The Unapologetic Mathematician

Mathematics for the interested outsider

Conservation of Electromagnetic Energy

Let’s start with Ampère’s law, including Maxwell’s correction:

\displaystyle\nabla\times B=\mu_0J+\epsilon_0\mu_0\frac{\partial E}{\partial t}

Now let’s take the dot product of this with the electric field:

\displaystyle E\cdot(\nabla\times B)=\mu_0E\cdot J+\epsilon_0\mu_0E\cdot\frac{\partial E}{\partial t}

On the left, we can run a product rule in reverse:

\displaystyle B\cdot(\nabla\times E)-\nabla\cdot(E\times B)=\mu_0E\cdot J+\epsilon_0\mu_0E\cdot\frac{\partial E}{\partial t}

Now, Faraday’s law tells us that

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

so we can write:

\displaystyle-B\cdot\frac{\partial B}{\partial t}-\nabla\cdot(E\times B)=\mu_0E\cdot J+\epsilon_0\mu_0E\cdot\frac{\partial E}{\partial t}

Let’s rearrange this a bit:

\displaystyle-\frac{1}{\mu_0}B\cdot\frac{\partial B}{\partial t}-\epsilon_0E\cdot\frac{\partial E}{\partial t}=\nabla\cdot\left(\frac{E\times B}{\mu_0}\right)+E\cdot J

The dot product of a vector field with its own derivative should look familiar; we can rewrite:

\displaystyle-\frac{\partial}{\partial t}\left(\frac{1}{2\mu_0}B\cdot B-\frac{\epsilon_0}{2}E\cdot E\right)=\nabla\cdot\left(\frac{E\times B}{\mu_0}\right)+E\cdot J

But now we should recognize almost all the terms in sight! On the left, we’re taking the derivative of the combined energy densities of the electric and magnetic fields:

\displaystyle U=\frac{\epsilon_0}{2}\lvert E\rvert^2+\frac{1}{2\mu_0}\lvert B\rvert^2

The second term on the right is the energy density lost to Joule heating per unit time. The only thing left is this vector field:

\displaystyle u=\frac{E\times B}{\mu_0}

which we call the “Poynting vector”. It’s really named after British physicist John Henry Poynting, but generations of students remember it because it “points” in the direction electromagnetic energy flows.

To see this, look at the final form of our equation:

\displaystyle-\frac{\partial U}{\partial t}=\nabla\cdot u+E\cdot J

On the left we have the rate at which the electromagnetic energy is going down at any given point. On the right, we have two terms; the second is the rate electromagnetic energy density is being lost to heat energy at the point, while the first is the rate electromagnetic energy is “flowing away from” the point.

Compare this with the conservation of charge:

\displaystyle-\frac{\partial\rho}{\partial t}=\nabla\cdot J

where the rate at which charge density decreases is equal to the rate that charge is “flowing away” through currents. The only difference is that there is no dissipation term for charge like there is for energy.

One other important thing to notice is what this tells us about our plane wave solutions. If we take such an electromagnetic wave propagating in the direction k and with the electric field polarized in some particular direction, then we can determine that

\displaystyle u=\frac{E\times B}{\mu_0}=\frac{\lvert E\rvert^2}{\mu_0c}k=\epsilon_0c\lvert E\rvert^2k

showing that electromagnetic waves carry electromagnetic energy in the direction that they propagate.

February 17, 2012 Posted by | Electromagnetism, Mathematical Physics | 3 Comments

Ohm’s Law

When calculating the potential energy of the magnetic field, we calculated the power needed to run a certain current around a certain circuit. Let’s look into that a little more deeply.

We start with Ohm’s law, which basically says that — as a first approximation — the electromotive force around a circuit is proportional to the current around it; push harder and you’ll move charge faster. As a formula:

\displaystyle V=IR

The electromotive force — or “voltage” — on the left is equal to the current around the circuit times the “resistance”. What’s the resistance? Well, here it’s basically just a constant of proportionality, which we read as “how hard is it to push charge around this circuit?”

