I am trying to convert Maxwell's equations from the well known differential form (found on Wikipedia Maxwell's equations) into scalar equations involving partial derivatives (more than four equations). The problem I'm having is finding the definitions of physical quantities. I'm not sure how charge density and current density are expressed in euclidean space. How are these expressed? What are Maxwell's Equations expressed in this way?

  • $\begingroup$ Just apply the definitions of divergence ($\bf{\nabla} \cdot$) and curl ($\bf{\nabla} \times$) to Maxwell's vector-differential equations. $\endgroup$ Mar 29, 2019 at 13:45
  • $\begingroup$ @ThomasFritsch I suspect that OP wants individual scalar equations (e.g., $\partial_xE_x=\text{something}$) & is questioning how that could be done. $\endgroup$
    – Kyle Kanos
    Mar 29, 2019 at 18:35
  • $\begingroup$ Is this helpful? zpenergy.com/downloads/Orig_maxwell_equations.pdf $\endgroup$
    – d_b
    Mar 29, 2019 at 18:38

2 Answers 2


I'm not sure how charge density and current density are expressed in euclidean space.

Charge density $\rho$ is just a scalar density (measured in Coulomb/m$^3$).
And current density $\vec{J}$ is a vector density (with cartesian components $J_x, J_y, J_z$, measured in Ampere/m$^2$).

What are Maxwell's Equations expressed in this way?

Maxwell's equations (written with vector-calculus) are given by $$ \begin{align} \vec{\nabla}\cdot\vec{B} &= 0 \tag{1a} \\ \vec{\nabla}\cdot\vec{E} &= \frac{\rho}{\epsilon_0} \tag{1b} \\ \vec{\nabla}\times\vec{E} + \frac{\partial\vec{B}}{\partial t} &= \vec{0} \tag{1c} \\ \vec{\nabla}\times\vec{B} - \frac{1}{c^2} \frac{\partial\vec{E}}{\partial t} &= \mu_0\vec{J} \tag{1d} \\ \end{align} $$

Using the definitions of divergence ($\vec{\nabla}\cdot$) and curl ($\vec{\nabla}\times$) it is straight-forward to write the above equations with the cartesian field components ($E_x,E_y,E_z,B_x,B_y,B_z$) in cartesian coordinates ($x,y,z$).

The scalar equations (1a) and (1b) each remain one equation.
The vector equations (1c) and (1d) each expand to three equations (for $x, y, z$ components). $$ \begin{align} \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} &= 0 \\ & \\ \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} &= \frac{\rho}{\epsilon_0} \\ & \\ \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} + \frac{\partial B_x}{\partial t} &= 0 \\ \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} + \frac{\partial B_y}{\partial t} &= 0 \\ \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} + \frac{\partial B_z}{\partial t} &= 0 \\ & \\ \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} - \frac{1}{c^2}\frac{\partial E_x}{\partial t} &= \mu_0 J_x \\ \frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x} - \frac{1}{c^2}\frac{\partial E_y}{\partial t} &= \mu_0 J_y \\ \frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y} - \frac{1}{c^2}\frac{\partial E_z}{\partial t} &= \mu_0 J_z \end{align} $$


If I understand you correctly,there are a couple ways you can proceed.

Method 1:

Maxwell's Equations in vector form already represent partial differential equations of the components of the Electric and Magnetic Fields. You can re-express the standard equations strictly in terms of those components.

This can be done most compactly using Einstein's Summation notation.Repeated indices in a sum are assumed to be summed over even without the usual $\Sigma$

In this way, the dot product of two vectora $\vec{A}$ and $\vec{B}$ is:


$$\nabla \cdot \vec{E}= \frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=\sum_{k=1}^3\frac{\partial}{\partial x_i}E_i=\frac{\partial}{\partial x^i}E_i$$

Where $i$ ranges from 1 to 3, $x_1=x, x_2=y, x_3=z$

You drop the sum easing the notation especially when you start using double sums.

In this way you can represent the curl of a vector field component by component:

$$ (\nabla \times \vec{B})_i=\epsilon_{ijk}\frac{\partial}{x_j}B_k$$

Putting it all together with Maxwell's equations, you get:

$$\frac{\partial}{\partial x_i}E_i=\rho/\epsilon_o$$ $$\frac{\partial}{\partial x_i}B_i=0$$ $$\epsilon_{ijk}\frac{\partial}{\partial x_i}E_i=-\frac{\partial}{\partial t}B_i$$ $$\epsilon_{ijk}\frac{\partial}{\partial x_i}B_i=\mu_0J_i+\frac{\partial}{\partial t}E_i$$

Method 2:

If you let $\vec{B}=\nabla \times \vec{A}$ and $\vec{E}=-\nabla V - \frac{\partial}{\partial}\vec{A}$ where $V$ is the scalar potential and $\vec{A}$ is the magnetic vector potential, and use the Lorentz Gauage, then you can rewrite Maxwell's equations in the form of 4 (net) equations.

$$\vec{E}=-\nabla V - \partial\vec{A}/\partial t$$ $$\nabla \cdot \vec{E}=\rho/\epsilon_0=-\nabla^2 V-\partial(\nabla \cdot \vec{A})\partial t$$

By the Lorentz Gauge, $\nabla \cdot \vec{A} = -\frac{1}{c^2}\frac{\partial}{\partial t}V$

Re arranging, we get:

$$\nabla ^2 V - \frac{1}{c^2}\frac{\partial ^2}{\partial t ^2}V=-\rho/\epsilon_0$$

There is also:

$$\nabla \times \vec{B}=\nabla \times (\nabla \times \vec{A})=\nabla(\nabla \cdot \vec{A}) - \nabla^2\vec{A}=\mu_0\vec{J}+\frac{1}{c^2}\frac{\partial}{\partial t}(-\nabla V - \partial \vec{A}/\partial t)$$


$$ \nabla(\nabla \cdot \vec{A} + \frac{1}{c^2}\partial V/ \partial t)-\nabla^2\vec{A}+\frac{1}{c^2}\frac{\partial ^2}{\partial t^2}\vec{A}= \mu_0\vec{J}$$

Where the far left term vanishes again the lorentz gauge. Then as in method one, this final equation can be represented component by component is talked about in method one.


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