# Vector calculus identity and electric/magnetic polarization

#### dRic2

Gold Member
Homework Statement
My professor told us to prove $$\int_{V} \mathbf x \rho_{b} d_3 \mathbf x = \int_{V} \mathbf P d_3 \mathbf x$$ using just the relation between bound charges and the electric dipole moment per unit volume.
Homework Equations
$$\rho_b = - \nabla \cdot \mathbf P$$
I spent a good amount of time thinking about it and in the end I gave up and asked to a friend of mine. He said it's a 1-line-proof: just "integrate by parts" and that's it. I'm not sure you can do that, so instead I tried using the identity:

to express the first term on the right-hand side (which is my $\mathbf x \nabla \cdot \mathbf P$) as a summation of all the other therms, but then I don't know how to get rid of all those integrals...

PS: I have to do the same thing for the magnetic field (prove that $\int_{V} \mathbf x × \mathbf J = \int_{V} \mathbf M$) but I hope that, once I get the trick, I can adapt it to the ather case.

Thanks
Ric

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#### timetraveller123

i think it might go something like first you might want to consider the x component of the integral
so
like this
$\int \vec r \nabla . \vec P d^3x|_x = \int x \nabla . \vec P d^3x$
then use divergence rules
$\nabla.(x\vec F) = x\nabla.\vec F + \nabla x . \vec F$

#### dRic2

Gold Member
Elaborating a little bit on what you suggested, can I draw the conclusion that $$\nabla \cdot ( \mathbf x ⊗ \mathbf P) = \mathbf x \nabla \cdot \mathbf P + \mathbf P \cdot \nabla \mathbf x$$
?
With $⊗$ I mean the outer product (or tensor product, I don't really know all this stuff ), basically $(\mathbf x ⊗ \mathbf P)_{ij} = x_iP_j$

#### timetraveller123

i am sorry i don't what is outer product(or tensor product) is either maybe the experts can help

#### Delta2

Homework Helper
Gold Member
I am not familiar with tensor product either, but I don't think we have to do with that, the hint at post #2 is good enough, we just need one more thing:
Is it given by the problem statement that $\vec{P}$ is tangential or zero at the boundary surface $\partial V$ of volume $V$?

#### dRic2

Gold Member
Is it given by the problem statement that →PP→\vec{P} is tangential or zero at the boundary surface ∂V∂V\partial V of volume VVV?
No. That's an other problem I'm trying to solve. I know $\int \nabla \cdot P d_3 x = 0$ because the medium is assumed to be overall neutral. But from that I don't know how to draw any significant conclusion.

BTW I asked about the tensor product because this and the other analogue for the magnetic field - which is far more difficult I think - are supposed to be exercises on vector calculus identities.

#### Delta2

Homework Helper
Gold Member
No. That's an other problem I'm trying to solve. I know $\int \nabla \cdot P d_3 x = 0$ because the medium is assumed to be overall neutral. But from that I don't know how to draw any significant conclusion.
well by divergence theorem we know that $\oint \vec{P}\cdot d\vec{S}=\int \nabla \cdot \vec{P} dV=0$ and if we can take advantage of some sort of symmetry (the following certainly holds if the volume V is spherical) we can conclude that $\vec{P}\cdot d\vec{S}=0$ which means that $\vec{P}$ is perpendicular to the normal, i.e tangential to the surface.

BTW I asked about the tensor product because this and the other analogue for the magnetic field - which is far more difficult I think - are supposed to be exercises on vector calculus identities.
I don't think its much more difficult, you just have to use the $$\nabla \cdot (x \times J)=(\nabla \times x)\cdot J-x\cdot (\nabla \times J)$$ identity for the divergence of cross product.

#### timetraveller123

Is it given by the problem statement that →PP→\vec{P} is tangential or zero at the boundary surface ∂V∂V\partial V of volume VVV?
well from what i have seen in some texts is that the trick is to integrate over a volume larger than the body itself this extra volume would not contribute to the volume integral as the P is zero and hence the surface integral is also zero

edit:
after some looking through i found that griffiths uses this trick when he derives ampere law from biot and savart law

would this apply here?

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#### Delta2

Homework Helper
Gold Member
well from what i have seen in some texts is that the trick is to integrate over a volume larger than the body itself this extra volume would not contribute to the volume integral as the P is zero and hence the surface integral is also zero

edit:
after some looking through i found that griffiths uses this trick when he derives ampere law from biot and savart law
View attachment 250812
would this apply here?
Yes I think this would work here too, however I feel there is something fishy about it, cant figure out what it is.

