Determining c from Maxwell's Equations

[Buckethead]

There are a few good youtube videos that calculate c from Maxwell's Equations, but the jist of it still hasn't gelled with me. In other words, I'm still scratching my head. I wonder if anyone could (without using math) explain to me how a velocity can be calculated from magnetic and electric field relationships. I think it has something to do with the change of the ratio of the field strengths of the two, but can you explain this so it makes sense? I recognize that in an electromagnetic wave when the mag field gets stronger the electric field gets weaker and these two oscillate, but how can a velocity be calculated from this? In other words, what is the logic behind this? Thanks.

 

[Nugatory]

It will make more sense after you've worked through a textbook like https://www.amazon.com/dp/0070048606/?tag=pfamazon01-20 (not a specific recommendation, just happens to be the one I learned from), which has many examples of how waves naturally arise in many different physical situations.

Loosely speaking, what's going is that the speed of a wave depends on two things: How much the thing that is waving resists being displaced and how rapidly it responds to the restoring force. Combine these two effects and you get a differential equation called the "wave equation" (google for that), and its solutions are waves traveling at a particular speed determined by the ratio of the two effects. Although there's no physical displacement and restoring force involved in electromagnetic radiation, the electric and magnetic fields have a similar relation under Maxwell's equations, so a similar differential equation appears, and its solutions are waves traveling at speed c.

 

[robphy]

I recognize that in an electromagnetic wave when the mag field gets stronger the electric field gets weaker and these two oscillate, but how can a velocity be calculated from this? In other words, what is the logic behind this? Thanks.

By the way, in the usual plane wave solution, the electric and magnetic fields are in phase... when one is at a maximum, minimum, or zero... the other is as well...

Also possibly useful reading: http://feynmanlectures.caltech.edu/II_18.html

 

[Buckethead]

Thank you for the responses and the book recommendation and the link. I see that in the Feynman lecture, in the 4th equation c^2 appears which I hadn't seen before. So the constant c was already known when the equations were written? So this would shed some light onto the reason why the equations lead to the conclusion why the velocity of light is equal to c, since c is already in the equations. I'll just have to accept that, but then I have to ask, how was c calculated (or rather without the math how was c logically arrived at?)
Thanks

 

[robphy]

In "SI" units, one had $\mu_0=\frac{1}{\epsilon_0 c^2}$ ( https://en.wikipedia.org/wiki/Vacuum_permeability ).
Rewrite everything replacing the $c^2$ with $\frac{1}{\mu_0\epsilon_0}$... and simplifying to get
(in integral form)

So... the intro-textbook story goes that
only after Maxwell's term was introduced did you have the combination $\mu_0\epsilon_0$.

 

[Nugatory]

So the constant c was already known when the equations were written? So this would shed some light onto the reason why the equations lead to the conclusion why the velocity of light is equal to c, since c is already in the equations...how was c calculated (or rather without the math how was c logically arrived at?)

It wasn't calculated, it was measured; you don't need to know what something is to measure its speed. Quite accurate measurements of the speed of light were being done years before Maxwell's equations were discovered.

Googling for "speed of light history" will find a surprising number of surprisingly early and good results. For example, Newton got within 15% in 1700, more than a century and a half before Maxwell. Considering what was available by way of clocks and time-measuring technology at the end of the 17th century, that's pretty impressive.

 

[Buckethead]

It wasn't calculated, it was measured; you don't need to know what something is to measure its speed. Quite accurate measurements of the speed of light were being done years before Maxwell's equations were discovered.

The speed of light was measured, but c and the speed of light are not the same thing, they just happen to have the same value. How was c determined? Or am I missing something? Do Maxwell's equations simply show that light travels at c or do the equations actually allow for the value of c to be determined. I'm assuming it's the first since c already shows up in the equations. Sorry for the elementary questions, but I'm just a hobbyist with a great interest in this.

 

[jtbell]

The speed of light can be measured without any reference to Maxwell's equations, using various optical methods, including Fizeau's toothed-wheel method and Foucault's rotating-mirror method:

https://en.wikipedia.org/wiki/Fizeau–Foucault_apparatus

These were done before Maxwell put forth his famous equations. (I did a version of Foucault's method as an undergraduate more than 40 years ago.)

Using his equations, Maxwell derived an equivalent to the relationship that robphy has already noted: $c = 1 / \sqrt {\mu_0 \varepsilon_0}$. At that time the quantities $\mu_0$ and $\varepsilon_0$ had been measured in magnetic and electrical experiments respectively. He plugged those values into the equation above, and got a speed which agreed pretty well with the measurements of $c$ noted above.

