# Homework Help: EM Waves in media

1. Nov 20, 2013

### CAF123

1. The problem statement, all variables and given/known data
The following dispersion relation is obtained by substituting a plane wave solution for $\mathbf{B}$ into $$\nabla^2 \mathbf{B} = \mu \epsilon \frac{\partial^2 \mathbf{B}}{\partial^2 t} + \mu \sigma \frac{\partial \mathbf{B}}{\partial t}:$$ $$\hat{k} = (\mu \epsilon w^2 + i\mu \sigma w)^{1/2}$$

1)Expand this to obtain the limiting expressions for the skin depth in poor and good conductors.
2)Show that in a good conductor the magnetic field lags the electric field by π/4.

2. Relevant equations
Binomial expansion

3. The attempt at a solution
1)$\hat{k}$ can be rexpressed as $\sqrt{\mu \epsilon w^2} \sqrt{\left(1+\frac{i\sigma}{\epsilon w}\right)}$ For a poor conductor, $\sigma/ \epsilon w \rightarrow 0$ so I can binomial expand. Doing this gives me a correct answer but when I consider expanding in $\epsilon w/ \sigma$ for a good conductor I don't. Is there a better way to do this expansion? Even though I obtained one correct answer, I feel that it is incorrect because of the imaginary unit. On an argand plane, i has magnitude 1, so via some 'hand-waved' argument overall $i \sigma/\epsilon w$ is <<1.

2)To derive this, we consider writing $E = E_o e^{i \delta_E}$ and $B = B_oe^{i\delta_B}$ where $\delta_{E,B}$ are characteristic skin depths of E and B (see Griffiths) but I am not sure how these come about.

2. Nov 20, 2013

### rude man

This belongs in the advanced physics forum, IMO.

3. Nov 20, 2013

### CAF123

I'll report it, and if the mentors agree maybe they'll move it. Do you have any thoughts on my questions?

Thank you.

4. Nov 20, 2013

### rude man

Well, first, I think k is mis-stated.
I think it's k = [jwμ(σ + jwε)]1/2
or k = [-w2με + jwμσ]1/2
which differs from what you were given by the sign of the 1st term (the one with ε in it.)

So you'll have to decide which to pick!

Regardless, k has a real and an imaginary component, and by definition,
skin depth δ = 1/(real part of k).

For good and mediocre conductors, the term containing ε can be disregarded, and you come up with an expression for δ that is a function of w, μ and σ. For really poor conductors I guess you have to include the ε term. I don't believe that is normally done.

In any case, this is just an exercise in finding the real and imaginary parts of k.

P.S. δ has to turn out to be a positive real number.

For part (b) you need to dig up the complex intrinsic impedance in a conducting medium. The formula for intrinsic impedance η in a conducting medium involves w, μ and k and reduces to the familiar √(μ/ε) if conductivity is zero, but I can't give you the formula. It's the result of a lot of derivation so if you're not given it in some part of your course then I have to assume you're supposed to derive it which is a lot of math I don't want to wade thru. Realize that k is complex which shows up as a phase shift in E vs. H.

5. Nov 21, 2013

### CAF123

It is stated correctly in my notes. I also checked Griffiths just to be sure and it is the same as in my notes.
Should that be 1/(Im(k)?

k = k + iκ and to obtain a form like this, I need to taylor expand k. I was going to binomial expand because the quantities σ/εω <<1 for a poor conductor and εω/σ << 1 for a good conductor. My question is when I expand for a poor conductor, I get the correct answer, but when I expand for the good conductor, I don't. Here is my working: $$\epsilon \omega/\sigma << 1 \Rightarrow \mathbf{k} = \sqrt{\frac{\epsilon \omega}{i \sigma} + 1} \sqrt{i} \sqrt{\sigma \mu \omega} \approx. \left(1+ \frac{1}{2} \frac{\epsilon \omega}{i \sigma}\right) \sqrt{\sigma \omega \mu} \sqrt{i}$$ Rexpressing $\sqrt{i} = \frac{1}{\sqrt{2}} \left(1+i\right)$ and collecting the imaginary part after simplification gives me an incorrect answer.

