12 ms·

There’s indeed something terribly wrong with the way they explain/teach electricity, to the point that I sometimes suspect that whoever is doing it are themselv

There’s indeed something terribly wrong with the way they explain/teach electricity, to the point that I sometimes suspect that whoever is doing it are themselves confused, whether they realize it or not. In particular, I am yet to find a lucid and intuitive explanation of why we need both E and D for the electric field and B and H for the magnetic field. (And no, talking about them being “differential forms” of various kinds is not helpful in most cases.) Also, the multitude of systems of units does not help learning, either.

dataflow 6y ago
E = q2 F basically motivates & defines the electric field from first principles by normalizing the force against the charge of the particle experiencing it. It relates the object to the properties of its surrounding space. D = q1 / 4 pi R^2 describes the effect the first particle has on space at the location of the second particle. It relates the subject to the properties of its surrounding space. And D = E epsilon of course tells you how those quantities relate to each other. It's basically the verb. Together these describe how two stationary particles interact with each other electrically. They link the subject and object together. Think of it as subject -> verb -> object. The motivation for all of is to decouple things so we can reason about them independently. Hopefully that's lucid and intuitive?
- Koshkin 6y ago
  You would think that (oh the curse of knowledge!), but there is still something that muddies the waters here. It is the missing relation between the physical electric field which is what the free charges actually generate and the (still mysterious) displacement field. This makes your explanation more abstract that it needs to be. The "D = E epsilon" that you mention in passing is in fact another source of confusion, because it is not clear whether E is what the free charges ("q1") generate or it is what acts on the probe ("q2").
frongpik 6y ago
I've been trying to understand the same thing: what those equations actually describe. So far I've come up with this analogy. The B field describes surface of the magnetic medium or substance of unknown origin, the E field is merely the apparent velocity of this surface. Electrons and other charged particles blindly run down the slopes of the magnetic medium. There's also this rot B and rot E part that creates vorticity, similar to what we see in liquids.
- fsh 6y ago
  This sounds very much like the aether hypothesis of the 19th century which was abandoned since it is incompatible with many experimental observations, such as the Michelson-Morley experiment. The modern point of view is that EM fields are fundamental and have no underlying mechanism such as an aether medium or something like that.
  - frongpik 6y ago
    Nah, that experiment disproved the "static background medium" within which the earth would be moving.
    - fsh 6y ago
      Sure, there are some loopholes like a dragged aether that could resolve the contradiction with Michelson-Morley. But you are going to have a very hard time explaining all the other tests of special relativity such as Ives-Stilwell type experiments.
effie 6y ago
> I am yet to find a lucid and intuitive explanation of why we need both E and D for the electric field and B and H for the magnetic field. E alone cannot describe fully state of dielectric, and B alone cannot fully describe state of magnetic medium. We need additional concepts, either charge/current densities, or polarization/magnetization, or D/H, in that order of importance. In modern understanding, D and H are auxiliary concepts that are useful mainly in special but important cases they were historically devised for. Let's take D for example. If free charge is absent, Maxwell's equations imply very nice equation div D = 0. The equation for E is more complicated (includes bound charge density that is hard to measure, hard to relate to other things in the model and so is usually preferred to be ignored). If in addition the medium is linear (a special but important case), D is a simple linear function of E ( there is no such simple relation to charge density or current density) so things such as the concept of EM energy in medium is possible, and wave propagation in crystals can be analyzed most elegantly. Sure, all could be analyzed in terms of E and rho and j which would be more "physical" but it is not the most elegant way to do it. It is like in mechanics: we can analyze complicated systems with constraints by writing down many equations involving constraint forces, but there are easier methods such as the Lagrangian.
cygx 6y ago
Note: In the following, I'll be using Lorentz-Heaviside units. For other unit systems, throw in factors of ε_0 or 4π as needed... The reason for the introduction of D is that even though a medium may have no net charge, it's still made from protons and electrons, meaning the charge density will be nonzero. We call this the density of 'bound' charges ρ_b. The medium is often modelled as consisting of tiny dipoles. If left to their own devices, the dipoles tend to align in such a way that the generated field largely cancels out. Now, let's add additional charges via a density of 'free' charges ρ_f, yielding a total charge density ρ = ρ_f + ρ_b. The bound charges will respond to the presence of the free charges, and the field they generate will no longer cancel out. We call this field -P, with P the electric polarization (the negative sign comes in through its definition as the dipole moment density of the dipole model). Defining the electric displacement field as D = E + P, Gauss's law div(E) = ρ can be decomposed into div(D) = ρ_f and div(P) = -ρ_b In many cases, the response of the medium will be approximately linear, and polarization will be proportional to the field generated by the free charges, P = η·D. Note that the definition of P was such that for positive η, the actual field will be opposing D. Equivalently, we can consider the response linear in the total field, P = χ·E if we introduce the electric susceptibility χ = η/(1-η). For the displacement field, we get D = E + P = (1 + χ)·E = ε·E with ε the permittivity. What's the point, you may ask? Simplification, as we can now just measure the linear response of the medium, instead of having to figure out the microscopic distribution of bound charges.
- Koshkin 6y ago
  This is a pretty standard treatment. What remains unclear is whether E is what the free charges generate (and what brings the mysterious D into existence), or it is what acts on the probe charge (which is calculated from D).
  - cygx 6y ago
    E is generated by all charges: div(E) = ρ = ρ_f + ρ_b D is generated by the free charges only: div(D) = ρ_f The quantity that makes an appearance in the Lorentz force law (hence acts on test charges) is E. Admittedly, things get muddled because the distinction E vs D does double duty: We also associate E with (specific) forces, and D with fluxes. Depending on your system of units, this necessitates the introduction of a conversion factor.

