The D field in Maxwell's equations

[QuasarBoy543298]

I recently saw that in the solution of a problem the following assumption was made - "there are no free charges in the problem, therefore the D field must be equal to 0 ". however if we use that logic to calculate the field of a polaraized sphere we get a wrong result (E=-P/e0 instead of E = -P/(3*e0)).
so my question is - what is wrong here? does it really true to treat D as the field that related only to the free charges? (assuming we work with linear dielectrics and curlP = 0)

 

[Charles Link]

$\nabla \cdot \vec{D}=\rho_{free}=0$ everywhere does not imply $\vec{D}=0$ everywhere. There are solutions with finite $\vec{D}$ that have $\nabla \cdot \vec{D}=0$. $\\$ The uniform $\vec{D}=\frac{2}{3} \vec{P}$ inside a dielectric sphere with uniform polarization $\vec{P}=P_o \hat{k}$ along with the field external to the sphere from the surface polarization charge $\sigma_p=\vec{P} \cdot \hat{n}$ is a good example of this. If you check these solutions, you will see $\nabla \cdot \vec{D}=0$ everywhere.

 

[QuasarBoy543298]

so what does D represents , and why is it useful? I thought that the whole point of defining D is to ignore the "matter effects" (at least in dielectrics) .

 

[Charles Link]

$D$ is a mathematical construction. It can be somewhat useful, but it doesn't make the polarization charge $\rho_p=-\nabla \cdot P$ disappear. Most of the time the electric displacement vector $D$ simplifies some of the algebra, but it really can't work magic. Polarization charge along with free charge is responsible for the electric field $E$, which is the physical entity, along with the polarization $P$ . The electric displacement $D$ is a mathematical construction, and is non-physical. $\\$ One thing you will note in these E&M problems is that picking the correct homogeneous solution is often necessary in solving the problem. For example, suppose we did have a $\rho_{free}$ in the problem. We begin with $\nabla \cdot D=\rho_{free}$, and can always write $D(\vec{x})=\frac{1}{4 \pi} \int \frac{\rho_{free}(\vec{x}')(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \, d^3x'$ as the inverse square law integral solution to the non-homogeneous differential equation. (In our case above, where we had $\rho_{free}=0$, your proposal that $D=0$ is the corresponding solution that would apply in the limit $\rho_{free} \rightarrow 0$ ). We must also consider solutions of the homogeneous equation $\nabla \cdot D=0$ to obtain the complete and correct solution. $\\$ You will find the necessary inclusion of a homogeneous solution to sometimes also be the case in magnetostatics, e.g. when considering solutions to the equation $\nabla \times H= J_{free}$, where $J_{free}$ is the current density for currents in conductors. For this curl operator type differential equation, the integral solution is of the Biot-Savart form. The solution that is obtained in this manner ignores the magnetic material with magnetization $M$ that can be present with contributions to $H$ from magnetic poles $\rho_m=-\nabla \cdot M$. The $H$ from them, (found with a similar integral inverse square law equation to the $D$ above with $\rho_m$ replacing $\rho$), obeys $\nabla \times H=0$. The correct homogeneous solution for $H$ must be included to have the complete solution for $H$. Once you have the correct $H$, the next step in the solution is to write $B=\mu_o H+M$ to get the result for $B$. $\\$ This covers more than your original question, but I think you are likely to also encounter the magnetostatic case, with a similar feature, in the near future. (In a similar manner to $D$, the "field" $H$ is also non-physical, and is a mathematical construction. $B$ and $M$ are the quantities that are physical entities).

 

[Delta2]

in my opinion both D and P are mathematical constructs that have some very good intuitive correspondence: loosely and intuitively speaking, D is the electric field that would exist if the free charge density was the only one that existed, and P is the electric field that would exist if the polarization charge density was the only one that existed.

But the electric field E which is the physical entity, again loosely and intuitively speaking is the sum of D and P ( to speak more accurately it is $\vec{E}=\frac{\vec{D}-\vec{P}}{\epsilon_0}$).

