Find the power radiated using the Poyting vector

[blueyellow]

Find the power radiated using the Poynting vector

Homework Statement

As a model for electric quadrupole radiation, consider two oppositely oriented oscillating electric dipoles, +p$_{0}$ cos $\omega$ t at z=+d/2, and -p$_{0}$ cos $\omega$ t at z=-d/2, hence separated by a distance d along the z-axis. In the radiation zone, the scalar and vector potentials are given by:

$\phi$=-($\mu$(subscript0) p(subscript 0) omega^2 d/4pi)((cos^2 theta)/r) cos [$\omega$ (t-r/c)]

A=-(mu(subscript 0) p(subscript 0) omega^2 d)/(4pi cr) cos $\theta$ cos [$\omega$ (t-r/c)] z-hat

a) Find the electric and magnetic fields, E and B.
b) Find the Poynting vector, N, and the power radiated, P= loop integral (subscript S) N.n-hat da

The Attempt at a Solution

a) I have done this but I am not writing all of it out.
b) N= E cross H
= [($\mu$(subscript 0) p(subscript0)^2 $\omega$^4 d^2)/($\mu$ *16 pi^2 c r^2)] cos^3 $\theta$ [[($\omega$^2)/(c^2)] sin^2 ($\omega$(t-(r/c))) - ($\omega$/c) (sin $\theta$)/($\mu$$\mu_{0}$ r) cos ($\omega$(t-(r/c)))
-2sin$\theta$ cos($\omega$(t-(r/c)))((p$_{0}$/$\mu$) [($\omega$^2 d)/(4pi c r)] cos$\theta$(($\omega$/c) sin($\omega$(t-(r/c)) - (sin$\theta$/$\mu$$\mu_{0}$ r) cos($\omega$(t-(r/c)))

But how does one work out the power radiated from this? My books and notes don't provide a simple answer. I am having trouble visualising the electric quadrupole and don't know what the unit vector is. And it has to be integrated over the area. The area of what? How can there be an area when it is two electric dipoles? Please help.

 

[vela]

You want to integrate over a closed surface, e.g. a sphere, to find the total power being radiated away. This is just like when you used the integral form of Gauss's law except instead of integrating the electric field, you're integrating the Poynting vector.

 

[blueyellow]

I found an easier way to do this. From my notes it says:

P=$\oint_{S}$ N.n-hat da
=(1/μ$_{0}$) $\int^{2pi}_{0}$$\int^{pi}_{0}$B$_{\phi}$E$_{\theta}$ r$^{2}$ sin$\theta$ d$\theta$ d$\phi$
=(1/μ$_{0}$) $\int^{2pi}_{0}$$\int^{pi}_{0}$ $\frac{\mu_{0}I_{0}\delta l}{4\pi}$ sin$\theta$ ($\frac{-i\omega}{rc}$+$\frac{1}{r^{2}}$exp(i(kr - omega*t)) $\frac{\mu_{0}I_{0}\delta l}{4\pi}$ sin$\theta$ sin $\theta$ (-i*omega/r) exp(i(kr - omega*t)) sin $\theta$ d$\theta$ d$\phi$

One problem: what do I do with the $\delta$l's? How are they to be gotten rid of? Do I just ignore them?

 

[Thaakisfox]

There is no dl at all. You didnt plug the proper fields in the calculation of the poynting vector. You should be calculating electric quadrupole radiation not magnetic dipole radiation. The fields you wrote for the poynting vector are the fields of a magnetic dipole.

Write the fields which u calculated for the electric quadrupole..

 

[blueyellow]

"Write the fields which u calculated for the electric quadrupole.."

I don't know how to do that. How do I go from a dipole field to a quadrupole field then? Multiply everything by 2? That doesn't seem right. Yes, I will look in my notes and books and see if they say anything, but I am asking just in case this turns out to be one of those complicated questions that require 'thinking' rather than bookwork.

 

[blueyellow]

Never mind, I just realized I copied down the equations for B and E from the question above the question I was supposed to be doing.