To be pedantically correct, I'd better note that physicists used different units for electric and magnetic quantities than the ones we use today, so the actual equation that Maxwell used was somewhat different. I once saw Maxwell's original description and numbers, but I don't have a link handy. Think of the above description as a "translation" of Maxwell's procedure into today's SI units.

 

[Buckethead]

Ah, wonderful. So c was calculated using this equation and the two noted measurements. So now the question is, was this equation for c a massaging of the two constants so that c could match the measured speed of light or was this equation put in this form for another reason and then the value c assigned to it without regard for the measured speed of light?

 

[jtbell]

The relationship $c = 1 / \sqrt{\mu_0 \varepsilon_0}$ can be derived from the fundamental electromagnetic equations (Maxwell's equations) that robphy showed on that t-shirt. Those equations in turn were based on the collective body of experiments on electricity and magnetism that had been done up to that time.

 

[Buckethead]

Interesting. The fact that they were able to derive c from the equations means that they needed to already know that there was a constant that they had to derive in the first place. I know this is somewhat diving into the mind of the physicists at that time, but why did they need to find a constant that they would call c? Was it because they suspected that the measured speed of light might somehow be related to Maxwell's equations and so they just said "let's call the measured speed of light c and then see if we can use Maxwell's equations to derive this constant"?

 

[jtbell]

I hope you can get the main point of the following even if you don't know any calculus or differential equations.

There is a "differential wave equation" which describes any kind of wave: $$\nabla^2 f = \frac 1 {c^2} \frac {\partial^2 f} {\partial t^2}$$ where $f$ is the quantity that is "waving" and $c$ is the speed of the wave. It can be shown that there is a set of solutions for Maxwell's equations in which the electric field $\vec E$ and the magnetic field $\vec B$ both also satisfy the following equations: $$\nabla^2 \vec E = \mu_0 \varepsilon_0 \frac {\partial^2 \vec E} {\partial t^2} \\ \nabla^2 \vec B = \mu_0 \varepsilon_0 \frac {\partial^2 \vec B} {\partial t^2}$$ which have the same form as the differential wave equation but with $\mu_0 \varepsilon_0$ in place of $\frac 1 {c^2}$. These solutions are therefore waves of $\vec E$ and $\vec B$ in which $c = 1 / \sqrt {\mu_0 \varepsilon_0}$.

I don't know whether Maxwell was looking specifically for this kind of solution to his equations, or whether he merely happened to stumble upon it.

It's worth noting that at that time, it was pretty well established that light is a wave of "something", based on optical experiments involving interference and diffraction (by Young, Fresnel, et al.)

 

[PeterDonis]

why did they need to find a constant that they would call c?

They didn't. Maxwell added the displacement current term to his equations because he thought it made them look more symmetrical. Then he discovered that with that extra term, the source-free equations had wave solutions. Then he used the known values of $\mu_0$ and $\epsilon_0$, derived from other experiments, to calculate the speed of the waves, and found out that it was the same as the speed of light, which had been measured in various ways by that time. He wasn't looking for any of this in advance; it just came out.

I don't know whether Maxwell was looking specifically for this kind of solution to his equations, or whether he merely happened to stumble upon it.

AFAIK he just stumbled upon it, as above.

 

[Buckethead]

Wow! This is amazing! First, thank you all very much for talking the time to answer my questions. I understand more now than months of doing google searches. You folks are all just fantastic!

So c was just the variable used back then to hold the speed of any wave, not just light and then when applied to electromagnetism, gives the modern constant c

I clearly see how you can make the ε and μ substitutions in Maxwell's equations with c from the wave equation and I can also imagine that once his two solutions were written out and being armed with the wave equation in his head, the substitution was an easy step to take. One thing I'm still unclear on is we have 2 variables in the wave equation when using this for light: c and the solution we are looking for (the ƒ part) (yes I know the speed of light was already measured, but we are trying to calculate it here so we are calling it an unknown). If we go to Maxwell's equations however we know μ and ε so we can solve the two equations.
Now that we are armed with the solutions for those two Maxwell equations we can then substitute c for ε and μ and then just solve for c. Is this right?

I'm right at the very edge of my understanding of this, but I'm getting there.