Is my expansion justified? The presence of i in the binomial expansion makes me unsure, but I argued that magnitude of i on an argand plane is 1 so this mulitiplies a quantity which is <<1, hence the expansion applies. But I am not sure.
I was wondering how the expressions E=E0expi(δE) and B=B0expi(δB) were derived. They are simple written down(as far as I can see) in Griffiths and in my notes.

6. Nov 21, 2013

### rude man

That is weird... I guess they must both be valid then. I know mine is.

No. The complex wave goes as exp(kx) so the real part is a decaying exponential while the imaginary part represents a constant-amplitude sine wave. The decaying part represents the skin depth.

If you start with either your or my expression for k, if σ >> wε then k = (jwμσ)1/2
so there must be something wrong with either your k or your expansion. I might question whether the formula

(1 + x)1/2 ~ (1 + x/2) for x << 1 is valid for complex x. In any case, you know the correct answer for k for large σ & I can't be of much more help here.
How is δ defined in your notes or Griffith? In my book (Skilling), for large σ, and at any frequency below about 1% of that of light, k = (1 + j)/δ where δ is skin depth. Then E ~ exp(+/-k). So δ ~ 1/√(ωμσ/2).

7. Nov 21, 2013

### rude man

What's the problem? If you let σ >> wε you get k = √(jwμσ) which is correct.

8. Nov 21, 2013

### CAF123

I am supposed to get to $$\delta = 1/\kappa = 1/Im(\mathbf{k}) \approx \left(\frac{2}{\mu \omega \sigma}\right)^{1/2}$$ I just don't see how this comes from what I showed previously. My book uses the notation k= k + iκ.
Edit: I see it now, thanks.

Last edited: Nov 21, 2013
9. Nov 21, 2013

### CAF123

To find the ratio of the amplitudes, I use the fact that $\hat{B_o} = \frac{\hat{k}}{w}\hat{E_o}$ and in the limit of a good conductor, I obtain the ratio to be $\sqrt{\sigma \mu i/w }$ Similarly, for the poor conductor, I get $\sqrt{\mu \epsilon}$. I think the latter one is correct, but I am not sure about the former. For a good conductor, $\hat{k} = \sqrt{\sigma \mu \omega} e^{i \pi/4}$ which when divided by w gives what I showed above. From another source, the correct answer is $\sqrt{\sigma \mu/w}$

Last edited: Nov 21, 2013
10. Nov 21, 2013

### rude man

That is not quite correct. There's a j missing somewhere!
That mistake I think carries on in your subsequent computations.

11. Nov 21, 2013

### CAF123

It is just quoted from my question sheet, except I used hats instead of tildas because I don't know the latex. $\hat{B_o}$ and $\hat{E_o}$ are inherently complex.

Also, I checked my notes (and Griffiths) again and they definitely define the skin depth to be the reciprocal of the imaginary part of the complex wavenumber. I guess our sources are using different notations.

12. Nov 21, 2013

### rude man

That's OK - I don't know latex at all! (I use bold for vectors & in this case complex numbers).

For conductors, they are actually the same: k = (1 + j)/δ. Our general expressions for k (see post 4) differ only in the ε terms. This term I believe is never applicable to "skin depth" since that implies current and therefore a conductor.

So probably no cause for losing sleep over it.

13. Nov 21, 2013

### rude man

My guess is that your question sheet is wrong. That missing j in the E/H (or B/E) ratio makes all the difference ... BTW hope you're tolerating my using j instead of i ...

14. Nov 21, 2013

### rude man

I get B/E = √(σμ/ωj). So we're off by 90 degrees since √j = exp(jπ/4)
and 1/√j = exp(-jπ/4). There is a π/4 phase difference between B and E in both cases! Just that you show B leading E & I show it lagging ... (and BTW the question agrees with me, for what that's worth ...). I can but return to my previous statement that your ratio is missing a j. I believe that accounts for our difference.

