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I can't understand the concept of the curl of an electromagnetic wave.

$$ \nabla \times E = -\frac{\partial \textbf{B}}{\partial t} $$

All of the examples I find show a current through a conductor, or that paddle wheel in water which I fail to see the distinction of that with an E-M wave. What I am trying to understand is the curl of say a laser beam light.

So lets say I have this sin wave which I plot in MATLAB which represents a section of a laser beam propagating through free space (We shall say it is the net product of $E_x$ and $E_y$):

laser radiation

How do you take the curl of it? Where do you take the curl of it. eg the whole beam, a certain section? Can I do this in MATLAB to see visualization of the concept of curl?

I mainly seek descriptive or pictorial answers.

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    $\begingroup$ The curl acts on three dimensional vector fields. So you would have to describe your laser beam in three dimensions instead of one to calculate its curl. $\endgroup$
    – M.Herzkamp
    Jun 18 '14 at 14:02
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    $\begingroup$ This article has a nice explanation of what a plane wave is and how to calculate its curl. $\endgroup$ Jun 18 '14 at 17:21
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    $\begingroup$ It is quite clear here that the questioner means that the E-field has the form E = E_0 sin(kz-omega t) in either the x or y directions. $\endgroup$
    – ProfRob
    Jun 19 '14 at 8:00
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Let us deal with this on two parts: (1) How do you take the curl?

The E-field you describe has the form ${\bf E} = E_0 \sin (kz-\omega t)\, {\bf i}$, where the latter unit vector assumes the wave is polarised along the x-axis, while the wave travels along the z-axis. (You can equally polarise it along the y-axis if you choose). If I define it in this way, then the E-field only has an x-component. i.e. $E_{y}=E_{z}=0$

The curl of the wave can be evaluated as described in the answer by JamalS, so in this case, as $E_{y}=E_{z}=0$, then the partial derivatives of these components are also zero and there are only two possible non-zero terms in the curl.

$$\nabla \times {\bf E} = {\bf j} \frac{\partial E_{x}}{\partial z} - {\bf k} \frac{\partial E_{x}}{\partial y}\, .$$ Because we chose to have the wave travelling along the z-axis then $E_{x}$ is not a function of $y$, so the second term is zero and $$\nabla \times {\bf E} = {\bf j} \frac{\partial E_{x}}{\partial z} = kE_{0} \cos (kz -\omega t)\, {\bf j}\, ,$$ perpendicular to the E-field and the direction of motion of the wave, but changing direction with time.

OK, that's the Maths, but (2) how to "visualise" or deduce without doing the Maths? I have to admit to struggling with this one. The paddle wheel analogy is always the one I use. A field with a non-zero curl will make the paddle wheel turn and the axis of rotation points in the direction of the curl.

If I assume that the MATLAB plot you show has z along the horizontal axis, then you can imagine the E-field as vertical arrows of size proportional to the E-field strength at that instant in time. Imagine placing a paddle wheel at some point along the axis. In general, the E-field on one side of the wheel will be of a different strength to the E-field on the other side of the wheel and hence it will rotate around an axis perpendicular to z. If we then roll the clock forward to some later instant of time, the situation could reverse with the E-field now stronger on the other side and the wheel rotates in the opposite direction - i.e. a cosinusoidally varying curl in a direction perpendicular to the wave motion and perpendicular to the E-field.

Here's an attempt to show this. The two plots show the instantaneous E-field strengths and directions at two instants of time. I add a paddle wheel which sits in the field at the horizontal positions shown. In the top plot the wheele rotates clockwise. Some time later the field has changed as shown in the bottom plot and the wheel would rotate anti-clockwise.

The E-field strength as a function of coordinate along the direction of wave motion at two instants in time. A paddle wheel inserted into the field (assuming the usual analogy between any vector field and a velocity field) would turn clockwise in the upper plot, anti-clockwise in the lower plot, about an axis perpendicular to the page.

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  • $\begingroup$ Thank you very much. The step by step explanation is great. There is one last thing confusing me. You say "paddle wheel turn and the axis of rotation points in the direction of the curl". If the paddle wheel was on the very peak of the wave, isn't its curl zero? Therefore isn't the derivative of the tangent to that slope (as Harold says), equal to zero and we simply get a cos wave? $\endgroup$ Jun 19 '14 at 14:39
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    $\begingroup$ Spot on. If you look at the mathematical derivation of the curl you will see it does end up being cosinusoidal. At the peak of the sine wave (i.e. at a time shortly before the second plot), the cosine is zero and there is no curl (at that instant of time). $\endgroup$
    – ProfRob
    Jun 19 '14 at 14:58
  • $\begingroup$ Thanks, but my confusion is that when I look at an electromagnetic wave schematic, the magnetic field is not zero when the sin wave is at its peak. The magnetic field is in phase at a 90$^\circ$ rotation along the direction of propagation axis... $\endgroup$ Jun 19 '14 at 15:17
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    $\begingroup$ The curl of the E-field depends on dB/dt according to Faraday's law, which is also a cosine function and zero at the same point. i.e. E and B are out of phase with their curls. $\endgroup$
    – ProfRob
    Jun 20 '14 at 7:18
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    $\begingroup$ Yes, you've got it. This applet falstad.com/emwave1 allows you to look at this stuff. You can set up a plane wave and then look at representations of E, B, dB/dt etc. at any moment in time or as an animation. $\endgroup$
    – ProfRob
    Jun 20 '14 at 15:02
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There are two ways of writing Maxwell's equations: the differential form (the one you adopted) in terms of gradients and densities, and the integral form in terms of integrals and total charges/currents.

The differential form is usually referred to as local, for obvious reasons: there is no information about the spatial extent of a source, so you don't know where its influence should start to wear off.

The curl involves derivatives, which, as I am sure you know, have a geometric meaning: they correspond to the slopes of the tangent lines to a point of the curve.

So, if you were to take the curl of a function, you would be calculating the slopes of the tangent lines at each point of it. This would result in a new, macroscopic (i.e. not local anymore) graph, which in this particular case would correspond to the time variation of the magnetic field. The key thing here is that you build a macroscopic field from local information only.

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  • $\begingroup$ Thank you. That clears a lot up for me. I wish I could accept more than one answer... $\endgroup$ Jun 19 '14 at 14:33
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Generally, the curl of a vector field $\vec{v}$ in $\mathbb{R}^3$ is given by,

$$\nabla \times \vec{v}= \left( \begin{matrix} \partial_y v_z -\partial_z v_y \\ \partial_z v_x - \partial_x v_z \\ \partial_x v_y -\partial_y v_x \end{matrix} \right)$$

which may be viewed mathematically as simply the cross product of the field with the operator, $\partial_i$. From a physical perspective, the curl describes the rotation of a field. If we drop a ball in the field, the angular speed of rotation is precisely half the curl at the point, i.e.

$$\omega = \frac{1}{2} \left( \nabla \times \vec{v}\right)$$

The rotation axis points in the direction of the curl of the vector field, at the point. Now, as you may know from vector calculus, Green's theorem is useful to compute 2D line integrals, and as such we can define a curl for a two-dimensional field, but it 'lives' in $\mathbb{R}^3$ nonetheless. Therefore, as long as you know what the values of all the $E_i$, you can compute the curl, even if one happens to be zero.

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  • $\begingroup$ Thanks for your input. I could not fully follow your answer though sorry. I am sure it is all correct I just really needed a purely physical explanation... $\endgroup$ Jun 19 '14 at 14:34

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