Tuesday, December 3, 2019

Unizor - Physics4Teens - Energy - Light as Quants





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

Quants of Light

By the end of 19th century the electrons, as carriers of electricity, were discovered by Sir Joseph Thomson in 1897, and many scientists experimented with electricity.

At that time the wave theory of light was dominant. It could explain most of experimental facts and was shared by most physicists.
However, there were some new interesting experimental facts that could not be easily explained in the framework of the classical wave theory of light as oscillation of electromagnetic field.

The photoelectric effect was one of such experimental facts that physicists could not fit into classical wave theory.

Consider a simple electrical experiment when two poles, positive and negative, positioned close to each other, are gradually charged. After a charge reaches certain value, a spark between these poles causes the discharge of electricity.

The electric charge was attributed to electrons with a negatively charged pole having more electrons than in an electrically neutral state and a positively charged one having less electrons than in an electrically neutral state.
The electric spark was the flow of excess electrons from a negative pole to a positive one, thus bringing them both to an electrically neutral state.

It was observed that, if the negative pole is lit by light, the discharge occurs earlier, with less amount of charge accumulated in the poles.
This was so-called photoelectric effect.

The experimental characteristics of the photoelectric effect were:
(a) if photoelectric effect is observed with specific frequency of the light, the number of electrons leaving the negative pole in a unit of time is proportional to intensity of the light;
(b) the speed of photo-electrons and, therefore, their kinetic energy do not depend on intensity of light, but on its frequency; higher intensity light produces more electrons, but their speed remains the same, while higher frequency light produces faster photo-electrons;
(c) for each material, used as a negative pole, there is minimal frequency of light necessary to initiate the photoelectric effect; high intensity or prolong time exposure to light of a lesser frequency do not produce photoelectric effect.

The explanation coming to mind within a framework of the wave theory of light would be as follows.
Light is oscillations of the electromagnetic field. Electrons, accumulated during the charging process, are vibrating more intensely as a result of the oscillations of the electromagnetic field of the light, whose energy is transformed to electrons, so photo-electrons leave the negative pole easier, thus facilitating an earlier discharge.

This would be a great explanation if not for a couple of contradictory facts.
The first contradiction is the property (b) of the photoelectric effect. According to the classical wave theory, the speed of photo-electrons must be dependent on the intensity of the light (amplitude of electromagnetic waves), which was not observed. And the (c) property is also unexplainable by classical wave theory, because, again, within a framework of the classical wave theory for any frequency we can find an intensity of light sufficient to "knock" out the electrons from the negative pole or keep the light of lesser intensity long enough time to transfer to electrons sufficient amount of energy to fly off the surface of the pole, which was not observed.

The explanation of these phenomena came with introduction of quants of light - a hypothesis offered by Planck and used by Einstein to explain the properties of photoelectric effect.

According to the explanation of photoelectric effect offered by Einstein, light propagates in space not as a continuous stream of waves of electromagnetic oscillations, but in small indivisible packets (quants of light or photons), separated in space and traveling along the same path, thus resurrecting the corpuscular theory, but without rejecting the electromagnetic origin of light.

The energy of each photon proportionally depends on the frequency of oscillation of electromagnetic field that carries the light, not on intensity of light, with intensity of light being just a measure of the number of photons passing through a point in space in a unit of time.

The energy of light is absorbed by electrons also in these photons. To break away the electron needs certain minimal energy.
If the energy of a single photon is sufficient to overcome the atomic forces that keep the electron inside the negative pole, this electron becomes a photo-electron and flies away to a positive pole.
If the energy of a photon is less than this minimal amount necessary to overcome the atomic forces, this energy is dissipated as heat, and no photo-electrons are produced.

Within the framework of this new quantum theory of light all the characteristics of the photoelectric effect can find their explanation.
Let's analyze them.

(a) If photoelectric effect is observed with specific frequency of the light, the number of electrons leaving the negative pole in a unit of time is proportional to intensity of the light;
Explanation: each photon has sufficient amount of energy to "knock" out an electron, and intensity is the number of such photons per unit of time.

(b) The speed of photo-electrons and, therefore, their kinetic energy do not depend on intensity of light, but on its frequency; higher intensity light produces more electrons, but their speed remains the same, while higher frequency light produces faster photo-electrons;
Explanation: speed of electrons and, therefore, their kinetic energy depend on the energy of a photon that "knocked" these electrons out, which, in turn, is proportional to the frequency of light, while intensity of light (the number of photons per unit of time) affects the number of photo-electrons produced by light, not their individual energy.

(c) For each material, used as a negative pole, there is minimal frequency of light necessary to initiate the photoelectric effect; high intensity or prolong time exposure to light of a lesser frequency do not produce photoelectric effect.
Explanation: the energy, needed by an electron to break away from atomic forces that keep it inside the material, obviously depend on the material; as soon as the light frequency is sufficient for one photon to carry that amount of energy, the photoelectric effect can start; photons of lesser level of frequency cannot "knock" out the electrons from the surrounding material, and the energy of the light is just dissipated as heat.

Thursday, November 14, 2019

Unizor - Physics4Teens - Energy of Light - Light as Waves





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

Light as Waves

Corpuscular theory of light, suggested by Newton, while adequately explaining such properties of light as traveling along a straight line and reflection off the mirror, has not been able to explain certain other properties of light observed by experimentators. In particular, interferencediffraction and polarization of light remained unexplainable in the framework of corpuscular theory.

Attempts to model the light as oscillation of waves were made by several scientists, but the main formulation and explanation of wave-like properties of light is mostly related to a brilliant English physician and physicist Thomas Young.
His famous double-slit experiment, that showed the interference of light passing through two small close to each other slits, convincingly proved that light has wave-like properties.

This experiment is extremely simple, anybody can reproduce it at home, its schema is below.


The light from a single monochromatic source equidistant from two slits goes through these slits and is projected on a screen. The picture on the screen is a sequence of light and dark stripes parallel to the slits. This is a picture of interference that can only be explained within a framework of waves.
Two light waves coming through two slits to a surface of a screen in phase (both at its top wave state or both at the bottom) increase each other, making bright stripes. Those that come in opposite phases (one at its top state, while another at the bottom) nullify each other, making dark stripes.



Let's interpret the interference of light from the wave theory viewpoint.

