defined. This "looks" rotational: the $\hat\theta$ vector points around the origin. However, the exact scaling of $1/r$ is just what it needs for the curl to vanish.

So, any naive "the curl tells you how twisty something looks" interpretation is wrong, because here is a thing which looks twisty but has no curl.

Better: the force-torque interpretation.

Suppose $\vec V$ represents a force field -- in the usual physics sense of a force defined at every point in space, please, not the sci-fi sense of an invisible wall.

Now suppose we put a little pinwheel inside the force field and have it feel those forces. Without loss of generality, let's write it as a little circle of radius $\epsilon$ in the $xy$-plane about the origin, the points $\delta\vec r(\theta) = [\epsilon \cos\theta, \epsilon \sin\theta, 0]$ for $0\le\theta\lt 2\pi.$ The force on any point of this circle is approximately: $$\vec F(\theta) \approx \vec V(0, 0, 0) + \frac{\partial \vec V}{\partial x}~\epsilon~\cos(\theta) + \frac{\partial \vec V}{\partial y}~\epsilon~\sin(\theta).$$ To first order in $\epsilon$, of course, the net force on this pinwheel is just $\vec V(0, 0, 0)$: the terms linear in $\epsilon$ vanish.

The net torque, however, does not vanish: $$\delta \vec\tau = \int_0^{2\pi} d\theta~ \delta\vec r(\theta)\times\vec F(\theta) \approx \epsilon \int_0^{2\pi} d\theta~\hat z~\big(\partial_x V_y \cos^2\theta - \partial_y V_x \sin^2\theta \big),$$with all the other terms disappearing as we integrate plain sines and cosines or occasionally $\sin\theta\cos\theta = \frac12 \sin(2\theta),$ all of which oscillate about $0$. Only $\cos^2$ and $\sin^2$ oscillate about a different average, namely $\frac12$. Performing the integral therefore leads to the small torque:$$\delta \vec\tau \approx \frac12~\epsilon ~\hat z \left(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\right).$$This is $\vec \tau = \epsilon~\hat z~(\hat z \cdot (\nabla \times \vec V)),$ so the proper extension of this to all coordinates is simply: at position $\vec r$ insert a pinwheel in the $\hat n$ direction, the torque per unit radius on that pinwheel is simply $\hat n \cdot (\nabla \times \vec V).$

So the curl of a force field is the torque-per-unit-radius on a small pinwheel, pointed in the direction of greatest torque.

Since a velocity field at small distances will generally influence particles by a linear drag $\vec F \propto \vec v$, this is also a good interpretation for velocity fields: insert a little pinwheel that you are holding fixed, and the curl tells you the torque that the fluid will start exerting on that pinwheel (which you will have to oppose to keep it fixed).

The $\hat \theta / r$ field, therefore, does not rotate a small object as it flows past it. You can roughly understand this by describing the pinwheel not as a little circle of radius $\epsilon$ but a little trapezoid $\delta r, \delta \theta.$ The fluid stays in contact with the outer surface for a length $\delta \theta~(r + \delta r)$ but is only in contact with the inner surface for the length $\delta \theta~ r.$ By making the force go like $\vec F = U_0~\hat \theta/r,$ the work done on the inner surface travelling around the loop is $-U_0~\delta\theta~r/r ~+~ U_0 \delta\theta (r + \delta r) / (r + \delta r) = 0,$ as it must be if the pinwheel doesn't want to rotate that way.

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    $\begingroup$ Of course I agree with all this. The usual way of addressing this is to use the example of the B-field outside an infinite current-carrying wire. That depends on $r^{-1}$ and must have a curl of zero by Ampere's law. That is the example I use to point out the dangers of just looking at the (curly) field lines and trying to decide. $\endgroup$ – Rob Jeffries Oct 10 '15 at 21:51
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Consider $\vec V$ to represent the velocity field of a fluid. We choose a point $x_0$. In a small region around this point we can approximate the vector field by $\vec V(\vec x_0 + \vec x) = \vec V(\vec x_0) + D\vec V(\vec x_0)x$, where $D\vec V(\vec x_0)$ represents the Jacobian matrix. We can decompose this derivative uniquely into the symmetric and antisymmetric part: $J := D\vec V(\vec x_0) = S + A$, where $A^T = -A$ and $S^T = S$. Explicitely, $A = \frac 1 2 (J - J^T)$ and $S = \frac 1 2 (J + J^T)$.

As the velocity fields generated by the symmetric and antisymmetric parts are simply superposed, we can consider their effects separately (and here we only consider the antisymmetric part).

Close to the point $x_0$, the antisymmetric part $A$ generates a velocity field: \begin{align*} \vec V_A(\vec x_0 + \vec x) &= \begin{pmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{pmatrix} \vec x = \vec \omega \times \vec x. \end{align*} This corresponds to the velocity field of a rigid body rotating with the angular velocity $\vec \omega$. (Note, that one simply cleverly names the components of $A$ by $\omega_i$ to get this result).

Writing out the elements of $A$ explicitely, one can trivially identify them with the components of $\nabla \times \vec V$ (e.g. $(\nabla \times \vec V)_1 = \partial_2 V_3 - \partial_3 V_2 = 2A_{23} = 2\omega_1$).

In conclusion, if $\vec V$ is a velocity field, then $\nabla \times \vec V(\vec x) = 2\vec \omega(\vec x)$ with the angular velocity $\vec \omega(\vec x)$ with respect to point $\vec x$.

A similar analysis of the symmetric part reveals that $\mathrm{Tr}(S) = \nabla \cdot V$ is related to the increased volume of the flow (i.e. to the increase of the volume of a small volume transported in the flow).

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$\nabla$ represent the pseudo-vector $(\frac\partial{\partial x},\frac\partial{\partial y})$ (in 2D). So :

  • $\nabla(v)$ is the gradient $(\frac{\partial v}{\partial x} , \frac{\partial v}{\partial y})$,
  • $\nabla\cdot v$ is the divergence $\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y}$,
  • $\nabla \times v$ is the curl $\frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y}$

So if $v$ is a rotating field $v(r,\theta) = r\dot{\theta}(-\sin(\theta),\cos(\theta))$ , i.e. $v(x,y)=\dot{\theta}(-y,x)$, you see that curl($v$) = $2\dot{\theta}$, with is (twice) the constante rate of rotation.

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  • $\begingroup$ completed up to rotation rate. $\endgroup$ – Fabrice NEYRET Oct 10 '15 at 18:02

protected by Qmechanic Oct 10 '15 at 17:55

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