#### timetraveller123

yes i have the feeling too something feels off about that

#### timetraveller123

i think what it is saying is that surface charges don't exist if we were to take the divergence with sensible results there should be a continuous function if i am not wrong. so maybe like what it says is just that since outside P is zero for the polarization to be continuous since the E field is then the P has to taper off to zero at surface ie no surface charge which makes sense because initially we assumed bound charges is only divergence of P hence only volume charge.
if not then we also have to include surface charge in the bound charge expression from the start i think.....

edit then in that we might have something like this

$-\int x \nabla.\vec P d^3x + \int x \vec P.da = \int \vec P d^3x - \int x \vec P.\vec da + \int x \vec P.da = \int \vec P d^3x\\$
since surface charge is $\vec P.\hat n$

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#### dRic2

Gold Member
@Delta2 but $J$ is the curl of $M$ not the divergence. So it is not that simple.

@timetraveller123 thanks for the insight. I'll think about it when I'll be back home :)

#### dRic2

Gold Member
Ok I got it for $P$, but the analogue for the magnetic field still eludes me...

#### timetraveller123

i am sorry i don't know how to do the second part breaking up into components doesn't work this is what i had so far but i am stuck half way through
$\int \vec r \times \vec J_b d^3x + \int \vec r \times \vec K_b d^2x\\ \int \vec r \times (\nabla \times \vec M) d^3x + \int \vec r \times (\vec M \times \hat n) d^2x\\ \int (\nabla(\vec r.\vec M) - \vec r.(\nabla \vec M) - \vec M )d^3x + \int (\vec M(\vec r.\hat n) - \hat n(\vec r. \vec M) )d^2x$
the first term of the first and last term of second integral cancel out from gradient divergence theorem
then
$\int (- \vec r.(\nabla \vec M) - \vec M )d^3x + \int (\vec M(\vec r.\hat n) d^2x$

i read that even for dyadics this theorem holds true similar to vector

$\int \nabla \vec M d^3x = \int \hat n \vec M d^2x$

but i am not what happens when dotted with r but even if those integrals cancel i still only have have -M integral the negative of what is required i am stuck here .

#### vanhees71

Gold Member
In my experience such questions are most easily solved using the Ricci calculus. Here we can use the simplified version for Euclidean vector analysis with all indices as lower indices. Just calculate $\partial_{i} (n_i x_j \rho)$, where $\vec{n}$ is an arbitrary constant unit vector, and use Gauss's integral theorem.

#### dRic2

Gold Member
In my experience such questions are most easily solved using the Ricci calculus. Here we can use the simplified version for Euclidean vector analysis with all indices as lower indices. Just calculate $\partial_{i} (n_i x_j \rho)$, where $\vec{n}$ is an arbitrary constant unit vector, and use Gauss's integral theorem.
Using that notation the first one should be $x_j \partial_i P_i = \partial_i (x_j P_i) - P_i \partial_i x_j$ and I get three equation for J=1,2,3; right ? Using the divergence theorem I get the same result because $P_i \partial_i x_j = P_j$ while the first term vanishes as explained by @timetraveller123.

But the second one still eludes me.

If I remember correctly what I did some days ago

$(x × div(J))_g = \epsilon_{ghi} x_h \epsilon_{jki} \partial_j J_k$

and using the property $\epsilon_{ghi} \epsilon_{jki} = \delta_{gj} \delta_{hk} - \delta_{gk} \delta_{jk}$ I should arrive to something like:

$x_i \partial_g J_i - x_i \partial_i J_g$

(sorry if I do not post the calculations but I need to go now, but I'm confident in the result I remember even though I'm very new to this kind of notation so it's very likely I'm wrong )

Anyway... I'm still stuck there.

edit: might be a sign error but the main problem remains

#### vanhees71

Gold Member
My idea was nonsense. It's much simpler. Just use
$$x_j \rho_b=-x_j \partial_k P_k=-\partial_k(x_j P_k)+P_k \partial_k x_j=-\partial_k(x_j P_k)+ P_j.$$
Now integrate this over $V$, use Gauss's theorem and assume that the entire chargd is completely contained in $V$.

#### dRic2

Gold Member
@vanhees71 but I already solved the first problem about electric polarization. It's the second that troubles me (the one about the magnetization vector)

BTW isn't your solution from problem #1 the same as mine and @timetraveller123's ?