First of all, if light is a wave, we have to determine the carrier of these waves. Until late 19th century physicists accepted a hypothesis that certain substance called ether (or aether) fills all the space and is a material carrier of waves of light and other waves of electromagnetic oscillations.
This hypothesis of ether was later on rejected, but for understanding of wave-like properties of light we can still think about ether as a substance, through which light travels as oscillations of this substance.
It might be considered similar to sound spreading inside a metal object, when you hit this metal object with another one. The crystal structure of the metal will start vibrating at the point of impact (the source of the sound) and these vibrations will spread throughout the whole object in all directions in a form of compression waves.
Notice that in this model the amplitude of vibrations is a characteristic of the strength of the sound and speed of propagation of sound waves depends on properties of material the object is made of.

If light is the waves of a carrier substance, we can talk about this wave's length (distance between two adjacent crests or two adjacent troughs), amplitude (maximum deviation from the neutral position) and speed of propagation (how far a crest moves in a unit of time).

If the speed of light propagation in the ether is c(m/sec) and the wave length is λ, it means that in a 1 sec the crest of a wave moves by a distance c meters. With a wave length λ meters there will be f=c/λ crests on this distance. So, the frequency of oscillation f, that is the number of oscillations per second, the wave length λ and the speed of propagation c are related by the equation
f = c/λ
or
λ = c/f
or
f·λ = c

Let's consider the double slit experiment and analyze what will be observed on the screen at the point equidistant from both slits. Two slits can be considered as two independent sources of light and, considering they are letting through the light emitted by a single source equidistant from them, we can assume that the waves of light are coming in-phase to these slits and go through them also in-phase. Then the two beams of light travel the same distance to a point on a screen that we have chosen as equidistant from the slits, arriving to this point also in-phase. At that point they are combined and intensity of the light is increasing at there. That's why the stripe on a screen in the middle of a picture is bright, it consists of points equidistant from both slits.

If, however, the distance these two beams of light have to travel towards the screen is different, the compounded effect of them, observed on a screen, will be dependent on whether the beams come to a screen in-phase or not.
More precisely, if the difference in the distance these beams travel towards a screen is a multiple of the length of a wave, they will come in-phase, and the overlapping beams will strengthen each other, there will be a bright spot on a screen.
If the difference in the distance equals to a multiple of wave length plus half a wave length, they will come out-of-phase, and the overlapping beams will nullify each other, there will be a dark spot on a screen.
In other cases the effect depends on how close the difference in distance is to a multiple of wave length or a multiple of wave length plus half a wave length with intermediary results.

So, the wave theory can explain the effect of interference. Other wave-related properties of light that can be explained within the framework of the wave theory will be presented in the Optics chapter of this course.

While the corpuscular properties of light do exist, they also can be explained within a framework of the wave theory and the energy of waves. That makes the wave theory of light more universal and more acceptable among physicists, though the auxiliary concept of ether has been totally rejected.

The concept of ether was rejected as there was no experiment that could prove its existence. Moreover, a famous experiment of Michelson-Morley has proven that the speed of light does not depend on the speed of the source of light, which cannot be explained from the position of ether as a medium, through which the light waves propagate.

There were some interesting observations of electric and magnetic properties, which related to an experimental fact that the speed of propagation of electric signals along wires is practically the same as the speed of light. This and other experiments prompted physicists to seek the explanation of the nature of light in the domain of electricity and magnetism.

James Maxwell came up with his famous equations that describe the electromagnetic field and concluded that light is the oscillations of electromagnetic field. There is no special medium, like ether, that carries the light as the sound is carried by the waves of compression of the medium. Instead, the electromagnetic field propagates through space, thus carrying the light.
The properties of electromagnetic field will be presented in the parts of this course dedicated to Electricity and Magnetism.

Tuesday, November 5, 2019

Unizor - Physics4Teens - Energy - Energy of Light - Light as Corpuscles





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

Light as Corpuscles

The nature of light was one of the topics of Newton's research. He experimented with light, made instruments to research the nature of light and formulated the first theory of light - corpuscular theory.

According to corpuscular theory, light is a bunch of very small particles emitted by a source of light and flying with a very high speed. When they enter our eye, we see the light as a result of bombarding of inner surface of an eye with these tiny particles of light that Newton called corpuscles.
Each light corpuscle has mass, speed, kinetic energy and trajectory, like any other material object.

Furthermore, Newton decided that there are different kinds of light corpuscles that cause our perception as different colors. He assumed that different colors of light corpuscles are due to their different sizes.

He also realized that the white color is a combination of different kinds of light corpuscles that can be separated into different individual colors. Thus, white light after going through a green glass becomes green, because a green glass separates different kinds of light corpuscles, letting through only those that are perceived by our eye as green.

According to Newton, light corpuscles are elastic, which explains perfectly their reflection from the mirror.

The effect of refraction of light, when it changes the direction going from one medium, like air, into another, like water, was explained by Newton as a result of changing the speed of propagation of light corpuscles, when they go from one medium into another.

The corpuscular theory of light was unable to explain the effect of diffraction, when the light seems bending around an edge of an obstacles or aperture.
Here is the picture formed on a screen by red light going through a small round hole.


The corpuscular theory of light was unable to explain the effect of interference, when the light going through two small holes positioned near each other forms a complex wave-like picture on the screen.


The corpuscular theory of light was unable to explain the effect of polarization, when the light consecutively going through two crystals of tourmaline changes its intensity from maximum to zero, depending on orientation of these crystals relatively to each other, when we rotate them around the axis coinciding with the direction of light.


These and some other difficulties in explanation of observed properties of light led to another theory - the wave theory of light.

Thursday, October 31, 2019

Unizor - Physics4Teens - Energy - Light as Energy Carrier





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

Light as Energy Carrier

Sun is a source of light and a source of heat. Just looking at light, we don't see energy it carries, but, touching a surface that was under Sun's light for some time, we feel its temperature, which is a measure of inner energy of the molecules inside the object. The object warms up under the Sun's light, which can only be explained by energy the light carries.

Let's conduct an experiment with light to prove that it carries energy.
On YouTube there are a few videos that show it, one of them demonstrates a radiometer.

The radiometer ("solar mill") looks like this:

Four small plates are arranged on a spinning wheel in the vacuum to avoid interference of air motion. Each plate has two surfaces, one silver and another black.
When light is directed on this "solar mill", black sides absorb light and are heated more than silver ones that reflect light.
As a result, the molecules near the black surface are moving more intensely than those near the surface of the silver side, thus pushing the silver side more and causing the rotation of the "solar mill".