But let’s dig in a bit more. A current doesn’t really flow around an infinitely-thin wire; it flows around a wire with some thickness. The thicker the wire is — the bigger its cross-sectional area — the easier it should be to push charge around, while the longer the circuit is, the harder. We’ll write down our resistance in the form

\displaystyle R=\eta\frac{l}{A}

where l is the length of the wire, A is its cross-sectional area, and \eta is a new proportionality constant we call “resistivity”. Putting this together with the first form of Ohm’s law we find

\displaystyle V=\eta\frac{l}{A}I

But look at this: the current is made up of a current density flowing along the wire, integrated across a cross-section. If the wire is running in the z direction and the current density in that direction is constantly J_z, then I=JA. Further — at least to a first approximation — the electromotive force is the z-component of the electric field E_z times the length l traveled in that direction.

Thus we conclude that E_z=\eta J_z. But since there’s nothing really special about the z direction, we actually find that

\displaystyle E=\eta J

which is Ohm’s law again, but now in terms of fields and current distributions.

But what about the power? We’ve got a battery pushing a current around a circuit and using power to do it; where does the energy go? Well, if we think about pushing little bits of charge around the wire, they’re going to hit parts of the wire and lose some energy in the process. The parts they hit get shaken up, and this appears as heat energy; the process is called “Ohmic” or “Joule” heating, the latter from Joule’s own experiments using a resistive wire to heat up a tub of water.

If we have a current I made up of N bits of charge q per unit time, then each bit takes an energy of qV to go around the circuit once. This happens N times per unit time, so the total power expenditure is

\displaystyle P=NqV=IV

just as we said last time. But now we can do the same trick as above and write

\displaystyle P=IV=(J\cdot E)Al

or

\displaystyle\frac{P}{Al}=E\cdot J

which measures the power per unit volume dissipated through Joule heating in the circuit.

February 16, 2012 Posted by | Electromagnetism, Mathematical Physics | 1 Comment

Energy and the Magnetic Field

Last time we calculated the energy of the electric field. Now let’s repeat with the magnetic field, and let’s try to be a little more careful about it since magnetic fields can be slippery.

Let’s consider a static magnetic field B generated by a collection of circuits C_i, each carrying a current I_i. Recall that Gauss’ law for magnetism tells us that \nabla\cdot B=0; since space is contractible, we know that its homology is trivial, and thus B must be the curl of some other vector field A, which we call the “magnetic potential” or “vector potential”. Now we can write down the flux of the magnetic field through each circuit:

\displaystyle\Phi_i=\int\limits_{S_i}B\cdot dS_i=\int\limits_{C_i}A\cdot dr_i

Now Faraday’s law tells us about the electromotive force induced on the circuit:

\displaystyle V_i=\frac{d\Phi_i}{dt}

This electromotive force must be counterbalanced by a battery maintaining the current or else the magnetic field wouldn’t be static.

We can determine how much power the battery must expend to maintain the current; a charge q moving around the circuit goes down by qV_i in potential energy, which the battery must replace to send it around again. If n such charges pass around in unit time, this is a work of nqV_i per unit time; since nq=I — the current — we find that the power expenditure is P_i=I_iV_i, or.

\displaystyle P_i=I_i\frac{d\Phi_i}{dt}

Thus if we want to ramp the currents — and the field — up from a cold start in a time T it takes a total work of

\displaystyle W=\sum\limits_{i=1}^N\int\limits_0^TI_i\frac{d\Phi_i}{dt}\,dt

which is then the energy stored in the magnetic field.

