EM field of dipole derivation from Green's function

In summary: I was going through some of the references you recommended and I think I understand it a little better now. In summary, the electric field due to a dipole is given by \mathbf{E}= [1-i(\omega/c) r]\frac{3 (\mathbf{p}\cdot\mathbf{r})\mathbf{r}-\mathbf{p} }{r^3}+(\omega/c)^2\frac{\mathbf{p}-(\mathbf{p}\cdot\mathbf{r})\mathbf{r}}{r} e^{i(\omega/c)r}
  • #1
krindik
65
1
Hi all,

I know that the electric field generated by a dipole is given by

[itex]\mathbf{E}= [1-i(\omega/c) r]\frac{3 (\mathbf{p}\cdot\mathbf{r})\mathbf{r}-\mathbf{p} }{r^3}+(\omega/c)^2\frac{\mathbf{p}-(\mathbf{p}\cdot\mathbf{r})\mathbf{r}}{r} e^{i(\omega/c)r}[/itex]
where [itex]\mathbf{p} [/itex] is the dipole's dipole moment proportional to [itex]e^{-i\omega t}[/itex].

I'm struggling to find out how this is derived from a Green's function approach. Can somebody help me with this or point me to somebook/reference that shows derivation?

Thanks in advance.

cheers,
Krindik
 
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  • #2
Some references that deal with deriving the dyadic Green's function are Jackson's "Classical Electrodynamics" and Chew's "Waves and Fields in Inhomogeneous Media."
 
  • #3
Hi,

Thanks for the response.
This is what I understood of its derivation. Hope u could clarify.

The electric field due to the dipole can be given by
[itex]\mathbf{E}(\mathbf{r}) = k^2 \mathbf{G}(\mathbf{r}, \mathbf{r}')\mathbf{p}(\mathbf{r'})[/itex] where [itex]\mathbf{G}(\mathbf{r}, \mathbf{r}')[/itex] is the dyadic Green's functions for the dipole source, which is located at [itex]\mathbf{r}'[/itex] and k is the wave number in the unbounded background.
[itex]\mathbf{G}(\mathbf{r}, \mathbf{r}')[/itex] can be determined from the scalar Green's function [itex]G_0(\mathbf{r}, \mathbf{r}')[/itex] by
[itex]\mathbf{G} = \left[\mathbf{I} + \frac{1}{k^2}\nabla \nabla \right]G_0(\mathbf{r}, \mathbf{r}')[/itex] where [itex]\mathbf{I}[/itex] is the unit dyad, and [itex]G_0 = \displaystyle\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}[/itex].


Is this correct? Really appreciate your response.

cheers,
Krindik
 
  • #4
krindik said:
Hi,

Thanks for the response.
This is what I understood of its derivation. Hope u could clarify.

The electric field due to the dipole can be given by
[itex]\mathbf{E}(\mathbf{r}) = k^2 \mathbf{G}(\mathbf{r}, \mathbf{r}')\mathbf{p}(\mathbf{r'})[/itex] where [itex]\mathbf{G}(\mathbf{r}, \mathbf{r}')[/itex] is the dyadic Green's functions for the dipole source, which is located at [itex]\mathbf{r}'[/itex] and k is the wave number in the unbounded background.
[itex]\mathbf{G}(\mathbf{r}, \mathbf{r}')[/itex] can be determined from the scalar Green's function [itex]G_0(\mathbf{r}, \mathbf{r}')[/itex] by
[itex]\mathbf{G} = \left[\mathbf{I} + \frac{1}{k^2}\nabla \nabla \right]G_0(\mathbf{r}, \mathbf{r}')[/itex] where [itex]\mathbf{I}[/itex] is the unit dyad, and [itex]G_0 = \displaystyle\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}[/itex].


Is this correct? Really appreciate your response.

cheers,
Krindik

Yep, that should be pretty much it. Actually you skipped the first step, the relationship between the electric field and the source currents is integrated over all space as
[tex]\mathbf{E}(\mathbf{r}) = C\int_V \overline{\mathbf{G}}(\mathbf{r},\mathbf{r}') \cdot \mathbf{J}(\mathbf{r}') d\mathbf{r}' [/tex]
I can't remember what the constant C is off hand. But in the case of the dipole, the current source is a point source and so we can drop the integration. Relating the resulting expression to what you have in the OP is a different story. They should be equivalent but it probably would require some work to show that. Chew's text doesn't derive the expression you gave originally but it's a good resource about deriving the dyadic Green's function and his chapter 2 applies it to dipoles and dipoles in layered media. Jackson's would probably work with an expression more closely related to what you gave. It's just that I am much more familiar with Chew's and Kong's texts than say Jackson.
 
  • #5
Thanks a lot Born2bwire, u were really helpful.
 

Related to EM field of dipole derivation from Green's function

1. What is an EM field of dipole?

An EM field of dipole refers to the electric and magnetic fields generated by a dipole, which is a pair of equal and opposite charges separated by a small distance. These dipoles can be found in various systems, such as atoms, molecules, and antennas, and their EM fields play a crucial role in many physical phenomena.

2. What is Green's function in relation to the EM field of dipole?

Green's function is a mathematical tool used to solve differential equations, such as the ones that describe the behavior of EM fields. In the context of the EM field of dipole, Green's function helps to derive the equations that describe the electric and magnetic fields produced by a dipole.

3. How is the EM field of dipole derived from Green's function?

The EM field of dipole can be derived from Green's function by using the Maxwell's equations, which relate the electric and magnetic fields to their sources, i.e. charges and currents. By applying Green's function to the differential equations in these equations, the resulting solutions can be used to determine the EM fields produced by a dipole.

4. What are the applications of the EM field of dipole derived from Green's function?

The derived equations for the EM field of dipole have various applications, including in the study of electromagnetic radiation, antenna design, and molecular interactions. These equations provide a powerful tool for understanding and predicting the behavior of EM fields in systems containing dipoles.

5. Is the EM field of dipole derived from Green's function an accurate representation of real-world systems?

While the equations derived from Green's function provide a useful approximation of the EM fields produced by dipoles, they may not accurately capture all the complexities of real-world systems. Factors such as material properties and environmental conditions can affect the behavior of EM fields, and these may not be fully accounted for in the derivation process.

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