So, there is no doubt that light carries energy. An important question is, how it does it.
We used to think about heat as the intensity of molecular motion. In case of light there is no such motion, light travels from Sun to Earth through vacuum.
Apparently, the situation is similar to gravity in a sense that gravity carries energy, but does not require any medium, like molecules, to carry it. Recall that we have introduced a concept of a field as a certain domain of space were forces exist and energy is present without any material substance. Somewhat similar situation is with light. Its nature is the waves of electromagnetic field.

Electromagnetic field is a completely different substance than gravitational field, but both are capable to carry energy without any material presence.

Classifying light as the waves of an electromagnetic field, we are opening the door to using this wave model for explanation of different kinds of light.
First of all, let's consider the light we see with a naked eye. The vision itself is possible only if the light carries some energy, that agitates some cells inside our eye, that, in turn, send an electric signal to a brain - one more argument toward a light as a carrier of energy.

Light that can be seen by a naked eye is called visible. But what about different colors that we can differentiate by an eye and different intensity of light that we view as "bright" or "faint"? The only explanation within a wave theory of light is that different intensities and colors that our eye sees are attributable to different kinds of waves of an electromagnetic field.

Any wave has two major characteristics: amplitude and frequency. This is similar to a pendulum, where amplitude is the maximum angle of deviation from a vertical and frequency is measured as a number of oscillations per unit of time. Light, as a wave, also has these two characteristics. The amplitude is an intensity of light, while frequency of the visible light is viewed as its color.

We all know that the photo laboratories, developing old fashioned films, are using rather faint red light during the developing process in order not to overexpose the film to light. The obvious reason is that red light carries less energy than white one, that is known to be a combination of many differently colored kinds of light. So, the color-defining frequency, as well as an intensity-defining amplitude of electromagnetic waves, determine the amount of energy carried by light.

Not only visible light is a manifestation of electromagnetic waves. From cosmic radiation called gamma rays to x-rays to ultra-violet light to visible light to infra-red light to microwave to radio waves - all are electromagnetic waves of different frequencies.

The unit of measurement of frequency is called Hertz, abbreviated as Hz with 1 Hz meaning 1 oscillation per second.

The range of frequencies of electromagnetic waves is from a few oscillations per second for very low frequency radio waves to 1024 oscillations per second for very high frequency gamma rays. The visible spectrum of frequencies is close to 1015 oscillations per second with the light perceived as red having a smaller frequency around 0.4·1015 Hz, followed in increasing frequency order by orange, yellow, green, blue and violet with a frequency around 0.7·1015 Hz.

Amount of energy carried by light depends on the amplitude and frequency of electromagnetic waves that constitute this light. Generally speaking, the higher the amplitude - the higher the energy is carried by light in a unit of time and, similarly, the higher the frequency - the higher the energy is carried by light in a unit of time.

Dependency of the energy carried by light on the frequency of electromagnetic waves that constitute this light is a more complex problem, that was solved in the framework of the Quantum Theory of light. According to Quantum Theory, electromagnetic waves propagate in packets called photons. Each photon carries an energy proportional to the frequency of waves that constitute this photon, and the amplitude of electromagnetic waves is, simply, a measure of the number of photons participating in these electromagnetic waves. That's why the picture of a light as a sinusoidal wave is a very simplified view on the nature of light as it is understood by contemporary physics.

Tuesday, October 15, 2019

Unizor - Physics4Teens - Energy - Gravitational Potential





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

Gravity Integration 4 -
Solid Sphere


1. Determine the potential of the gravitational field of a uniform solid sphere at any point outside it.

Let's establish a system of coordinates with a sphere's center at the origin of coordinates and X-axis going through a point of interest P, where we have to determine the gravitational potential.

Assume that the sphere's radius is R and the mass is M. Then its volume is 4πR³/3; and the mass density per unit of volume is ρ=3M/(4πR³).
Assume further that X-coordinate of a point P, where we want to calculate the gravitational potential, is H, which is greater than the radius of a sphere R.
If, instead of a sphere, we had a point mass M concentrated in its center at point O(0,0,0), its gravitational potential at a point P would be
V0(H) = −G·M/H
(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point P, and the field performs this work for us, so we perform negative work).

As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.

To calculate a gravitational potential of a solid sphere at point P on the X-axis, let's divide it into infinite number of infinitesimally thin concentric spherical shells, all centered at the origin of coordinates and use the results presented in the previous lecture about spherical shell.

As the variable of integration we will chose a radius of a spherical shell r that varies from 0 to R. Its outside surface area is 4πr², its thickness is dr and, therefore, its volume is 4πr²·dr.
This allows us to calculate the mass of this spherical shell using the volume and mass density calculated above.
dm = ρ·4πr²·dr =
= 3M·4πr²·
dr/(4πR³) =
= 3M·r²·
dr/

The formula for gravitational potential of a spherical shell, derived in the previous lecture was V=−G·M/H, where G is a gravitational constant, M is a mass of a spherical shell and H is a distance from a center of a shell to a point of interest.
In case of a solid sphere divided into infinite number of infinitesimally thin concentric spherical shells the distance H remains the same. So, all we have to do is to substitute the mass in the formula for a shell with the variable mass of a shells we divided our solid sphere and to integrate by variable radius r:
V(H) = −(G/H)[0;R]dm =
= −(G/H)[0;R]3M·r²·
dr/R³ =
= −
[3M·G/(H·R³)][0;R]r²·dr
The indefinite integral of  is r³/3, which gives the value of the integral
[0;R]r²·dr = R³/3 − 0 = R³/3
Therefore, finally,
V(H) = −G·M/H

Remarkably, the formula is exactly the same as if the whole mass was concentrated in the center of a sphere, the same as in case of a spherical shell.

It's a justification for considering gravitational potential of space objects in many cases to be approximately equivalent to a potential of point-masses of the same mass concentrated in one point, the center of mass of an object.

In theory, this result was easily predictable. The gravitational potential of each spherical shell is the same as if its mass is concentrated at its center. All shells are concentric, therefore the masses of all of them are concentrated in the origin of coordinates and can be added together, since the gravitational potential is additive. Thus, we come to the same value of gravitational potential of a sphere, as if its total mass is concentrated in one point - its center.