 

[phyzguy]

I think it was the other way around. Nobody knew that light was electromagnetic waves before Maxwell. Maxwell inserted the displacement current, and then found that it led to differential equations for waves that propagated at the speed $$\sqrt{\mu_0 \epsilon_0}$$. He knew the value of these two constants, so he calculated the speed of these waves and found it to be equal to the speed of light. One can imagine the stroke of insight when he realized that light is these electromagnetic waves. Nobody knew that before then. Quoting from his paper, he said,

"We can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena. "

To me this is one of the great unifications in the history of physics.

 

[Nugatory]

So c was just the variable used back then to hold the speed of any wave, not just light and then when applied to electromagnetism, gives the modern constant c...

The history is more haphazard than that.
- $c$ was used for the speed of light, a measured quantity.
- $\mu_0$ and $\epsilon_0$ were also measured, by studying the behavior of electrical and magnetic fields.
- Maxwell discovered his equations of electrodynamics, which naturally incorporated the empirically determined constants $\mu_0$ and $\epsilon_0$ because they're part of the behavior or electrical and magnetic fields.
- Maxwell found that his equations predicted waves traveling with speed $1/\sqrt{\mu_0\epsilon_0}$.
- Remarkably, when he cqlculated that value, it turned out to be equal to $c$, the measured speed of light, so he could assert that the radiation his equations predicted was light and he could write $c=1/\sqrt{\mu_0\epsilon_0}$.

 

[jtbell]

"We can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena."

I was thinking of that quote, although I didn't remember the exact words. With those words I was able to find the original source:

https://en.wikisource.org/wiki/Page:Philosophical_magazine_23_series_4.djvu/38

 

[Mister T]

I wonder if anyone could (without using math) explain to me how a velocity can be calculated from magnetic and electric field relationships.

As others have told you, Maxwell determined that electric fields and magnetic fields can each generate the other in a way that allows them to move from one place to another. Based on the known strengths of these fields, and the way they interact, he was able to calculate the speed at which they should move. The speed he calculated matched the speed that others had measured for light. He then concluded that light must be this propagating combination of electric and magnetic fields.

 

[vanhees71]

Well, this debate shows why I abhorr the SI units in theoretical electrodynamics. It confuses things in introducing unnecessary asymmetries into the equations of the field components, $F_{\mu \nu}$ (or equivalently $(\vec{E},\vec{B})$ in the (1+3) notation). The most natural units are rationalized Gaussian units (also known as Heaviside-Lorentz units). In these units $\vec{E}$ and $\vec{B}$ have the same dimension, and the fundamental Maxwell equations read
$$\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,$$
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
These are the microscopic Maxwell equations with $j^{\mu}=(c \rho,\vec{j})$ the complete charge-current density.

To derive electromagnetic waves for the free fields, i.e., $\rho=0$ and $\vec{j}=0$, take the curl of the first equation
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\frac{1}{c} \partial_t \vec{\nabla} \times \vec{B}=0.$$
Now use the 3rd equation with $\vec{j}=0$, and you get
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\frac{1}{c^2} \partial_t^2 \vec{E}=0.$$
Further you can work in Cartesian coordinates and then simplify the first term to
$$\vec{\nabla} (\vec{\nabla} \cdot \vec{E})-\Delta \vec{E} + \frac{1}{c^2} \partial_t^2 \vec{E}=0.$$
Using the fourth Maxwell equation with $\rho=0$ finally gives the wave equation
$$\left (\frac{1}{c^2} \partial_t^2-\Delta \right) \vec{E}=0.$$
This shows that there are solutions for $\vec{E}$ describing plane waves,
$$\vec{E}(t,\vec{x})=\vec{E}_0 \cos(\omega t-\vec{k} \cdot \vec{x}),$$
where
$$\omega=c|\vec{k}|,$$
i.e., a wave with phase velocity $c$, i.e., the parameter $c$ in the Maxwell equation is the phase velocity of electromagnetic waves. The value of $c$ was known at Maxwell's times from electrostatic and magnetostatic (static!) measurements, and thus the Maxwell equations lead to a prediction (a) of the existence of electromagnetic waves and (b) a strong hint that light is an electromagnetic wave, because $c$ was pretty much the value measured by optical means.

