Electric Field

[Little Bang]

Does anyone know of experiments to test what happens to an electric field when it is squeezed with another electric field?

[Erasmus00]

Electric fields don't interact with each other, so I don't see what you mean by one field squeezing another.

[sanctus]

Erasmus, this might be a stubid question. But photons being the carriers of EM-force and photons interact does this not imply also that two EM-fields interact?

[Erasmus00]

Photons don't interact with other photons, there are no photon/photon couplings. Photons can interact through a fermion loop, but thats a 4th order effect.

[sanctus]

Ok, fell into the trap that if typical electron positron annihilation exists giving 2 photons also the reverse is possible...but then it is just a quantum fluctuation of 1 photon. I.e.:

 

[math] e^-+e^+ \to \gamma + \gamma[/math]

but not [math] \gamma +\gamma \to e^- +e^+[/math]

 

but just [math] \gamma \to e^- +e^+[/math]

 

Thanks!

[Erasmus00]

[imath] e^+ + e^- -> \gamma \gamma [/imath] is actually photons interacting with (virtual) electrons and positrons. It isn't a photon/photon coupling.

[lawcat]

Electric field is a charge differential field, or potential field. In other words, there has to be 1) a charge, 2) another charge, and 3) a "measurable" difference in charge between the two. Then you have an Electric field.

 

An Electric Field is a "vector" force field. Thus, an electric field is charge measurable and a vector field.

 

Now to your question. the answer yes, of course there have been experiments. Any time there are three separate charges present, you have three measurable vector fields and their interactions, or resulting field. You can measure the field, and you can calculate it through vector analysis.

 

Further, all modern physics experiments and equations are based on this measurability of electric fields, and its vector modeling. If you look at Schrodinger, LaPlace operator, Klein Gordon, Maxwell, symmetry cancellation, all of those use vector calculus, the del operator, the partial derivative etc. So what happens when two fields collide can be modeled in vector calculus, and you get vector additions, and curvatures, etc.

 

Edit: Every paper that deals with atoms or molecules, or anything that has to do with electricty describes interaction between electric fields, starting with one of the most famous papers, Bohrs model of atom.

[Little Bang]

An excellent post lawcat. If we look at a collision between two protons, each traveling at say seventy percent of C with respect to each other, wouldn't the resulting collision collapse the electric field between them to a very high field density? If that is the case would we be able to calculate the field density at one nanosecond before the actual contact?

[lawcat]

I can't answer your question with much confidence because I am not a physicist. I am an electrical engineer. But you are describing proton-proton fusion. I am unsure if 70% c velocity carries enough energy for complete fusion, such as that which occurs in the Sun, a star. But here is a good starting point for your research. Proton-proton fusion. It appears that if the fusion is complete, energy or velocity is sufficient for complete fusion, then one of the fields collapses, one of the protons becomes a neutron. Also, I am sure that this type of fusion has been discussed ad nauseum here at hypography, so some of the older mods may find a quick answer for you.

 

As far as field density, there is no such thing. There is electric field intensity, and electric flux density. Field intensity depends on the medium. Flux density does not. Field is charge measurable. Flux is not directly measurable, but can be calculated or indirectly measured beacuse the flux = charge. Field intensity depends on the distance between charges and the medium, and does not depend on the time. So yes we can determine field intensity at one nanosecond before the collision as long as we know the distance between charges at one nanosecond before the collision. Flux density can be linear, surface, or volume density, and that can be calculated from the Field intensity and the area of interest.

[Qfwfq]

Does anyone know of experiments to test what happens to an electric field when it is squeezed with another electric field?

They say ouch because being squeezed together hurts them! :cocktail:

 

The EM field is almost perfectly linear because, as already said here, photons don't interact directly with each other; the indirect coupling between them is very very weak. Consequently, with electric fields the superposition principle is quite exact for practicle purposes.

 

It appears that if the fusion is complete, energy or velocity is sufficient for complete fusion, then one of the fields collapses, one of the protons becomes a neutron.