Let's analyze the force of gravity, acting on a probe object of a mass m at a point of interest on the distance H from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:
F(H) = m·dV(H)/dH =
= G·m·M/H²

which is a well known Newton's Law of Gravitation.

2. Determine the potential of the gravitational field of a uniform solid sphere of radius R at a point inside it at distance r from a center.

For this problem, as in the Problem 1 above, we will need a mass density per unit of volume ρ=3M/(4πR³).

Assume that a probe object is a distance r from a center of sphere, which is less than the radius of a sphere R. Let's calculate the gravitational potential V(r) of the combination of two separate sources - the solid sphere of radius r with a probe object on its surface and a thick empty spherical object between a surface of a sphere of the radius r and surface of a sphere of the radius R.

The gravitational potential of a uniform solid sphere of radius r on its surface is discussed above as a Problem 1. To use the results of this problem, we need a mass M1(r) of a source of gravity and the distance of a point of interest from a center H.
The mass is
M1(r) = ρ·4πr³/3 = M·r³/
The distance form a center is
H = r
The gravitational potential on the surface of this solid sphere of the radius r equals to
V1(r) = −G·M·r²/R³

Consider now the second source of gravity - a thick empty sphere between the radiuses r and R.
As in Problem 1 above, we will divide a thick empty sphere into an infinite number of concentric infinitesimally thin spherical shells of a variable radius x and thickness dx.
The mass of each shell is
dm(x) = ρ·4πx²·dx =
= 3M·4πx²·
dx/(4πR³) =
= 3M·x²·
dx/

The potential inside such an infinitesimally thin spherical shell of radius x is, as we know from a previous lecture, constant and equals to
dV(x) = −G·dm(x)/x =
= −3G·M·x·
dx/

To get a full potential inside such a thick empty sphere we have to perform integration of this expression from x=r to x=R.
V2(r) = (−3G·M/R³)[r;R]dx =
= −3G·M·(R²−r²)/(2R³)


The total potential at distance r from a center equal to sum of two potentials calculated above
V(r) = V1(r) + V2(r) =
= −G·M·r²/R³ −
−3G·M·(R²−r²)/(2R³) =
= −G·M·(3R²−r²)/(2R³)


On the outer surface of this sphere, when r=R, the above formula converts into the one derived in Problem 1:
V(R) = −G·M·/R

In the center of a solid sphere, when r=0, the potential is
V(R) = −(3/2)·G·M·/R

Let's analyze the force of gravity, acting on a probe object of a mass m at a point of interest on the distance r from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:
Case inside a solid sphere:
F(r) = m·dV(r)/dr =
= G·m·M·r/R³

So, as we move from a center of a solid sphere (r=0) towards its outer surface (r=R), the force is linearly growing from zero at the center to G·m·M/R² at the end on the surface.
Case outside a solid sphere (using the results of Problem 1 above for V(H)=−G·M/H, where H=r is greater than R):
F(r) = m·dV(r)/dr =
= G·m·M/r²

So, as we move from a surface of a solid sphere (r=R) outwards to infinity, increasing r, the force is decreasing inversely to a square of a distance from the center from G·m·M/R² to zero at infinity.

It's quite interesting to graph the force of gravitation as a function of a distance of a probe object from a center of a solid sphere. We have two different functions that represent this force, one inside and one outside the surface of a sphere.
The graph looks like this:





Monday, October 7, 2019

Unizor - Physics4Teens - Energy - Gravitational Potential - Thin Spheric...





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

Gravity Integration 3 -
Thin Spherical Shell


1. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point outside it.

Let's establish a system of coordinates with a spherical shell's center at the origin of coordinates and X-axis going through a point of interest P, where we have to determine the gravitational potential.

Assume that the sphere's radius is R and the mass is M. Then its surface is 4πR² and the mass density per unit of surface area is ρ=M/(4πR²).
Assume further that X-coordinate of a point P, where we want to calculate the gravitational potential, is H, which is greater than the radius of a spherical shell R.


If, instead of a spherical shell, we had a point mass M concentrated in its center at point O(0,0,0), its gravitational potential at a point P would be
V0 = −G·M/H
(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point P, and the field performs this work for us, so we perform negative work).

As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.

To calculate a gravitational potential of an infinitesimally thin spherical shell at point P on the X-axis, let's divide a spherical shell into infinite number of infinitesimally thin rings that are parallel to the YZ-plane and, therefore, perpendicular to X-axis, that goes through a center of each ring.

The angle φ from X-axis (that is, from OP) to a radius from an origin of coordinates to any point on a ring will be our variable of integration.
Then the radius of a ring will be
r(φ) = R·sin(φ)
The distance from the origin of coordinates to a center of a ring is R·cos(φ).
The area of a ring between angles φ and φ+dφ will be equal to the product of infinitesimal width of a ring dφ and its circumference 2πR·sin(φ)
Therefore, the mass of a ring will be
dm(φ) = ρ·2πR²·sin(φ)·dφ =
= M·2πsin(φ)·
dφ/(4π) =
= M·sin(φ)·
dφ/2


Knowing the mass of a ring dm(φ), its radius r(φ) and the distance from the ring's center to point of interest P, that is equal to H−R·cos(φ), we can use the formula of the ring's potential from a previous lecture
V = −G·M /R²+H²
substituting
dV(φ) instead of V
dm(φ) instead of M
H−R·cos(φ) instead of H
r(φ) instead of R

Therefore,
dV(φ) = −G·dm(φ) /r²(φ)+[H−R·cos(φ)]² =
= −G·M·sin(φ)·
dφ /2√R²+H²−2R·H·cos(φ)


Now all we need is to integrate this by φ in limits from 0 to π.
Substitute
y = √R²+H²−2R·H·cos(φ)
Incidentally, the geometric meaning of this value is the distance from point of interest P to any point on a ring for a particular angle φ.
Then
dy = R·H·sin(φ)·dφ /R²+H²−2R·H·cos(φ)

The limits of integration for φ from 0 to π in terms of y are from |H−R| (which, for our case of point P being outside the sphere, equals to H−R) to H+R.

In terms of y
dV(y) = −G·M·d/(2R·H)
which we have to integrate by y from H−R to H+R.

Simple integration of this function by y on a segment [H−R;H+R] produces −G·M·y/(2R·H) in limits from H−R to H+R:
V = −G·M·(H+R)/(2R·H) +
+ G·M·(H−R)/(2R·H) =
= −G·M/H


Remarkably, it's exactly the same gravitational potential, as if the whole mass was concentrated in a center of a spherical shell, as noted above as V0.