This expression doesn’t depend on exactly how the field turns on, so let’s say the currents ramp up linearly:

\displaystyle I_i(t)=I_i(T)\frac{t}{T}

and since the fluxes are proportional to the currents they must also ramp up linearly:

\displaystyle\Phi_i(t)=\Phi_i(T)\frac{t}{T}

Plugging these in above, we find:

\displaystyle W=\sum\limits_{i=1}^N\int\limits_0^TI_i(T)\Phi_i(T)\frac{t}{T^2}\,dt=\frac{1}{2}\sum\limits_{i=1}^NI_i(T)\Phi_i(T)

Now we can plug in our original expression for the flux:

\displaystyle W=\frac{1}{2}\sum\limits_{i=1}^NI_i\int\limits_{C_i}A\cdot dr_i

This is great. But to be more general, let’s replace our currents with a current distribution:

\displaystyle W=\frac{1}{2}\int\limits_{\mathbb{R}^3}A\cdot J\,dV

Now we can use Ampère’s law to write

\displaystyle\begin{aligned}W&=\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}A\cdot(\nabla\times B)\,dV\\&=\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}B\cdot(\nabla\times A)-\nabla\cdot(A\times B)\,dV\\&=\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}B\cdot B\,dV-\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}\nabla\cdot(A\times B)\,dV\end{aligned}

We can pull the same sort of trick last time to make the second integral go away; use the divergence theorem to convert to

\displaystyle\frac{1}{2\mu_0}\lim\limits_{R\to\infty}\int\limits_{S_R}(A\times B)\cdot dA

and take the surface far enough away that the integral becomes negligible. We handwave that A\times B falls off roughly as the inverse fifth power of R, while the area of S_R only grows as the second power, and say that the term goes to zero.

So now we have a similar expression as last time for a magnetic energy density:

\displaystyle u_B=\frac{1}{2\mu_0}\lvert B\rvert^2

Again, we can check the units; the magnetic field has units of force per unit charge per unit velocity:

\displaystyle\frac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}

while the magnetic constant has units of henries per meter:

\displaystyle\frac{\mathrm{m}\cdot\mathrm{kg}}{\mathrm{C}^2}

Putting together an inverse factor of the magnetic constant and two of the magnetic field and we get:

\displaystyle\frac{\mathrm{C}^2}{\mathrm{m}\cdot\mathrm{kg}}\frac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}\frac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}=\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{m}^3\cdot\mathrm{s}^2}=\frac{\mathrm{J}}{\mathrm{m}^3}

or, units of energy per unit volume, just like we expect for an energy density.

February 14, 2012 Posted by | Electromagnetism, Mathematical Physics | 4 Comments

Energy and the Electric Field

Okay, now let’s consider the electric field from the perspective of energy. We have an idea that this might be interesting because we know that the field produces a force, and forces and energies interact in interesting ways.

So recall that if we have a “test charge” q at a point p in an electric field E it experiences a force F=qE(p). As we saw when discussing Faraday’s law, for a static electric field we can write E=-\nabla\phi for some “electric potential” function \phi. Thus we can also write F=-\nabla U for the potential energy function U=q\phi.

Now, say the field is generated by a charge distribution \rho; how much potential energy is contained in the force the field exerts on the little bit of charge at p? We count U=\rho(p)\phi(p), but this is too much — half of it is due to the rest of the distribution acting on the bit of charge at r and half of it comes from r acting back. We can thus find the total potential energy by integrating

\displaystyle U=\frac{1}{2}\int\limits_{\mathbb{R}^3}\rho(p)\phi(p)\,d^3p

Now, Gauss’ law tells us that \rho=\epsilon_0\nabla\cdot E, so we substitute:

\displaystyle U=\frac{1}{2}\int\limits_{\mathbb{R}^3}\epsilon_0(\nabla\cdot E)\phi\,dV

Next we use a form of the product rule — \nabla\cdot(fV)=(\nabla f)\cdot V+f(\nabla\cdot V) — and run it backwards to write:

\displaystyle\begin{aligned}U&=\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}\nabla\cdot(\phi E)-(\nabla\phi)\cdot E\,dV\\&=\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}\nabla\cdot(\phi E)\,dV+\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}(-\nabla\phi)\cdot E\,dV\\&=\frac{\epsilon_0}{2}\lim\limits_{R\to\infty}\int\limits_{B_R}\nabla\cdot(\phi E)+\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}E\cdot E\,dV\end{aligned}

where we evaluate the first integral over space by evaluating it over the solid ball of radius R and taking the limit as R goes off to infinity. The divergence theorem says we can write:

\displaystyle\begin{aligned}U&=\frac{\epsilon_0}{2}\lim\limits_{R\to\infty}\int\limits_{S_R}\phi E\cdot dA+\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}E\cdot E\,dV\\&=\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}E\cdot E\,dV\end{aligned}

where, as usual, we have taken the charge distribution to be compactly supported, so as our sphere gets large enough, the potential energy \phi goes to zero. Yes, this is very hand-wavy, but this is how the physicists do it.

