Momentum of Light: UNIZOR.COM - Physics 4 All - Waves - Electromagnetic ...

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

Momentum of Light

In this lecture we will deal with electromagnetic field in vacuum with a flat wave front consisting of sinusoidal monochromatic (same frequency) synchronous (same phase) oscillations, which we will simply refer to as light.

We know that light carries energy.
More precisely, as we have established earlier (lectures Electric Field Energy and Magnetic Field Energy of topic Energy of Waves of the current part Waves of the course), the total electric + magnetic energy density of electromagnetic field (that is, an amount of energy per unit of volume) is
P_E+M = ½·[ε·E²+(1/μ)·B²]
where E(t,x,y,z) is an intensity of electric component and B(t,x,y,z) is a magnetic component's intensity of the field.

In our case of monochromatic sinusoidal oscillations in vacuum this energy density expression is simplified to (see lecture Electric Flux Density, where we used B=E/c relation)
P_E+M = ε₀·E²

When the light hits an absorbing surface of some object, this energy is transferred into this object causing moving its electrons, heating and other manifestations, even mechanical movement of an object.
Let's explain the nature of this transfer of energy.

On the picture above the light propagates along X-axis, electric component E of the electromagnetic field oscillates along Y-axis and magnetic component B oscillates along Z-axis.

As vector E oscillates up and down, negatively charged electrons on an object's surface move down and up.
The Lorentz force on these electrons caused by their movement in the magnetic field B pushes them forward along X-axis.
When the direction of vector E changes to opposite, the direction of electrons' movement and vector B also change to opposite. As a result, the Lorentz force will still push the electrons forward along X-axis in the same direction.

So, electrons inside the surface layer of an object are always pushed in the same direction with pulsating force, exerting a radiation pressure on the entire object.
Let's analyze the quantitative characteristic of this pressure.

Assume the flat surface of some object is perpendicular to the direction of light propagation and its area is A.
Let the density of electric charge on this surface be σ, so the total charge on this surface is q=σ·A.

The sinusoidal electric component E(t) acts on this charge q, causing electrons to move up and down with velocity v(t).
The Lorentz force on this charge consists of electric component moving electrons along Y-axis and magnetic component pressing electrons perpendicularly to their velocity vector along X-axis.
The total force on the electrons in vector form is, therefore,
F = q·E + q·(v⨯B)

Since only B component pushes electrons in the direction of light propagation and, as we mentioned above, B = E/c, the force that pushes object along the X-axis is
F_x(t) = q·v(t)·E(t)/c
Notice that v(t) is the speed of electrons along the Y-axis, the same axis the electric component of the field E(t) acts.

Let's deviate for a moment from the electromagnetic field and consider classical Newtonian Mechanics.
Recall the Newton's Second Law of Mechanics connecting the force F, mass m and acceleration a:
F(t) = m·a(t)

Since acceleration a(t) is a derivative of speed v(t) by time, the above can be transformed into a relation between an impulse of force and an increment of an object's momentum
F(t) = m·[dv(t)/dt] =
= d[m·v(t)]/dt
from which follows
F(t)·dt = d[m·v(t)] = dp(t)
where p(t) is a momentum of an object.

Using the above relation between an increment of impulse F(t)·dt and an increment of momentum dp(t), we can state that during an infinitesimal time from t to t+dt the force F_x(t) gives an object a push forward in the X-direction quantified as impulse F_x(t)·dt which is converted into an increase of momentum of an object p_x(t):
F_x(t)·dt = dp_x(t)

Using the formula for F_x(t) above, the expression for an increment of a momentum of an object is
dp_x(t) = (1/c)·q·v(t)·E(t)·dt

Let's analyze and interpret the right side of the equation above.

First of all, a product of a charge q and the intensity of an electric component E(t) is the electric force the field exerts upon a charge:
F_e(t) = q·E(t)

Secondly, expressing the speed v(t) as a derivative of the distance of electrons' movements along Y-axis, that is v(t)=dy(t)/dt, we can write F_e(t)·v(t) = F_e(t)·dy(t)/dt =
= dW(t)/dt
where W(t) is work performed by the field to move electrons along Y-axis.

