Dirac Delta source for vectorial equation

In summary, the conversation discusses the use of Maxwell's equations and the Lorentz' gauge to obtain a vector wave equation. The first step for the solution is to consider a point source, which leads to the use of a Green function and a scalar function. The Helmholtz equation applies to each component of the vector field separately, and the retarded Green's function is typically used for waves irradiated outwards from the sources.
  • #1
EmilyRuck
136
6
Hello!
By manipulating Maxwell's equation, with the potential vector [itex]\mathbf{A}[/itex] and the Lorentz' gauge, one can obtain the following vector wave equation:

[itex]∇^2 \mathbf{A}(\mathbf{r}) + k^2 \mathbf{A}(\mathbf{r}) = -\mu \mathbf{J}(\mathbf{r})[/itex]

The first step for the solution is to consider a point source [itex]- \delta (\mathbf{r} - \mathbf{r}')[/itex]. But what about its meaning in a vector form?
A vector is a 1-dimensional object in the space; a point is a 0-dimensional object, according to linear algebra. How can we relate a 0-dimensional object to a vector context?
With such a source, the unknown function is a Green function and the previous equation becomes

[itex]∇^2 G(\mathbf{r}, \mathbf{r}') + k^2 G(\mathbf{r}, \mathbf{r}') = - \delta (\mathbf{r} - \mathbf{r}')[/itex]

but now we don't have a vector function [itex]\mathbf{A}(\mathbf{r})[/itex] any more, but just a scalar function [itex]G(\mathbf{r}, \mathbf{r}')[/itex]. Why?
This is like saying that, when the source is a point, the vector potential [itex]\mathbf{A}(\mathbf{r})[/itex] becomes scalar...?!
Thank you for having read,

Emily
 
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  • #2
The Helmholtz equation holds for each (Cartesian!) component of the vector field separately. So as soon as you have the Green's function for the scalar wave function you can apply it to each component.

BTW: Usually one needs the retarded Green's function, which gives you waves irradiated outwards from the sources. So you need to find a
[tex]G(\vec{r},\vec{r}') \propto \frac{\exp(+\mathrm{i}k|\vec{r}-\vec{r}'|)}{|\vec{r}-\vec{r}'|}.[/tex]
 
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Related to Dirac Delta source for vectorial equation

1. What is the Dirac Delta source for vectorial equation?

The Dirac Delta source for vectorial equation is a mathematical function that represents an infinitely narrow, infinitely tall spike at a specific point in space. It is often used in physics and engineering to model point sources of energy or forces.

2. How is the Dirac Delta source represented in equations?

The Dirac Delta source is typically represented as δ(x), where x is the variable representing the point in space where the source is located. It can also be represented as δ(x-x0), where x0 is the location of the source.

3. What is the physical interpretation of the Dirac Delta source?

The physical interpretation of the Dirac Delta source is that it represents a point source of energy or force. This means that at the specific point in space where the source is located, there is an infinite amount of energy or force acting in a specific direction.

4. How is the Dirac Delta source used in vectorial equations?

In vectorial equations, the Dirac Delta source is often used to model point sources of energy or forces that act on a system. It is typically used in conjunction with other equations and boundary conditions to solve for the behavior of the system.

5. Are there any limitations to using the Dirac Delta source?

Yes, there are limitations to using the Dirac Delta source. It is a mathematical idealization and does not have a physical counterpart in the real world. Additionally, it can only be used to model point sources and not distributed sources of energy or forces.

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