 

[Thaakisfox]

no, the quadrupole is the next term in the multipole expansion of the fields. After the dipole comes the quadrupole in the expansion.

 

[blueyellow]

So

P=$\oint$N.n-hat da
=$\int$ div N d(tau)
=$\int$ $\frac{\partial N_{z}}{\partial z}$ d(tau)

but N has no z component so the integral goes to zero.

That the power seems zero doesn't seem right.

 

[blueyellow]

I sort of ignored the n-hat because I didn't see any other way to go about doing the question. What would be the direction of n-hat anyway? All we know about the quadrupole by considering two dipoles is:

+p$_{0}$ cos ωt at z=d/2
-p$_{0}$ cos ωt at z=-d/2

so n-hat could be pointing in either the r-hat or theta-hat direction. Does the n-hat need to be taken into account to do the question properly?

 

[blueyellow]

Previously I made a mistake while posting.

N= E cross H
= [($\mu$(subscript 0) p(subscript0)^2 $\omega$^4 d^2)/($\mu$ *16 pi^2 c r^2)] cos^3 $\theta$ [[($\omega$^2)/(c^2)] sin^2 ($\omega$(t-(r/c))) - ($\omega$/c) (sin $\theta$)/($\mu$$\mu_{0}$ r) cos ($\omega$(t-(r/c)))
-2sin$\theta$ cos($\omega$(t-(r/c)))((p$_{0}$/$\mu$) [($\omega$^2 d)/(4pi c r)] cos$\theta$(($\omega$/c) sin($\omega$(t-(r/c)) - (sin$\theta$/$\mu$$\mu_{0}$ r) cos($\omega$(t-(r/c)))] z-hat

I had missed the z-hat out at the end.

 

[blueyellow]

I changed it from cylindrical to spherical coordinates. I found that in spherical coordinates it only has a rho-component. So only $\frac{1}{\rho^{2}}$ $\frac{\partial}{\partial\rho}$ ($\rho^{2}$ N$_{\rho}$) matters when trying to work out div N.

But N$_{\rho}$ has no $\rho$ terms in it. So div N=0. This is the same as the result I got when I did it in cylindrical coordinates.

I spoke to someone else who is doing the same homework and he also doesn't think it makes sense for the power radiated to be zero. Is the power radiated really zero? If so, why? If not, is there some sort of trick that will help one to answer the question? Thanks if anyone replies.

 

[vela]

What did you get for E and B?

 

[blueyellow]

E=-grad$\phi$
=$\frac{\partial\phi}{\partial r}$r-hat+$\frac{1}{r}$$\frac{\partial\phi}{\partial\theta}$θ-hat +$\frac{\partial\phi}{\partial z}$z-hat
=[$\frac{-\mu_{0}p_{0}\omega^{2}d}{4\pi}$$\frac{cos^{2}θ}{r}$($\frac{\omega}{c}$sin($\omega$(t-$\frac{r}{c}$))+$\frac{-\mu_{0}p_{0}\omega^{2}d}{4\pi}$(-2sinθ $\frac{cosθ}{r}$)cos($\omega$(t-$\frac{r}{c}$))]r-hat+[$\frac{\mu_{0}p_{0}\omega^{2}d}{4r^{2}\pi}$sinθ cosθ cos(ω(t-$\frac{r}{c}$))]θ-hat
=$\frac{\mu_{0}p_{0}\omega^{2}d cosθ}{4\pi r}(-cosθ \frac{\omega}{c}sin(\omega$(t-$\frac{r}{c}$)))r-hat+($\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r^{2}}$sinθcosθcos($\omega(t-\frac{r}{c}$)))θ-hat

B=$\nabla$$\times$A

=$\frac{1}{r}$$\frac{\partial A_{z}}{\partialθ}$r-hat -$\frac{\partial A_{z}}{\partial r}$theta-hat
=$\frac{\mu_{0}p_{0}\omega^{2}d sinθ}{4\pi cr^{2}}$sinθcos(ω(t-r/c)))r-hat+($\frac{\mu_{0}p_{0}\omega^{2}d sinθ}{4\pi cr}$cosθ($\frac{ω}{c}$sin(ω(t-r/c))-$\frac{sinθ}{r}$cos(ω(t-r/c)))θ-hat

Have I gone wrong somewhere?

