Find the power radiated using the Poyting vector

[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.

 

[Antiphon]

The radiation zone is where the transverse field dominate over any radial components. It is defined to be R = 2D^2/lambda where D is the diameter of the source and lambda is the wavelength.

 

[blueyellow]

Is z-hat in spherical coordinates just r-hat*cosθ?

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

?

 

[blueyellow]

So if

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

and dropping the (1/(r^2)) terms for the radiation zone it is:

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))r-hat

And

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

so

dA/dt=$\frac{-\mu_{0}p_{0}\omega^{3}d}{4\pi cr}$cos$^{2}$θsin($\omega$(t-r/c))r-hat

This means that E=-grad $\phi$-dA/dt=0

This can't be right, please help.

 

[blueyellow]

The radiation zone is where the transverse field dominate over any radial components. It is defined to be R = 2D^2/lambda where D is the diameter of the source and lambda is the wavelength.

Thanks, but how would I apply this to the question?

So for the question R=(2(d^2))/(2pi/omega).

Does this have any implications on how I should do the question/approximate?

Thanks if you reply.

 

[blueyellow]

I just realized, I made a mistake while expressing z-hat in spherical coordinates.

https://www.physicsforums.com/showthread.php?t=126702 - this told me that to find out what z-hat is in spherical coordinates, one has to do grad(z)=grad(r cos theta)=z-hat.

Please tell me if what they say is wrong. I have found no other textbook or website that will tell me how to express z-hat in spherical coordinates.

 

[vela]

You could do it that way, but there are other ways. For example, you should know that because $\hat{r}$, $\hat{\theta}$, and $\hat{\phi}$ form an orthogonal basis, you have that $\hat{z} = (\hat{z}\cdot\hat{r})\hat{r} + (\hat{z}\cdot\hat{\theta})\hat{\theta} + (\hat{z}\cdot\hat{\phi})\hat{\phi}$. Presumably, you have expressions for $\hat{r}$, $\hat{\theta}$, and $\hat{\phi}$ in terms of the Cartesian unit vectors, so that expression should be trivial to work out.

In any case, in 9.1.2, Griffiths mentions what $\hat{z}$ is in terms of the spherical unit vectors. Have you taken the time to read and understand what he did in that section? If you understand what he did there, this problem is very straightforward.