## Tuesday, November 28, 2023

### Newton+Maxwell=mc2:UNIZOR.COM - Relativity 4 All - Conservation

Notes to a video lecture on UNIZOR.COM

Newton + Maxwell = m·c²

In this lecture we will use the laws of classic Newtonian mechanics and the properties of light we discussed in the chapter Waves - Electromagnetic Field Waves of this course to derive the famous Einstein's equation E=m·c².

In particular, we will use the results presented in the lecture Momentum of Light of the above mentioned chapter of this course, which we recommend to review prior to studying materials presented in this lecture.

In the Momentum of Light lecture we have derived the relationship between the amount of work W performed by light moving an object with absorbing surface and the momentum p lost by this light during this process of light absorption:
p = W/c (c is the speed of light)
which, basically, means that the energy E contained in certain amount of light and this light's momentum are related as p=E/c.

Now consider the following thought experiment.
Initially, two identical objects, α and β, of mass M each are at rest and positioned on a distance 2L from each other along X-axis with coordinates −L for α and L for β.
We assume that no external forces are acting on these objects.

Since no external forces act on these objects and they are at rest, the total momentum of motion is zero and the center of mass is at X-coordinate x=0.

At some moment an object α sends a short light signal towards object β.
As we know, this light carries some energy E and momentum p=E/c.

For this totally closed isolated system the total momentum must be preserved.
In particular, it means that the center of mass should stay at the original location x=0.

It also necessitates that immediately after issuing a light signal object α must acquire a momentum −p to neutralize the momentum p of the light signal, that is, object α recoils after issuing this light signal.

This momentum of object α means that it has acquired some speed −V (the value V is presumed positive) and moves towards negative direction of the X-axis, while a light signal with speed c moves towards object β in the positive direction of the X-axis.

If light, together with its energy, does not carry some portion of object α's mass, the center of mass would shift with object α's movement, which should not be the case.
So, light signal carries not only some energy E and momentum p=E/c, but also some part of object α's mass m, leaving object α with mass M−m.

But, according to classic Newtonian Mechanics, if the light signal carries mass m moving with speed c, its momentum must be p=m·c (mass times speed of light).
Since, as we stated above, p=E/c, we have a relationship between light signal's mass and energy E it carries:
p=m·c = E/c
from which follows
E = m·c²

The last equation means that energy E lost by an object α by issuing a light signal equals to its lost mass m times a square of the speed of light.

We came to the same Einstein's relationship between mass and energy using a combination of the Maxwell theory of electromagnetic field, which gave us a relationship between light energy E and its momentum p as p=E/c, and Newtonian Mechanics with its definition of a momentum p as mass m times speed that in the case of light equals to c.

Let's continue our analysis of this experiment and find the behavior of all components involved in it.

Obviously, the light signal moves with speed c from α towards β.
It's initial coordinate is x=−L, so its movement can be described as
xsig(t) = −L + c·t
for all t from 0 to T=2L/c - the moment light hits object β.
Then it's absorbed by object β and disappears as an independent entity (see below the analysis of movement of β object.

Object α is losing mass m and moves to the negative direction of X-axis with speed Vα.
To conserve the momentum, this speed must satisfy the equation
(M−m)·Vα = m·c
Therefore,
Vα = m·c/(M−m)
and the equation of motion for α object is
xα(t) = −L − m·c·t/(M−m)

Object β stays in place at coordinate x=L up until it gets hit by a light signal at time T=2L/c.
Then β gets hit by a light signal and absorbs it energy, momentum and mass.
From the Law of Conservation of Momentum follows that the momentum of an object that absorbed the light signal, increasing its mass by m, should be equal to a momentum of this signal:
(M+m)·Vβ = m·c
Therefore,
Vβ = m·c/(M+m)
and the equation of motion for β object, considering it starts by time T later than α, is
xβ(t) = L + m·c·(t−T)/(M+m)

Let's check if the center of mass is not changing the position.
There are two stages of this experiment: before the light signal hit β object and after.

1. For time t less than T we have three objects - α, light signal and β.
Their combined "mass times X-coordinate" is
(M−m)·xα(t)+m·xsig(t)+M·L =
= −L·(M−m) − m·c·t −
− L·m + m·c·t + M·L = 0

as it should.

2. For time t equal or greater than T we have only two objects - α and β.
Their combined "mass times X-coordinate" is
(M−m)·xα(t) + (M+m)·xβ(t) =
= −L·(M−m) − m·c·t +
+ (M+m)·L+m·c·t−m·c·2L/c =0

as it should.

As we see, the center of mass is always at x=0, which is expected from a closed system.

## Friday, November 17, 2023

### 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
PE+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)
PE+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·(vB)

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
Fx(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 Fx(t) gives an object a push forward in the X-direction quantified as impulse Fx(t)·dt which is converted into an increase of momentum of an object px(t):
Fx(t)·dt = dpx(t)

Using the formula for Fx(t) above, the expression for an increment of a momentum of an object is
dpx(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:
Fe(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 Fe(t)·v(t) = Fe(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
dpx(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.
dpx(t) = (1/c)·dW(t)
(c) an object absorbing light absorbs its energy and momentum in the same proportion

Incidentally, since
Fx(t)·dt = dpx(t)
we can express the force exerted by light on an object fully absorbing this light as
Fx(t) = dpx(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)=Fx(t)/A on the left side and energy flux density divided by the speed of light on the right side:
Fx(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
Pabs(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
Pref(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.

## Sunday, November 5, 2023

### 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
PE+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)·(EB)

Using the Poynting vector, we have represented the rate of change of electromagnetic energy flux as
PE+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, PE+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 with E²/c² in the expression for PE+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
PE+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.