# 5.1. Maxwell’s Equations¶

## 5.1.1. Electromagnetic Field¶

The electromagnetic field is fully described by a vector field called the 4-potential . It has four components that we can label any way we want, the traditional way is to use:

where is called the electrostatic scalar potential, is called the vector potential and is the speed of light. The Lagrangian density for the free (noninteracting) field is:

The Lagrangian for a (charged) particle is:

it produces the following charge density:

The interaction between the charged particle (or in general any charged body) with some charge density and the electromagnetic field is given by the Lagrangian density (this follows from Local Gauge Invariance):

where:

All together, the Lagrangian of a charged particle and an electromagnetic field is:

(5.1.1.1)¶

Note that:

There are several approaches how to obtain the above Lagrangian from some other assumptions, but ultimately the exact form of the Lagrangians has to be given by experiment. This Lagrangian is our only assumption and we derive everything else from it.

The Euler-Lagrange equations for the electromagnetic field (in terms of and ) are:

(5.1.1.2)¶

Equations for the particle (in terms of and ) are:

(5.1.1.3)¶

Where is called the electromagnetic field strength tensor:

The only way to measure the electric field is through its interaction with the charge particle. As such, the actual physical field (that can be measured) is , which is invariant under any gauge transformation:

where is a gauge function:

In other words, two different related by the gauge transformation represent the exact same physical electromagnetic field (as given by the field tensor). As such, we can modify the Lagrangian by applying the gauge transformation to the field : this changes the equations of motion for the field (thus the numerical values for will be different), but doesn’t change the equation of motion for the particle, so the change will not have any physical effect (cannot be measured).

By choosing as a solution to the equation , we get:

So for any 4-potential we can find such that the transformed 4-potential obeys the Lorenz gauge condition .

In order to obtain a gauge invariant Lagrangian, we need to express it using using the following identity:

The 4-divergence doesn’t change Euler-Lagrange equations, so we can ignore it. We can see, that in the Lorenz gauge the term (which is gauge invariant) simplifies to the term in the Lagrangian (5.1.1.1). The gauge invariant Lagrangian is:

(5.1.1.4)¶

The E.-L. equation for the particle doesn’t change, the equation for the field becomes:

(5.1.1.5)¶

Which in Lorenz gauge simplifies to equation (5.1.1.2). In order to write equations of motion in terms of only, we need another equation for it:

(5.1.1.6)¶

We used the fact, that the partial derivatives are symmetric in the indices and while is antisymmetric.

## 5.1.2. Maxwell’s Equations¶

Maxwell’s equations are the equations for the electromagnetic field in terms of the physical field strengh tensor, equations (5.1.1.5) and (5.1.1.6):

The field strength tensor is antisymmetric, so it has 6 independent components (we use metric tensor with signature -2):

There is freedom in how we label the components. The standard way is to express them using physical fields and that are introduced by:

or in components:

Comparing to the above, we get:

In particular:

so we get:

In terms of and fields, the Maxwell’s equations become:

In Lorenz gauge, the equation for the 4-potential is (5.1.1.2):

The solution to this equation is:

For scalar potential () we get:

(5.1.2.1)¶

And for vector potential () we get:

(5.1.2.2)¶

## 5.1.3. Lorentz Force¶

The equation for the charge particle (5.1.1.3) is:

In components:

Using coordinate time and coordinates instead of the proper time and 4-vector , we need to rewrite the action:

where is the Lagrangian expressed in coordinates and (and thus is not Lorentz invariant):

the particle’s canonical momentum is:

where is the kinetic momentum. Euler-Lagrange equations are:

For continuous case (current), the force due to the magnetic field is:

## 5.1.4. Hamiltonian¶

Expressing in terms of we get:

The system of equations was solved for using the code (in there , and ):

```
>>> from sympy import var, solve
>>> var("P1 P2 P3 m c v1s v2s v3s")
(P1, P2, P3, m, c, v1s, v2s, v3s)
>>> vs = v1s+v2s+v3s
>>> solve([P1**2*(1-vs/c**2) -v1s*m**2,
... P2**2*(1-vs/c**2) -v2s*m**2,
... P3**2*(1-vs/c**2) -v3s*m**2], [v1s, v2s, v3s])
{v1s: P1**2*c**2/(P1**2 + P2**2 + P3**2 + c**2*m**2),
v2s: P2**2*c**2/(P1**2 + P2**2 + P3**2 + c**2*m**2),
v3s: P3**2*c**2/(P1**2 + P2**2 + P3**2 + c**2*m**2)}
```

And the absolute value was removed by using the fact, that has the same sign as which follows from the second equation.

The Hamiltonian is:

## 5.1.5. Electromagnetic Stress Tensor¶

The stress tensor is calculated from the Lagrangian:

using the Noether formula:

We raise the index:

This tensor is not symmetric under the exchange of the indices. To make it symmetric, we add a total derivative term , where is antisymmetric in its first two indices. This guarantees that so that the new stress energy tensor is still conserved. We choose and get:

where we used .

