Is it possible to describe the physical meaning of Maxwell's equations and show how they lead to electromagnetic wave, with little involvement of mathematics ?

  • 1
    $\begingroup$ Hi user38578, and welcome to Physics Stack Exchange! These kinds of questions are better if you just ask what you want to know, rather than asking for a book. People will still recommend resources in their answers as they see fit. I've edited accordingly, but if you really must have a resource rather than a direct answer you can edit the question to explain why. $\endgroup$ – David Z Jan 30 '14 at 7:49
  • $\begingroup$ What about them do you find unphysical? Is it the roles of $E$ and $B$ themselves, or the vector calculus involved? $\endgroup$ – Robert Mastragostino Jan 30 '14 at 8:08
  • $\begingroup$ In the beginning, there was light. And then there was Maxwell who was able to formulate a complete and symmetric theory of electricity and magnetism which predicted electromagnet waves and the speed of light. The following is the tensor packaging of Maxwell's Equations in Minkowski space: $\partial_{[\mu}F_{\nu \lambda}=0$, and $\partial_{\mu}F^{\mu \nu}=\mu_{0}J^\nu$. And no it's not possible - the description is inversely proportional to your knowledge of the subject - of which we know nothing. To understand the physical meaning, go camping for a week without any electronic devices. $\endgroup$ – Cinaed Simson Jul 22 at 1:02

Is it possible to describe the physical meaning of Maxwell's equations and show how they lead to electromagnetic wave, with little involvement of mathematics ?


But don't be put off, I'll go through the mathematics slow.

Starting with Maxwell's equations. Maxwell's equations are relatively old and don't take into account quantum mechanics and are used (like Newtons law of gravity) because it gets the right answer (pretty much) the better quantum field theory (General relativity for gravity) is too maths heavy to be practical for most uses.

$$ \nabla \cdot \vec E = \frac{\rho}{ \epsilon} $$

$$ \nabla \cdot \vec B = 0 $$

$$ \nabla \times \vec E = - \frac {\partial \vec B}{\partial t} $$

$$ \nabla \times \vec B = \mu \vec J + \epsilon \mu \frac {\partial \vec E}{\partial t} $$


$\vec B$ is magnetic field

$\vec E$ is electric field

The arrows $\vec E$ and $\vec B$ indicate that this quantity is a vector and points (flows) in a given direction

$\rho$ is the charge density (amount of charge)

$\vec J$ is current (density)

$\epsilon , \mu$ are just constants

$\nabla$ (upside down triangle [aka del]) on its own is the gradient operator (just a mathematical thing [that does stuff to functions]) (it describes how something changes in space)

$\nabla \cdot$ (up-side-down triangle followed by a dot [aka del dot]) is an The divergence operator (how something changes in space [in an expanding or contracting kind of way])

$\nabla \times$ (up-side-down triangle followed by a multiplied by "x" [aka del cross]) is the curl operator (how something changes in space [in an turning kind of way])

$\frac{\partial}{\partial t}$ is the time derivative operator (how something changes in time). Let's start with the easiest one:

$$ \nabla \cdot \vec B = 0 $$

This just says the divergence of the magnetic field is zero always no matter what. This means that the magnetic $\vec B$ field can be thought of as a tank of water where it can move but never will there be bubbles of no water or areas of higher density water. (unlike air which can compress if you push it hard enough)

Which can also be written as:

$$ \iint_{S} \vec B \cdot d \vec s = 0 $$

Which says the same thing in a different way, that if you add up $\iint$ all of the $\vec B$ field escaping through the tiny bits of area $d\vec s$ of a closed surface $S$ then they will add to zero. So basically if some $\vec B$ field enters at one point it has to leave again at another point.


$$ \nabla \cdot \vec E = \frac{\rho}{ \epsilon} $$

This says a very similar thing: if there is no charge then the $\vec E$ field is divergent free i.e. and in-compressible fluid. However if there is a charge e.g. a proton then it acts like the end of a hose pipe and $\vec E$ "fluid" bursts out in all directions (referred to as a source). If there is a negative charge then the $\vec E$ "fluid" gets sucked in (referred to as a sink). This can be re written as:

$$ \iint_{S} \vec E \cdot d\vec s = \frac {1}{\epsilon} \iiint_{V}\, \rho dv $$

This way of saying it means that if you add up $\iint$ all of the $E$ field escaping through the tiny bits of area $d\vec s$ of a closed surface $S$ then they will be equal to what you get if you add up $\iiint$ the amount of charge density $\rho$ times all of the little volumes $dv$ inside the volume $V$ of the surface $S$.

So basically you can tell the number of hose ends within an area by the amount of water that flows from that area. If there are no charge (or the same number of sucking hoses as blowing hoses) then this adds to zero.


$$ \nabla \times \vec E = - \frac {\partial \vec B}{\partial t} $$

This says that if $\vec B$ changes in time $\dfrac {\partial \vec B}{\partial t}$ then it will cause $\vec E$ to move in space $\nabla \times$ i.e. that a changing magnetic field will cause an electric field which if it were in a wire we would call a voltage.

And finally:

$$ \nabla \times \vec B = \mu \vec J +\epsilon \mu \frac {\partial \vec E}{\partial t} $$

This has three parts so lets break it down. Note that $\vec J$ appears for the first time, this is current density (the movement of charge carriers).

If there are no charge carriers (i.e. no protons and electrons) like the way there are no (Maybe there are I'm still waiting on you LHC) magnetic charge carriers (called mono-poles) Then this law looks very like the one previous one.


$$ \vec J = \vec 0 $$


$$ \nabla \times \vec B = \epsilon \mu \frac {\partial \vec E}{\partial t} $$

Looks a lot like:

$$ \nabla \times \vec E = - \frac {\partial \vec B}{\partial t} $$

This is actually the one of the conditions for light to propagate.