Anyway, what does this tell us? It means that a static electric field contains energy with a density

\displaystyle u_E=\frac{1}{2}\epsilon_0\lvert E\rvert^2

which we can integrate over any region of space to find the electrostatic potential energy contained in the field.

We can also check the units here; the electric field has units of force per unit charge:

\displaystyle\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{C}\cdot\mathrm{s}^2}

while the electric constant has units of farads per meter:

\displaystyle\frac{\mathrm{s}^2\cdot\mathrm{C}^2}{\mathrm{m}^3\cdot\mathrm{kg}}

Putting these together — two factors of E and one of \epsilon_0 we find the units:

\displaystyle\frac{\mathrm{s}^2\cdot\mathrm{C}^2}{\mathrm{m}^3\cdot\mathrm{kg}}\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{C}\cdot\mathrm{s}^2}\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{C}\cdot\mathrm{s}^2}=\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{m}^3\cdot\mathrm{s}^2}=\frac{\mathrm{J}}{\mathrm{m}^3}

Joules per cubic meter — energy per unit of volume, just as we’d expect for an energy density.

February 14, 2012 Posted by | Electromagnetism, Mathematical Physics | 5 Comments

Polarization of Electromagnetic Waves

Let’s look at another property of our plane wave solutions of Maxwell’s equations. Specifically, we’ll assume that the electric and magnetic fields are each plane waves in the directions k_E and k_B, repectively:

\displaystyle\begin{aligned}E(r,t)&=\hat{E}(k_E\cdot r-ct)\\B(r,t)&=\hat{B}(k_B\cdot r-ct)\end{aligned}

We can take these and plug them into the vacuum version of Maxwell’s equations, and evaluate them at (r,t)=(0,0):

\displaystyle\begin{aligned}k_E\cdot\hat{E}'(0)&=0\\k_E\times\hat{E}'(0)&=c\hat{B}'(0)\\k_B\cdot\hat{B}'(0)&=0\\k_B\times\hat{B}'(0)&=-\frac{1}{c}\hat{E}'(0)\end{aligned}

The first equation says that \hat{E}'(0) is perpendicular to k_E, but the second equation implies, in part, that \hat{B}'(0) is also perpendicular to k_E. Similarly, the third and fourth equations say that both \hat{E}'(0) and \hat{B}'(0) are perpendicular to k_B, meaning that k_E and k_B either point in the same direction or in opposite directions. We can always pick our coordinates so that k_E points in the direction of the z-axis and \hat{E}'(0) points in the direction of the x-axis; then \hat{B}'(0) points in the direction of the y-axis. It’s then straightforward to check that k_B=k_E rather than k_B=-k_E. Of course, it’s possible that \hat{E}'(0) — and thus \hat{B}'(0) also — is zero; in this case we can just pick some different time at which to evaluate the equations. There must be some time for which these values are nonzero, or else \hat{E} and \hat{B} are simply constants, which is a pretty vacuous solution that we’ll just subtract off and ignore.

The upshot of this is that E and B must be plane waves traveling in the same direction. We put this back into our assumption:

\displaystyle\begin{aligned}E(r,t)&=\hat{E}(k\cdot r-ct)\\B(r,t)&=\hat{B}(k\cdot r-ct)\end{aligned}

and then Maxwell’s equations imply

\displaystyle\begin{aligned}k\cdot\hat{E}'&=0\\k\times\hat{E}'&=c\hat{B}'\\k\cdot\hat{B}'&=0\\k\times\hat{B}'&=-\frac{1}{c}\hat{E}'\end{aligned}

where these are now full functions and not just evaluations at some conveniently-chosen point. And, incidentally, the second and fourth equations are completely equivalent. Now we can see that \hat{E}' and \hat{B}' are perpendicular at every point. Further, whatever component either vector has in the k direction is constant, and again we will just subtract it off and ignore it.

