## Saturday, October 21, 2023

### Electromagnetic Energy Continuity: UNIZOR.COM - Physics4Teens - Waves - ...

Notes to a video lecture on http://www.unizor.com

Electromagnetic Energy Continuity

Let us recall the definition of the divergence of a time-dependent vector field V(x,y,z,t) in three-dimensional Cartesian space (X,Y,Z), with three components Vx(x,y,z,t), Vy(x,y,z,t) and Vz(x,y,z,t) along each space coordinate presented in the lecture Divergence of the topic Electromagnetic Field Waves of the Waves part of this course:

divV(x,y,z,t) = · V(x,y,z,t) =
=
Vx(x,y,z,t)/x +
+
Vy(x,y,z,t)/y +
+
Vz(x,y,z,t)/z

In particular, in that lecture we used an example of magnitude and direction of air blown by the wind in a unit of time, a time-dependent vector field V(x,y,z,t), and its relationship with air density, a time-dependent scalar field ρ(x,y,z,t).

The final conclusion we came up with was an equation that connects the divergence of the above vector field at any point and time with the rate of change of the air density there as follows
· V(x,y,z,t) = −ρ(x,y,z,t)/t

Constant Flux Density

Consider a trivial example.
If oil of density 800 kg/m³ uniformly moves along a pipeline with speed 6 m/s and the pipe cross section is 0.2 m², the rate of oil flow is
q = 800 kg/m³ · 6 m/s · 0.2 m²

In this case the flux density j, by definition, is the rate of oil flow per unit of time per unit of area of a pipe cross section, which equals to
j = 800 kg/m³ · 6 m/s =
= 4,800 (kg/s)/m²

For an area A = 0.2 m² of a pipe cross section the total rate of flow of oil through a pipe in a unit of time is q = j·A:
q = 4,800 (kg/s)/m² · 0.2 m² =
= 960 kg/s

Flux Density

For any such "flowing" substance at any point inside this substance we define flux density vector field (usually denoted j) as an amount of this substance "flowing" through a unit of area perpendicular to its direction during a unit of time or, in other words, the rate of change of the density of this substance at a given point.

Then, given some surface and knowing the value of flux density at each point of this surface, we can evaluate the quantity of a substance "flowing" through this surface per unit of time.

In the oil flow of the example above the flux density vector at any point inside a pipe at any moment in time is a vector directed along a pipe of a magnitude j=4,800(kg/s)/m² independent of space position inside a pipe and time.
Assuming the pipe is stretched along X-axis, this flux density vector is
j(x,y,z,t) = {j,0,0}
and its divergence is
·j(x,y,z,t) = 0
because all components of the flux density vector are constants.
Therefore, according to the formula about relationship between divergence and density
the derivative of oil density by time is zero, which means that the density of oil is constant.

The above examples of an air flow because of a wind or oil flow in a pipe can be generalized to other measurable substances that can "flow" - number of molecules, mass, electric charge, momentum, energy, etc.

Assume, ρ(x,y,z,t) represents a time-dependent density in space of some substance.
Further assume that a time-dependent vector field j(x,y,z,t) represent the flux density of this substance, that is an amount of substance going through a unit of area perpendicular to the flux vector in a unit of time.

Then an equation similar to the above air flow equation
· j(x,y,z,t) = −ρ(x,y,z,t)/t
constitutes the continuity equation for this substance.

Electric Charge Flux Density

Let's examine the flow of electricity in terms of the flux density, which somewhat similar to an air wind.

Assume, we have certain time-dependent continuous distribution of electric charge in space with charge density ρ(x,y,z,t).

The amount of electricity going through some area in a unit of time is an electric current
I = dq/dt
Therefore, the electric current density (electric current per unit of area) is
J = dI/dA
where A is a perpendicular to electric current unit area, through which the current I is flowing.