Another way to derive the stress energy tensor is from general relativity using the formula:

So we write the action:

And vary with respect to :

And we get:

## 5.1.6. Examples¶

### Coulomb Law¶

Maxwell’s equations in Lorenz gauge (5.1.1.2):

have the solution for the scalar potential (5.1.2.1):

Assuming :

Assuming the vector potential is time independent, we get for the electric field:

If the charge distribution can be approximated by an infinitely-narrow wire with linear charge density , we get:

and:

#### Example: Straight Wire¶

Let’s assume infinite straight wire with constant linear charge density :

For :

We can also calculate the scalar potential as follows:

Note that in the radial direction (let’s set for example ) the result is scale (translation) invariant, i.e. .

In order to calculate with , we need to regularize it first. Cutoff regularization is:

where is the regulator and also an auxiliary scale. In this regularization, we lost the translational symmetry. The physical quantities don’t depend on in the limit :

and

Dimensional regularization expresses the integral in the dimension as follows:

Here is the regulator and is the auxiliary scale. This regularization preserves the translational symmetry. Now we can renormalize the integral. The minimal subtraction (MS) renormalization is:

Another option is the modified minimal subtraction () renormalization is:

Once we choose a renormalization scheme, we can calculate the electric field as follows:

and the potential difference as:

In agreement with the previous result. The final results don’t depend on the auxiliary scale and we are not doing any limits.

### Biot-Savart Law¶

Maxwell’s equations in Lorenz gauge (5.1.1.2):

have the solution for the vector potential (5.1.2.2):

Assuming :

The magnetic field is then:

If the current can be approximated by an infinitely-narrow wire, we get:

and:

#### Example: Straight Wire¶

Let’s assume infinite straight wire carrying constant current :

Where we used the value of the folowing integral:

For :

#### Example: Circular Loop¶

Let’s assume a circular loop:

Due to the symmetry of the problem, we can set :

In the last equation we used the fact, that is odd and is even on the interval . For we get:

#### Helmholtz Coil¶

Helmholtz coil is a set of two circular loops of radius , that are apart, where . Let’s calculate the magnetic field on the axis. Magnetic field of the first coil is (see the previous example):

Second coil is positioned above the first one:

The total magnetic field is:

The field in the middle:

For we get:

where the magnitude of is:

For and turns we get the magnitude of the field as (we use SI units, so is in and in tesla):

Code:

```
>>> from math import pi, sqrt
>>> "%e" % (8*4*pi*1e-7*130 / (5*sqrt(5)*0.15))
'7.792861e-04'
```

Equation of motion for an electron in this field is:

The general solution is:

So the electron is moving in a circle with a center , depends on the initial direction of the velocity and is the magnitude of the initial velocity. There can also be a possible movement in the direction, but for the following initial conditions there is none:

Then we get:

So the radius of the circle is . Let the electrons by accelerated by the electric potential :

So the initial velocity is:

and we get for the radius:

from which the electron charge versus mass ratio is:

For , , , , we get:

Code:

```
>>> from math import pi
>>> r = 0.15
>>> N = 130
>>> V = 300
>>> R = 0.05
>>> I = 1.48
>>> mu0 = 4*pi*1e-7
>>> "%e" % (125 * V * r**2 / (32 * mu0**2 * R**2 * N**2 * I**2))
'1.804238e+11'
```

Reference value is:

Code:

```
>>> e = 1.6021766e-19
>>> c = 299792458
>>> eV = e
>>> KeV = 1e3 * eV
>>> m = 510.998910 * KeV / c**2
>>> m
9.109382795192204e-31
>>> "%e" % (e / m)
'1.758820e+11'
```

or even simpler (we do not actually need the value of the electron charge ):

```
>>> c = 299792458
>>> KeV = 1e3
>>> m = 510.998910 * KeV / c**2
>>> "%e" % (1/m)
'1.758820e+11'
```

We can use the experimental value to calculate the electron rest mass energy:

### Ampère’s Force Law¶

The force on a wire 1 due to a magnetic field of a wire 2 is:

Where is the magnetic field produced by the wire 2. Combining these two equations we get:

#### Parallel Straight Wires¶

We calculate the force between two parallel straight infinite wires:

Where we used the value of the folowing integral:

As such, the direction of the force on the first wire (at coordinates going in the direction) will be to the left and the force per unit length is:

Because the second wire is at the coordinates and the force on the first wire is in the direction , the force between the wires is attractive, as long as and have the same sign (either both currents go up, or both down) and repulsive if and have opposite signs.

Let , , then the force is attractive and (we also use ):

#### Perpendicular Straight Wires¶

We calculate the force between two perpendicular straight infinite wires:

The integral is an odd functin of , so it is zero. We used the value of the folowing integral (but in fact it is already seen before this integral is needed that the double integral must be zero):

As such, there will be no net force.