As the wave propagates in the direction of k, the electric and magnetic fields move around in the plane perpendicular to k. If we pick our z-axis in the direction of k, we can write \hat{E}=\hat{E}_x\hat{i}+\hat{E}_y\hat{j} and \hat{B}=\hat{B}_x\hat{i}+\hat{B}_y\hat{j}. Then the second (and fourth) equation tells us

\displaystyle\hat{E}_x'\hat{j}-\hat{E}_y'\hat{i}=c\hat{B}_x'\hat{i}+c\hat{B}_y'\hat{j}

That is, we get two decoupled equations:

\displaystyle\begin{aligned}\hat{E}_x'&=c\hat{B}_y'\\\hat{E}_y'&=-c\hat{B}_x'\end{aligned}

This tells us that we can break up our plane wave solution into two different plane wave solutions. In one, the electric field “waves” in the x direction while the magnetic field waves in the y direction; in the other, the electric field waves in the y direction and the magnetic field waves in the -x direction.

This decomposition is the basis of polarized light. We can create filters that only allow waves with the electric field oriented in one direction to pass; generic waves can be decomposed into a component waving in the chosen direction and a component waving in the perpendicular direction, and the latter component gets destroyed as the wave passes through the Polaroid filter — yes, that’s where the company got its name — leaving only the light oriented in the “right” way.

As a quick, familiar application, we can make glasses with a film over the left eye that polarizes light vertically, and one over the right eye that polarizes light horizontally. Then if we show a quickly-alternating series of images, each polarized with the opposite axis, then they will be presented to each eye separately. This is the basis of the earliest modern stereoscopic — or “3-D” — glasses, which had the problem that if you tilted your head the effect was first lost, and then reversed as your neck’s angle increased. If you’ve been paying attention, you should be able to see why.

February 10, 2012 Posted by | Electromagnetism, Mathematical Physics | 4 Comments

The Propagation Velocity of Electromagnetic Waves

Now we’ve derived the wave equation from Maxwell’s equations, and we have worked out the plane-wave solutions. But there’s more to Maxwell’s equations than just the wave equation. Still, let’s take some plane-waves and see what we get.

First and foremost, what’s the propagation velocity of our plane-wave solutions? Well, it’s c for the generic wave equation

\displaystyle\frac{\partial^2F}{\partial t^2}-c^2\nabla^2F=0

while our electromagnetic wave equation is

\displaystyle\begin{aligned}\frac{\partial^2E}{\partial t^2}-\frac{1}{\epsilon_0\mu_0}\nabla^2E&=0\\\frac{\partial^2B}{\partial t^2}-\frac{1}{\epsilon_0\mu_0}\nabla^2B&=0\end{aligned}

so we find the propagation velocity of waves in both electric and magnetic fields is

\displaystyle c=\frac{1}{\sqrt{\epsilon_0\mu_0}}

Hm.

Conveniently, I already gave values for both \epsilon_0 and \mu_0:

\displaystyle\begin{aligned}\epsilon_0&=8.85418782\times10^{-12}\frac{\mathrm{F}}{\mathrm{m}}&=8.85418782\times10^{-12}\frac{\mathrm{s}^2\cdot\mathrm{C}^2}{\mathrm{m}^3\cdot\mathrm{kg}}\\\mu_0&=1.2566370614\times10^{-6}\frac{\mathrm{H}}{\mathrm{m}}&=1.2566370614\times10^{-6}\frac{\mathrm{m}\cdot\mathrm{kg}}{\mathrm{C}^2}\end{aligned}

Multiplying, we find:

\displaystyle\epsilon_0\mu_0=8.85418782\times1.2566370614\times10^{-18}\frac{\mathrm{s}^2}{\mathrm{m}^2}=11.1265006\times10^{-18}\frac{\mathrm{s}^2}{\mathrm{m}^2}

which means that

\displaystyle c=\frac{1}{\sqrt{\epsilon_0\mu_0}}=0.299792457\times10^9\frac{\mathrm{m}}{\mathrm{s}}=299\,792\,457\frac{\mathrm{m}}{\mathrm{s}}

And this is a number which should look very familiar: it’s the speed of light. In an 1864 paper, Maxwell himself noted:

The agreement of the results seems to show that light and magnetism are affections of the same substance, and that light is an electromagnetic disturbance propagated through the field according to electromagnetic laws.

Indeed, this supposition has been borne out in experiment after experiment over the last century and a half: light is an electromagnetic wave.

February 9, 2012 Posted by | Electromagnetism, Mathematical Physics | 4 Comments

The Electromagnetic Wave Equations

Maxwell’s equations give us a collection of differential equations to describe the behavior of the electric and magnetic fields. Juggling them, we can come up with other differential equations that give us more insight into how these fields interact. And, in particular, we come up with a familiar equation that describes waves.

Specifically, let’s consider Maxwell’s equations in a vacuum, where there are no charges and no currents:

\displaystyle\begin{aligned}\nabla\cdot E&=0\\\nabla\times E&=-\frac{\partial B}{\partial t}\\\nabla\cdot B&=0\\\nabla\times B&=\epsilon_0\mu_0\frac{\partial E}{\partial t}\end{aligned}

Now let’s take the curl of both of the curl equations:

\displaystyle\begin{aligned}\nabla\times(\nabla\times E)&=-\frac{\partial}{\partial t}(\nabla\times B)\\&=-\frac{\partial}{\partial t}\left(\epsilon_0\mu_0\frac{\partial E}{\partial t}\right)\\&=-\epsilon_0\mu_0\frac{\partial^2 E}{\partial t^2}\\\nabla\times(\nabla\times B)&=\epsilon_0\mu_0\frac{\partial}{\partial t}(\nabla\times E)\\&=\epsilon_0\mu_0\frac{\partial}{\partial t}\left(-\frac{\partial B}{\partial t}\right)\\&=-\epsilon_0\mu_0\frac{\partial^2 B}{\partial t^2}\end{aligned}

We also have an identity for the double curl:

\displaystyle\nabla\times(\nabla\times F)=\nabla(\nabla\cdot F)-\nabla^2F

But for both of our fields we have \nabla\cdot F=0, meaning we can rewrite our equations as

\displaystyle\begin{aligned}\frac{\partial^2 E}{\partial t^2}-\frac{1}{\epsilon_0\mu_0}\nabla^2E&=0\\\frac{\partial^2 B}{\partial t^2}-\frac{1}{\epsilon_0\mu_0}\nabla^2B&=0\end{aligned}

which are the wave equations we were looking for.

February 7, 2012 Posted by | Electromagnetism, Mathematical Physics | 2 Comments

Deriving Physics from Maxwell’s Equations

It’s important to note at this point that we didn’t have to start with our experimentally-justified axioms. Maxwell’s equations suffice to derive all the physics we need.

In the case of Faraday’s law, we’re already done, since it’s exactly the third of Maxwell’s equations in integral form. So far, so good.

Coulomb’s law is almost as simple. If we have a point charge q it makes sense that it generate a spherically symmetric, radial electric field. Given this assumption, we just need to calculate its magnitude at the radius r. To do this, set up a sphere of that radius around the point; Gauss’ law in integral form tells us that the flow of E out through this sphere is the total charge q inside. But it’s easy to calculate the integral, getting

\displaystyle4\pi r^2\lvert E(\lvert r\rvert)\rvert=\frac{q}{\epsilon_0}

or

\displaystyle\lvert E(\lvert r\rvert)\rvert=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

which is the magnitude given by Coulomb’s law.