dataflow 6y ago

E = q2 F basically motivates & defines the electric field from first principles by normalizing the force against the charge of the particle experiencing it. It relates the object to the properties of its surrounding space. D = q1 / 4 pi R^2 describes the effect the first particle has on space at the location of the second particle. It relates the subject to the properties of its surrounding space. And D = E epsilon of course tells you how those quantities relate to each other. It's basically the verb. Together these describe how two stationary particles interact with each other electrically. They link the subject and object together. Think of it as subject -> verb -> object. The motivation for all of is to decouple things so we can reason about them independently. Hopefully that's lucid and intuitive?

Koshkin 6y ago
You would think that (oh the curse of knowledge!), but there is still something that muddies the waters here. It is the missing relation between the physical electric field which is what the free charges actually generate and the (still mysterious) displacement field. This makes your explanation more abstract that it needs to be. The "D = E epsilon" that you mention in passing is in fact another source of confusion, because it is not clear whether E is what the free charges ("q1") generate or it is what acts on the probe ("q2").

frongpik 6y ago

I've been trying to understand the same thing: what those equations actually describe. So far I've come up with this analogy. The B field describes surface of the magnetic medium or substance of unknown origin, the E field is merely the apparent velocity of this surface. Electrons and other charged particles blindly run down the slopes of the magnetic medium. There's also this rot B and rot E part that creates vorticity, similar to what we see in liquids.

fsh 6y ago
This sounds very much like the aether hypothesis of the 19th century which was abandoned since it is incompatible with many experimental observations, such as the Michelson-Morley experiment. The modern point of view is that EM fields are fundamental and have no underlying mechanism such as an aether medium or something like that.
- frongpik 6y ago
  Nah, that experiment disproved the "static background medium" within which the earth would be moving.
  - fsh 6y ago
    Sure, there are some loopholes like a dragged aether that could resolve the contradiction with Michelson-Morley. But you are going to have a very hard time explaining all the other tests of special relativity such as Ives-Stilwell type experiments.

effie 6y ago

> I am yet to find a lucid and intuitive explanation of why we need both E and D for the electric field and B and H for the magnetic field. E alone cannot describe fully state of dielectric, and B alone cannot fully describe state of magnetic medium. We need additional concepts, either charge/current densities, or polarization/magnetization, or D/H, in that order of importance. In modern understanding, D and H are auxiliary concepts that are useful mainly in special but important cases they were historically devised for. Let's take D for example. If free charge is absent, Maxwell's equations imply very nice equation div D = 0. The equation for E is more complicated (includes bound charge density that is hard to measure, hard to relate to other things in the model and so is usually preferred to be ignored). If in addition the medium is linear (a special but important case), D is a simple linear function of E ( there is no such simple relation to charge density or current density) so things such as the concept of EM energy in medium is possible, and wave propagation in crystals can be analyzed most elegantly. Sure, all could be analyzed in terms of E and rho and j which would be more "physical" but it is not the most elegant way to do it. It is like in mechanics: we can analyze complicated systems with constraints by writing down many equations involving constraint forces, but there are easier methods such as the Lagrangian.

cygx 6y ago

Note: In the following, I'll be using Lorentz-Heaviside units. For other unit systems, throw in factors of ε_0 or 4π as needed... The reason for the introduction of D is that even though a medium may have no net charge, it's still made from protons and electrons, meaning the charge density will be nonzero. We call this the density of 'bound' charges ρ_b. The medium is often modelled as consisting of tiny dipoles. If left to their own devices, the dipoles tend to align in such a way that the generated field largely cancels out. Now, let's add additional charges via a density of 'free' charges ρ_f, yielding a total charge density ρ = ρ_f + ρ_b. The bound charges will respond to the presence of the free charges, and the field they generate will no longer cancel out. We call this field -P, with P the electric polarization (the negative sign comes in through its definition as the dipole moment density of the dipole model). Defining the electric displacement field as D = E + P, Gauss's law div(E) = ρ can be decomposed into div(D) = ρ_f and div(P) = -ρ_b In many cases, the response of the medium will be approximately linear, and polarization will be proportional to the field generated by the free charges, P = η·D. Note that the definition of P was such that for positive η, the actual field will be opposing D. Equivalently, we can consider the response linear in the total field, P = χ·E if we introduce the electric susceptibility χ = η/(1-η). For the displacement field, we get D = E + P = (1 + χ)·E = ε·E with ε the permittivity. What's the point, you may ask? Simplification, as we can now just measure the linear response of the medium, instead of having to figure out the microscopic distribution of bound charges.

Koshkin 6y ago
This is a pretty standard treatment. What remains unclear is whether E is what the free charges generate (and what brings the mysterious D into existence), or it is what acts on the probe charge (which is calculated from D).
- cygx 6y ago
  E is generated by all charges: div(E) = ρ = ρ_f + ρ_b D is generated by the free charges only: div(D) = ρ_f The quantity that makes an appearance in the Lorentz force law (hence acts on test charges) is E. Admittedly, things get muddled because the distinction E vs D does double duty: We also associate E with (specific) forces, and D with fluxes. Depending on your system of units, this necessitates the introduction of a conversion factor.