 

[Charles Link]

in my opinion both D and P are mathematical constructs that have some very good intuitive correspondence: loosely and intuitively speaking, D is the electric field that would exist if the free charge density was the only one that existed, and P is the electric field that would exist if the polarization charge density was the only one that existed.

But the electric field E which is the physical entity, again loosely and intuitively speaking is the sum of D and P ( to speak more accurately it is $\vec{E}=\frac{\vec{D}-\vec{P}}{\epsilon_0}$).

I would like to make a correction to this. $P$ is a physical entity, and the polarization charge density is given by $\rho_p=-\nabla \cdot P$. From this $E_p(x)=\frac{1}{4 \pi \epsilon_o} \int \frac{\rho_p(x')(x-x')}{|x-x'|^3} \, d^3x'$. The construction $D=\epsilon_o E+P$ shows what it is comprised of upon taking the divergence, ( $\nabla \cdot$ ), on both sides of the equation. $E_p$ along with $P$ is the important parameter. The polarization $P$ is not an electric field and does not behave like an electric field, even though the equation $D=\epsilon_o E+P$ might suggest that it does.

 

[Delta2]

I don't understand Charles, isn't $E_p=P$?. Anyway viewing polarization as a physical process or physical phenomenon ofcourse it is something real. And the polarization vector tell us how strong is polarization so yes it has a real physical meaning afterall.

 

[Charles Link]

I don't understand Charles, isn't $E_p=P$?. Anyway viewing polarization as a physical process or physical phenomenon ofcourse it is something real. And the polarization vector tell us how strong is polarization so yes it has a real physical meaning afterall.

The electric field that results from polarization charges that occur as a result of polarization $P$ depends on the geometry. In some cases, such as for a uniformly polarized sphere, it is simply the result of polarization charges on the surface where polarization surface charge density $\sigma_p=P \cdot \hat{n}$. For the spherical geometry, $E_p=-\frac{1}{3} \frac{P}{\epsilon_o}$ everywhere inside the sphere, as the OP mentioned in his first post. (Since $P$ is uniform inside the sphere, $- \nabla \cdot P=0$, so there is zero polarization charge density inside the uniformly polarized sphere).

 

[Delta2]

The electric field that results from polarization charges that occur as a result of polarization $P$ depends on the geometry. In some cases, such as for a uniformly polarized sphere, it is simply the result of polarization charges on the surface where polarization surface charge density $\sigma_p=P \cdot \hat{n}$. For the spherical geometry, $E_p=-\frac{1}{3} \frac{P}{\epsilon_o}$ everywhere inside the sphere, as the OP mentioned in his first post. (Since $P$ is uniform inside the sphere, $- \nabla \cdot P=0$, so there is zero polarization charge density inside the uniformly polarized sphere).

Judging from the integral equation for $E_p(x)$ you give at post #6 and from $\nabla\cdot P=-\rho_p$ it follows that $E_p(x)=-\frac{P}{\epsilon_0}$ . But regardless of what D, P are exactly and formally, I believe that using our loose intuition the "loose" truth about D and P is as I say in first paragraph of post #5. But then again, yes the "loose" truth can be source of problems and that's why this thread was created ...

 

[Charles Link]

Judging from the integral equation for $E_p(x)$ you give at post #6 and from $\nabla\cdot P=-\rho_p$ it follows that $E_p(x)=-\frac{P}{\epsilon_0}$ (the $4\pi$ factor just changes units). But regardless of what D, P are exactly and formally, I believe that using our loose intuition the "loose" truth about D and P is as I say in first paragraph of post #5. But then again, yes the "loose" truth can be source of problems and that's why this thread was created ...

In general $E_p \neq -\frac{P}{\epsilon_o}$. The equality does in fact hold for a dielectric slab that is uniformly polarized perpendicular to the slab, but for a uniformly polarized sphere $E_p=-\frac{P}{3 \epsilon_o}$. Most of the time, the integral solution for $E_p$ is too mathematically difficult to be readily evaluated, and alternative mathematical techniques, such as the Legendre polynomial method, need to be employed to determine $E_p$.

