Charge density using the Dirac delta funciton

Click For Summary
The Dirac delta function is utilized in electromagnetism to represent charge density as a point charge, expressed mathematically as ρ = qδ(x - r(t)). This formulation ensures that the charge is localized at a specific point in space, with r(t) denoting the electron's trajectory. The delta function is defined such that its integral over all space equals one, effectively modeling an infinite spike at the charge's location while maintaining a total charge of q. Additionally, the current density can be represented as j(t, x) = qṙ(t)δ(x - r(t)), linking charge density and motion. This framework allows for the rigorous mathematical treatment of point charges and instantaneous changes in physical systems.
doggydan42
Messages
169
Reaction score
18
Currently, I am reading this article which introduces electromagnetism.

It gives a function for the charge density as: $$\rho = q\delta(x-r(t))$$
The paper states that "the delta-function ensures that all the charge sits at a point," but how does it do that? Also, if ##r(t)## is the trajectory of the electron, what is its purpose in the equation?
 
Physics news on Phys.org
The Dirac delta function is not a simple function but rather a more abstract mathematical object. If you consider all of the ways you can linearly map functions to numbers (called a linear functional), a great number of the more useful ones can be expressed as integrals with respect to another function.
\phi_w: f\mapsto a = \int_{-\infty}^\infty f(x)w(x) dx
(read ##\mapsto## as "is mapped to")
The choice of weighting function or distribution ##w## determines a large class of these linear functionals but not all. For example you cannot express the linear mapping that takes a function to its value at zero (or some other fixed point). ## f\mapsto f(0)##
There is no actual function but we can invent a notation, which looks like function notation but only has meaning as a place holder inside these functional integrals.

By definition:
\int_{-\infty}^\infty f(x) \delta(x) dx = f(0)
We can then assert that this place holder obeys the same rules a functions inside of integrals with respect to change of variable, rules of differentiation and so on.
We can thus shift the variable of integration to get:
\int_{-\infty}^\infty f(x) \delta(x-a) dx = f(a)
This is what we have to do to give the Dirac delta function rigorous meaning mathematically. In practice you can think of it as a "function" that is infinite at one point, zero everywhere else, and is normalized to integrate to 1. It is often expressed as the limit of e.g. Gaussian distribution functions in the limit as the standard deviation goes to zero.

It is of course very handy in physics applications where you are working in a paradigm of charge or mass densities but want to introduce a point mass or point charge. Another application is instantaneous changes in velocity where one can assert the applied force and thus the acceleration took the form of a delta function (of time).

In your case the delta function ## \delta(x - r(t))## is zero everywhere except where ##x=r(t)## at which point you have the infinite spike representing the charge density. Since again the total integral of the delta function is 1, multiplying by ##q## scales it to have total charge ##q##.
 
  • Like
Likes dextercioby
The correct equation is
$$\rho(t,\vec{x})=q \delta^{(3)}[\vec{x}-\vec{r}(t)],$$
where ##\vec{r}(t)## is the trajectory of the electron written in a fixed (inertial) reference frame.

The current density is
$$\vec{j}(t,\vec{x}) = q \dot{\vec{r}}(t) \delta^{(3)}[\vec{x}-\vec{r}(t)].$$
It is a very good exercise to prove that ##(\rho,\vec{j})## is a Minkowski four-vector field!

Hint: Write the expressions in terms of an integral over the proper time of the electron!
 
Thread 'Inducing EMF Through a Coil: Understanding Flux'
Thank you for reading my post. I can understand why a change in magnetic flux through a conducting surface would induce an emf, but how does this work when inducing an emf through a coil? How does the flux through the empty space between the wires have an effect on the electrons in the wire itself? In the image below is a coil with a magnetic field going through the space between the wires but not necessarily through the wires themselves. Thank you.

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 1 ·
Replies
1
Views
646