1
$\begingroup$

I take two identical lasers, point them at each other from a large distance in space, and fire a pulse from each. The pulse is $N$ wavelengths long. The pulses intersect at the mid point and, for a while, a standing wave is formed by their superposition, and in that standing wave are a series of stationary points at which the E and H fields are 0. There will be $N$ such points, some longer lived than others.

However, both pulses are carrying energy, in the E and H fields, and that energy arrives at the final destination, so energy is being transferred, in the form of E and H fields, through points that never have an E or H field.

I can do maths, but I'm looking for some sort of physical explanation of what is actually happening to allow this. Can anyone point me in the right direction?

I've found a related question which doesn't have answers I can accept (or, for that matter, an accepted answer)

How do traveling waves pass through a standing wave node, if the node doesn't move?

and this one which has the right title but isn't the question I'm asking

How does the energy in a standing wave travel beyond a node?

This last one doesn't answer the question - the accepted answer instead explains how an additional wave can travel through a pre existing standing wave - I want to know how the two waves forming the standing wave are propagating energy through the nodes.

$\endgroup$
4
  • $\begingroup$ Calculate the sum of all amplitudes at a node. That's all that matters. $\endgroup$ Dec 21 '20 at 13:54
  • $\begingroup$ The sum of all amplitudes at the node is 0, because otherwise it wouldn't be a node? The question you just linked shows how an additional wave can pass through a standing wave and I linked to it already - sorry :/ $\endgroup$ Dec 21 '20 at 14:11
  • $\begingroup$ Your question is answered in the linked items. If you add a wave of instantaneous amplitude +X to another wave of instantaneous amplitude -X, you see zero net amplitude. equal amounts of energy are passing in opposite directions. I don't know how to make that any clearer than others already have. $\endgroup$ Dec 21 '20 at 16:02
  • $\begingroup$ I completely understand the maths I can create a computer simulation that animates the waves and shows me it, what I don't get is how energy is transported past a point at which no energy is ever present. $\endgroup$ Dec 21 '20 at 19:05
1
$\begingroup$

The pulses intersect at the mid point and, for a while, a standing wave is formed by their superposition, and in that standing wave are a series of stationary points at which the E and H fields are 0.

Unfortunately, your assumption about how this works is incorrect. It is best to work through the math on problems like this. For clarity I will use natural units ($c=\mu_0=\epsilon_0=1$) and a frequency such that $\omega=k=1$. Let $$\vec E_1 = (0,A,0) \cos(x-t)$$ $$\vec B_1=(0,0,A) \cos(x-t)$$ and let $$\vec E_2=(0,A,0) \cos(-x-t)$$ $$\vec B_2 = (0,0,-A) \cos(-x-t)$$ and let $$\vec E_3 = \vec E_1 + \vec E_2$$ $$\vec B_3 = \vec B_1 + \vec B_2$$

So the first fields represent a plane wave traveling in the $+x$ direction, the second fields represent a plane wave traveling in the $-x$ direction, and the third fields represent the standing wave formed by their sum. I leave it as an exercise to show that all three pairs of fields satisfy Maxwell's equations.

Note that by simplifying we obtain $$\vec E_3 = (0,2A,0) \cos(t) \cos(x)$$ $$\vec B_3 = (0,0,2A)\sin(t) \sin(x)$$ So although it is correct that $\vec E_3$ has a series of stationary points at which it is always 0 (nodes) and it is also correct that $\vec B_3$ has a series of nodes, the nodes for $\vec E_3$ are different from the nodes for $\vec B_3$. There are no points that are always zero for both fields and there are also no times when both fields are zero everywhere.

To understand the flow of energy in this field configuration we can calculate the energy density and the Poynting vector. $$\vec S_3 = \vec E_3 \times \vec B_3 = (A^2,0,0) \sin(2t) \sin(2x)$$ $$u_3=\frac{1}{2}(E_3^2+B_3^2)=A^2\left(\cos^2(t-x)+\cos^2(t+x)\right)$$

However, both pulses are carrying energy, in the E and H fields, and that energy arrives at the final destination, so energy is being transferred, in the form of E and H fields, through points that never have an E or H field.

This is partly correct and partly incorrect. There are points that are nodes for the E field and different points that are nodes of the B field. There are no points are nodes for both the E and B field.

However, there are indeed nodes for the Poynting vector $\vec S_3$. The nodes of $\vec S_3$ include all of the nodes of $\vec E_3$ and of $\vec B_3$. Any point where $\vec S = \vec 0$ means that electromagnetic energy is not flowing through that point. So electromagnetic energy never flows through the nodes of $\vec S_3$ at any time. Instead, by looking at $u_3$ and $\vec S_3$ we see that EM energy merely oscillates back and forth between the nodes. There is no transfer of energy through the nodes of $\vec S_3$.

Just because energy arrives at the final destination does not imply that energy goes through any of the nodes of $\vec S_3$. Energy doesn't get some cosmic ID tag, so there is no sense in which you can say that the energy leaving one laser must be the same energy as the energy arriving at the other laser and therefore no justification for the assumption that energy must flow through the nodes. Energy is conserved, and energy conservation does not require it to flow through the nodes.

$\endgroup$
2
  • 1
    $\begingroup$ I missed that the H fields were out of phase, which was silly of me. As for the poynting vector, which makes my question valid again (phew) I kind of could tag the energy with an ID, so long as I insisted it "bounced" off the nodes, which is an amusing idea, but I will accept "it doesn't need to" as the real answer. Thankyou. $\endgroup$ Dec 21 '20 at 23:46
  • $\begingroup$ @AndyNewman you are very welcome! $\endgroup$
    – Dale
    Dec 21 '20 at 23:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.