 

[Delta2]

In general $E_p \neq -\frac{P}{\epsilon_o}$. The equality does in fact hold for a dielectric slab that is uniformly polarized perpendicular to the slab, but for a uniformly polarized sphere $E_p=-\frac{P}{3 \epsilon_o}$. Most of the time, the integral solution for $E_p$ is too mathematically difficult to be readily evaluated, and alternative mathematical techniques, such as the Legendre polynomial method, need to be employed to determine $E_p$.

Hmmm from what you saying at post #2, I infer that $E_D=\frac{2}{3}\frac{P}{\epsilon_0}$, $E_p=-\frac{P}{\epsilon_0}$ and $E=E_D+E_p=-\frac{P}{3\epsilon_0}$. Where do I go wrong?

 

[Charles Link]

Hmmm from what you saying at post #2, I infer that $E_D=\frac{2}{3}\frac{P}{\epsilon_0}$, $E_p=-\frac{P}{\epsilon_0}$ and $E=E_D+E_p=-\frac{P}{3\epsilon_0}$. Where do I go wrong?

$E_p=-\frac{P}{3 \epsilon_o}$. The evaluation of the integral of $\sigma_p=P \cdot \hat{n}=P \cos(\theta)$to compute $E_p$ at the center of the polarized sphere is relatively straightforward. $E_{p \, center}=-\frac{1}{4 \pi} \frac{P}{\epsilon_o} \iint \cos(\theta) \cos(\theta) \sin(\theta) \, d \theta \, d \phi$ where the extra $\cos(\theta)$ is to pick off the z-component. The result uses the inverse square law and the $a^2$ in the denominator cancels the $a^2$ of the $dA=a^2 \sin(\theta) \, d \theta \, d \phi$. Evaluating the integral, with $\theta$ from $0$ to $\pi$ gives: $E_p=\frac{1}{2}\frac{P}{\epsilon_o}[\frac{1}{3}\cos^3(\theta)|_0^{\pi}]=-\frac{1}{3}\frac{P}{\epsilon_o}$. $\\$ Meanwhile $E_{total}=E_A +E_p$ where $E_A$ is any externally applied electric field. In this case we assumed it to be zero, with a spontaneous polarization $P$. There is no $E_D$. The $D$ is defined by $D=\epsilon_o E+P$. For this problem, $D=\frac{2}{3}P$ everywhere inside the sphere. The OP was thinking, (incorrectly), that the $D$ might be zero because $\nabla \cdot D=0$ everywhere. $\\$ We just evaluated the $E_p$ at the center of the sphere, but more detailed calculations do show it is indeed uniform everywhere inside the sphere.

 

[kuruman]

The electric displacement $D$ is a mathematical construction, and is non-physical.

Yet, the normal component of $\vec D$ is continuous across a dielectric interface as is the tangential component of $\vec E.~$ Is the electric field more physical than the electric displacement? I don't think so. Arguably, the electric field is also a mathematical construct in terms of a force as in $\vec E=\vec F/q.~$ So now we have a force field which itself is a mathematical construct to help relate, through space, the acceleration of an object to its cause.

 

[Charles Link]

Yet, the normal component of $\vec D$ is continuous across a dielectric interface as is the tangential component of $\vec E.~$ Is the electric field more physical than the electric displacement? I don't think so. Arguably, the electric field is also a mathematical construct in terms of a force as in $\vec E=\vec F/q.~$ So now we have a force field which itself is a mathematical construct to help relate, through space, the acceleration of an object to its cause.

The equation $\nabla \cdot E=\frac{\rho_{total}}{\epsilon_o}$ is very straightforward. The equation $\nabla \cdot D=\rho_{free}$ suggests there is something, which I claim is simply a man-made construction, that allows $D$ to sort out the difference between electrical charges that are of the "free" variety, as opposed to being "polarization" charges.