Substituting all the obtained equalities into the formula for an increment in momentum of an object above, we obtain
dp_x(t) = (1/c)·[dW(t)/dt]·dt =
= (1/c)·dW(t)

The equation above relates increment of momentum of an object in the direction of the light propagation (X-axis) and amount of work done by this light.

The Law of Conservation of Energy then dictates that the light by its electric intensity component E(t) has performed this work by transferring its own energy to move the electrons on an object's surface along Y-axis.
That, in turn, caused these electrons to push along X-axis because of action of light's magnetic intensity component B(t).
Finally, this caused a movement of an object along X-axis and increased its momentum in the direction of propagation of light.
The amount of this lost energy of light, divided by the speed of light, equals to an increment of a momentum of a movement of an object along X-axis.

But there is also the Law of Conservation of a Momentum. Therefore, that increment of a momentum of an object must be equal to a decrement in light's momentum.
That means:
(a) light carries an energy and a momentum
(b) when light is completely absorbed by a surface of some object, its energy and momentum are transferred to this object and the decrement of light's momentum equals to the decrement of light's energy divided by the speed of light.
dp_x(t) = (1/c)·dW(t)
(c) an object absorbing light absorbs its energy and momentum in the same proportion

Incidentally, since
F_x(t)·dt = dp_x(t)
we can express the force exerted by light on an object fully absorbing this light as
F_x(t) = dp_x(t)/dt =
= (1/c)·dW(t)/dt

The expression dW(t)/dt represents amount of energy carried by light in a unit of time.

Dividing both sides by the area A on which light falls, we will get a radiation pressure P(t)=F_x(t)/A on the left side and energy flux density divided by the speed of light on the right side:
F_x(t)/A = (1/c)·[dW(t)/dt]/A

As presented in the previous lecture Electromagnetic Energy Flux Density, the energy flux density can be expressed as Poynting vector S(t)=(1/μ)·E(t)⨯B(t).
Therefore, in terms of Poynting vector, a radiation pressure on an object fully absorbing the light is
P_abs(t) = S(t)/c

Everything above was about the case of full absorption of the light by an object (the object fully absorbing the light is called blackbody).

Let's assume now, we have a fully reflective object.
That means that, if an incident light had momentum p, the reflected light will have momentum −p.

From the Law of Conservation of Momentum follows that, if the momentum of the light was p and became −p, the increment of the momentum of an object should be 2p.

That means, the force of radiation pressure in case of fully reflecting object should be twice as strong as in case of fully absorbing object, that is
P_ref(t) = 2·S(t)/c
So, a light reflective object feels twice as strong radiation pressure than a light absorbing object.

The following experiment confirms this.

Four small squares with one side black and another being a reflective mirror are arranged like a propeller that can freely rotate on a needle inside a sphere with vacuum inside.

As soon as light falls on this propeller, it starts rotating because the mirror side has more radiation pressure than the black one.

E-M Energy Flux Density: UNIZOR.COM - Physics 4 Teens - Waves - Electrom...

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

Electromagnetic
Energy Flux Density

As we have established earlier (lectures Electric Field Energy and Magnetic Field Energy of topic Energy of Waves of the current part Waves of the course), the total electric + magnetic energy density of electromagnetic energy (that is, an amount of energy per unit of volume) is
P_E+M = ½·[ε·E²+(1/μ)·B²]
where E(t,x,y,z) is an intensity of electric component and B(t,x,y,z) is a magnetic component's intensity of the field.

In the previous lecture Energy Continuity of the current topic Electromagnetic Field Waves we discussed a concept of electromagnetic field energy flux density vector, which, based on a concept of energy continuity, we have identified as Poynting vector
S = (1/μ)·(E⨯B)

Using the Poynting vector, we have represented the rate of change of electromagnetic energy flux as
−∂P_E+M /∂t = ∇·S
In this form this expression consttutes the continuity equation of the flux density rate of change of electro-magnetic energy.