To get the Biot-Savart law, we can use Ampère’s law to calculate the magnetic field around an infinitely long straight current I. We again argue on geometric grounds that the magnitude of the magnetic field should only depend on the distance from the current and should point directly around the current. If we set up a circle of radius r then, the total circulation around the circle is, by Ampère’s law:

\displaystyle2\pi r\lvert B(\lvert r\rvert)\rvert=\mu_0I

or

\displaystyle\lvert B(\lvert r\rvert)\rvert=\frac{\mu_0}{2\pi}\frac{I}{r}

Now, we can compare this to the last time we computed the magnetic field of the straight infinite current by integrating the Biot-Savart law directly and got essentially the same answer.

Finally, we can derive conservation of charge from Ampère’s law, with Maxwell’s correction by taking its divergence:

\displaystyle\nabla\cdot(\nabla\times B)=\mu_0\nabla\cdot J+\epsilon_0\mu_0\frac{\partial}{\partial t}(\nabla\cdot E)

The quantity on the left is the divergence of a curl, so it automatically vanishes. Meanwhile, Gauss’ law tells us that \epsilon\nabla\cdot E=\rho, so we conclude

\displaystyle0=\mu_0\left(\nabla\cdot J+\frac{\partial\rho}{\partial t}\right)

or

\displaystyle\nabla\cdot J+\frac{\partial\rho}{\partial t}=0

which is the “continuity equation” expressing the conservation of charge.

The importance is that while we originally derived Maxwell’s equations from four experimentally-justified laws, those laws are themselves essentially derivable from Maxwell’s equations. Thus any reformulation of Maxwell’s equations is just as sufficient a basis for all of electromagnetism as our original physical axioms.

February 3, 2012 Posted by | Electromagnetism, Mathematical Physics | 3 Comments

Maxwell’s Equations (Integral Form)

It is sometimes easier to understand Maxwell’s equations in their integral form; the version we outlined last time is the differential form.

For Gauss’ law and Gauss’ law for magnetism, we’ve actually already done this. First, we write them in differential form:

\displaystyle\begin{aligned}\nabla\cdot E&=\frac{1}{\epsilon_0}\rho\\\nabla\cdot B&=0\end{aligned}

We pick any region V we want and integrate both sides of each equation over that region:

\displaystyle\begin{aligned}\int\limits_V\nabla\cdot E\,dV&=\int\limits_V\frac{1}{\epsilon_0}\rho\,dV\\\int\limits_V\nabla\cdot B\,dV&=\int\limits_V0\,dV\end{aligned}

On the left-hand sides we can use the divergence theorem, while the right sides can simply be evaluated:

\displaystyle\begin{aligned}\int\limits_{\partial V}E\cdot dS&=\frac{1}{\epsilon_0}Q(V)\\\int\limits_{\partial V}B\cdot dS&=0\end{aligned}

where Q(V) is the total charge contained within the region V. Gauss’ law tells us that the flux of the electric field out through a closed surface is (basically) equal to the charge contained inside the surface, while Gauss’ law for magnetism tells us that there is no such thing as a magnetic charge.

Faraday’s law was basically given to us in integral form, but we can get it back from the differential form:

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

We pick any surface S and integrate the flux of both sides through it:

\displaystyle\int\limits_S\nabla\times E\cdot dS=\int\limits_S-\frac{\partial B}{\partial t}\cdot dS

On the left we can use Stokes’ theorem, while on the right we can pull the derivative outside the integral:

\displaystyle\int\limits_{\partial S}E\cdot dr=-\frac{\partial}{\partial t}\Phi_S(B)

where \Phi_S(B) is the flux of the magnetic field B through the surface S. Faraday’s law tells us that a changing magnetic field induces a current around a circuit.

A similar analysis helps with Ampère’s law:

\displaystyle\nabla\times B=\mu_0J+\epsilon_0\mu_0\frac{\partial E}{\partial t}

We pick a surface and integrate:

\displaystyle\int\limits_S\nabla\times B\cdot dS=\int\limits_S\mu_0J\cdot dS+\int\limits_S\epsilon_0\mu_0\frac{\partial E}{\partial t}\cdot dS

Then we simplify each side.