#### vanhees71

Gold Member
Hm, despite a factor of 2 that's an identity. The magnetic moment of a current distribution is
$$\vec{\mu}=\frac{1}{2} \int_V \mathrm{d}^3 x \vec{x} \times \vec{j}(\vec{x}).$$
You can derive this using the vector potential in Coulomb gauge, [CORRECTED by a before missing factor $1/c$!!!]
$$\vec{A}(\vec{x})=\int_{V} \mathrm{d}^3 x \frac{\vec{j}(\vec{x}')}{4 \pi c |\vec{x}-\vec{x}'|}$$
and expanding the integrand in powers of $|\vec{x}'|/\vec{x}|$.

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#### timetraveller123

hi @vanhees71 oh wow i completely forgot about the relation but what about the factor of 1/2

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#### vanhees71

Gold Member
This calculation is a bit more cumbersome. I'm not so familiar with the Homework forum. So I'm not sure, whether it's allowed to post too much details. On the other hand, it's already a quite long thread, and it may help in this case.

The calculation I have in mind in #19 is to use the Cartesian multipole expansion. The idea behind this is that $V$ is the volume around some finite (charge and) current distribution, and we look at the fields far away. Say that $V$ is completely contained in a sphere of radius $R$. Then we look for $r=|\vec{x}| \gg R$, i.e., $r'=|\vec{x}'|\ll r$.

Then we can Taylor expand the denominator of the integrand around $\vec{x}'=0$:
$$\frac{1}{|\vec{x}-\vec{x}'|}=\frac{1}{\sqrt{r^2+r^{\prime 2}-2 \vec{x} \cdot \vec{x}'}} = \frac{1}{r} \, \frac{1}{\sqrt{1+\epsilon}}$$
with
$$\epsilon=\frac{r^{\prime 2}}{r^2} - 2 \frac{\vec{x} \cdot \vec{x}'}{r^2}.$$
Note that the 1st piece is of order $\mathcal{O}[(r'/r)^2]$ and the 2nd or order $\mathcal{O}(r'/r)$ which makes the Cartesian multipole expansion a bit cumbersome, but for the first few terms it's fine. Now we have
$$\frac{1}{\sqrt{1+\epsilon}}=(1+\epsilon)^{-1/2}=1-\frac{1}{2} \epsilon + \mathcal{O}(\epsilon^2).$$
Now $\mathcal{O}(\epsilon^2)=\mathcal{O}[(r'/r)^2]$, and we want the expansion up to order $\mathcal{O}(r'/r)$, and thus we finally use
$$\frac{1}{|\vec{x}-\vec{x}'|}=\frac{1}{r} \left [1+\frac{\vec{x} \cdot \vec{x}'}{r^2} + \mathcal{O}[(r'/r)^2 \right].$$
Now we use these two terms to approxmate $\vec{A}(\vec{x})$. The first term gives
$$\vec{A}_0(\vec{x})=\frac{1}{4 \pi c r} \int_{V} \mathrm{d}^3 x' \vec{j}(\vec{x}').$$
Now comes a trick similar to the one used for the calculation of $\vec{P}$ in the electrostatic case. We note that
$$\partial_b (x_a j_b)=j_a + x_a \partial_b j_b=j_a + x_1 (\vec{\nabla} \cdot \vec{j})=j_a,$$
because due to charge conservation we must have $\vec{\nabla} \cdot \vec{j}=0$ in magnetostatics. Thus the integrand for $\vec{A}_0$ is a total divergence and thus
$$\int_V \mathrm{d}^3 x' j_a(\vec{x}')=\int_V \mathrm{d}^3 x' \partial_b' (x_a' j_b)=\int_{\partial V} \mathrm{d}^2 \vec{f}_b' x_a' j_b(\vec{x}')=0,$$
the latter equation holds, because the boundary of $V$ is assumed to be completely outside of the charge distribution. This reflects the wellknown fact that there are no magnetic monopoles and thus no magnetic monopole field.

