Reactive power with electromagnetic sources in free space

[EmilyRuck]

Good morning,
in circuit theory I know that reacting power arise from phasors and represents a power which can't be used, because not delivered to any load, but continuously flows back and forth between the load and the generator with a zero mean during one period.
I can't understand very well, anyway, the meaning of this power in a field context, with electromagnetic sources in free space.
Let's consider a hertzian dipole, which has several fields component with several dipendence from the distance $r$ (in spherical coordinates).

$H_{\varphi} = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{jk}{r} + \frac{1}{r^2} \right) \sin \theta \\ E_r = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{2 \eta}{r^2} + \frac{2}{j \omega \epsilon r^3} \right) \cos \theta \\ E_{\theta} = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{j \omega \mu}{r} + \frac{\eta}{r^2} + \frac{1}{j \omega \epsilon r^3} \right) \sin \theta$

Components proportional to $1/r^2$ and $1/r^3$ are called near field components; components proportional to 1/r are called far-field components. $I_0$ is the phasor of the current in the dipole and $h$ its length.

I calculate the power as flux of the Poynting vector through a surface $S$:

$P = \displaystyle \oint_S \mathbf{E} \times \mathbf{H}^* \cdot d\mathbf{S}$

where the dipole is in the origin and $S$ is a sphere with radius $r$ centered in the origin too.

I find for the power $P$ a real part, which is the power that goes away from the dipole, and an imaginary part, which is the reactive power and is only stored near the dipole. In the waveguide theory, reactive power is carried by modes that cannot propagate, that is, modes that have a purely imaginary propagation constant: they attenuate exponentially along the waveguide. But now the reactive power is carried by a field that propagate, because it has the $e^{-jkr}$ term like the "far field" components: how can it is possible?
Moreover, what about the meaning of this reactive power? Books say that it is due to the $1/r^2$ and $1/r^3$ components, which are related to *static* fields. So, should I state that reactive power is that carried from static fields? A static field stores an energy and I can *use* this energy if I place a charge in the field, because the field will move the charge, making a work: but this is a sort of active power, a suitable power, isn't it?
Thank you for having read and sorry for my confusion.
Bye!

Emily

 

[sophiecentaur]

I feel that the term 'reactive power' is not strictly appropriate because no energy is being transferred. I think that the term 'reactive energy' is a better one because it is an energy build up between the inductive and capacitive elements in the structure. It takes a finite time, after switch-on for the waves / oscillating fields to build up - this can be either an antenna or a transmission line feeding into a load, where there are mismatches along the way. When the dipole (or any other antenna structure) is at resonance, the reactive energy is maximum and the power source 'sees' just a resistance (the radiation resistance).

 

[EmilyRuck]

it is an energy build up between the inductive and capacitive elements in the structure.

Thank you! What "elements" can you consider as inductive/capacitive in the free space?

It takes a finite time, after switch-on for the waves / oscillating fields to build up

And I suppose this time is necessary to "charge" the inductors and/or the capacitors.

Emily

 

[sophiecentaur]

Thank you! What "elements" can you consider as inductive/capacitive in the free space?



And I suppose this time is necessary to "charge" the inductors and/or the capacitors.

Emily

The are no 'elements' in free space but there is a characteristic impedance for free space of 377Ω.
For a line, Z₀ =√(L/C)

The time I was referring to is the time for the waves to reach a steady state of energy flow in a transmission line - taking into account any reflections there might be at interfaces, due to mis-matches. The 'standing waves' need to establish themselves and this will take a number of journeys along the section of line. The 'charge / discharge' times of reactive elements is included in the transmission equations, I think.

I don't know how relevant this is, actually, to your post, now I think about it.

 

[EmilyRuck]

The are no 'elements' in free space but there is a characteristic impedance for free space of 377Ω.
For a line, Z₀ =√(L/C)

Ok, that's right, we can establish an analogy between transmission lines and free space.

I don't know how relevant this is, actually, to your post, now I think about it.

This is not really about my topic, but is anyway a useful in-depth analysis.
Now I would like to know more about the question: reactive power is that carried from static fields? I observed that a static field stores an energy and I can *use* this energy if I place a charge in the field, because the field will move the charge, making a work. Which is the difference between the energy provided by a static field and the energy provided by an electro-magnetic field in dynamic conditions?

Emily

 

[sophiecentaur]

I don't think the term "reactive" applies to static fields, does it? For a static situation, you will either have a reactance of zero or infinity. It's more 'potential energy' that is involved in that case, I think. Also, does a 'static field' actually carry energy (implies from place to place?)
But in your OP, you are referring to changing fields - EM waves associated with a Hertzian dipole. I think my post about transmission lines is, in fact, relevant in as far as the quadrature E and H fields in the near field region must take time to build up to the steady state. These fields store Energy and not Power (you don't store power, by definition because power is rate of energy transfer). Because they exist at a distance from the dipole, they take time to establish themselves. The in-phase components of the field are the ones involved with radiating power. (I don't think this is being too pedantic, is it?)

Which is the difference between the energy provided by a static field and the energy provided by an electro-magnetic field in dynamic conditions?

I don't think there needs to be any difference - after all, a very slowly varying field is indistinguishable from a static field so where would you draw the line, apart from where QM becomes relevant?

 

[EmilyRuck]

I don't think the term "reactive" applies to static fields, does it? For a static situation, you will either have a reactance of zero or infinity.

You're right. But if we substitute "reactive" with "storing and giving back energy", we can talk also about static fields.

Also, does a 'static field' actually carry energy (implies from place to place?)

A charge in the space is subjected to the field and the field can make (theoretically) everywhere a work on this charge: this is a sort of transfer of energy from the source of the fields to the place where the work is made, so, yes, I think from place to place. But the difference with the dynamic case is that this energy, stored by the field, is used only in the volume where the field lies, whereas the active energy "moves away" from the hertzian dipole. Is it right?

Because they exist at a distance from the dipole, they take time to establish themselves.

This is more and more similar to the transmission line theory with a reactive load: here the reactive load is space and there is a sort of standing wave with one end in the source and the other end in the surrounding space, until the near-field components are significant.

The in-phase components of the field are the ones involved with radiating power. (I don't think this is being too pedantic, is it?)

No, I completely agree!

I don't think there needs to be any difference - after all, a very slowly varying field is indistinguishable from a static field so where would you draw the line, apart from where QM becomes relevant?

My line was $\omega = 0$ for static fields and simply $\omega \neq 0$ for dynamic ones. But now it is not important, because you gave me an excellent point of view: "a very slowly varying field is indistinguishable from a static field". Thank you!

 

[sophiecentaur]

So we're all happy - excellent!