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As we know the diffraction pattern from a circular hole looks something like

enter image description here

Intuitively I would think the "bessel function like" pattern on the right would be due to poynting flux through the screen but every single resource (text book and papers) that I have seen seems to indicate that the bessel function is due to the electric field flux through the back screen. For instance, from these notes

https://www.ece.rutgers.edu/~orfanidi/ewa/ch18.pdf

enter image description here

I thought it is the poynting vector that carries energy and thus illuminates the screen. Or is it the electric field vector that excites the material in the back screen creating the pattern seen?

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It is the electric field.

Whilst you may have two electromagnetic waves where $\vec{N}_1 = \vec{E}_1 \times \vec{H}_1$ and $\vec{N}_2 = \vec{E}_2 \times \vec{H}_2$, it is not necessarily true that $\vec{N}_{\rm tot} = \vec{N}_1 + \vec{N}_2$.

Instead you should write $$\vec{N}_{\rm tot} = (\vec{E}_1 + \vec{E}_2) \times (\vec{H}_1 + \vec{H}_2)$$ because it is the fields, not the Poynting vector, that obey the principle of superposition.

A pair of oscillating Poynting vectors representing electromagnetic waves will not add to give you a zero intensity in a diffraction pattern unless both Poynting vectors are zero at the same point in space at all times. But this isn't the case. The Poynting vector from a point light source will oscillate in time between zero and some maximum

It is the fields, which can be positive or negative, that sum together to produce a net zero. That's why you work out the fields and then work out their intensity.

To put it yet another way, whilst it is rather hard to arrange for $\Sigma_i E_i^2 =0$ at any point for all $t$, $(\Sigma_i E_i)^2 =0$ certainly is possible.

If you re asking what is observed, rather than what "creates" the diffraction pattern, then it is the (time-averaged) intensity of the light, or equivalently, the energy density of the electromagnetic fields, that are important, and they are proportional to $(\Sigma_i E_i)^2$ or the Poynting vector associated with the vector sum of the electric fields.

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  • $\begingroup$ Thank you for your response/perspective. Of course when calculating the entire Poynting field we have to account for cross terms but I don't see how that indicates that it is the electric field that produces the diffraction pattern as opposed to the Poynting vector. If possible can you elaborate? Thank you in advanced. $\endgroup$ – user112202 Jun 16 '20 at 17:48
  • $\begingroup$ @user112202 What do you mean by "produces the diffraction pattern"? It is the fields that interfere constructively and destructively. $\endgroup$ – ProfRob Jun 16 '20 at 17:52
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    $\begingroup$ I think I understand what you're saying........because it is the electric fields that obey the principle of superposition (i.e. interferes constructively and destructively) it must the the relevant field that causes the diffraction pattern (after all as you said the diffraction pattern must occur from constructive and destructive interference which only the electric and magnetic fields can do). Do I understand correctly? $\endgroup$ – user112202 Jun 16 '20 at 18:00
  • $\begingroup$ @user112202 Yes, that is my view. $\endgroup$ – ProfRob Jun 16 '20 at 18:05
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The underlying level of reality is at present modeled by quantum mechanics. In this diffraction pattern built one photon at a time, it is the energy carried by the photons that creates the pattern:

singlphotdbls

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

This video of the interference of two beams shows that the pattern does not depend on the existence of the screen. The patterns are recorded by quantum mechanical transitions in energy levels of the atoms and molecules of the screen.

Edit after reading other answers:

It can be proven mathematically that the classical electromagnetic field is built up by a large number of individual photons. One would expect that it is the Poynting vector at the classical level, that generates the patterns on the screens, but the definition of the Poynting vector:

poynting

where :

Emagnitude

is on averages, it has no sinusoidal dependence, and I will agree with the other answers that it is the sinusoidal instantaneous dependence of the electric field that generates the patterns in the classical analysis. Still, I believe that the quantum mechanical level has the best model, since it is the sinusoidal dependence of the probability of locating a photon at (x,y,z,t), $Ψ^*Ψ$ ,that is varying and generating patterns in space time. See figure 1 here.