Let's examine the correspondents of both expressions, P_E+M and S, in a simple case of flat sinusoidal monochromatic electromagnetic waves.

In the lecture E-M Waves Amplitude of the current topic we came up with a simple relationship between electric and magnetic components of the plane electromagnetic waves in vacuum.

If a sinusoidal in time (t) electromagnetic field propagates along Z-axis with speed c according to these equations
E(t,z) = E₀·sin(ω·(t−z/c))
B(t,z) = B₀·sin(ω·(t−z/c))
where electrical component E oscillates along X-axis and magnetic component B oscillates along Y-axis, then
E₀ /c = B₀
and, therefore,
E(t,z)/c = B(t,z)

It's quite appropriate now to check if our formula for Poynting vector as an energy flux density checks in this simple case.

Replacing B² with E²/c² in the expression for P_E+M and taking into consideration that the speed of light c in terms of electrical permittivity ε and magnetic permeability μ is expressed as c=√1/(ε·μ) (see lecture Speed of Light of the current topic), the equation for total energy density can be simplified as
P_E+M = ½·[ε·E²+(1/μ)·E²/c²] =
= ½·[ε·E²+(1/μ)·E²·ε·μ] =
= ε·E²

Consider a unit area of 1m² and light flowing perpendicularly through it with speed c.
During the unit time interval of 1s the light going through this area will fill the volume
1(m²)·c(m/s)·1(s) = c(m³)

So, the amount of electromagnetic energy flowing through a unit area during a unit of time (that is, energy flux density j) is this volume times the calculated above density of the energy ε·E²
j = c·ε·E²
The above is the magnitude of an electromagnetic energy flux density vector directed along Z-axis.

In our special case vectors E (oscillating along X-axis) and B (oscillating along Y-axis) are perpendicular to each other.
Therefore, the magnitude of their vector product equals to a product of their magnitudes and the magnitude of Poynting vector is
|S| = (1/μ)·|E|·|B| =
= (1/μ)·E·B = (1/μ)·E·E/c =
= (1/μ)·E²/c
Since c²=1/(ε·μ) and 1/μ=c²·ε,
|S| = c·ε·E²
which is exactly the same as j above calculated based on energy density.

The direction of Poynting vector is perpendicular to both electrical and magnetic components and, therefore, is along Z-axis.

As we see, Poynting vector in this special case fully corresponds in magnitude and direction to the energy flux density obtained by direct calculation based on energy density and speed of light.

This confirms (at least, in this simple case) that Poynting vector correctly represents the electromagnetic energy flux density.

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 V_x(x,y,z,t), V_y(x,y,z,t) and V_z(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) =
= ∂V_x(x,y,z,t)/∂x +
+ ∂V_y(x,y,z,t)/∂y +
+ ∂V_z(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 P_E and magnetic P_B energy densities of electromagnetic field with electric intensity E and magnetic intensity B:
P_E = ½ε·E²
P_M = 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,
P_E+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
P_E+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:
∂P_E+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
∂P_E+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) = ∇·(P⨯Q)
(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 B⨯E:
∂P_E+M /∂t = (1/μ)·∇·(B⨯E)

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

−∂P_E+M /∂t = (1/μ)·∇·(E⨯B)

The equation above implies that vector
S = (1/μ)·(E⨯B)
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.

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) = {V_x,V_y,V_z}
using unit vectors i, j and k along the coordinate axes:
∇⨯V =
= (∂V_z/∂y − ∂V_y/∂z)·i +
+ (∂V_x/∂z − ∂V_z/∂x)·j +
+ (∂V_y/∂x − ∂V_x/∂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 E_x(t,z)≠0,
∇⨯E =
= (∂E_z/∂y − ∂E_y/∂z)·i +
+ (∂E_x/∂z − ∂E_z/∂x)·j +
+ (∂E_y/∂x − ∂E_x/∂y)·k =
= (∂E_x/∂z)·j

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

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

Hence,
∂E_x/∂z = −∂B_y/∂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

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.

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


Answer

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.

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
E_before = 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, E_after = E_before.
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.