As I said, the introduction of a fourth unit for charge (or as done officially in the SI electrich current) obscures this very clear picture a bit. In fact in the modern way to build the SI units it becomes very clear that $\epsilon_0$ and $\mu_0$ are just conversion factors from the arbitrary SI units to the more natural units (in the extreme you work in Planck units, where everything is expressed in dimensionless numbers ;-)): For the mechanical+electromagnetic part of physics the SI units are based on relativity, and there is one fundamental parameter, which is the speed of light, $c$. Consequently $c$ is no longer measured but fixed exactly to define the unit of length (metre) in terms of the unit of time (second). Indeed, given relativity it's unnatural to measure space and time with different units, and that's why $c$ is just a conversion factor between metres and light seconds. This fixes the kinematical units of space and time. As a third unit for mechanics you need also a unit for mass, which is the kg in the SI.

In principle there's no need for a third unit of electric charge or current, and that's how the Gaussian or Heaviside-Lorentz units are built. Nevertheless, it leads to a bit of strange units for charges and currents with non-integeger powers of the base units for the electric charge and current. That's why in the SI they introduce the Ampere as a fourth unit. This requires a conversion factor, the "permeability of the vacuum", $\mu_0=4 \pi 10^{-7} \text{N}/\text{A}^2$, where
$1 \text{N}=1 \text{kg} \text{m}/\text{s}^2$ is the unit of force (Newton), and $1 \text{A}$ is the unit of the electromagnetic current, formally defined through the force per unit length of two infinitely long very thin wires in parallel at a distance of $1 \text{m}$. Consequently (and since idiosyncrazily the units of the electric and the magnetic field components are also measured in different units!) you also need another conversion factor, the permittivity of the vacuum, defined by the Maxwell equation
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
In SI units the Maxwell equations lead to
$$c^2=\frac{1}{\mu_0 \epsilon_0},$$
which together with the definition of the speed of light through fixing the metre by the definition of the second and $\mu_0$ through the definition of the Ampere leads to the definition of $\epsilon_0$. So indeed $\mu_0$ and $\epsilon_0$ are merely conversion factors from the artificial units of the SI to the more natural Gaussian or (even better) rationalized Gaussian units used earlier.

 

[Battlemage!]

Well, this debate shows why I abhorr the SI units in theoretical electrodynamics. It confuses things in introducing unnecessary asymmetries into the equations of the field components, $F_{\mu \nu}$ (or equivalently $(\vec{E},\vec{B})$ in the (1+3) notation). The most natural units are rationalized Gaussian units (also known as Heaviside-Lorentz units). In these units $\vec{E}$ and $\vec{B}$ have the same dimension, and the fundamental Maxwell equations read
$$\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,$$
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
These are the microscopic Maxwell equations with $j^{\mu}=(c \rho,\vec{j})$ the complete charge-current density.

To derive electromagnetic waves for the free fields, i.e., $\rho=0$ and $\vec{j}=0$, take the curl of the first equation
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\frac{1}{c} \partial_t \vec{\nabla} \times \vec{B}=0.$$
Now use the 3rd equation with $\vec{j}=0$, and you get
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\frac{1}{c^2} \partial_t^2 \vec{E}=0.$$
Further you can work in Cartesian coordinates and then simplify the first term to
$$\vec{\nabla} (\vec{\nabla} \cdot \vec{E})-\Delta \vec{E} + \frac{1}{c^2} \partial_t^2 \vec{E}=0.$$
Using the fourth Maxwell equation with $\rho=0$ finally gives the wave equation
$$\left (\frac{1}{c^2} \partial_t^2-\Delta \right) \vec{E}=0.$$
This shows that there are solutions for $\vec{E}$ describing plane waves,
$$\vec{E}(t,\vec{x})=\vec{E}_0 \cos(\omega t-\vec{k} \cdot \vec{x}),$$
where
$$\omega=c|\vec{k}|,$$
i.e., a wave with phase velocity $c$, i.e., the parameter $c$ in the Maxwell equation is the phase velocity of electromagnetic waves. The value of $c$ was known at Maxwell's times from electrostatic and magnetostatic (static!) measurements, and thus the Maxwell equations lead to a prediction (a) of the existence of electromagnetic waves and (b) a strong hint that light is an electromagnetic wave, because $c$ was pretty much the value measured by optical means.

