# Can current density $J$ for a thin wire by written in terms of the current $I$ and Dirac delta function?

If there is a thin wire with current $$I$$ flowing through it, could I write the current density at all points in space of a horizontal 2D slice of the wire as $$I \cdot \delta^2(\vec r)$$ ?

I'm a bit new to the idea of the Dirac delta function, but as I understand, it can be used to model point sources. The area under the curve of the function is always 1. I'm not sure how to write it in 2-Dimensions so i'll write it as $$\delta^2(\vec{r})$$ where $$\vec r$$ is a 2D point in space.

Relationship between current & current density:

$$\iint_A{J\cdot dA}=I$$

Volume under the Dirac delta function is 1:

$$\iint_A I\cdot\delta^2(\vec r)\cdot dA=I\iint_A\delta^2(\vec r)\cdot dA=I$$

Setting the two equal to each other:

$$\iint_A{J\cdot dA}=\iint_A I\cdot\delta^2(\vec r)\cdot dA$$

$$J=I\cdot\delta^2 (\vec r)$$

Is this logic correct? The equality seems to work for calculating the $$B$$-Field around a long thin wire using Ampere's Law but I'm not sure if the math behind it is correct or if I'm just getting the right result by misunderstanding the math.

• Look alright. But be aware that there will be a Jacobian factor when transform the current density to cylindric coordinate.
– ytlu
Apr 13, 2021 at 4:06
• Not a duplicate, but very closely related: Dirac Delta Magnetic field. In particular, see the comment on my answer there. Nov 19, 2021 at 11:08
• Assuming that your current density is homogeneous (which is rarely the case), this can be done. Dec 21, 2021 at 10:35

You are thinking in the right direction, although it is not quite clear what $$\delta^2(\vec{r})$$ exactly means.
The current density of a current $$I$$ through a thin wire flowing along the $$z$$-axis can be written as $$\vec{J}(\vec{r})=I\delta(x)\delta(y) \vec{e}_z$$ where $$\vec{e}_z$$ is the unit vector in $$z$$-direction. You can check this by verifying $$\iint_A\vec{J}\cdot d\vec{A}=\begin{cases} I \quad\text{ , if the area A includes the point (x,y)=(0,0)} \\ 0 \quad\text{ , if the area A doesn't include the point (x,y)=(0,0)} \end{cases}$$
I am also trying to define a current density with delta function for a filament, wondering if it is correct. The definition is $$\vec{j} = \frac{I}{\pi} \cdot \frac{\delta(\rho)}{\rho}\cdot \vec{e_z}$$ where $$\rho$$ is a radius in cylindrical coordinates. Thus if we calculate the current, it is $$current = \int \vec{j}d\vec{S} = \int 2\pi I \cdot \frac{\delta(\rho)}{\rho}\cdot \vec{e_z} \cdot d\vec{S}$$ where the surface element is equal to $$d\vec{S}=\vec{e_z}\rho d\rho d\phi$$ thus the current is equal to $$\int \frac{I}{\pi} \cdot \frac{\delta(\rho)}{\rho}\cdot \vec{e_z} \cdot \vec{e_z}\rho d\rho d\phi = \frac{I}{\pi} \int_{0}^{\infty} \delta(\rho) d\rho \int_{0}^{2\pi}d\phi = \frac{I}{\pi}\frac{1}{2}2\pi=I$$