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Electromagnetism - intrinsic impedance and poynting vector

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erece
#1
Aug22-12, 02:41 AM
P: 72
1. I know that 1/2 of the real part of the complex poynting vector gives the average power flow per unit area. But what is the significance of its imaginary part ?
2. What is the significance of the complex intrinsic impedance ? For lossless medium (i.e. with no ohmic losses) it is real , but in circuit theory real impedance signifies ohmic losses. Then how the concept differs in both cases ? And also tell the significance of both the real and imaginary part of complex intrinsic impedance ?
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gnurf
#2
Aug22-12, 03:23 AM
P: 330
Quote Quote by erece View Post
1. I know that 1/2 of the real part of the complex poynting vector gives the average power flow per unit area. But what is the significance of its imaginary part ?
2. What is the significance of the complex intrinsic impedance ? For lossless medium (i.e. with no ohmic losses) it is real , but in circuit theory real impedance signifies ohmic losses. Then how the concept differs in both cases ? And also tell the significance of both the real and imaginary part of complex intrinsic impedance ?
I can't do these questions proper justice, so I'll leave them for someone else, but maybe this quote sheds some light on your first question:
The imaginary part of the Poynting vector gives you information about evanescent (i.e. non-propagating) fields. If you calculate the Poynting vector for a plane wave, you will find the imaginary part is zero. However, when the Poynting vector is calculated from the fields near complex structures, you will often find a non-zero imaginary part of the Poynting vector. This indicates there are resonant, non-propagating fields at that location. This can be useful information in some situations. For example, it's best to keep the PML absorbing boundary conditions far enough away from the structures so the evanescent fields are zero near the PML boundaries.

Source: http://docs.lumerical.com/en/mode/us...ng_vector.html
Perhaps you can tell me what to make of thatómy EM books are covered in metaphorical dust.
yungman
#3
Aug22-12, 03:34 AM
P: 3,883
Let me try as see whether this help:

[tex] e^{jx}=\cos x +j\sin x\; \Rightarrow \; \cos x =Re [e^{jx}][/tex]

cos x is not a complex, it has no imaginary part. It is just the real part of the [itex] e^{jx}\;[/itex]. there is no jsin x.

Remember for EM propagates in z direction

[tex]\vec E_{(z,t)}=\hat x E_0 e^{-\alpha z} \cos(\omega t -\beta z) [/tex]

This can be represented by

[tex]\vec E_{(z,t)}=Re[\tilde E_{(z)}e^{j\omega t}] \;\hbox { Where }\;\tilde E_{(z)}= \hat x E_0 e^{-\alpha z} e^{-j\beta z} [/tex]

You can write out the [itex] \vec H_{(z,t)}\; [/itex] the same way.

Remember these are not complex number. They are just the real part of a complex representation, not that they are a complex number.

Poynting vector is defined as:

[tex] \vec E_{(z,t)}\times \vec H_{(z,t)} [/tex]

Which is also a real number, not a complex number. You need to throw out the complex part as the reason given above.......That both E and H are real number represented only by the real part of the complex number. There is no imaginary part. It is just easier to do the calculation in complex form and throw away the imaginary part.

And me too, I have to dust up my note book to give you this. I have to review my notes on lossless and lossy dielectric before I get back to you on the second part.

yungman
#4
Aug22-12, 04:01 AM
P: 3,883
Electromagnetism - intrinsic impedance and poynting vector

OK regarding to complex intrinsic impedance. You start out from the homogeneous wave equation:

[tex] \nabla ^2 \vec E -\mu σ \frac{\partial \vec E}{\partial t}-\mu\epsilon \frac{\partial^2\vec E}{\partial t^2}= 0 \;\hbox { which in phasor form becomes }\; \nabla ^2 \vec E - j\omega\mu σ \tilde E + \omega^2 \mu\epsilon \tilde E = \nabla ^2 \tilde E +\omega^2\mu(\epsilon -j\frac {\sigma}{\omega}) \tilde E = 0[/tex]

You let [itex] \epsilon_c=(\epsilon -j\frac {\sigma}{\omega}) [/itex] so the wave equation becomes:

[tex]\nabla ^2 \tilde E + \omega^2\mu \epsilon_c \tilde E=0[/tex]

The intrinsic impedance:

[tex] η_0=\sqrt{\frac{\mu}{\epsilon_c}}[/tex]

So the impedance is complex. For lossless dielectric, [itex]\sigma=0\;[/itex]. The impedance become real.

Remember this is characteristic impedance, unless it is lossy, there is no conduction.

I am skipping a lot of steps as it is quite long in both of your question.
erece
#5
Aug22-12, 04:49 AM
P: 72
so for lossless medium intrinsic impedance is real, but real imedance signifies pure resistance and losses . So how this is possible in lossless medium ?
gnurf
#6
Aug22-12, 05:46 AM
P: 330
Quote Quote by erece View Post
so for lossless medium intrinsic impedance is real, but real imedance signifies pure resistance and losses . So how this is possible in lossless medium ?
I suspect your confusion is caused by you thinking about this as being the same as the electrical impedance of e.g. a wire. It's not. Also, I really need to get out of this thread. Good luck.
Enthalpy
#7
Aug22-12, 11:08 AM
P: 661
The real caracteristic impedance does mean a real power as in a circuit, but this power is transmitted further, not dissipated.

Fun: is this power is light from a star that disappeared a billion years ago, we can still detect it, which means the energy has been stored since it was emitted: vacuum stores energy. I fact, the energy of a vacuum insulated capacitor is stored in the vacuum, not in the electrodes. Same for a coil.
yungman
#8
Aug22-12, 11:12 AM
P: 3,883
Quote Quote by erece View Post
so for lossless medium intrinsic impedance is real, but real imedance signifies pure resistance and losses . So how this is possible in lossless medium ?
This is not real resistance, it is from the wave equation defined as intrinsic impedance η. The resistance part is only in the imaginary part with σ. For lossless dielectric ( air and others), σ=0 and the imaginary term disappeared and the resistance is infinite.

It is important to know intrinsic impedance is not resistance in electrical means, that's the reason I pulled out the wave equation. It is really a definition. For example, for vacuum:

[tex]η_0=\sqrt{\frac{\mu_0}{\epsilon_0}}=377Ω[/tex]

This don't mean you can measure the vacuum and see 377Ω. The kind of question you asked is not easily explained by common sense terms. From what I study so far, Electromagnetics and Maxwell's equations cannot be easily explained in common sense terms that you can like tell a story. It started out as an observation and the original Maxwell's equation came out. Then the whole vector calculus was invented to help explain the Maxwell's equations. That's the reason it is in vector calculus form. Explanations in English usually are an interpretation of the calculus equations..........or at least I am not good enough to explain it better. That's the reason I have to pull out those fancy equations to talk.

I am not an expert in EM theory, If anyone have better explanation, I would love to hear. This is not an easy subject, I studied three books, worked through most of the problems in three separate times and I still have more questions than answers. It is like peeling an onion, one layer at a time.
erece
#9
Aug22-12, 11:23 AM
P: 72
got that
thanks
Enthalpy
#10
Aug22-12, 11:34 AM
P: 661
The caracteristic impedance is the ratio between E and H for a plane wave; it's real when E and H are in phase. If you understand transmission lines first (U and I in phase) then free-space propagation is easier, to my taste at least.


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