As I said, the introduction of a fourth unit for charge (or as done officially in the SI electrich current) obscures this very clear picture a bit. In fact in the modern way to build the SI units it becomes very clear that $\epsilon_0$ and $\mu_0$ are just conversion factors from the arbitrary SI units to the more natural units (in the extreme you work in Planck units, where everything is expressed in dimensionless numbers ;-)): For the mechanical+electromagnetic part of physics the SI units are based on relativity, and there is one fundamental parameter, which is the speed of light, $c$. Consequently $c$ is no longer measured but fixed exactly to define the unit of length (metre) in terms of the unit of time (second). Indeed, given relativity it's unnatural to measure space and time with different units, and that's why $c$ is just a conversion factor between metres and light seconds. This fixes the kinematical units of space and time. As a third unit for mechanics you need also a unit for mass, which is the kg in the SI.

In principle there's no need for a third unit of electric charge or current, and that's how the Gaussian or Heaviside-Lorentz units are built. Nevertheless, it leads to a bit of strange units for charges and currents with non-integeger powers of the base units for the electric charge and current. That's why in the SI they introduce the Ampere as a fourth unit. This requires a conversion factor, the "permeability of the vacuum", $\mu_0=4 \pi 10^{-7} \text{N}/\text{A}^2$, where
$1 \text{N}=1 \text{kg} \text{m}/\text{s}^2$ is the unit of force (Newton), and $1 \text{A}$ is the unit of the electromagnetic current, formally defined through the force per unit length of two infinitely long very thin wires in parallel at a distance of $1 \text{m}$. Consequently (and since idiosyncrazily the units of the electric and the magnetic field components are also measured in different units!) you also need another conversion factor, the permittivity of the vacuum, defined by the Maxwell equation
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
In SI units the Maxwell equations lead to
$$c^2=\frac{1}{\mu_0 \epsilon_0},$$
which together with the definition of the speed of light through fixing the metre by the definition of the second and $\mu_0$ through the definition of the Ampere leads to the definition of $\epsilon_0$. So indeed $\mu_0$ and $\epsilon_0$ are merely conversion factors from the artificial units of the SI to the more natural Gaussian or (even better) rationalized Gaussian units used earlier.

All I read in that was "idiosyncrazily." Hehe just kidding. Brilliant word! But seriously, that was a very informative post. Thank you. Together with Dale and a few others the last few days you guys have given me a lot of insight.

OP thanks for your question. This has been an informative thread so far.

 

[phyzguy]

Two comments to vanhees71's post. First, this whole way of looking at things was only possible after it was understood that light consists of electromagnetic waves. Nobody would have written $$\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0$$ before Maxwell's key insights. Second, while I agree that Gaussian units make it easier to see the relationships, try doing real-world calculations with them. If you are trying to calculate the capacitance of a parallel plate capacitor, or the mutual inductance of a transformer, you will quickly move to using SI units.

 

[jtbell]

Also consider Coulomb's law for the electric field, the Biot-Savart law for the magnetic field, and the Lorentz force law, in Gaussian units: $$d \vec E = \frac {dq} {r^2} \hat r \\
d \vec B = \frac 1 c \frac {I d \vec l \times \hat r} {r^2} \\
\vec F = q \left( \vec E + \frac 1 c \vec v \times \vec B \right)$$ In Gaussian units, Coulomb's law and the electric force law have no extra constants because they effectively define the unit of charge in terms of the force between two charges. But who would have thought (before Maxwell's work) of defining the magnetic field in such a way as to put the constant $\left( \frac 1 c = \right) 3.33 \times 10^{-9}~\rm{s/m}$ in front of both the Biot-Savart law and the magnetic force law?

Hindsight is 20/20; foresight, not so good.

 

[vanhees71]

Well, that's like the endless battle whether vi or emacs is the one and only editor, but why should it make a big difference in the derivation of the capacitance of a plate capacitor whether I use HL or SI units? Neglecting edge effect you use Gauss's law with the ansatz $E=\text{const}$ leading to (in SI units) $$E=Q/A\epsilon_0 \epsilon_r=U/d=Q/Cd,$$
i.e.,
$$C=\epsilon_0 \epsilon_r A/d.$$
In HL units you just need to set $\epsilon_0=1$.

 

[phyzguy]

Right. So in HL units, if I make a capacitor with a silicon dioxide dielectric 10 microns thick, what does the area need to be to give me 1 microfarad of capacitance?

 

[vanhees71]

In HL units the capacitance is measured in cm not in Farad ;-)), but that's really a useless discussion. Whenever you need a result in the SI, it's no problem at all to transform a calculation from any other system of units to the SI units.