We want to prove that Maxwell's equations have wave solutions. We can prove this by producing the wave equation:

$$ \frac {\partial^2 \vec A}{\partial x^2}= \frac{1}{v^2}\frac {\partial^2 \vec A}{\partial t^2} $$

Or in 3 dimensions:

$$ \nabla^2 \vec A = \frac{1}{v^2}\frac {\partial^2 \vec A}{\partial t^2} $$

Where $\nabla^2 $ is the 3D analog of $\frac {\partial^2}{\partial x^2}$ i.e. $\frac {\partial^2}{\partial x^2}+\frac {\partial^2}{\partial y^2}+\frac {\partial^2}{\partial z^2}$

Any system that obeys the above wave equation behaves with wavelike properties. To prove that Maxwell's equations have these solutions we also have to let:

$$ \rho=0 $$


$$ \nabla \cdot \vec E = 0 $$

We then take the curl of both sides of: $\nabla \times \vec E = - \frac {\partial \vec B}{\partial t}$

To get:

$$ \nabla \times \nabla \times \vec E = -\nabla \times \frac {\partial \vec B}{\partial t} $$

A theorem in calculus that states that the curl of a curl of a field can be written as:

$$ \nabla \times \nabla \times \vec A = \nabla (\nabla \cdot \vec A) - \nabla^2 \vec A $$


$$ \nabla \times \nabla \times \vec E = \nabla (\nabla \cdot \vec E) - \nabla^2 \vec E $$


$$ \nabla \cdot \vec E = 0 $$


$$ \nabla \times \nabla \times \vec E = - \nabla^2 \vec E $$

Replacing that in $\nabla \times \nabla \times \vec E = -\nabla \times \frac {\partial \vec B}{\partial t}$

$$ - \nabla^2 \vec E = -\nabla \times \frac {\partial \vec B}{\partial t} $$

Cancelling the minus sign. Switching $-\nabla \times$ and $\frac {\partial}{\partial t}$

$$ \nabla^2 \vec E = \frac {\partial}{\partial t} \nabla \times \vec B $$

From Maxwell's equations we know that:

$$ \nabla \times \vec B = \epsilon \mu \frac {\partial \vec E}{\partial t} $$

Replacing that we get

$$ \nabla^2 \vec E = \frac {\partial}{\partial t} (\epsilon \mu \frac {\partial \vec E}{\partial t}) $$


$$ \nabla^2 \vec E =\epsilon \mu \frac {\partial^2\vec E}{\partial t^2} $$

Comparing this to our original wave equation.

$$ \nabla^2 \vec A = \frac{1}{v^2}\frac {\partial^2 \vec A}{\partial t^2} $$

We see that $\vec E$ corresponds to $\vec A$

And $\epsilon \mu$ corresponds to $\frac{1}{v^2}$

So the speed at which this wave travels is $v = \frac{1}{\sqrt{\epsilon \mu}}$ This speed is dubbed $c$

This derivation can be repeated for the magnetic field $\vec B$ to get that in wave solutions the two field always come in pairs perpendicular to each other and to the direction that they are going.

So basically changing electric field creates changing magnetic field which creates a changing electric field etc. This also tells us the speed of light $c$ is equal to $c = \dfrac{1}{\sqrt{\epsilon \mu}}$

Note: Einstein thought that it was interesting not what terms show up in the equation of the speed of light but the term missing- that this speed of light is independent of how fast you are travelling, and he thought that everybody regardless of speed would measure this speed to be the same.

Next we let $\vec J = \vec 0$

This time we are going to let $\vec E$ be constant in time so that

$$ \dfrac {\partial \vec E}{\partial t} = \vec 0 $$.

In this case:

$$ \nabla \times \vec B = \mu \vec J $$

The current (density) $\vec J$ creates a magnetic field (that changes in space but not time) so since this magnetic field does not chance in time it cannot create the $\vec E$ field from equation:

$$ \nabla \times \vec E = - \dfrac {\partial \vec B}{\partial t} $$

This is how electro-magnets work. The current (density) $\vec J$ creates a magnetic field around the wire, this magnetic field can be intensified by having the wire in loops like a solenoid the "soft" iron core of the solenoid helps the magnetic fluid to flow by changing the values of (μ) to make it easier to channel through it instead of the air.

Back to:

$$ \nabla \times \vec B = \mu \vec J +\epsilon \mu \dfrac {\partial \vec E}{\partial t} $$

If $\vec J$ is net zero over time so the charge carriers only move back and forth. So we ignore $\vec J$ (but charge carriers exist this time)

$$ \nabla \times \vec B = \epsilon \mu \dfrac {\partial \vec E}{\partial t} $$

It is possible for the change in electric field caused by the changing moving electrons to create a changing moving magnetic field. (Since the rate of change of the electric field is not constant [they are not just getting faster and faster they are getting faster then slower then faster {sorry if that's kinda confusing}]).

So changing $\vec E$ field creates changing $\vec B$ field creates changing $\vec E$ field on the far side of the Transformer core (it also creates loops of current [Eddie currents] inside the core that is why the core is made of layers instead of one solid core because we don't want current there it would generate heat and waste energy.)

It is possible to define a displacement current $\vec J_D$ that has the same units of $\vec J$ but also takes into account the energy that flows through the capacitor or transformer.


protected by Qmechanic Jul 14 '14 at 14:06

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