 

[Delta2]

$E_p=-\frac{P}{3 \epsilon_o}$. The evaluation of the integral of $\sigma_p=P \cdot \hat{n}=P \cos(\theta)$to compute $E_p$ at the center of the polarized sphere is relatively straightforward. $E_{p \, center}=-\frac{1}{4 \pi} \frac{P}{\epsilon_o} \iint \cos(\theta) \cos(\theta) \sin(\theta) \, d \theta \, d \phi$ where the extra $\cos(\theta)$ is to pick off the z-component. The result uses the inverse square law and the $a^2$ in the denominator cancels the $a^2$ of the $dA=a^2 \sin(\theta) \, d \theta \, d \phi$. Evaluating the integral, with $\theta$ from $0$ to $\pi$ gives: $E_p=\frac{1}{2}\frac{P}{\epsilon_o}[\frac{1}{3}\cos^3(\theta)|_0^{\pi}]=-\frac{1}{3}\frac{P}{\epsilon_o}$. $\\$ Meanwhile $E_{total}=E_A +E_p$ where $E_A$ is any externally applied electric field. In this case we assumed it to be zero, with a spontaneous polarization $P$. There is no $E_D$. The $D$ is defined by $D=\epsilon_o E+P$. For this problem, $D=\frac{2}{3}P$ everywhere inside the sphere. The OP was thinking, (incorrectly), that the $D$ might be zero because $\nabla \cdot D=0$ everywhere. $\\$ We just evaluated the $E_p$ at the center of the sphere, but more detailed calculations do show it is indeed uniform everywhere inside the sphere.

ok you are correct here but I believe we get a different result than $-\frac{P}{\epsilon_0}$ because we deal with a surface polarized charge density.
If we have a volume polarized charge density then the solution of $\nabla\cdot P=-\rho_p$ using green's function and given that $\nabla\times P=0$ is $P=-\frac{1}{4\pi}\int\frac{\rho_p(\vec{r'})(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3}d^3r'$ which is same as the equation you give for $E_p$ in post #6.

 

[kuruman]

The equation $\nabla \cdot E=\frac{\rho_{total}}{\epsilon_o}$ is very straightforward. The equation $\nabla \cdot D=\rho_{free}$ suggests there is something, which I claim is simply a man-made construction, that allows $D$ to sort out the difference between electrical charges that are of the "free" variety, as opposed to being "polarization" charges.

I don't disagree with you. One equation hides polarization in $\rho_{total}$ and the other in $\vec D$. It's a matter of preference. For me it is easier to find $\vec D$ and then use the constitutive equations to sort out $\vec E$, $\vec P$ and the bound charges. Yes, it is easier to conceptualize the electric field as arising from free and bound charges, but it is not very useful when one knows the free charge but not the bound charge. However, I prefer not to think of the electric field as "more physical" than the electric displacement. I would argue that the "more physical" field is $\vec D$ and that $\vec E$ is actually $\vec D$ in the special case where there is no polarizable matter present.

 

[Charles Link]

ok you are correct here but I believe we get a different result than $-\frac{P}{\epsilon_0}$ because we deal with a surface polarized charge density.
If we have a volume polarized charge density then the solution of $\nabla\cdot P=-\rho_p$ using green's function and given that $\nabla\times P=0$ is $P=-\frac{1}{4\pi}\int\frac{\rho_p(\vec{r'})(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3}d^3r'$ which is same as the equation you give for $E_p$ in post #6.

Your calculation of $P$ from the differential equation $\nabla \cdot P=-\rho_p$ is, what I would call, quite clever. I did a similar thing one time with the corresponding magnetostatic equation, $\nabla \cdot M=-\rho_m$, using a uniformly magnetized cylinder of finite length, and was very puzzled why this computation did not give me the original magnetization $M$. You will find the same kind of thing in the computation for $P$.$\\$ e.g. Take a cylinder of finite length of uniform polarization $P$ pointing along its axis. There are two pole endfaces that get polarization surface charge density. Clearly the $P$ that your calculation gives from this does not generate the original distribution of $P$. Your solution is missing the correct homogeneous solution to $\nabla \cdot P=0$, that, when added to your solution, will give the uniform distribution for $P$ that you started with.