Now comes the 2nd term,
$$\vec{A}_1(\vec{x})=\frac{1}{4 \pi c r^3} \int_{V} \mathrm{d}^3 x' (\vec{x} \cdot \vec{x}') \vec{j}(\vec{x}').$$
So we need the integral
$$M_{ab}=\int_{V} \mathrm{d}^3 x' x_a' j_b(\vec{x}').$$
To bring this in the usual form of a magnetic moment we use again the trick with the integral, but only to the symmetric part of the tensor, i.e.,
$$\frac{1}{2} (M_{ab}+M_{ba})= \frac{1}{2} \int_V \mathrm{d}^3 x' (x_a' j_b+x_b j_a).$$
Now we note that, again using $\partial_c' j_c=0$,
$$\partial_c' (x_a' x_b' j_c)=x_a' j_b + x_b' j_a$$
and that thus the symmetric part of $M_{ab}$ vanishes. Thus we have
$$M_{ab}=\frac{1}{2}(M_{ab}+M_{ba})+\frac{1}{2}(M_{ab}-M_{ba})=\frac{1}{2} (M_{ab}-M_{ba})$$
and thus finally
$$A_{1b}(\vec{x})=\frac{1}{4 \pi c r^3} \frac{1}{2} x_a (M_{ab}-M_{ba}).$$
Now
$$\frac{1}{2} x_a (M_{ab}-M_{ba})=\frac{1}{2} \int_{V} \mathrm{d}^3 x' x_a (x_a' j_b-x_b' j_a).$$
Now we can write
$$x_a' j_b-x_b' j_a=\epsilon_{abc} (\vec{x}' \times \vec{j})_c$$
and thus
$$A_{1b}(\vec{x})=\frac{1}{4 \pi c r^3} x_a \epsilon_{abc} \frac{1}{2} \int_V \mathrm{d}^3 x' (\vec{x}' \times \vec{j})_c=\frac{1}{4 \pi c r^3} (\vec{M} \times \vec{x})_b$$
with
$$\vec{M}=\frac{1}{2} \int_V \mathrm{d}^3 x' \vec{x}' \times \vec{j}(\vec{x}').$$
The leading term of the vector potential in our multpole expansion thus is
$$\vec{A}_{1}=\frac{\vec{M} \times \vec{x}}{4 \pi c r^3}.$$
You can prove by direct calculation that
$$\vec{B}_1=\vec{\nabla} \times \vec{A}_1=\frac{1}{4 \pi c} \left [\frac{3 \vec{x} (\vec{x} \cdot \vec{M})}{r^5}-\frac{\vec{m}}{r^3} \right].$$
Comparing this with the electric dipole field you see that this is of the analogous form and thus that $\vec{M}$ is the magnetic dipole moment of the current distribution.

NB: I've used Heaviside-Lorentz units. You get the result in SI units by just substituting $1/(4 \pi c)$ in the formulae for $\vec{A}$ and $\vec{B}$ by $\mu_0/(4 \pi)$.

#### dRic2

Gold Member
Sorry if I'm not answering but I had a full day. I hope to get back at it during the weekend. thanks a lot for the help!

#### timetraveller123

This calculation is a bit more cumbersome. I'm not so familiar with the Homework forum. So I'm not sure, whether it's allowed to post too much details. On the other hand, it's already a quite long thread, and it may help in this case.

The calculation I have in mind in #19 is to use the Cartesian multipole expansion. The idea behind this is that $V$ is the volume around some finite (charge and) current distribution, and we look at the fields far away. Say that $V$ is completely contained in a sphere of radius $R$. Then we look for $r=|\vec{x}| \gg R$, i.e., $r'=|\vec{x}'|\ll r$.

Then we can Taylor expand the denominator of the integrand around $\vec{x}'=0$:
$$\frac{1}{|\vec{x}-\vec{x}'|}=\frac{1}{\sqrt{r^2+r^{\prime 2}-2 \vec{x} \cdot \vec{x}'}} = \frac{1}{r} \, \frac{1}{\sqrt{1+\epsilon}}$$
with
$$\epsilon=\frac{r^{\prime 2}}{r^2} - 2 \frac{\vec{x} \cdot \vec{x}'}{r^2}.$$
Note that the 1st piece is of order $\mathcal{O}[(r'/r)^2]$ and the 2nd or order $\mathcal{O}(r'/r)$ which makes the Cartesian multipole expansion a bit cumbersome, but for the first few terms it's fine. Now we have
$$\frac{1}{\sqrt{1+\epsilon}}=(1+\epsilon)^{-1/2}=1-\frac{1}{2} \epsilon + \mathcal{O}(\epsilon^2).$$
Now $\mathcal{O}(\epsilon^2)=\mathcal{O}[(r'/r)^2]$, and we want the expansion up to order $\mathcal{O}(r'/r)$, and thus we finally use
$$\frac{1}{|\vec{x}-\vec{x}'|}=\frac{1}{r} \left [1+\frac{\vec{x} \cdot \vec{x}'}{r^2} + \mathcal{O}[(r'/r)^2 \right].$$
Now we use these two terms to approxmate $\vec{A}(\vec{x})$. The first term gives
$$\vec{A}_0(\vec{x})=\frac{1}{4 \pi c r} \int_{V} \mathrm{d}^3 x' \vec{j}(\vec{x}').$$
Now comes a trick similar to the one used for the calculation of $\vec{P}$ in the electrostatic case. We note that
$$\partial_b (x_a j_b)=j_a + x_a \partial_b j_b=j_a + x_1 (\vec{\nabla} \cdot \vec{j})=j_a,$$
because due to charge conservation we must have $\vec{\nabla} \cdot \vec{j}=0$ in magnetostatics. Thus the integrand for $\vec{A}_0$ is a total divergence and thus
$$\int_V \mathrm{d}^3 x' j_a(\vec{x}')=\int_V \mathrm{d}^3 x' \partial_b' (x_a' j_b)=\int_{\partial V} \mathrm{d}^2 \vec{f}_b' x_a' j_b(\vec{x}')=0,$$
the latter equation holds, because the boundary of $V$ is assumed to be completely outside of the charge distribution. This reflects the wellknown fact that there are no magnetic monopoles and thus no magnetic monopole field.

