# The D Vector

1. Feb 20, 2007

### Swapnil

Why is D called the "electric flux density vector"?

Last edited: Feb 20, 2007
2. Feb 20, 2007

### Tom Mattson

Staff Emeritus
Never heard of the "electrostatic flux density vector". Are you perhaps referring to the electric displacement field? It also goes by the name $\vec{D}$, and it is the solution to Gauss' law:

$$\vec{\nabla}\cdot\vec{D}=\rho$$

3. Feb 20, 2007

### robphy

"electric flux density vector" is appropriately named because its divergence [in Tom Mattson's reply] is the electric-flux per unit volume.

4. Feb 20, 2007

### Tom Mattson

Staff Emeritus
Ah well, in that case back to Swapnil's question:

In addition to being the solution to Gauss' law, which I posted earlier, we can approach $\vec{D}$ a different way. Let's start with an electric field $\vec{E}$ in space in which there are free charges. It satisfies the equation $\nabla\cdot\vec{E}=\rho/\epsilon_0$ (which is just Gauss' law under these special circumstances). Now bring into the space a polarizable material. The electrons are pushed around to create another charge density, called the bound charge density $\rho_B$, which gives rise to a polarization density $\vec{P}$. $\vec{P}$ is the solution to the differential equation $\nabla\cdot\vec{P}=-\rho_B$.

So now Gauss' law reads as follows.

$$\nabla\cdot\vec{E}=\frac{\rho + \rho_B}{\epsilon_0}$$
$$\epsilon_0\nabla\cdot\vec{E}=\rho+\rho_B$$
$$\nabla\cdot\epsilon_0\vec{E}=\rho-\nabla\cdot\vec{P}$$
$$\nabla\cdot(\epsilon_0\vec{E}+\vec{P})=\rho$$

Now define the operand of the divergence operator to be $\vec{D}=\epsilon_0\vec{E}+\vec{P}$, and you've got it.

Last edited: Feb 20, 2007
5. Feb 20, 2007

### ObsessiveMathsFreak

The units of D are C/m^2, Columbs per meter squared. D is a measure of columbs per unit area and as such is most appropriately integrated over areas. In vector notation, this is not explicitly clear. However, in differential forms notation, D is explicitly a two-form, which means the only thing you can do with it is integrated it over 2D manifolds, i.e. surfaces. In Geometric Calculus notation, D is a bivector, or directed area, so the distinction is explicit here also.

D is a flux because it is a measurement of how much "charge" is "flowing" through a surface. Of course charges are actually moving, so I should really say it's a measure of the density of the electric field lines through that surface. Where the electric field lines bunch up, the electric flux density is high, and where they are spare the flux density is low. Of course, implicitly I mean this to be the case for surfaces that the field lines are perpendicualar to, but in vector notation D is considered apart from the surfaces you integrate it over.

6. Feb 20, 2007

### leright

Because when you integrate the D-field over a closed surface the number that results is the net OUTWARD flux. So, D is therefore called the electric flux density. It is the flux per unit surface area.

If there is a charge enclosed by the surface of integration then you will get a net outward flux through the surface. The net outward flux is equal to the charge enclosed.

7. Feb 20, 2007

### Swapnil

Thanks guys. Now I know.

BTW, is there something wrong with calling D the "electric field strength" (just like we call H the "magnetic field strength")?

Last edited: Feb 20, 2007
8. Feb 21, 2007

### arunma

To be honest, I never heard about this name until I read this thread. I'm aware that H is sometimes called the "magnetic field strength." But I remember my professors and textbook authors insisting that all of the names people have invented for these two vectors are pointless, and that we should simply call them D and H. My guess is that the letters are in greater popular use than these funny names. IOW, we probably shouldn't be naming the vectors when we write papers.

9. Feb 21, 2007

### ObsessiveMathsFreak

E and H are fields. (V/m, A/m)

D and B are fluxes. (C/m^2, Wb/m^2)

10. Feb 21, 2007

### robphy

In my opinion, one should use the "operational" units [or better, the arguably more fundamental differential-form] as a guide to naming these quantities.

Via differential forms [https://www.amazon.com/Tensor-Analysis-Physicists-Second-Schouten/dp/0486655822", etc...]:

• D, a [twisted] two-form with units Coulomb/m^2 [as mentioned above],
appears in Gauss' {Electric] Law and Ampere-Maxwell's Law
(some names: electric flux-density, electric induction, [di-]electric displacement, electric excitation)
• B, a two-form with units Weber/m^2,
appears in Gauss' [Magnetic] Law and in the Lorentz Force Law
(some names: magnetic flux-density, magnetic field induction, magnetic field strength*less common)
• E, a one-form with units of Volt/m,
appears in Faraday's Law and in the Lorentz Force Law
(some names: electric field strength, electric field intensity)
• H, a [twisted] one-form with units of Ampere/m,
appears in Ampere-Maxwell's Law
(some names: magnetic field strength, magnetic field intensity, magnetic excitation)
These differential-forms have a natural geometric interpretation and visualization.

It seems that, historically, there has been some conflicting naming conventions regarding H and B.