According to your link, the proton becomes a neutron emitting a positron and a neutrino. I wouldn't really say its field collapses.

 

When two like charges are brought together, the field along the conjoining line isn't greater than when they are separated. The field around them, yes, the nearer they get and the more they are like a single charge equal to their sum.

[lawcat]

I wouldn't really say its field collapses.

.

 

Let me explain and see if you can agree.

 

The electric field E is proportional to charge Q, and inversly proportional to the square of the distance r. E = Q / 4 * pi * r^2 * permitivity = k * (Q/r^2) = E

 

When two protons approach each other, there are two fields, each attributable to each proton. So, field one, E1 = kQ1/r^2. And field two, E2 = kQ2/r^2. The two fields are vector force fields, and the force is repulsive for each field due to the action between like charges.

 

Just before the collision, the distance r is so small, the field intensity E approaches infinity. However, that unstable structure can not last, and one of the protons becomes a neutron. Since neutron has no charge, Q = 0, then E2 = 0 * k/r^2 = 0. The field E2 collapses, but the collapsed field emits a positron and a neutrino.

[Qfwfq]

Just before the collision, the distance r is so small, the field intensity E approaches infinity.

Uhm, which E? E1 or E2? Each of these depends on distance from the respective charge, neither of them depends on the position of the other charge.

 

Now, by the time you get to a distance of about 1 fermi from the centre of a proton you can no longer consider it a point charge and hence the Coulomb potential packs its bags and steps out, the corrected potential won't approach infinity.

 

However, that unstable structure can not last, and one of the protons becomes a neutron. Since neutron has no charge, Q = 0, then E2 = 0 * k/r^2 = 0. The field E2 collapses, but the collapsed field emits a positron and a neutrino.

No, these are emitted by some dumb quark in one of the protons and this is why that one becomes a neutron. Charge is conserved. It's a weak interaction process, just like beta decay in which a neutron becomes a proton emitting eletron and neutrino. You could almost say the two cases are the opposite of each other, only that the lepton is outgoing in both.

[lawcat]

Uhm, which E? E1 or E2? Each of these depends on distance from the respective charge, neither of them depends on the position of the other charge.

Both E1 and E2. Neither E field depends on the other charge, but the force the field exerts, the E intensity, depends on the distance between charges.

 

Now, by the time you get to a distance of about 1 fermi from the centre of a proton you can no longer consider it a point charge

I was not taking into consideration that one proton can get inside the other.

 

No, these are emitted by some dumb quark in one of the protons and this is why that one becomes a neutron. Charge is conserved. It's a weak interaction process, just like beta decay in which a neutron becomes a proton emitting eletron and neutrino. You could almost say the two cases are the opposite of each other, only that the lepton is outgoing in both.

 

I don't know about all that. I was just considering the interaction between fields as charges approach each other.

 

Back to OP, another simple example of the electric fields interacting is two magnets. The repelling force that we feel or observe between two magnets is the electric field intensity.

[Qfwfq]

Both E1 and E2. Neither E field depends on the other charge, but the force the field exerts, the E intensity, depends on the distance between charges.

I've never called the force the field intensity. It's always been F = qE in my books.

 

I was not taking into consideration that one proton can get inside the other.

They can even go right through each other! Mainly, you were reasoning as if all the charge of each were at one surface point. Even if they were rigid balls, with charge density fixed and evenly spread around the surface, the maximum force would be when they reach contact and it would be given by setting the denominator to the sum of the radii.

 

I was just considering the interaction between fields as charges approach each other.

The superposition principle is what makes sense. It isn't an actual interaction, only a matter of linearity, the combined effect of both on whatever they might act upon. Each of the two fields essentially minds its own business.

 

Back to OP, another simple example of the electric fields interacting is two magnets. The repelling force that we feel or observe between two magnets is the electric field intensity.

Well, it's a bit more complicated than that...

[lawcat]

I agree. No real disagreement, 'cept for semantics.