 

[blueyellow]

Sorry, I think I know where I have gone wrong now. E is not -grad phi in this case.

 

[vela]

What were your results in spherical coordinates? You probably don't want to use cylindrical coordinates here.

Even if you do, you have to remember to rewrite the potentials in terms of the cylindrical coordinate $(r, \theta, z)$ first because $r$ and $\theta$ aren't defined the same way in cylindrical and spherical coordinates.

Since you're looking for the fields in the radiation zone, where r is large, you can drop contributions proportional to 1/r². The 1/r terms will dominate.

 

[blueyellow]

So I have to find E and B in spherical coordinates? I can't just find them in cylindrical coordinates, and then convert them to spherical when I find N? Because as I said, when I convert the result I got for N into spherical coordinates, and then I do the integral, it just goes to zero.

I have realized more now that the question is about a quadrupole, but it is asking you to consider two dipoles. Does that mean I have to multiply everything I did by two?

What is a 'radiation zone'?
Thanks if anyone explains.

 

[vela]

Despite the presence of $\hat{z}$, the expressions for the potentials are, I'm pretty sure, in spherical coordinates. (You should check that.) So using spherical coordinates is the path of least resistance. Also, the end result won't and, more important, can't matter on which coordinate system you choose.

What textbook are you using? For terms like radiation zone, you should look those up in your text rather than asking here. You want to learn how to find that kind of basic information on your own.

 

[blueyellow]

I am using Griffiths. It does not explain what radiation zone is. It just highlights it in bold. It says 'we are interested in fields that survive at large distances from the source, in the so called radiation zone'.

'large distances'?
That means the radiation zone is everywhere? If the radiation zone is everywhere, and there is no difference between one place and another with regards to whether it is in the radiation zone or not, then why is it mentioned in the question?

 

[blueyellow]

I get the feeling that 'radiation zone' is just one of those things that get mentioned, but you don't really need to pay much attention to those words to do the question...

 

[vela]

I probably have an older edition of Griffiths, so the section number might be slightly off. In my copy, section 9.1.2 goes over a similar problem for the radiation from a single dipole. You want to do the same thing starting with the potentials you've been given.

"large distances" doesn't mean everywhere. There are two lengths in the problem. One of them is d, the distance between the dipoles, and the other one is the wavelength $\lambda = 2\pi c/\omega$. In the radiation zone, r>>d and $r>>\lambda$.

 

[blueyellow]

Thanks.

I realized that E=-grad(phi) - $\frac{\partial A}{\partial t}$

$\frac{\partial A}{\partial t}$
=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi cr}$(cosθ)$\omega$sin($\omega$(t-r/c))z-hat

E=$\frac{\mu_{0}p_{0}\omega^{2}d cosθ}{4\pi r}(-cosθ \frac{\omega}{c}sin(\omega$(t-$\frac{r}{c}$)))r-hat+($\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r^{2}}$sinθcosθcos($\omega(t-\frac{r}{c}$)))θ-hat-$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi cr}$(cosθ)$\omega$sin($\omega$(t-r/c))z-hat

N= E cross H
= [($\mu$(subscript 0) p(subscript0)^2 $\omega$^4 d^2)/($\mu$ *16 pi^2 c r^2)] cos^3 $\theta$ [[($\omega$^2)/(c^2)] sin^2 ($\omega$(t-(r/c))) - ($\omega$/c) (sin $\theta$)/($\mu$$\mu_{0}$ r) cos ($\omega$(t-(r/c)))
-2sin$\theta$ cos($\omega$(t-(r/c)))((p$_{0}$/$\mu$) [($\omega$^2 d)/(4pi c r)] cos$\theta$(($\omega$/c) sin($\omega$(t-(r/c)) - (sin$\theta$/$\mu$$\mu_{0}$ r) cos($\omega$(t-(r/c)))] z-hat
+($\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r^{2}}$sinθcosθcos($\omega$(t-r/c)))$\frac{p_{0}\omega^{2}d}{\mu 4\pi cr}$cos($\frac{\omega}{c}$sin($\omega$(t-r/c))-$\frac{sinθ}{\mu\mu_{0}r}$cos($\omega$(t-r/c)))r-hat
-$\frac{\mu_{0}p_{0}\omega^{2}d sinθ}{4\pi r^{2}}$cos($\omega$(t-r/c))(cosθ +$\frac{1}{c\mu\mu_{0}}$)θ-hat