It's a justification for considering gravitational potential of space objects in many cases to be approximately equivalent to a potential of point-masses of the same mass concentrated in one point, the center of mass of an object.

2. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point inside it.

Using the same notation as in the previous case, this problem requires the distance from a point of interest P to a center of a spherical shell O to be less than the radius R of a spherical shell.
Doing exactly the same manipulation and substitution
y = √R²+H²−2R·H·cos(φ)
we see that the only difference from the previous case is in the limits of integration in terms of y.
The limits of integration for φ from 0 to π in terms of y are from |H−R| (which, in this case of point P being inside the sphere, equals to R−H) to H+R.

Integration by y on a segment [R−H;H+R] produces −G·M·y/(2R·H) in limits from R−H to H+R:
V = −G·M·(H+R)/(2R·H) + G·M·(R−H)/(2R·H) =
= −G·M/R


Remarkably, it's constant and is independent of the position of point P inside a spherical shell.

We have mentioned in the earlier lecture on gravitational field that in one dimensional case the gravitational force is a derivative of gravitational potential by distance from the source of gravity times mass of a probe object:
F(r)=G·M·m /r²=m·dV(r)/dr

The fact that the gravitational potential is constant and, therefore, its derivative is zero, signifies that there is no force of gravity inside a spherical shell. The forces of gravity from all directions nullify each other.

An intuitive explanation of this is in the fact that, if you consider any conical surface with a vertex at point P inside a sphere, cutting pieces of spherical shell's surface in both directions, the areas of the pieces will be proportional to a square of a distance from point P, while the gravitational forces produced by these pieces of surface are inversely proportional to a square of a distance from point P, thus both forces from opposite ends of a cone are equal in magnitude and opposite in direction, thus nullify each other.



Friday, October 4, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...





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

Gravity Integration 2

1. Determine the potential of the gravitational field of an infinitely thin uniform solid ring at any point on the line perpendicular to a plane of the ring and going through its center.

Let's establish a system of coordinates with a ring in the XY-plane with its center coincided with the origin of coordinates. Then the Z-axis is the line, on which we have to determine the gravitational potential of the ring.
Assume that the ring's radius is R and the mass is M, so the density of mass per unit of length is ρ=M/(2πR).
Assume further that the Z-coordinate of a point P, where we want to calculate the gravitational potential, is H.

If, instead of a ring, we had a point mass M concentrated in its center at point (0,0,0), its gravitational potential at a point P would be
V0 = −G·M/H
(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point P, and the field performs this work for us, so we perform negative work).

Since the mass in our case is distributed along the circumference of a ring, and every point on a ring is on a distance r=√R²+H² from point P, which is further from this point than the center of a ring, the gravitational potential of a ring at point P will be smaller.

As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.
Therefore, to calculate a gravitational potential of a ring, we can divide it into infinite number of infinitesimal pieces, calculate the gravitational potential of each piece at the point of interest P and integrate all these potentials.

Let's choose an angle from the positive direction of the X-axis to a point on a ring as the main integration variable φ∈[0;2π]. Its increment dφ gives an increment of the circumference of a ring
dl = R·dφ
The mass of this infinitesimal segment of a ring is
dm = ρ·dl = M·R·dφ /(2πR) = M·dφ /(2π)

The distance from this infinitecimal segment of a ring to a point of interest P is independent of variable φ and is equal to constant r=√R²+H².

Therefore, gravitational potential of an infinitecimal segment of a ring is
dV = −G·d/r = −G·M·dφ /(2π√R²+H²)

Integrating this by variable φ on [0;2φ], we obtain the total gravitational potential of a ring at point P:
V = [0;2π]dV = −[0;2π]G·M·dφ /(2π√R²+H²)
Finally,
V = −G·M /R²+H²

2. Determine the potential of the gravitational field of an infinitely thin uniform solid disc at any point on the line perpendicular to a plane of the disc and going through its center.

Let's establish a system of coordinates with a disc in the XY-plane with its center coincided with the origin of coordinates. Then the Z-axis is the line, on which we have to determine the gravitational potential of the ring.
Assume that the ring's radius is R and the mass is M, so the density of mass per unit of surface is ρ=M/(πR²).
Assume further that the Z-coordinate of a point P, where we want to calculate the gravitational potential, is H.

Let's split our disc into infinite number of infinitely thin concentric rings of radius from x=0 to x=R of width dx each and use the previous problem to determine the potential of each ring.

The mass of each ring is
dm(x) = ρ·2πx·dx
This gravitational potential of this ring at point P, according to the previous problem, is
dV(x) = −G·dm(x) /x²+H² =
= −G·ρ·2πx·
d/x²+H² =
= −G·M·2πx·
d/(πR²√x²+H²) =
= −G·M·2x·
d/(R²√x²+H²)


To determine gravitational potential of an entire disc, we have to integrate this expression in limits from x=0 to x=R.
V = [0;R]dV(x) = −k·[0;R]2x·d/x²+H²
where k = G·M /

Substituting y=x²+H² and noticing the dy=2x·dx, we get
V = −k·[H²;H²+R²] d/y
The derivative of y is /(2√y) Therefore, the indefinite integral of /y is
2√y + C

Finally,
V = −k·(2√H²+R²−2H) = −2G·M·(√H²+R²−H) /

Tuesday, September 24, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...





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

Gravity Integration 1

Determine the potential of the gravitational field of an infinitely thin solid rod at any point outside of it.

Let's establish a system of coordinates with a rod and a point mass lying in the XY-plane with the rod on the X-axis with one end at point A(a,0) and another at point B(b,0).
Assume that the rod's length is L=b−a and the mass is M, so the density of mass per unit of length is ρ=M/L.
Assume further that the coordinates of a point P, where we want to calculate the gravitational potential, are (p,q).

If, instead of a rod, we had a point mass M concentrated in the midpoint of a rod at point ((a+b)/2,0), its gravitational potential at a point (p,q) would be
V0=G·M/r
where r is the distance between the midpoint of a rod and a point of measurement of gravitational potential P:
r = {[(p−(a+b)/2]2 + q2}1/2

Since the mass in our case is distributed along the rod, the gravitational potential will be different.