 

[Charles Link]

Just an additional comment on this: Starting with the equation $B=\mu_oH+M$, (which is really left unproven in most E&M textbooks. I proved this independently a number of years after I had my college E&M courses=see https://www.overleaf.com/read/kdhnbkpypxfk for a detailed proof. The paper uses c.g.s. units, but the proof would be identical using MKS.), some interesting calculations can be done: Taking divergence of both sides and setting $\nabla \cdot B=0$, we get $\nabla \cdot (\mu_o H)=-\nabla \cdot M$. One might think that this implies $\mu_o H=-M$, but for reasons discussed above, this is not the case. $\\$ Similarly, in cases of no free electric charge, where $\nabla \cdot D=0$, this does not imply that $\epsilon_o E=-P$. (Starting with $D=\epsilon_o E+P$).$\\$ One further comment is I don't recall the E&M instructors emphasizing this and other finer points, with the result that E&M was a very difficult subject in the way it was presented. In hindsight, I think they could have made it a lot easier, if the professors would have shown us a few more of the finer details as well as a couple of the more basic concepts in a manner that was easier to understand. $\\$ e.g. In the problem of a polarized sphere, the simpler approach is the one the OP has taken, where the polarization is spontaneous and not caused by an external electric field. If an external electric field is introduced, and the dielectric has a susceptibility $\chi$, so that $P=\epsilon_o \chi E$, the problem, with the solution being of the self-consistent variety, becomes much more complex. In the case of the polarized sphere, it would be beneficial to first present the simpler case, (of a spontaneous polarization), to illustrate the concepts.

 

[Delta2]

Just an additional comment on this: Starting with the equation $B=\mu_oH+M$, (which is really left unproven in most E&M textbooks. I proved this independently a number of years after I had my college E&M courses=see https://www.overleaf.com/read/kdhnbkpypxfk for a detailed proof. The paper uses c.g.s. units, but the proof would be identical using MKS.), some interesting calculations can be done: Taking divergence of both sides and setting $\nabla \cdot B=0$, we get $\nabla \cdot (\mu_o H)=-\nabla \cdot M$. One might think that this implies $\mu_o H=-M$, but for reasons discussed above, this is not the case. $\\$ Similarly, in cases of no free electric charge, where $\nabla \cdot D=0$, this does not imply that $\epsilon_o E=-P$. (Starting with $D=\epsilon_o E+P$).

I think I ll have to disagree after having a careful look at Helmholtz Decomposition Theorem https://en.wikipedia.org/wiki/Helmholtz_decomposition and specifically the section https://en.wikipedia.org/wiki/Helmholtz_decomposition#Fields_with_prescribed_divergence_and_curl.
For the static case where we have $\nabla \times E=\nabla \times B=0$ taking the divergence of $B=\mu_oH+M$ (1) we get
$\nabla \cdot (\mu_o H)=-\nabla \cdot M$ (2) and taking the curl of (1) we get
$\nabla \times (\mu_o H)=-\nabla \times M$ (3).

(2) and (3) tell us that the fields $-M$, and $\mu_o H$ have the same curl and the same divergence, hence from Helmholtz decomposition theorem and with the assumption that they satisfy some additional smoothness and vanishing conditions they are uniquely determined, i.e they are equal. And similar result we can get for D and P. So at least for the static case I can't agree with you on this. BUT it might be the case that due to irregular boundary conditions, the smoothness and vanishing conditions required by Helmholtz theorem are not satisfied, so you might be right for such cases afterall.

 

[Charles Link]