div N=$\frac{1}{r}$$\frac{\partial}{\partial r}$(r N$_{r}$)+$\frac{1}{r}$$\frac{\partial N_{\theta}}{\partial\theta}$+0

"$\frac{1}{r}$$\frac{\partial}{\partial r}$(r N$_{r}$)"

Is there an easier way to differentiate N$_{r}$ with respect to r?

N$_{r}=$($\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r^{2}}$sinθcosθcos($\omega$(t-r/c)))$\frac{p_{0}\omega^{2}d}{\mu 4\pi cr}$cos($\frac{\omega}{c}$sin($\omega$(t-r/c))-$\frac{sinθ}{\mu\mu_{0}r}$cos($\omega$(t-r/c)))

and I tried doing product rule differentiation for that - but then I realized I could not do that easily because there were too many parts to deal with. Please help.

 

[blueyellow]

Never mind. I think I can just about manage to do the differentiation of N (subcript r). But it is so hideous to do that I am worried that I made a mistake previously. And one has to do the integral of div N. And that will look hideous, but I'm guessing it will be fine if it turns out to be an odd integral.

 

[vela]

I wouldn't use the divergence theorem.

 

[blueyellow]

How else is one meant to do the integral then please? Because it has to be integrated over the area, and the three components of N suggest that it should be intgerated over the volume. So I thought, use divergence theorem to change the area integral into a volume integral. That's the only theorem I know that allows one to change from an area integral to a volume integral.

 

[blueyellow]

And I can't think of any way that I would be able to use Stoke's theorem.

 

[vela]

I cleaned up your expressions for dA/dt and E:
\begin{align*}
\frac{\partial \mathbf{A}}{\partial t} &= \frac{\mu_0 p_0 \omega^3 d}{4\pi c}\frac{\cos \theta}{r} \sin[\omega(t-r/c)] \hat{\mathbf{z}} \\
\mathbf{E} &= -\frac{\mu_0 p_0 \omega^3 d}{4\pi c} \frac{\cos^2 \theta}{r} \sin[\omega(t-r/c)] \hat{\mathbf{r}} +
\frac{\mu_0 p_0 \omega^2 d}{4\pi} \frac{\sin \theta \cos\theta}{r^2} \cos[\omega(t-r/c)] \hat{\theta} -
\frac{\mu_0 p_0 \omega^3 d}{4\pi c} \frac{\cos\theta}{r} \sin[\omega(t-r/c)] \hat{\mathbf{z}}
\end{align*}First, it appears you didn't calculate $\nabla\phi$ correctly, so you need to correct that. Second, you want to express $\hat{\mathbf{z}}$ in terms of the spherical unit vectors $\hat{\mathbf{r}}$ and $\hat{\theta}$. Finally, use the assumption of being in the radiation zone to simplify the expression.

The expression you used for the curl
$$ \mathbf{B}= \nabla \times \mathbf{A} = \frac{1}{r}\frac{\partial A_z}{\partial\theta} \hat{\mathbf{r}} - \frac{\partial A_z}{\partial r}\hat{\theta} $$is for cylindrical coordinates, but the vector potential is written using spherical coordinates. Again, you need to express $\hat{\mathbf{z}}$ in terms of the spherical unit vectors $\hat{\mathbf{r}}$ and $\hat{\theta}$ so you can identify what $A_r$ and $A_\theta$ are, and then find the magnetic field using the correct expression for the curl in spherical coordinates. You will find there is only a $\hat{\phi}$ component. Then, again, use the assumption that you're in the radiation zone to simplify the expression.