The electric current density is an example of a flux density applied to electricity. Therefore, according to the same divergence theorem, the following continuity equation holds:
· J(x,y,z,t) = −ρ(x,y,z,t)/t

The equation above represents a local law of conservation of electric charge.
The charge does not appear from nowhere, neither it disappear without a trace, but gradually increases or decreases with electric current density delivering additional charge into an area or taking charge from an area.

This law is stronger than a statement that the total amount of charge in a closed space remains constant because the latter does not exclude instantaneous transfer of charge from one point to another within a closed space without any trace in between.

As such, the equation above represents a continuity of electric charge.

Electromagnetic
Field Flux Density

The next step after analyzing an electric current flux density and its relation to electric charge continuity and local law of conservation is to discuss electromagnetic field energy flux density and corresponding continuity and local conservation law.

The first difficulty is to realize that electromagnetic field is not similar to air or electric charge - those are real material substances, while electromagnetic field is not "material" in the same sense as air molecules or electrons.

Nevertheless, electromagnetic field carries energy (see lectures Electric Field Energy and Magnetic Field Energy in the topic Energy of Waves of the Waves part of this course), electromagnetic field energy is transferred in space with some speed (with speed of light in vacuum), so it must have a flux.

So, our task is to express the electromagnetic field energy flux density vector in terms of characteristics that define this field - time-dependent vector fields E(x,y,z,t) of electric intensity and B(x,y,z,t) of magnetic intensity.

In the above mentioned lectures we have derived two formulas for electric PE and magnetic PB energy densities of electromagnetic field with electric intensity E and magnetic intensity B:
PE = ½ε·E²
PM = ½(1/μ)·B²
where ε is electric permittivity and μ is magnetic permeability of the medium where electromagnetic field exists (correspondingly, ε0 and μ0 for vacuum).

The total energy density of electromagnetic field is, therefore,
PE+M = ½·[ε·E²+(1/μ)·B²]
Strictly speaking, we have to use vectors E and B instead of scalars E and B. Then the energy density would be
PE+M =
= ½·
[ε·(E·E)+(1/μ)·(B·B)]

Here are the steps that we will follow to accomplish our task of expressing the energy flux density vector field S(x,y,z,t) in terms of vector fields E(x,y,z,t) of electric intensity and B(x,y,z,t) of magnetic intensity.
.

The continuity equation relates the flux density with the rate of change of the density of whatever substance we talk about.
In our case we are analyzing the electromagnetic energy as such a substance and know how its density expressed in terms of intensities of electromagnetic field E and B.

Using the Maxwell equations, we can express the rate of E and B change (that is, their time derivative) in terms of divergence of B and E (see the Faraday Law as the equation #3 and Amper-Maxwell Law as the equation #4).

Finally, comparing the expression for the rate of change of energy density in term of divergence of E and B with an expression in terms of divergence of energy flux density vector, we will express the energy flux density vector in terms of vectors E and B.

To determine the rate of electromagnetic field energy density change, we have to differentiate the above expression for the energy density in terms of E and B by time:
PE+M /t =
= ε·(E·
E/t)+(1/μ)·(B·B/t)

Let's use the Faraday's Law (Maxwell equation #3 in this course) to express B/t in terms of E
E = −B/t

Let's use the Amper-Maxwell Law (Maxwell equation #4 in this course) to express E/t in terms of B, assuming a simple case of the vacuum (no electric current within a field)
B = μ·ε·E/t

Now the rate of change of the energy density looks like
PE+M /t =
= ε·(E·(
B))/(μ·ε) −
− (1/μ)·(B·(
E)) =
= (1/μ)·
[E·(B)−B·(E)]

At this point we can use a vector identity
Q·(∇⨯P)−P·(∇⨯Q) = ∇·(PQ)
(see the proof of this identity in Problem 3 of Vector Field Identities notes from topic Electromagnetic Field Waves of the Waves part of this course)

Using this identity for
E=Q and
B=P,
we obtain an expression for the rate of change of the energy density in terms of a divergence of vector BE:
PE+M /t = (1/μ)··(BE)

Standard continuity equation has negative rate of change equated to divergence of the flux density.
Therefore, we can rewrite the above equation in a standard form, changing the sign on the left and order of vectors on the right, getting

PE+M /t = (1/μ)··(EB)

The equation above implies that vector
S = (1/μ)·(EB)
represent the electromagnetic field energy density rate of change.
This vector S is called Poynting vector.