Some groupings to note [as mentioned above]:
• flux-densities are associated with the 2-forms: D and B
field-intensities are associated with 1-forms: E and H
• forces are associated with E and B
sources are associated with D and H (hence, "excitations")
• upon making the usual [Hodge-dual] association with these differential-forms and vectors, E and D-vector are "polar vectors", whereas B and H are "axial vectors"

It would be nice if these names [including those from the more-standard textbooks (Jackson, etc...) ] were surveyed and all sorted out.

Last edited by a moderator: May 2, 2017
11. Feb 21, 2007

### marcusl

Here's a short survey from an earlier thread

More on whether to call H or B the magnetic field:

As for "getting it all sorted out", it didn't happen during the first 150 years of E&M theory. Maybe the next 150? :rofl:

Last edited by a moderator: Apr 22, 2017
12. Feb 21, 2007

### Swapnil

Wow! I didn't know that such a simple question would have such an interesting answer!

BTW, robphy the way you describe E, H, B, & D as geometric objects with "one/two-form;" what branch of mathematics would you need to study in order to become familiar with these types of terminologies?

13. Feb 21, 2007

### robphy

tensor calculus and differential geometry.
consult the references I linked in my previous post

14. Feb 21, 2007

### arunma

:yuck:

I've had a fair amount of experience with both, and I can say with scientific certainty that tensor calculus is the devil!

Seriously though, thanks for the links. To be honest I never even paid attention to the units of D and H. Now I know why they're useful for finding the field inside a material.

15. Feb 21, 2007

### Meir Achuz

Your units for D and B are for flux densities.

16. Feb 22, 2007

### shungmunga

Hi. I am reading the text by Jackson that was mentioned by marcusl in this thread and am having trouble understanding the nature of the polarisation P and electric displacement D vectors. I only want to go so far as these fundamental concepts in knowing electrodynamics and I would greatly appreciate someone clarifying my misunderstandings. In my understanding, “flux” is only a mathematical concept meaning the surface integral of a vector field. For the case of an electric field, with coulomb’s constant and the charge taken outside the integral, the integral becomes that of the solid angle subtended by the surface, which for an enclosing surface is $4\pi$. So Gauss’ law says that the flux is directly proportional to the charge enclosed, which has the differential form:
$$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_{0}}$$
This would mean that determining “flux density”, which is the area density of flux, undoes the process of integrating the field over the surface area and is exactly the same as the “field”, just a different word.

If it were taken as given that $\vec{D}=\varepsilon\vec{E}$, then not only the $4\pi$ would cancel but the permittivity as well, leaving:
$$\nabla\cdot\vec{D}=\rho$$

My problem is that I don’t have anywhere near the maths ability or the motivation to understand how to get a polarisation vector from the dipole moments of the molecules of a dielectric and then the associated function for potential but in Jackson’s book, its got:
$$\nabla\cdot\vec{E}=\frac{1}{\varepsilon_{0}}(\rho-\nabla\cdot\vec{P})$$
and with P assumed directly proportional to E, this does in fact lead to $\vec{D}=\varepsilon\vec{E}$.

I think that E could rightly be called both the “flux density” and the “field”, but some sources, including lehright in this thread, claim that D can be called the “flux density” because it is a measure of the net outward flux. To me, this implies that D is a measure of the equivalent field in the material summing the effects of E and the polarised molecules, and everyone says that polarisation reduces the effect of E rather than aiding it. If:
• D is the net outward flux and
• Polarisation does reduce the effect of E and
• P does equal 0xeE (0 and xe being positive numbers)
then why is the plus a plus in:
$$\vec{D}=\varepsilon_{0}\vec{E}+\vec{P}$$
when polarisation apparently reduces E and doesn’t aid it. Or alternatively, why are $\varepsilon_{0}$ and $\chi_{e}$ positive.
On the other hand, Jackson says that “the fundamental fields are E and B… The derived fields, D and H are introduced as a matter of convenience” which I’m guessing means that D does not have the physical significance of an equivalent field and shouldn’t be called flux density either.

I have the same problem in magnetostatics with magnetisation M where:
$$\vec{H}=\frac{1}{\mu_{0}}\vec{B}-\vec{M}$$
when paramagnets and ferromagnets with positive susceptibility supposedly increase the field instead of reduce it. So my question is what is the physical meaning of vectors P, M, D and H? Also, what is the point of the permittivity number if it just cancels with that in the denominator of Coulomb’s constant when multiplying it with E? Someone please help show me where I’m going wrong.

17. Feb 22, 2007

### Meir Achuz

You might do better with "Classical Electromagnetism" by Franklin which answers your question in chapter 6.

18. Feb 22, 2007

### shungmunga

sounds good but i may not be able to get hold of the book so i was wondering if you could give me the jist of it, mainly whether of not D and H are fields equivalent to the net effects of E and P and B and M respectively.

19. Feb 23, 2007

### Meir Achuz

The main point is that SI units completely confuse the connection between
E,D,P,B,H,M. You just have to look at any text that uses gaussian units.
In gaussian units, E is the field due to all charges (free and bound).
D is the field due to only the free charges.
P is the dipole moment per unit volume of the matter.
-div P gives the bound charge density in the matter,
and P_n at the surface is bound surface charge.
If D=E+4pi P, then div D=4pi rho_{free}, and div E=4pi rho_{free+bound}