15. Nov 21, 2013

### CAF123

The relation I posted can be derived from Maxwells third eqn in media $\nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t$. It is also the same relation in Griffiths. Does your book say otherwise?

16. Nov 21, 2013

### rude man

What exact relation again please? We've got a bunch here ...

17. Nov 21, 2013

### CAF123

Sorry, I mean to say $\hat{B_o} = \frac{\hat{k}}{w}\hat{E_o}$ is quoted in Griffiths too. I derived it in an earlier part by taking the curl of $E = \hat{E_o} \exp(\hat{k}z-wt)$ and the time derivative of $B = \hat{B_o} \exp(\hat{k}z-wt)$ and using MIII.

18. Nov 21, 2013

### vanhees71

Could it be that you differences are just due to the different conventions of the Fourier transformation concerning the sign in the exponential. I've the impression that, e.g., in electrical engineering people prefer the opposite sign than the physicists. The latter is the one in Griffiths. E.g., for the magnetic field you write (also using the convention to put all factors involving $2 \pi$ in the Fourier transformation to the frequency-wavenumber integrals
$$\vec{B}(t,\vec{x})=\frac{1}{(2 \pi)^4} \int \mathrm{d} \omega \int \mathrm{d}^3 \vec{k} \tilde{\vec{B}}(\omega,\vec{k}) \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x}).$$
Now you can use the correspondences between the time-space and the frequency-wavenumber domain in the in-medium Maxwell equations:
$$\partial_t \rightarrow -\mathrm{i} \omega, \quad \vec{\nabla} \leftrightarrow \mathrm{i} \vec{k}.$$

Then from your wave equation for the magnetic field in the orignal posting you get the dispersion relation
$$-\vec{k}^2=-\mu \epsilon \omega^2 - \mathrm{i} \mu \sigma \omega$$
or
$$\vec{k}^2=\mu \epsilon \omega^2 + \mathrm{i} \mu \sigma \omega.$$
Note that in general $\mu$, $\epsilon$, and $\sigma$ are frequency dependent (for usual media) and also $\vec{k}$ dependent (e.g., for plasmas).

Now if you consider a plane wave, i.e., a field of the form
$$\vec{B}=\vec{B}_0 \exp(-\mathrm{i} \omega t + \mathrm{i} k z),$$
you get
$$k^2=\mu \epsilon \omega^2 + \mathrm{i} \mu \sigma \omega,$$
and it makes sense to take the square root. You have to choose the sign of the square root such that the imaginary part becomes positive so that the wave is attenuating in positive $z$ direction, i.e., the direction of wave propagation rather than exponentially increasing.

19. Nov 21, 2013

### rude man

We are only indirectly quibbling over the correct expression of k. For a good conductor we have the same k, yet we get a different phase relationship between the B and the E vectors. And the posed question agrees with me, which is that the B vector lags the E vector.

How about you giving us your B/E phase relationship derivation? The more, the merrier ...

20. Nov 21, 2013

### rude man

We can compare derivations. Let propagation direction = x, E direction = y, then B direction is z. Let z = unit vector in z direction. Remember that k is a complex scalar and not the unit vector in the z direction.

Define
E = electric field,
E0 = E(t=0) so is not a function of t;
E0y is the y component of E0
Em = constant = the value of E0y at the origin and is not a function of x, y, z or t.

And same convention for B with y replaced by z etc.

Now,
curl E0 = -jwB0
= z ∂E0y/∂x (plane wave)
= z ∂/∂x Emexp(-kx)
= - z k Emexp(-kx)
So jwB0 = z k Emexp(-kx)

so Bz = k/jw Ey
or the B field lags the E field by π/4 since k = √(jμσω).

So clearly η = Ey/Bz = jw/k.

Maybe you can find where our conventions differ, or whatever. Probably some stupid variation. But I do think you shoud finally get the same phase relationship between E and B that I did.

