Which component of an electromagnetic wave is responsible for visible effect?

[nipiano93]

Which component E or B of an electromagnetic wave is responsible for visible effect?

 

[Meir Achuz]

It is chiefly B that induces currents in your retina.

 

[Bill_K]

It is chiefly B that induces currents in your retina.

Is that really correct? Everything I read says it is simply the energy in the EM wave that excites a photosensitive chemical.

 

[Meir Achuz]

The energy "excites" a chemical by the mechanism of the B field inducing currents.

 

[nasu]

Do you have any reference about it? I find this very interesting.

 

[nipiano93]

Thank you clem! I think you are right.
@nasu
I think I can explain. The objects are visible to us because of currents induced in our retina. As the magnetic field is responsible for induction of current, it is the magnetic field that is responsible for the visibility of stuff around us.

 

[Bill_K]

Again I'd ask for a reference if you think this explanation is correct. Here's what I'm finding. It's rather complex!

The outermost neurons in the retina are arranged in three successive layers: photoreceptors (rods and cones), bipolar cells and and ganglion cells. Photoreceptor cells communicate with the bipolar cells through a continuous release of glutamate. Photoreceptors contain a chemical called retinal. When light strikes a photoreceptor, what happens is that the energy in the photon raises retinal to an isomeric state. This reduces the cell's release of glutamate, which in turn affects the bipolar cells lying behind it.

Bipolar cells are the second neuron layer. They come in two types: polarizing and depolarizing, arranged in a center-surround pattern. They have different responses to the glutamate reduction. This causes a voltage gradient to occur between the two types of cells, and this can be detected by the presence of a retinal current.

Bipolar cells in turn transmit signals to the ganglion cells, leading eventually to the brain.

So indeed retinal currents exist, but they are an indirect effect of the light reception, not in any way related to the B field of the photon.

 

[Born2bwire]

It was my recollection that the reception was a quantum mechanical effect from absorbing the energy of the photon. I too would also like to see a reference.

 

[nasu]

Thank you clem! I think you are right.
@nasu
I think I can explain. The objects are visible to us because of currents induced in our retina. As the magnetic field is responsible for induction of current, it is the magnetic field that is responsible for the visibility of stuff around us.

OK, i did not say that is impossible or that there is no way to imagine a mechanism responsible for this.
But explaining something that could happen does not mean that it really does.
And not only magnetic field is responsible for producing currents. The electric field can do it even better. So if you are referring to migration of electrons along a molecule or similar things, this is more likely to be produced by the electric field.
What currents are you actually talking about? In what kind of "circuit"?

So my question was if you have a reference or you just imagined it as plausible.

 

[Meir Achuz]

Disclaimer: I am not an authority on the retina (except for having an occasional tear), and don't know any useful reference. My original answer really applies for a good conductor (which I maybe mistakenly considered the retina to be) where B is much larger than E. I hope more expert witnesses can resolve this discussion.

 

[Born2bwire]

Disclaimer: I am not an authority on the retina (except for having an occasional tear), and don't know any useful reference. My original answer really applies for a good conductor (which I maybe mistakenly considered the retina to be) where B is much larger than E. I hope more expert witnesses can resolve this discussion.

Generally though, when we treat a perturbation problem we treat the force from the electric field as first order and the magnetic field as second order. This is due to the fact that the magnetic field only operates on moving charges. Personally though, I would not say that one or the other is more important here. At very low frequencies we can make quasi-static approximations and decouple the fields. But at optical frequencies, I would normally view it as a quantum problem. Only for something like nanoscale objects would I think that you could treat it as a classical problem (like we do with plasmon resonances in nanoscale materials like gold spheres). But under those conditions we still treat it as an electromagnetic wave and I think it is a bit misleading to think of the effect being predominantly one or the other since they always work in tandem.

 

[Drakkith]

Can I assume that the question can be boiled down to:

Which field is involved when a photon is absorbed?

 

[Claude Bile]

Echoing Born2bwire's sentiment; we cannot isolate the E or B fields at optical frequencies, thus we are forced to conclude that both fields play a role in optical absorption.

Claude.

 

[nipiano93]

All right everyone, this was my own question and so I don't have any reference. Firstly, let us think about this in a different manner. I now believe my previous reasoning was incorrect. Let us modify the question to avoid any biological explanation (it's physicsforums after all. Consider the case of a camera. What is it that enables the camera to record images? (I mean which field).

 

[nipiano93]

Echoing Born2bwire's sentiment; we cannot isolate the E or B fields at optical frequencies, thus we are forced to conclude that both fields play a role in optical absorption.

Claude.