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  • $\begingroup$ Thank you very much for your response. I wish I could upvote it but I have less than 15 reputation points. I would also think it is the Poynting vector that creates the pattern but strangely all resources I've come across seem to say that is the electric field.....which makes no sense to me (see the second picture I posted). Thank you again. $\endgroup$ – user112202 Jun 16 '20 at 17:29
  • $\begingroup$ The fact that waves are modeled sinusoidaly, and that mathematically one could get the electric field to reproduce the patterns, cannot explain the energy deposited on the screens , imo. The pattern is made by interference at the quantum level of the wavefunctions, at the quentum level. at the classical level it is again interference of the sinusoid waves describing the electric field. $\endgroup$ – anna v Jun 16 '20 at 17:42
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Something has to detect the radiation. It could be a detector, or it could be light reflected from the screen into your eye. Whatever it is, the predominant mechanism for light to interact with matter is electrical. It could be absorption in a CCD pixel, or reflection due to induced electric dipoles, but in any case electric dipole intereactions are overwhelmingly the dominant mechanism. I would say that the Poyning vector provides us with the rate that interactions take place, but the mechanism itself is electrical.

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  • $\begingroup$ Thank you for your perspective, I will look more into it. If I may ask a further question. Let's say we have a normally incident wave through an aperture in the xy plane and the far screen is parallel to this plane. Is it the electric flux through the far screen (i.e. perpendicular to the far screen) that causes the electrical interactions or the components parallel to the screen that cause the electrical interactions. I would think it would be the former given the second picture I posted in the OP. $\endgroup$ – user112202 Jun 16 '20 at 17:38
  • $\begingroup$ It's both, in general. The electric field polarizes the medium of the screen, and it depends on the molecular structure at the surface as to whether or not the perpendiuclar component contributes. I can't think of a situation where the perpendicular component would not be allowed. The symmetry breaking at the surface almost guarantees that the perp. component will contribute. Of course this is only for off-axis directions. Exactly on axis there is no perpendicular component. $\endgroup$ – garyp Jun 17 '20 at 4:31
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I think that we can use the model of the work done by a plane wave (even if for Fresnel difraction the waves are not treated as planes) to get the conclusion that the physical effects comes form the E-field:

The work done in a charge of the screen depends on the Lorentz forces and its displacement: $$\Delta W = q(\mathbf E + \mathbf v \times \mathbf B).\Delta \mathbf x$$

Considering the wave propagating in the $x$ direction, there is no components of $\mathbf E$ and $\mathbf B$ in this direction:

$$\Delta W = q((0,E_y,E_z) + (v_yB_z - v_zB_y , -v_xB_z , v_xB_y)).(\Delta x,\Delta y,\Delta z) =>$$

$$\Delta W = q((v_yB_z - v_zB_y)\Delta x + (E_y – v_xB_z)\Delta y + (E_z + v_xB_y)\Delta z)$$

Dividing both sides by the volume and by a small time interval we get density of energy per time. Taking the limit when the time interval and volume tends to zero:

$$\frac{\partial W_v}{\partial t} = \rho( v_xv_yB_z - v_xv_zB_y + E_yv_y – v_xv_yB_z + E_zv_z + v_xv_zB_y)$$ =>

The $\mathbf B$ components vanish, and only $\mathbf E$ does work on the charges. $$\frac{\partial W_v}{\partial t} = \rho(E_yv_y + E_zv_z) = \rho \mathbf {v.E}$$

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If you're using an energy detector, it isn't either/or. The energy (Poynting) flux determines the intensity pattern on the detector. Electromagnetic interference determines the pattern in the Poynting flux. To calculate the pattern in the plane wave approximation, you may take the Poynting flux as proportional to the square of the electric field magnitude. Textbooks tend to skip over this, jumping from field to detected pattern.

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The diffraction pattern observed in the photos is due the power distribution in the light's EM field. The power density is proportional to the square of the amplitude of the E field. So it may be more appropriate to say that the diffraction pattern is created by variations in the square of the E field.

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  • $\begingroup$ Combining $E^2$ from different points on the source cannot be responsible for "creating the pattern seen", since $\Sigma_i E_i^2 \neq 0$ at any point. $\endgroup$ – ProfRob Jun 17 '20 at 7:59
  • $\begingroup$ Not sure I understand your objection. You're right of course that E^2 is not what is combined: E from different points is combined vectorially. However, what we see or detect is E^2 at each point on the screen/detector. We can't see the E vector. $\endgroup$ – S. McGrew Jun 17 '20 at 12:19

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