 

[blueyellow]

I found out what E and B were in spherical coordinates. E has r and theta components. B only has a phi component. And to do E cross H, I converted everything to Cartesian coordinates, because apparently there is no easy way to take the cross product in Spherical coordinates.

So N= E cross H has x, y and z components, and the expression is really long and complicated.

To convert it back to spherical coordinates would give an even more long and complicated expression, which I wouldn't know how to integrate.

I could leave it in Cartesian coordinates and integrate it, but what would be the limits of integration? From 0 to r for x, y and z?

How do I consider the fact that it is a quadrupole to be considered as two dipoles? Do I have to integrate it and then times the result by 2?

And does the information given in the question that z=d/2 and z=-d/2 need to be used anywhere.

I'm really confused so I'd appreciate it if you could help please.

 

[blueyellow]

I forgot, I would still have to do div N before I integrate it. And div N wouldn't work because div N in Cartesian gives 0.

 

[blueyellow]

The theta-component of E has a (1/(r^2)) in front of it - so can I just approximate and ignore the theta component because it has an (1/(r^2)) in front of it?

Sorry I'm editing this post because I just realized it was said previously that 1/r^2 terms can be dropped

 

[blueyellow]

Sorry I got stuck again.

I worked out E and H to be:

E=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r}$[cosθ $\frac{\omega}{c}$sin($\omega$(t-r/c))-cos($\omega$(t-r/c))-$\frac{\omega}{c}$sin($\omega$(t-r/c))]r-hat

H=$\frac{p_{0}\omega^{2}d}{\mu 4\pi cr^{2}}$cos($\omega$(t-r/c))$\phi$-hat

N=$\frac{\mu_{0}p^{2}_{0}\omega^{4}d}{\mu 16\pi^{2} cr^{3}}$cosθ sinθ [cosθ $\frac{\omega}{c}$sin($\omega$(t-r/c))-cos($\omega$(t-r/c))-$\frac{\omega}{c}$sin($\omega$(t-r/c))]cos($\omega$(t-r/c))θ-hat

But I am having trouble integrating doing the integral of N.n with respect to the area. I have tried to do this for hours, but I don't know what to do, because if I try to integrate it with respect to r it doesn't quite work because I do integration by parts and it goes around in circles. And I still don't know how to do the integral without using divergence theorem. Please help.

 

[blueyellow]

Never mind. I think I figured it out now.

 

[vela]

Sorry I got stuck again.

I worked out E and H to be:

E=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r}$[cosθ $\frac{\omega}{c}$sin($\omega$(t-r/c))-cos($\omega$(t-r/c))-$\frac{\omega}{c}$sin($\omega$(t-r/c))]r-hat

H=$\frac{p_{0}\omega^{2}d}{\mu 4\pi cr^{2}}$cos($\omega$(t-r/c))$\phi$-hat

Neither of those are correct.

N=$\frac{\mu_{0}p^{2}_{0}\omega^{4}d}{\mu 16\pi^{2} cr^{3}}$cosθ sinθ [cosθ $\frac{\omega}{c}$sin($\omega$(t-r/c))-cos($\omega$(t-r/c))-$\frac{\omega}{c}$sin($\omega$(t-r/c))]cos($\omega$(t-r/c))θ-hat

But I am having trouble integrating doing the integral of N.n with respect to the area. I have tried to do this for hours, but I don't know what to do, because if I try to integrate it with respect to r it doesn't quite work because I do integration by parts and it goes around in circles. And I still don't know how to do the integral without using divergence theorem. Please help.

 

[blueyellow]

Really? I think I did them really carefully.

E=-grad $\phi$ -dA/dt

grad $\phi$=(d$\phi$/dr]r-hat +(1/r)(d$\phi$/dθ)θ-hat

=$\frac{-\mu_{0}p_{0}\omega^{2}d}{4\pi}$cos$^{2}$θ($\frac{-1}{r}$$\frac{-\omega}{c}$sin($\omega$(t-r/c))+$\frac{1}{r^{2}}$cos($\omega$(t-r/c)))r-hat
+$\frac{1}{r}$($\frac{-\mu_{0}p_{0}\omega^{2}d}{4\pi r}$cos($\omega$(t-r/c))(-sin 2θ))θ-hat

Oh, I think I realized my mistake now. It was some mistake I made with factorising with brackets that made me think that one of my (1/r^2) terms was an (1/r) term.