Last edited: Nov 21, 2013
21. Nov 22, 2013

### CAF123

I thought the fact that $\mathbf{k} \cdot \mathbf{r} = \hat{k}z$ in the argument of the exp implied that the direction of propagation was in z direction?

22. Nov 22, 2013

### CAF123

When I do the derivation, the i cancels on each side of the eqn.

23. Nov 22, 2013

### vanhees71

$$\vec{E}=E_0 \vec{e}_x \exp[-\mathrm{i} \omega t+\mathrm{i} k z]. \qquad (1)$$
We can choose our time origin such that $E_0 \in \mathbb{R}$.

Then the magnetic field is determined by Faraday's Law,
$$\partial_t \vec{B}=-\vec{\nabla} \times \vec{E}.$$
With the plane-wave ansatz
$$\vec{B}=B_0 \vec{e}_y \exp[-\mathrm{i} \omega t+\mathrm{i} k z]$$
$$-\mathrm{i} \omega B_0 = -\mathrm{i} E_0 k \; \Rightarrow \; B_0=\frac{k}{\omega} B_0.$$
As detailed in my previous posting the square root in the complex dispersion relation has to be taken such that the imaginary part of $k$ is positive, i.e.,
$$k=k_R+\mathrm{i} k_I, \quad k_R,k_I > 0.$$
That means that the phase shift $\phi$ between $B_0=\hat{B}_0 \exp(\mathrm{i} \phi)$ and $E_0=\hat{E}_0$ is given by
$$\phi=\arccos \left (\frac{k_I}{\sqrt{k_{R}^2+k_{I}^2}} \right )>0.$$
This means that the magnetic field's phase stays always behind the electric field's (because in my convention we have a time dependence $\propto \exp(-\mathrm{i} \omega t)$.

From
$$k^2=k_R^2-k_I^2 + 2 \mathrm{i} k_R k_I=\mu \epsilon \omega^2+\mathrm{i} \mu \sigma \omega$$
you find
$$k_R=\sqrt{\frac{\mu \epsilon}{2}} \omega \left [ \left (1+\frac{\sigma^2}{\epsilon^2 \omega^2} \right )^{1/2}+1 \right ]^{1/2},$$
$$k_I=\sqrt{\frac{\mu \epsilon}{2}} \omega \left [ \left (1+\frac{\sigma^2}{\epsilon^2 \omega^2} \right )^{1/2}-1 \right ]^{1/2}.$$
Thus we have
$$\cos \phi=\frac{k_R}{\sqrt{k_R^2+k_I^2}}=\frac{1}{\sqrt{2}} \sqrt{\frac{1+\sqrt{1+\frac{\sigma^2}{\epsilon^2 \omega^2}}}{\sqrt{1+\frac{\sigma^2}{\epsilon^2 \omega^2}}}}.$$
This goes to 1 for $\sigma \rightarrow 0$, i.e., for a non-conducting material the phase shift between the magnetic and the electric field $\phi=0$ and it goes to $1/\sqrt{2}$, i.e., $\phi=\pi/4$ for $\sigma \rightarrow \infty$ for an ideal conductor.

This discussion can be much simplified by realizing that $0 \leq \phi \leq \pi/4$ and thus one can write
$$\tan \phi=\frac{k_I}{k_R}$$
and then using
$$\tan(2 \phi)=\frac{2 \tan \phi}{1-\tan^2 \phi}=\frac{\sigma}{\epsilon \omega}.$$
Thus you see that $\phi$ is monotonously growing with increasing $\sigma$ starting at $\phi=0$ for an insulator and ending at $\phi=\pi/4$ [corrected!] for an indeal conductor.

Last edited: Nov 22, 2013
24. Nov 22, 2013

### rude man

The problem did not define the direction of propagation. I chose the x direction.

25. Nov 22, 2013

### rude man

It is getting a bit monotonous at that.

Anyway, you got a lagging B as I and the problem statement did.