\displaystyle\int\limits_{\partial S}B\cdot dr=\mu_0I_S+\epsilon_0\mu_0\frac{\partial}{\partial t}\Phi_S(E)

where \Phi_S(E) is the flux of the electric field E through the surface S, and I_S is the total current flowing through the surface S. Ampère’s law tells us that a flowing current induces a magnetic field around the current, and Maxwell’s correction tells us that a changing electric field behaves just like a current made of moving charges.

We collect these together into the integral form of Maxwell’s equations:

\displaystyle\begin{aligned}\int\limits_{\partial V}E\cdot dS&=\frac{1}{\epsilon_0}Q(V)\\\int\limits_{\partial V}B\cdot dS&=0\\\int\limits_{\partial S}E\cdot dr&=-\frac{\partial}{\partial t}\Phi_S(B)\\\int\limits_{\partial S}B\cdot dr&=\mu_0I_S+\epsilon_0\mu_0\frac{\partial}{\partial t}\Phi_S(E)\end{aligned}

February 2, 2012 Posted by | Electromagnetism, Mathematical Physics | 6 Comments

Maxwell’s Equations

Okay, let’s see where we are. There is such a thing as charge, and there is such a thing as current, which often — but not always — arises from charges moving around.

We will write our charge distribution as a function \rho and our current distribution as a vector-valued function J, though these are not always “functions” in the usual sense. Often they will be “distributions” like the Dirac delta; we haven’t really gotten into their formal properties, but this shouldn’t cause us too much trouble since most of the time we’ll use them — like we’ve used the delta — to restrict integrals to smaller spaces.

Anyway, charge and current are “conserved”, in that they obey the conservation law:

\displaystyle\nabla\cdot J=-\frac{\partial\rho}{\partial t}

which states that the mount of current “flowing out of a point” is the rate at which the charge at that point is decreasing. This is justified by experiment.

Coulomb’s law says that electric charges give rise to an electric field. Given the charge distribution \rho we have the differential contribution to the electric field at the point r:

\displaystyle dE(r)=\frac{1}{4\pi\epsilon_0}\rho\frac{r}{\lvert r\rvert^3}dV

and we get the whole electric field by integrating this over the charge distribution. This, again, is justified by experiment.

The Biot-Savart law says that electric currents give rise to a magnetic field. Given the current distribution J we have the differential contribution to the magnetic field at the poinf r:

\displaystyle dB(r)=\frac{\mu_0}{4\pi}J\times\frac{r}{\lvert r\rvert^3}dV

which again we integrate over the current distribution to calculate the full magnetic field at r. This, again, is justified by experiment.

The electric and magnetic fields give rise to a force by the Lorentz force law. If a test particle of charge q is moving at velocity v through electric and magnetic fields E and B, it feels a force of

\displaystyle F=q(E+v\times B)

But we don’t work explicitly with force as much as we do with the fields. We do have an analogue for work, though — electromotive force:

\displaystyle\mathcal{E}=-\int\limits_CE\cdot dr

One unexpected source of electromotive force comes from our fourth and final experimentally-justified axiom: Faraday’s law of induction

\displaystyle\mathcal{E}=\frac{\partial}{\partial t}\int\limits_\Sigma B\cdot dS

This says that the electromotive force around a circuit is equal to the rate of change of magnetic flux through any surface bounded by the circuit.

Using these four experimental results and definitions, we can derive Maxwell’s equations:

\displaystyle\begin{aligned}\nabla\cdot E&=\frac{1}{\epsilon_0}\rho\\\nabla\cdot B&=0\\\nabla\times E&=-\frac{\partial B}{\partial t}\\\nabla\times B&=\mu_0J+\epsilon_0\mu_0\frac{\partial E}{\partial t}\end{aligned}

The first is Gauss’ law and the second is Gauss’ law for magnetism. The third is directly equivalent to Faraday’s law of induction, while the last is Ampère’s law, with Maxwell’s correction.

February 1, 2012 Posted by | Electromagnetism, Mathematical Physics | 10 Comments

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