Now comes the 2nd term,
$$\vec{A}_1(\vec{x})=\frac{1}{4 \pi c r^3} \int_{V} \mathrm{d}^3 x' (\vec{x} \cdot \vec{x}') \vec{j}(\vec{x}').$$
So we need the integral
$$M_{ab}=\int_{V} \mathrm{d}^3 x' x_a' j_b(\vec{x}').$$
To bring this in the usual form of a magnetic moment we use again the trick with the integral, but only to the symmetric part of the tensor, i.e.,
$$\frac{1}{2} (M_{ab}+M_{ba})= \frac{1}{2} \int_V \mathrm{d}^3 x' (x_a' j_b+x_b j_a).$$
Now we note that, again using $\partial_c' j_c=0$,
$$\partial_c' (x_a' x_b' j_c)=x_a' j_b + x_b' j_a$$
and that thus the symmetric part of $M_{ab}$ vanishes. Thus we have
$$M_{ab}=\frac{1}{2}(M_{ab}+M_{ba})+\frac{1}{2}(M_{ab}-M_{ba})=\frac{1}{2} (M_{ab}-M_{ba})$$
and thus finally
$$A_{1b}(\vec{x})=\frac{1}{4 \pi c r^3} \frac{1}{2} x_a (M_{ab}-M_{ba}).$$
Now
$$\frac{1}{2} x_a (M_{ab}-M_{ba})=\frac{1}{2} \int_{V} \mathrm{d}^3 x' x_a (x_a' j_b-x_b' j_a).$$
Now we can write
$$x_a' j_b-x_b' j_a=\epsilon_{abc} (\vec{x}' \times \vec{j})_c$$
and thus
$$A_{1b}(\vec{x})=\frac{1}{4 \pi c r^3} x_a \epsilon_{abc} \frac{1}{2} \int_V \mathrm{d}^3 x' (\vec{x}' \times \vec{j})_c=\frac{1}{4 \pi c r^3} (\vec{M} \times \vec{x})_b$$
with
$$\vec{M}=\frac{1}{2} \int_V \mathrm{d}^3 x' \vec{x}' \times \vec{j}(\vec{x}').$$
The leading term of the vector potential in our multpole expansion thus is
$$\vec{A}_{1}=\frac{\vec{M} \times \vec{x}}{4 \pi c r^3}.$$
You can prove by direct calculation that
$$\vec{B}_1=\vec{\nabla} \times \vec{A}_1=\frac{1}{4 \pi c} \left [\frac{3 \vec{x} (\vec{x} \cdot \vec{M})}{r^5}-\frac{\vec{m}}{r^3} \right].$$
Comparing this with the electric dipole field you see that this is of the analogous form and thus that $\vec{M}$ is the magnetic dipole moment of the current distribution.

NB: I've used Heaviside-Lorentz units. You get the result in SI units by just substituting $1/(4 \pi c)$ in the formulae for $\vec{A}$ and $\vec{B}$ by $\mu_0/(4 \pi)$.
no you misunderstood me i know the derivation for m what i am asking is

$\int \vec M d^3x = \vec m =\frac{1}{2} \int \vec r \times \vec J d^3x$
doesn't that leave a factor of half unaccounted for the qn asks to show integral of M is the integral without the 1/2 so that is what i was asking

#### vanhees71

Gold Member
But that's the standard definition of magnetization $\vec{M}$ and magnetic moment (including the factor 1/2). Where is the original question from? Maybe there's another context, where the factor 1/2 shouldn't be there? For the standard theory, you can also look at Wikipedia:

https://en.wikipedia.org/wiki/Magnetic_moment#Amperian_loop_model

#### dRic2

Gold Member
Sorry, I forgot the 1/2 in the original post

"Vector calculus identity and electric/magnetic polarization"

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