#### Infinitely Long Wire and a Square Loop¶

We calculate the net force on a square loop with current of side , whose center is far from an infinitely long wire with current :

The wire has coordinates and the magnetic field from it is (see the example above):

The four sides of the loop are ():

and the differentials are:

The net force on the loop is:

### Magnetic Dipole¶

### Bar Magnet¶

A good model of a bar magnet of the length and width is a combination of two magnetic monopoles (that sit inside the magnet, so one cannot actually see them, just their behavior outside the magnet):

where:

The magnetic moment vector is:

and its magnitude then is:

The permeability is:

For a typical bar magnet, we have for example:

The unit of is Tesla: .

### Bar Magnet in a Coil¶

We throw a magnet through a coil and calculate the voltage on the coil. We use two model of the bar magnet: a magnetic dipole and two monopoles apart.

Geometry:

Field of the dipole:

we will need:

and

Field of two monopoles:

we will need:

and

Now we can calculate the voltage:

for the dipole we get

For two monopoles we get

For the dipole, the function

has a maximum and minimum for:

with the max value:

Code:

```
>>> from sympy import var, solve, S, refine, Q
>>> var("a z")
(a, z)
>>> f = z / (a**2+z**2)**(S(5)/2)
>>> solve(f.diff(z), z)
[-a/2, a/2]
>>> f.subs(z, a/2)
16*sqrt(5)*a/(125*(a**2)**(5/2))
>>> refine(f.subs(z, a/2), Q.positive(a))
16*sqrt(5)/(125*a**4)
```

So the maximum voltage is:

If we drop the magnet from height above the coil into it, then its speed will be in the middle of the coil, when . Then:

And we get for the voltage dependence for dipole:

The time difference between the maximum and minimum is the time difference between and , so:

The total flux doesn’t depend on the particular dependence of and :

For the voltage dependence of two monopoles, we get:

The total flux doesn’t depend on the particular dependence of and :

Note that in the limit , we get the magnetic moment and the last formula for two monopoles flux becomes the dipole flux.

As a particular example, consider a coil with loops, , , . Then the total flux from the second peak is:

Code:

```
>>> from math import pi, sqrt
>>> mu0 = 4*pi*1e-7
>>> cm = 0.01
>>> Q_m = 43.
>>> d = 1.8*cm
>>> a = 1.4*cm
>>> N = 500
>>> -N*mu0*Q_m*d/sqrt(a**2+d**2)
-0.02132647889395681
```

For a single loop with we get:

and for a single loop with we get:

Code:

```
>>> a = 1.25*cm
>>> -mu0*Q_m*d/sqrt(a**2+d**2)
-4.438304942066266e-05
>>> a = 1.8*cm
>>> -mu0*Q_m*d/sqrt(a**2+d**2)
-3.820879326816195e-05
```

### RC Circuit¶

Let’s consider resistor (with voltage ) and capacitor (with voltage and current ) in a series. Voltage on the battery is , then the equation for the circuit is:

with initial condition . We differentiate it:

and the initial condition follows from the first equation . The solution is:

Now we calculate the charge (using the initial condition for the charge above for the lower bound of the integral):

The voltage on the resistor is:

The voltage on the capacitor is:

Half life of the capacitor is defined as the time so that the charge is half of the total charge, and we get:

# 5.2. Semiconductor Device Physics¶

In general, the task is to find the five quantities:

where () is the electron (hole) concentration, () is the electron (hole) current density, is the electric field.

And we have five equations that relate them. We start with the continuity equation:

where the current density is composed of electron and hole current densities:

and the charge density is composed of mobile (electrons and holes) and fixed charges (ionized donors and acceptors):

where and is the electron and hole concetration, is the net doping concetration ( where is the concentration of ionized donors, charged positive, and is the concentration of ionized acceptors, charged negative) and is the electron charge (positive). We get:

Assuming the fixed charges are time invariant, we get:

where is the net recombination rate for electrons and holes (a positive value means recombination, a negative value generation of carriers). We get the carrier continuity equations:

(5.2.1)¶

Then we need material relations that express how the current is generated using and and . A drift-diffusion model is to assume a drift current () and a diffusion (), which gives:

(5.2.2)¶

where , , , are the carrier mobilities and diffusivities.

Final equation is the Gauss’s law:

(5.2.3)¶

## 5.2.1. Equations¶

Combining (5.2.2) and (5.2.1) we get the following three equations for three unknowns , and :

And it is usually assumed that the magnetic field is time independent, so and we get:

(5.2.1.1)¶

These are three nonlinear (due to the terms and ) equations for three unknown functions , and .

### Example 1¶

We can substract the first two equations and we get:

and using and , we get:

So far we didn’t make any assumptions. Most of the times the net doping concetration is time independent, which gives:

Assuming further , we just get the equation of continuity and the Gauss law:

Finally, assuming also that that doesn’t depend on time, we get:

## 5.2.2. Example 3¶

Let’s calculate the 1D pn-junction. We take the equations (5.2.1.1) and write them in 1D for the stationary state ():

We expand the derivatives and assume that and is constant:

and we put the second derivatives on the left hand side:

(5.2.2.1)¶

now we introduce the variables :

and rewrite (5.2.2.1):

So we are solving the following six nonlinear first order ODE:

(5.2.2.2)¶