What's more interesting is, how Maxwell came to the conclusion that light is an electromagnetic wave. As I said before, he used his newly found equations (in a form almost incomprehensible to us nowadays since we are used to the modern vector calculus rather than writing out everything in not very intuitively named components or even in terms of quaternions) to derive that the speed within the then used system of units. At these times they used a different unit of charge for electrostatics than for magnetostatics, and there was a conversion factor of dimension speed. This factor was determined by Kohlrausch by first charging a capacitor to some voltage (measured in electrostatic units) and then discharging it via a galvanometer to measure the charge (in magnetostatic units). This enabled Kohlrausch to give a value for $c$, which was not too bad, and it was pretty close to the value for the speed of light measured in several ways optics wise.

 

[haushofer]

I hope you can get the main point of the following even if you don't know any calculus or differential equations.

There is a "differential wave equation" which describes any kind of wave: $$\nabla^2 f = \frac 1 {c^2} \frac {\partial^2 f} {\partial t^2}$$ where $f$ is the quantity that is "waving" and $c$ is the speed of the wave. It can be shown that there is a set of solutions for Maxwell's equations in which the electric field $\vec E$ and the magnetic field $\vec B$ both also satisfy the following equations: $$\nabla^2 \vec E = \mu_0 \varepsilon_0 \frac {\partial^2 \vec E} {\partial t^2} \\ \nabla^2 \vec B = \mu_0 \varepsilon_0 \frac {\partial^2 \vec B} {\partial t^2}$$ which have the same form as the differential wave equation but with $\mu_0 \varepsilon_0$ in place of $\frac 1 {c^2}$. These solutions are therefore waves of $\vec E$ and $\vec B$ in which $c = 1 / \sqrt {\mu_0 \varepsilon_0}$.

Not any kind of wave, but relativistic ones :P Non-relativistically you necessarliy have to leave out the time derivative due to Galilean boosts.

 

[bryanso]

Now use the 3rd equation with →j=0j→=0\vec{j}=0, and you get

→∇×(→∇×→E)+1c2∂2t→E=0.​

Hi, I hope someone can explain this mystery to me. When we set j to zero, why doesn't the displacement current term also vanish? There is no charge, no current, hence no E field, right?

 

[pervect]

There are a few good youtube videos that calculate c from Maxwell's Equations, but the jist of it still hasn't gelled with me. In other words, I'm still scratching my head. I wonder if anyone could (without using math) explain to me how a velocity can be calculated from magnetic and electric field relationships. I think it has something to do with the change of the ratio of the field strengths of the two, but can you explain this so it makes sense? I recognize that in an electromagnetic wave when the mag field gets stronger the electric field gets weaker and these two oscillate, but how can a velocity be calculated from this? In other words, what is the logic behind this? Thanks.

Well, the short version is that Maxwell's equations reduce to the wave equation, and when you solve the wave equation the speed of light in SI units is $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

Reducing Maxwell's equations to the wave equation is math, so is solving the wave equation - though I'd say the second step is easier than the first. I'm not sure how to give a more detailed explanation without math. It seems to me that a more detailed explanation will involve some math, it's just a question of what sort of math you're comfortable with.

 

[Ibix]

Hi, I hope someone can explain this mystery to me. When we set j to zero, why doesn't the displacement current term also vanish? There is no charge, no current, hence no E field, right?

Well, $j$ has to be non-zero somewhere - in the emitting antenna. But after that the wave is self-propagating. Normally when deriving the wave equation we don't discuss the source. We just show that a propagating wave is a valid solution to Maxwell's equations.

 

[robphy]

The displacement current is not really a current like the electric current $I=\int\vec j \cdot d\vec A$.
While Electric Fields are typically introduced as being "produced by charges", it turns out that (when we learn about Faraday's Law) you can have electric fields without have any electric charges around.

In fact, as has been suggested, a propagating electromagnetic wave (made by electric and magnetic fields) [in vacuum] has no charge or current sources around... just empty space.

 

[vanhees71]

Hi, I hope someone can explain this mystery to me. When we set j to zero, why doesn't the displacement current term also vanish? There is no charge, no current, hence no E field, right?

The displacement current is no current. Its name is from a highly complicated and utterly wrong early mechanical model invented by Maxwell to "derive" is famous equations. Maxwell himself later abandoned all these mechanical models and came to the conclusion that the electromagnetic field is a fundamental entity with no need for any mechanical models.

 

[bryanso]

vanhees71, Thank you very much. This is the most intuitive explanation. I will reread displacement current with this history perspective in mind.