Poynting vector represents the direction and magnitude of the electromagnetic field energy flow.
It's perpendicular to both electric and magnetic intensity vectors.

The equation above represents the continuity of electromagnetic field energy and local Energy Conservation Law.
Not only the total amount of energy in a closed system remains constant, but also the energy movement is continuous, it gradually flowing from one location to its immediate neighborhood.

## Friday, October 13, 2023

### Amplitude of Electromagnetic Waves: UNIZOR.COM - Physics 4 Teens - Waves...

Notes to a video lecture on http://www.unizor.com

Electromagnetic Waves Amplitude

In the previous lecture Speed of Light of the current topic Electromagnetic Field Waves we discussed the relationship between electric permittivity of a medium ε (ε0 for vacuum), its magnetic permeability μ (μ0 for vacuum) and the speed of light in this medium (in vacuum speed of light is c and 1/c² equals to ε0·μ0).

In the same lecture we derived an expression for electric component of the simplest electromagnetic field oscillations in vacuum - monochromatic plane waves - as a function of time t and distance z from the source of oscillations along the Z-axis (direction of propagation of waves)
E(t,z) = E0·sin(ω·(t−z/c))
where we assumed that the electric component E of an electromagnetic field regularly oscillates in the direction of X-axis, magnetic component B regularly oscillates in the Y-direction, electromagnetic waves propagate along Z-axis with speed c with angular frequency ω.

The expression for magnetic component B(t,z) of an electromagnetic field oscillations is similar and its derivation follows along exactly the same steps as the one above leading to the same formula:
B(t,z) = B0·sin(ω·(t−z/c))

In this lecture we will derive a simple relationship between amplitudes of electric (E0) and magnetic (B0) amplitudes of the simplest electromagnetic field oscillations - monochromatic plane waves.

The third Maxwell equation that relates induced electric field to a changing magnetic field - the Faraday's Law - is

E = −B/t

Recall the definition of a vector (cross) product of a pseudo-vector
∇={∂/∂x,∂/∂y,∂/∂z}
by any vector
V(x,y,z) = {Vx,Vy,Vz}
using unit vectors i, j and k along the coordinate axes:
V =
= (
Vz/y − Vy/z)·i +
+ (
Vx/z − Vz/x)·j +
+ (
Vy/x − Vx/y)·k

(you can refresh this in the lecture "Curl in 3D" of this chapter of a course on UNIZOR.COM)

Considering a special characteristics of vector E during the simplest oscillations of electromagnetic field under consideration with only Ex(t,z)≠0,
E =
= (
Ez/y − Ey/z)·i +
+ (
Ex/z − Ez/x)·j +
+ (
Ey/x − Ex/y)·k =
= (
Ex/z)·j

Similarly, considering a special characteristics of vector B with only By(t,z)≠0,
B/t = −(By/t)·j

Therefore, the third Maxwell equation in this case of a simple electromagnetic field takes the form
(Ex/z)·j = −(By/t)·j

Hence,
Ex/z = −By/t

Let's use the expressions for E(t,z) and B(t,z) as sinusoidal functions above (we dropped subscripts for brevity, as other components of the field electric and magnetic intensities are zero).
E(t,z) = E0·sin(ω·(t−z/c))
B(t,z) = B0·sin(ω·(t−z/c))

These functions should satisfy the above differential equation for the Faraday's Law.
E/z = −B/t

Therefore,
E/z=E0·cos(ω·(t−z/c))·(−ω/c)
B/t = −B0·cos(ω·(t−z/c))·ω
from which follows the relationship between amplitudes of electric and magnetic components of these simple electromagnetic field oscillations

E0 /c = B0

## Wednesday, October 11, 2023

### Doppler Effect for Sound: UNIZOR.COM - Physics 4 Teens - Waves - Waves i...

Notes to a video lecture on http://www.unizor.com

Doppler Effect for Sound Waves

In this lecture we will consider longitudinal sound waves in the air and dependence of their frequency at the source of sound with a perceived frequency by an observer moving relatively to this source with a speed less than the speed of sound waves propagation.