So E should be:

E=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r}$[cosθ $\frac{\omega}{c}$sin($\omega$(t-r/c))-$\frac{\omega}{c}$sin($\omega$(t-r/c))]r-hat

and I had made a mistake because I had left out the sin theta in H.

H should be:

H=$\frac{p_{0}\omega^{2}d}{\mu 4\pi cr^{2}}$sinθ cos($\omega$(t-r/c))$\phi$-hat

Is that correct now? Please tell me if I'm wrong and had made a stupid mistake or a major error.

 

[blueyellow]

I thought that to translate from cylindrical to spherical coordinates:

r=$\sqrt{s^{2}+z^{2}}$

s=0

so r=z

So A$_{r}$=A$_{z}$

So

A=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi cr}$cosθ cos($\omega$(t-r/c))r-hat

θ=arctan (s/z)=arctan 0= 0
=arccos (z/r)=arccos(z/z)=arccos 1=0
$\phi$=$\phi$=0

So since A only has an r component, and dA$_{r}$/d$\phi$=0

only (1/r)(-dA$_{r}$/dθ)$\phi$-hat matters while doing the curl of A.

So, B=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi cr^{2}}$sinθ cos($\omega$(t-r/c))$\phi$-hat

And H= what I said it equals in the previous post, because H=B/(mu*mu0).

Right? Please tell me if I has made a mistake somewhere, such as when converting from cynlindrical to spherical.

 

[vela]

Really? I think I did them really carefully.

E=-grad $\phi$ -dA/dt

grad $\phi$=(d$\phi$/dr]r-hat +(1/r)(d$\phi$/dθ)θ-hat

=$\frac{-\mu_{0}p_{0}\omega^{2}d}{4\pi}$cos$^{2}$θ($\frac{-1}{r}$$\frac{-\omega}{c}$sin($\omega$(t-r/c))+$\frac{1}{r^{2}}$cos($\omega$(t-r/c)))r-hat
+$\frac{1}{r}$($\frac{-\mu_{0}p_{0}\omega^{2}d}{4\pi r}$cos($\omega$(t-r/c))(-sin 2θ))θ-hat

Oh, I think I realized my mistake now. It was some mistake I made with factorising with brackets that made me think that one of my (1/r^2) terms was an (1/r) term.

So E should be:

E=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r}$[cosθ $\frac{\omega}{c}$sin($\omega$(t-r/c))-$\frac{\omega}{c}$sin($\omega$(t-r/c))]r-hat

The gradient looks fine now, but your dA/dt must be wrong.

I thought that to translate from cylindrical to spherical coordinates:

r=$\sqrt{s^{2}+z^{2}}$

s=0

so r=z

So A$_{r}$=A$_{z}$

So

A=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi cr}$cosθ cos($\omega$(t-r/c))r-hat

θ=arctan (s/z)=arctan 0= 0
=arccos (z/r)=arccos(z/z)=arccos 1=0
$\phi$=$\phi$=0

So since A only has an r component, and dA$_{r}$/d$\phi$=0

only (1/r)(-dA$_{r}$/dθ)$\phi$-hat matters while doing the curl of A.

So, B=$\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi cr^{2}}$sinθ cos($\omega$(t-r/c))$\phi$-hat

And H= what I said it equals in the previous post, because H=B/(mu*mu0).

Right? Please tell me if I has made a mistake somewhere, such as when converting from cynlindrical to spherical.

This is completely wrong. Except for $\hat{\mathbf{z}}$, the vector potential is already written in spherical coordinates. As I said back in post #26, all you have to do is rewrite $\hat{\mathbf{z}}$ in terms of the unit vectors for spherical coordinates.