As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.
Therefore, to calculate a gravitational potential of a rod, we can divide it into infinite number of infinitesimal pieces, calculate the gravitational potential of each piece at the point of interest P and integrate all these potentials.

Consider a picture below (we recommend to save it locally to see in the bigger format).

As a variable, we will use an X-coordinate of a point on a rod Q and calculate the gravitational potential at point of interest P(p,q) from an infinitesimal segment of a rod of the length dx around point Q(x,0).
Knowing that, we will integrate the result by x on a segment [a;b] to get the gravitational potential of the rod.

The infinitesimal segment of a rod dx, positioned around a point Q(x,0), has an infinitesimal mass dm that can be calculated based on the total mass of a rod M and its length L=b−a as
dm = M·d/L

The gravitational potential of this segment depends on its mass dm and its distance r(x) to a point of interest P(p,q).
dV = G·d/r(x)
Obviously,
r(x) = [(p−x)2+q2]1/2
Combining all this, the full gravitational potential of a rod [a;b] of mass M at point P(p,q) will then be
V(p,q) = abd/r(x) = abG·M·dx/{[(p−x)2+q2]1/2}

We can use the known indefinite integral
d/(t²+c²) = ln|t+√(t²+c²)|

Let's substitute in the integral for gravitational potential t=x−p.
Then
V(p,q) = G·M·d/[t2+q2]1/2
where integration is from t=a−p to t=b−p.
V(p,q) = (G·M/L)·[ln|b−p+√(b−p)²+q²| − ln|a−p+√(a−p)²+q²|]
where L = b−a

Since the difference of logarithms is a logarithm of the result of division,
V(p,q) = G·M·ln(R) /L
where
L = b−a and
R = |b−p+√(b−p)²+q²/ |a−p+√(a−p)²+q²|

Friday, September 20, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Energy...





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

Gravitational Energy Conservation

While moving an object from a distance r1 to a distance r2 from the center of gravity, the gravitational field has performed certain work W[r1,r2], spending certain amount of energy. Since energy must be conserved, it should materialize in some other way.

Indeed, the kinetic energy of a probe object at the end of its movement from point {r1,0,0} to {r2,0,0} must be equal to the work performed by the field.

We have positioned our probe object at point {r1,0,0} without any initial speed, that is Vr1=0. Therefore, the kinetic energy Kr1 at this initial point is zero.
At the end of a motion at point {r2,0,0} the speed Vr2 must have such a value that the kinetic energy Kr2 would be equal to work W[r1,r2] performed by the field.
Kr2 = m·V²r2 /2 = W[r1,r2]

From this equation, knowing how to calculate the work performed by a gravitational field (see the previous lecture), we can find a speed of a probe object at the end of its motion from point {r1,0,0}, where it was at rest, to point {r2,0,0}:
m·V²r2 /2 =
(G·M /r2 − G·M /r1
)·m
r2 = 2·G·M·(1/r2 − 1/r1)

In a particular case, when r1=∞ and r2=r, that is a probe object falls with no initial speed from the infinitely long distance from a source of gravity to a point at distance r from it, the formula is simplified:
r = 2·G·M /r

We would like to warn against falling into a point-mass that is a source of gravity, when the final distance from it is zero, that is r=0 in the above equation. It obviously produces infinite speed and infinite kinetic energy, which does not correspond to reality. The most important reason for this deviation from the reality is our assumption about a source of gravity to be a point-mass. Real objects have certain non-zero dimensions. For example, in case of a gravitational field around our planet should not be analyzed by this formula for values of r less than the radius of Earth.

Back to energy conservation.
The potential energy of an object is a measure of work that it can do, if left alone, that depends on a position of an object relative to other objects and such properties as its mass. Actually, these two parameters are the only ones needed to calculate the potential energy of a probe object in a gravitational field, provided we know everything about the field.

As we know (see the previous lecture), amount of work we need to move a probe object of mass m from an infinite distance to a distance r from a source of gravity equals to
Wr = −G·M·m /r
It's negative from our external to the gravitational field viewpoint, because we don't actually perform work, the field performs it for us. So, from the external viewpoint, the field gives certain energy to external object by performing some work on it, similar to a person, pushing the cart, spends energy, transferring it to a cart.

In this expression, skipping over the universal gravitational constant G, components M (mass of a source of gravitational field) and r (distance from the center of the gravitational field) characterize the gravitational field, while m (mass of a probe object) characterizes the object, whose potential energy we measure.

This energy is transferred to a probe object as its potential energy. If an object is not moving from this position, because some force holds it there, it retains this potential energy. As soon as there is no force holding it there, it will start moving towards the source of gravity, losing its potential energy and gaining the kinetic energy because it will move faster and faster.

As an example, let's calculate the kinetic energy and final speed of a small asteroid, free falling on the surface of the Moon, assuming the Moon is the only source of gravity in the Universe.
The Universal Gravitational Constant is
G=6.67408·10−11,
its units are m3·kg−1·sec−2.
The mass of the Moon is M=7.34767309·1022 kg.
The radius of the Moon is r=1.7371·106 m.
Let's assume that an asteroid falling on the Moon is relatively small one, say, m=50 kg.

According to the formula above, the gravitational field of the Moon did the work that equals to
W ≅ 6.67408·10−11·7.34767309·1022·50 /(1.7371·106) ≅ 141,151,800 (joules)
let's check the units to make sure we get joules, the units of work
m3·kg−1·sec−2·kg·kg·m−1 = kg·m2·sec−2 = N·m = J

The final speed V can be calculated by equating this amount of work and kinetic energy of an asteroid:
V2 ≅ 2·6.67408·10−11·7.34767309·1022 /(1.7371·106) ≅ 5646072
let's check the units to make sure we get the square of speed units
m3·kg−1·sec−2·kg·m−1 = m2·sec−2 = (m/sec)2

From this the speed of an asteroid falling from infinity onto Moon's surface is
V ≅ √5646072 ≅ 2376 (m/sec)
or about 2.4 km/sec.

Incidentally, this is the so-called escape speed from the Moon, the initial speed needed for an object to leave the gravitational field of the Moon. A stone, thrown perpendicularly to the surface of the Moon with an initial speed less than that will go for certain distance away from the Moon, but then it will be brought back by the Moon's gravitation. Only if the initial speed is equal or exceeds the one above, the distance an object will go will be infinite, that is the object will leave the gravitational field of the Moon.