Obviously, the analysis is applicable to other types of waves and media where these waves propagate.

We will consider the situation when the source of sound is at some fixed position and an observer is moving to or from this source with certain speed v.

The case when observer is at rest and the source of sound moves to or from an observer is no different than the above.

The case when both the source of sound and an observer are moving is just a combination of the above, as only the relative speed of an observer relative to a source of sound is really important.

For simplicity, we will consider only a one-dimensional case when both a source of a sound and an observer are always on the X-axis of some Cartesian coordinate system.
More complicated case of three-dimensional movement involves more calculations but is not important from the conceptual viewpoint.

Let's assume that a source of sound is at position x=0 and the speed of sound waves propagation in the air is u.

Let's assume further that the frequency of oscillations produced by a source of sound is f0 and waves propagate along X-axis in both direction with the same speed u mentioned above.

Then the standard relationships among parameters of oscillation are:
period T0 = 1/f0
wavelength λ0 = u·T0 = u/f0
frequency f0 = u/λ0

Case 1 - Observer moves towards a source of sound with speed v.

Let's take some relatively long time interval t substantially greater than a period of oscillations T0.
If an observer stands still (v=0) during this time interval t, considering the period of oscillations is T0, the number of waves passing the observer would be N0=t/T0.

If during time t there are N0 waves passing an observer, the perceived frequency of sound would be
N0 /t = (t/T0)/t = 1/T0 = f0
which is the original frequency of sound produced by a source.

So, a standing still observer perceives the same sound frequency as was emitted by a source of sound.

Assume now that an observer at an initial distance a (greater than zero) is moving with speed v (less than the speed of sound u) towards the source of sound.

Again, let's choose some time interval t, not very small (substantially larger than period of oscillations T0) but not too long, so an observer will not reach the source of sound (that is, t should be less that a/v).

If an observer is moving towards a source of sound with speed v during time t, he will cover the distance l=v·t and will come to a point at distance a−l from the source of sound.

Let's calculate how many waves of sound will pass an observer during this time interval.
In other words, how many waves will fall into a zone of perception of an observer.

The first wave to cross will be where an observer started, that is at distance a from the source.
Obviously, all waves that are already in the interval from distance a−l to distance a will be crossed.

In addition, all those sound waves, moving with speed u and positioned at the beginning of time closer to a source of sound at a distance from a−l−u·t to a−l by the time t will come to a distance of a−l or larger from the origin of sound, thereby crossing an interval where an observer can hear them.

Therefore, all waves that at time t=0 are at a distance from a source of sound from a−l−u·t to a will cross ways with an observer during time t.
The number of these waves is
N = (l+u·t)/λ0

If an observer hears N sound waves during time t, the perceived frequency of sound is
f = N/t = (u·t+l)/(λ0·t) =
= (u+v)/λ0 = (u/λ0)·(1+v/u) =
= f0·(1+v/u)

As we see, the perceived frequency f by an observer moving towards a fixed in the medium source of sound is greater than the frequency emitted by this source f0 by a factor 1+v/u, where u is the speed of sound emitted by a source and v is the speed of an observer.

Case 2 - Observer moves away from a source of sound with speed v.

Now the zone of an observer's perception is from a distance a from a source of sound to distance a+l, where l=v·t.

Initially, in this zone of perception there were l/λ0 waves of sound. But, as the time goes and an observer gets closer to the end of his zone of perception at distance a+l, all these waves will escape this zone because they move quicker than an observer.

