Electromagnetism - intrinsic impedance and poynting vector

In summary, the real and imaginary parts of the complex intrinsic impedance provide different information. The real part is about evanescent fields, while the imaginary part is about resonant fields. The real part is not a complex number, while the imaginary part is.
  • #1
erece
72
0
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 ?
 
Engineering news on Phys.org
  • #2
erece said:
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/user_guide_understanding_poynting_vector.html

Perhaps you can tell me what to make of that—my EM books are covered in metaphorical dust.
 
Last edited by a moderator:
  • #3
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.
 
Last edited:
  • #4
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.
 
Last edited:
  • #5
so for lossless medium intrinsic impedance is real, but real imedance signifies pure resistance and losses . So how this is possible in lossless medium ?
 
  • #6
erece said:
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.
 
  • #7
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.
 
  • #8
erece said:
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.
 
Last edited:
  • #9
got that
thanks
 
  • #10
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.
 

1. What is the intrinsic impedance of a medium?

The intrinsic impedance, also known as characteristic impedance, is a property of a medium that describes its resistance to the flow of electric and magnetic fields. It is represented by the symbol Z and is defined as the ratio of the magnitude of the electric field to the magnitude of the magnetic field in the medium. The unit of intrinsic impedance is ohms (Ω).

2. How is the intrinsic impedance related to the speed of electromagnetic waves?

The intrinsic impedance of a medium is directly proportional to the speed of electromagnetic waves in that medium. This means that as the intrinsic impedance increases, the speed of electromagnetic waves also increases. This relationship is described by the equation Z = μ0c, where μ0 is the permeability of free space and c is the speed of light.

3. What is the Poynting vector in electromagnetism?

The Poynting vector, represented by the symbol S, is a mathematical quantity that describes the direction and magnitude of the flow of electromagnetic energy in a given point in space. It is defined as the cross product of the electric field vector and the magnetic field vector. The direction of the Poynting vector is perpendicular to both the electric and magnetic field vectors.

4. How is the Poynting vector related to the intensity of electromagnetic waves?

The Poynting vector is directly proportional to the intensity of electromagnetic waves. This means that as the Poynting vector increases, the intensity of electromagnetic waves also increases. The relationship is described by the equation S = 1/μ0 * E x B, where E is the electric field and B is the magnetic field.

5. What is the significance of the intrinsic impedance and Poynting vector in electromagnetism?

The intrinsic impedance and Poynting vector are important concepts in electromagnetism as they help us understand the behavior of electromagnetic waves in different media. The intrinsic impedance helps us calculate the reflection and transmission of electromagnetic waves at the interface of two different media, while the Poynting vector helps us understand the direction and intensity of energy flow. These concepts are crucial in the design and analysis of various electrical and electronic systems.

Similar threads

Replies
14
Views
2K
Replies
4
Views
888
Replies
14
Views
1K
  • Electrical Engineering
Replies
5
Views
994
  • Electrical Engineering
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
758
  • Other Physics Topics
Replies
2
Views
4K
Replies
1
Views
566
  • Classical Physics
Replies
16
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
767
Back
Top