Let's do similar calculations for the Earth, using the same assumptions, the same asteroid and the same units of measurement.
M = 5.972·1024 kg
r = 6.371·106 m
W ≅ 6.67408·10−11·5.972·1024·50 /(6.371·106) ≅ 3,128,049,424 (joules)

V2 ≅ 2·6.67408·10−11·5.972·1024 /(6.371·106) ≅ 125121977 (m/sec)2
From this the speed of an asteroid falling from infinity onto Earth's surface is
V≅√125121977≅11186 m/sec
or about 11.2 km/sec.

This is also the escape speed needed to fly away from Earth's gravitational field.

Monday, September 16, 2019

Unizor - Physics4Teens - Energy - Gravitational Field - Problems





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

Problems on Gravity

Problem 1
Gravitational potential of a spherical gravitational field around a point-mass M at a distance r from it is defined as the work performed by gravity to bring a probe object of a unit mass from infinity to this point and is expressed as
Vr = −G·M /r
Why is this formula independent of trajectory of a probe object or its exact final position relative to the point-mass M, but only on a distance itself from the source of gravity?

Solution
Any movement can be represented as infinitely many infinitesimal displacements, combined together into a trajectory.
In our three-dimensional world the force and an infinitesimal displacement of a probe object are vectors, so the infinitesimal work dW performed by the force of gravity F during the movement of a probe object, described by the infinitesimal displacement dS, is a scalar product of these two vectors:
dW = F·dS
Note that the vector of gravitational force F is always directed towards the source of gravity.
Since a displacement vector dS can be represented as a sum of radial (towards the source of gravity) dSr and tangential (perpendicular to radius) dSt components, the above expression for a differential of work can be written as
dW = F·(dSr + dSt) =
F·
dSr + F·dSt
The second component in the above expression is a scalar product of two perpendicular vectors and is equal to zero. That's why we can completely ignore tangential movements, when calculating the work done by a central gravitational field, as not contributing to the amount of work. The total amount of work will be the same as if our probe object moved along a straight line towards the source of gravity and stopped at a distance r from it.

Problem 2
Given two point-masses of mass M each, fixed at a distance 2R from each other.
Prove that the gravitational potential of a gravitational field produced by both of them at each point on a perpendicular bisector between them equals to a sum of individual gravitational potentials of these point-masses at this point, as if they were the only source of gravitation. In other words, prove that gravitational potential is additive in this case.

Solution
Let's draw a diagram of this problem (you can download it to display in a bigger format).

Our two point-masses are at points A and B, the probe object is at point D on a perpendicular bisector of a segment AB going through point C.
The force of gravity towards point A is a segment DE, the force of gravity towards point B is a segment DF.
We will calculate the potential of a combined gravitational field of two point-masses at point D, where the probe object is located.
Let's assume that the segment CD equals to h.
The magnitude of each gravitational force equals to
F = G·M·m /(h2+r2)
Represent each of these forces as a sum of two vectors, one (green on a drawing) going vertically along the bisector CD, another (red) going horizontally parallel to AB.
Vertical components of these two forces will add to each other, as equal in magnitude and similarly directed downwards on a drawing, while horizontal ones will cancel each other, as equal in magnitude and opposite in direction to each other. So, the combined force acting on a probe object is a sum of vertical components of gravitational forces with a magnitude
Ftot = 2·G·M·m·sin(φ)/(h2+r2)
Since sin(φ) = CD/AD,
sin(φ) = h /[(h2+r2)1/2]
Ftot = 2·G·M·m·h /(h2+r2)3/2
If the gravitational field pulls a probe object along the perpendicular bisector of a segment AB from infinity to a distance h from the segment, the magnitude of a combined force of gravity, as a function of a distance from the segment x is changing, according to a similar formula:
Ftot(x) = 2·G·M·m·x /(x2+r2)3/2
To calculate work performed by a gravitational field pulling a probe object from infinity to height h above the segment AB, we have to integrate
Wtot = [∞;h]Ftot(x)·dx
It's supposed to be negative, since the direction of a force is opposite to a positive direction of the coordinate axis, we will take it into account later.
Wtot = 2GMm·x·d/(x2+r2)3/2
(within the same limits of integration [∞;h])
This integral can be easily calculated by substituting
y=x2+r2,
2·x·dx = dy,
infinite limit of integration remaining infinite and the x=h limit transforming into y=h2+r2. Now the work expression is
Wtot = G·M·m·y−3/2·dy
with limits from y=∞ to y=h2+r2.
The indefinite integral (anti-derivative) of y−3/2 is −2·y−1/2.
Therefore, the value of integral and the work are
Wtot = −2·G·M·m·(h2+r2)−1/2
For a unit mass m=1 this work is a gravitational potential of a combined gravitational field produced by two point-masses on a distance h from a midpoint between them along a perpendicular bisector
Vtot = −2·G·M·(h2+r2)−1/2
At the same time, the gravitational potential of a field produced by each one of the point-masses, considered separately, equals to
Vsingle = −G·M·(h2+r2)−1/2
As we see, the gravitational potential of two point-masses equals to a sum of gravitational potential of each of them, considered separately.
IMPORTANT NOTE
With more cumbersome calculations this principle can be proven for any two (not necessarily equal) point-masses at any point in space (not necessarily along the perpendicular bisector). This principle means that gravitational potential is additive, that is the gravitational potential of any set of objects at any point in space equals to sum of their individual gravitational potentials.

Problem 3
Express mass M of a spherical planet in terms of its radius R and a free fall acceleration g on its surface.

Solution
Let m be a mass of a probe object lying on a planet's surface.
According to the Newton's 2nd Law, its weight is
P = m·g
According to the Universal Law of Gravitation, the force of gravitation between a planet and a probe object is
Fgravity = G·M·m /R2
Since the force of gravitation is the weight Fgravity = P,
m·g = G·M·m /R2
from which
M = g·R2 /G

Problem 4
Express gravitational potential VR of a spherical planet on its surface in terms of its radius R and a free fall acceleration g on its surface.

Solution
From the definition of a gravitational potential on a distance R from a source of gravity
VR = −G·M /R
Using the expression of the planet's mass in terms of its radius R and a free fall acceleration g on its surface (see above),
M = g·R2 /G
Substituting this mass into a formula for potential,
VR = −G·g·R2 /(G·R) = −g·R


Tuesday, September 10, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravitational Field



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

Gravitational Field

Studying forces, we have paid attention to a force of attraction, that exists between any material objects, the force of gravity.
For example, if a comet from outer space flies not far from a Sun, it is attracted by Sun and changes its straight line trajectory.