@Delta² At least I got you thinking. The computation of $H$ for a cylinder of finite length of uniform magnetization is relatively straightforward. Magnetic pole density is $\rho_m=-\nabla \cdot M$. This means you get a surface charge $\sigma_m=M \cdot \hat{n}$ on the end faces, and no other $\rho_m$. The inverse square law applies to the results to compute $H$, and it is quite apparent that $H \neq \frac{-M}{\mu_o}$. (Geometrically, $H$ is much different from the uniform $M$ that we need). The $H$ doesn't have any homogeneous solution in this case, but the $M$ does. $\\$ (The goal here is to recover $M$ from the differential equation $\nabla \cdot M=-\mu_o \nabla \cdot H$ ). $\\$ The homogeneous solution for $M$ that will give the correct result, (from the equation $\nabla \cdot M=0$), consists of a uniformly magnetized cylinder of infinite length, along with (edit: it needs to be the $-\mu_o H$ from) two cylinders of semi-infinite length of magnetization $-M$ that are just outside the region of the cylinder of length $L$ that when superimposed will leave just the cylinder of length $L$ of magnetization $M$. $\\$ I leave it to you to work out the details at the end faces. Everywhere else, these magnetized portions obey $\nabla \cdot M=0$ , and these are what is needed to recover the original magnetization of length $M$ after the solution $M=-\mu_o H$ from the inhomgeneous equation, $\nabla \cdot M=-\mu_o \nabla \cdot H$, is employed. Note that the solution $M=-\mu_o H$ (look closely at the spatial extent of $H$) extends outside the physical region of the magnetized cylinder. (I'm assuming here the Green's function solution to $M$, which will yield $M=-\mu_o H$).

 

[Delta2]

I believe when we have surface charge densities the smoothness conditions required by Helmholtz decomposition theorem break down. Specifically I believe for the example you give the divergence of M becomes infinite at the surface of the end faces of the cylinder.

 

[Charles Link]

Hmmmm, I believe when we have surface charge densities the smoothness conditions required by Helmholtz decomposition theorem break down. Specifically I believe for the example you give the divergence of M becomes infinite at the surface of the end faces of the cylinder.

I'm also ironing out some of the details of this. Given $\nabla \cdot M=-\mu_o \nabla \cdot H$, and using the known $H$ and putting it in there basically leaves you with $\nabla \cdot M=0$ except at the end faces. Thereby, basically to recover $M$ we are solving a homogeneous differential equation, except for sources at the end faces. Recovering $M$ is not the simplest of operations, and it is completely different from $- \mu_o H$. $\\$ Edit: See also the "edit" in the middle paragraph of post 20 that I just added. I think that's the missing piece. Notice that has $\nabla \cdot H=0$ so if we let $M=-\mu_o H$, it also has $\nabla \cdot M =0$. Anything that has $\nabla \cdot M=0$ is fair game for this solution. And note that the function $M$ covers an infinite spatial extent here, even though the $M$ we are trying to recover has a finite spatial extent.

 

[DrDu]

I just want to note that both magnetisation and polarisation may be derived from $\partial_\mu \Pi^{\mu \nu} =j^\nu$, where $\Pi^{\mu \nu}$ is an antisymmetric tensor (the polarisation magnetisation tensor) with $\Pi^{0 i} = P_i$ and $\Pi ^{ij} =\epsilon^{ijk} M_k$ and $j$ is the charge current four vector. The definition is not unique as there are non-trivial solutions of the homogeneous equation $\partial_\mu \Pi^{\mu \nu} =0$. The condition on $j^\mu$ that $\partial_\mu j^\mu=0$ (charge conservation) is guaranteed to hold if j is derived from P.
Especially in optics, the freedom in the choice of $\Pi$ is used to make M=0, so that all material effects are subsumed in P.

 

[Charles Link]

@Delta² and @saar321412 I would like to return to the case of the polarized sphere: $\\$ It would seem logical, though incorrect, that since $\nabla \cdot D=0$, so that $\epsilon_o \nabla \cdot E=-\nabla \cdot P$, that $E_p=-\frac{P}{\epsilon_o}$ would be the solution. (both inside and outside the sphere).$\\$ One major problem with this solution though is that in the region where $P=0$ (outside of the sphere), it necessarily has $E=0$ outside of the sphere. A surface distribution of polarization charge of the form $\sigma_p=P_o \cos(\theta)$ does not lend itself to $E$ being zero everywhere outside the sphere. $\\$ Again, the best answer I have for why this doesn't work is, because of homogeneous solutions that may be differ between the $E$ and the $P$, the differential equation $\epsilon_o \nabla \cdot E=-\nabla \cdot P$ does not imply that $\epsilon_o E=-P$.