Instead, some waves that are behind an observer (closer to a source of sound) will be able to be in the zone of perception and be heard by an observer because they will overcome him by moving faster.

The first wave an observer perceives is the one at his initial position at distance a from a source of sound.

The last wave he perceives is the one that by time t is at the end of his perception zone at distance a+l from the source, that is that at time t=0 was at distance a+l−u·t from a source.

Therefore, all waves that at time t=0 are at the distance from a+l−u·t to a will be perceived by an observer moving from distance a to distance a+l from a source.

The interval from a+l−u·t to a is u·t−l long.
It contains all the waves heard by an observer while he moves from distance a to a+l.
The number of these waves is
N = (u·t−l)/λ0
The perceived frequency of sound is
f = N/t = (u·t−l)/(λ0·t) =
= (u−v)/λ0 = (u/λ0)·(1−v/u) =
= f0·(1−v/u)

As we see, the perceived frequency f by an observer moving away from a fixed in the medium source of sound is smaller than the frequency emitted by this source f0 by a factor 1−v/u, where u is the speed of sound emitted by a source and v is the speed of an observer.

Incidentally, this formula shows why we consider only a case of a speed of an observer to be smaller than a speed of sound. If that condition is not met, we would have a negative frequency, so formula would not be correct.

## Monday, October 9, 2023

### Energy-Momentum Ratio: UNIZOR.COM - Relativity 4 All - Conservation - Pr...

Notes to a video lecture on UNIZOR.COM

Problem on Relativistic Momentum and Energy

Recall an effect of photoelectricity discussed in Waves - Photoelectricity chapter of UNIZOR.COM course Physics 4 Teens - a ray of light, directed towards a metal plate, kicks electrons from a metal surface.

Electrons, kicked out of metal by a ray of light, have energy and momentum.
Therefore, according to the laws of conservation of energy and momentum, light must carry energy and momentum.

In the previous two lectures of this chapter of the course we have derived formulas for relativistic momentum p and relativistic kinetic energy K of an object. These formulas depend on the object's rest mass m0 and its speed u:
p =
 m0·u √1−u²/c²
= γ·m0·u
K =
 m0·c² √1−u²/c²
− m0·c² =
= m0·c²(γ−1)
where c is the speed of light in vacuum and γ is Lorentz factor
γ =
 1 √1−u²/c²

The problem to apply these formulas to light is that, on one hand, light has zero rest mass m0=0 and, on the other hand, Lorentz factor for light (u=c) is not defined since it turns into division by zero.

In spite of all these difficulties, here is a problem.

Problem

What is the ratio of relativistic kinetic energy of light to its momentum K/p?

Solution

Obviously, we cannot substitute light characteristics m0=0 and u=c directly into formulas for energy and momentum because of undefined Lorentz factor.
Instead, let's consider an object of some rest mass m0 and very high speed u.
Then for this object
K/p = (c²/u)·
 γ−1 γ
or
K/p = (c²/u)·(1−1/γ)

Now we can substitute parameters for light:
u = c
1/γ = √1−c²/c² = 0

The result of this substitution is
K/p = c

For light in vacuum the ratio of relativistic kinetic energy to its momentum is
K/p = c
Consequently, the momentum of light can be expressed in terms of its kinetic energy
p = K/c

Historical Note

The formula for ratio of energy of a photon to its momentum was derived from Maxwell equations long before the Theory of Relativity was invented by Einstein.
Compton effect and Poynting Theorem were the bases for this derivation.
We have derived this ratio from relativistic standpoint without directly resorting to Maxwell equations and properties of electromagnetic waves.

## Tuesday, October 3, 2023

### E=m·c² -- Total Relativistic Energy: UNIZOR.COM - Relativity 4 All - Con...

Notes to a video lecture on UNIZOR.COM

Relativistic Total Energy

Recall an expression for a kinetic energy K of an object of mass m0 moving with speed u derived in the previous lecture Relativistic Kinetic Energy of this course:
K =
 m0·c² √1−u²/c²
− m0·c²
where c is the speed of light in vacuum.