I think I disagree. The fields can ofcourse not be isolated, but I think that both fields play separate roles. For example, take the case of polarisation. Mainly electric field is responsible for polarisation because of its larger amplitude (E=B/c).

 

[Naty1]

This is one of the most erratic discussions I've come across...perhaps because the question is not clearly defined.

Photometry is the science of the measurement of light, in terms of its perceived brightness to the human eye.[1] It is distinct from radiometry, which is the science of measurement of radiant energy (including light) in terms of absolute power

;

http://en.wikipedia.org/wiki/Photometry_(optics )


Follow links here for film versus electronic image production...

Traditional cameras capture light onto photographic film or photographic plate. Video and digital cameras use an electronic image sensor, usually a charge coupled device (CCD) or a CMOS sensor to capture images which can be transferred or stored in a memory card or other storage inside the camera for later playback or processing.

 

[nipiano93]

Sorry I meant E=Bc.
@naty1 Look there has to be something that is responsible for the visibility effect. And now I think that Electric field is responsible because of its larger magnitude. ( It will have a greater impact on the photographic plate).

 

[Claude Bile]

I think I disagree. The fields can ofcourse not be isolated, but I think that both fields play separate roles. For example, take the case of polarisation. Mainly electric field is responsible for polarisation because of its larger amplitude (E=B/c).

Polarisation simply refers to the orientation of the E field; which is by convention. One could equally define polarisation in terms of the orientation of the B field; in fact, this is done in waveguides where the E and B fields are not mutually perpendicular. The relative amplitudes of E and B are due to convention also, specifically the choice of units used.

The point remains, you cannot separate the effect of E and B fields at optical frequencies, either in the classical or quantum regime.

Claude.

 

[nipiano93]

@claude bile
Hmm... strange! In my school syllabus, it is told to us that electric field is primarily responsible for polarisation. Anyways, then could you please sum up all this to give a final answer to my original question?

 

[jtbell]

Consider the case of a camera. What is it that enables the camera to record images? (I mean which field).

With both digital sensors and photographic film (silver halide grains), as with the eye, it is the absorption of individual photons by atoms or molecules that produces the image. At the level of individual photons it is not meaningful to speak in terms of the classical E and B fields.

 

[nasu]

@claude bile
Hmm... strange! In my school syllabus, it is told to us that electric field is primarily responsible for polarisation.

Does it even make sense to say "responsible for polarization" in respect to electric field?
The polarization describes the specific orientation of the field (electric, magnetic), displacement, etc in a wave.
Responsible for a specific polarization of the wave is the nature of the source and the eventual media and interfaces the wave had to go thorough.

 

[Born2bwire]

@claude bile
Hmm... strange! In my school syllabus, it is told to us that electric field is primarily responsible for polarisation. Anyways, then could you please sum up all this to give a final answer to my original question?

It doesn't make much sense to say that it is responsible. It is just taken as the definition really. When we speak of polarization we primarily talk about the alignment of the electric field but we can similarly discuss in respects to the magnetic field. In fact, in antenna research we often talk of the E plane polarization and the H plane polarization (the planes along which the electric and magentic fields are polarized).

Now often in a perturbation sense we take the electric field to be first order and the magnetic field to be second order. This is because the electric field acts directly on the charges, whether stationary or moving. The magnetic field acts only on moving charges and thus if we start with a stationary system the magnetic field does not play a part until we account for the acceleration from the electric field. But perturbation analysis requires a classical treatment and it also assumes a weak field interaction. This is something that we would do say in the analysis of a plasma. I do not think that it applies here.

 

[Naty1]

Look there has to be something that is responsible for the visibility effect

Visibility is due to the INTENSITY (energy or power) of an electromagnetic waves (and of course it's frequency or wavelength). A very low power distant light, say a star, even if in the visible spectrum can't be seen because there are too few photons to be "visible"..that is detected...

Light (an electromagnetic wave) is a transverse wave which means E and B oscillate perpendicular to the direction of energy transfer. At one moment at one point E is maximum and a moment later B is maximum...In a waveguide or fiber optic cable for example energy periodically surges back and forth between the center and the outer edges of the confiniing wall...For brief instants NO energy is flowing (whuich flows overall in the direction of propagation) because the field configurations are momentarily stable. (This description is based on my old copy of Hallidayand Resnick, PHYSICS FOR STUDENTS OF SCIENCE AND ENGINEERING.)

E and B are two sides of the same phenomena...they are a package deal, I don't see how you can split them apart...

I see it as analogous to frequency and wavelength...which carries the energy, the frequency or the wavelength? The question makes no sense because those "components" can't be separated...they also are two sides of the same phenomena...

 

[nipiano93]

thank you everyone. I got it.