So, kinetic energy is represented as a difference between two other expressions.
Naturally, it's tempting to consider these other two expressions as representing some type of energy themselves.

Consider an equivalent to above formula
m0·c² + K =
 m0·c² √1−u²/c²
It says that a sum of kinetic energy (K) and something else (m0·c²) equals that same something else increased by a familiar Lorentz γ-factor.
Logically speaking, that something else must be some form of energy because, when we add two physical variables, they must be of the same kind and units.

Moreover, the right side of this equation that expresses something else multiplied by Lorentz γ-factor depends on the speed of object u and becomes equal to something else when speed equals to zero.

Obvious conclusion is that something else (m0·c²) represents some form of energy that is concentrated in an object at rest in a frame of reference associated with it, while the right side of a formula is the result of increasing this rest energy by an amount of kinetic energy (K) developed by this object because of its motion.

In other words, an expression
E =
 m0·c² √1−u²/c²
represents the total energy of a moving object that is equal to a sum of its energy at rest and kinetic energy of its motion.

Let's analyze this from the viewpoint of the Law of Conservation of Energy.

Assume an object is at rest in an inertial frame of reference associated with it, and its mass is M.
Assume further that at some moment of time this object splits in two equal parts flying in the opposite to each other directions with speed u.

The total energy of this system before a split was only the rest energy of our object, that is
Ebefore = M·c²
The total energy after a split is a sum of two equal amounts of energy possessed by two parts of our initial object.
If we assume that the rest mass of each part is M/2, we'll face an increase of the total energy of the system without any action from outside, which contradicts the Law of Conservation of Energy.

What follows is a conclusion that the mass of each of two parts must be less than M/2 to compensate an increase in kinetic energy of these parts.

Let's assume that the rest mass of each part of an initial object is m (less than M/2).
Than the rest energy of each part would be m·c² and kinetic energy of each part would be
Kpart =
 m·c² √1−u²/c²
− m·c²
The total energy of each part would be
Epart =
 m·c² √1−u²/c²
And the total energy of a system of two moving parts after a split would be
Eafter =
 2m·c² √1−u²/c²
According to the Law of Energy Conservation, Eafter = Ebefore.
Therefore,
M·c² =
 2m·c² √1−u²/c²
This allows to calculate the ratio of mass decrease 2m/M as a function of speed u:
2m/M = √1−u²/c²

What remains to be proven in this logical train of thoughts is that the total relativistic energy is invariant relatively to Lorentz transformations of coordinates from one inertial reference frame to another.

To accomplish that, we will relate the total relativistic energy to the relativistic momentum and use the corresponding results obtained for the relativistic momentum in UNIZOR.COM - Relativity 4 All - Conservation - Momentum lecture of this part of a course.

Recall the definition of the relativistic momentum
p =
 m0·u √1−u²/c²
where m0 is the rest mass of an object and u is its speed (for simplicity, we assume one-dimensional motion, so speed and momentum are scalars, but in three-dimensional vector form the formula is the same).

The formula for the total relativistic energy discussed above is
E =
 m0·c² √1−u²/c²
Simple algebraic transformations of the above two expressions will lead us to the following equality
E2 − p2·c2 = m02·c4
or
E2 = p2·c2 + m02·c4
or
E2 = (p·c)2 + E02
where E0 is a rest energy the object has just because it has certan mass.
The above equality is called energy - momentum relation in Special Theory of Relativity.

In particular, for objects with no rest mass (like photons of light) with m0=0 the following equation between the total energy and momentum is held
E = p·c

As we see, the total relativistic energy of an object of a rest mass m0 can be expressed in terms of its relativistic momentum, which is invariant relative to Lorentz transformation, and constants that are independent of the frame where an object is observed (its rest mass and the speed of light in vacuum).

Therefore, the total relativistic energy is invariant relative to Lorentz transformation, as it should be.