In Mechanics we used to see the force as something between the objects touching each other, like a man pushing a wagon. In case of gravity the force obviously exists, but it acts on a distance, in "empty" space.
In Physics this concept of force acting on a distance is described by a term field. Basically, field is the area in space where some force acts on all objects or only on objects that have specific property. The force in this case depends on a point in space and an object that experiences this force and, as a result of the action of force, changes its movement.

Gravitational field exists around any material object (the source object of a field) and acts as an attraction towards this source object, experienced by any other material object (probe object) positioned in this field.
As described in the "Gravity, Weight" chapter of "Mechanics" part of this course, the magnitude of the gravitational force F is proportional to a product of masses of a source object and a probe object, M and m, and it is inversely proportional to a square of a distance r between these objects:
F = G·M·m /
where G - a constant of proportionality, since the units of force (N - newtons) have been defined already, and we want to measure the gravitational force in the same units as any other force.

The direction of the gravitational force acting on a probe object is towards the source object.

Let's return to our example of a comet flying not far from the Sun and, being attracted to the Sun, changing its trajectory. Obviously, to change the trajectory, some energy must be spent. So, we conclude that gravitational field has certain amount of energy at each point that it spends by applying the force onto a probe object.

To quantify this, assume that the source of gravity is a point mass M fixed at the origin of coordinates. Position a probe object of mass m at coordinates {r1,0,0} and let it go. The force of gravity will cause the motion of this probe object towards the center of gravity, the origin of coordinates, so the movement will be along the X-axis. Let the ending position of the probe object be {r2,0,0}, where r2 is smaller then r1. Let x be a variable X-coordinate (distance to the origin).

According to the Universal Law of Gravitation, the force of attraction of a probe object towards the source of a gravitational field at distance x from the origin equals to
F = −G·M·m /
where minus in front of it signifies that this force is directed opposite to increasing the X-coordinate.
This force causes the motion and, therefore, does some work, moving a probe object from point {r1,0,0} to point {r2,0,0} along the X-axis. To calculate the work done by this variable force, we can integrate dx from x=r1 to x=r2:
W[r1,r2] = [r1,r2]dx =
= −[r1,r2]G·M·m·
dx /x² =
= G·M·m /x
|[r1,r2] =
= G·M·m /r2 − G·M·m /r1 =
= (G·M /r2 − G·M /r1
)·m

The expression
V(r) = −G·M/r
is called gravitational potential.
It's a characteristic of a gravitational field sourced by a point mass M at a distance r from a source.
It equals to work needed by external forces to bring a probe object of mass m=1 to a point at distance r from a source of the field from infinity.
Indeed, set m=1, r1=∞ and r1=r in the above formula for work W[r1,r2] and take into consideration that gravitational field "helps" external forces to move a probe object, so the external forces spend negative amount of energy.

Using this concept of gravitational potential V(r), we can state that, to move a probe object of a unit mass from distance r1 relative to a source of gravitational field to a distance r2 relative to its source in the gravitational field with gravitational potential V(r), we have to spend the amount of energy equal to V(r1)−V(r2).
For a probe object of any mass m this amount should be multiplied by m.
If r2 is greater than r1, that is we move a probe object further from the source of gravity, working against the gravitational force, this expression is positive, we have to apply effort against the force of gravity. In an opposite case, when r2 is smaller than r1, that is we move closer to a source of gravity, the gravitational force "helps" us, we don't have to apply any efforts, and our work is negative.

Therefore, an expression EP=m·V(r) represents potential energy of a probe object of mass m at a distance r from a source of a gravitational field with gravitational potential V(r).

A useful consequence from a concept of a gravitational potential is that the force of gravity can be expressed as the derivative of the gravitational potential.
F = G·M·m /r² = m·dV(r)/dr
which emphasizes the statement that the gravitational potential is a characteristic of a field itself, not its source.
We, therefore, can discuss gravitational field as an abstract concept defined only by the function called gravitational potential.

Sunday, August 18, 2019

Unizor - Physics4Teens - Energy - Energy of Nucleus - Fusion



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



Nucleus Fusion



Fusion is a nuclear reaction, when light nuclei are brought together and combined into a heavier ones.

The reason for this reaction to release the energy is the difference
between amount of energy needed to overcome the repulsion between nuclei
because they have the same positive electric charge (this energy is
consumed by fusion) and the potential energy released by strong forces, when the formation of a combined nucleus occurs (this energy is released by fusion).

The former is less than the latter.



When the light nuclei are fused into a heavier one, the excess of potential energy of strong forces, released in the process of fusion,
over the energy needed to squeeze together protons against their
repulsion is converted into thermal and electromagnetic field energy.



Analogy to this process can be two magnets separated by a spring.


The magnets represent two separate protons, the magnetic force of attraction between them represents the strong force
that is supposed to hold the nucleus together, when these particles are
close to each other, the spring represents the electrical repulsive
force between them, acting on a larger distance, as both are positively
charged.

It's known that magnetic force is inversely proportional to a square of a
distance between objects, while the resistance of a spring against
contraction obeys the Hooke's Law and is proportional to the length of
contraction.

On the picture magnets are separated. To bring them together, we have to
spend certain amount of energy to move against a spring that resists
contraction. But the magnetic attraction grows faster then the
resistance of the spring, so, at some moment this attraction will be
greater than the resistance of a spring. At this moment nothing would
prevent magnets to fuse.



As is in the above analogy, if we want to fuse two protons, we have to bring them together sufficiently close for strong forces to overtake the repulsion of their positive charges.



Consider the following nuclear reaction of fusion.

One nucleus of hydrogen isotope deuterium 1H2 with atomic mass 2 contains one proton and one neutron.

One nucleus of hydrogen isotope tritium 1H3 with atomic mass 3 contains one proton and two neutrons.

If we force these two nuclei to fuse, they will form a nucleus of helium 2He4 and releasing certain amount of energy:

1H2 + 1H3 = 2He4 + 0n1



It's not easy to overcome the repulsion of protons. High temperature and
pressure, like in the core of our Sun, are conditions where it happens.
On Earth these conditions are created in the nuclear bomb, using the
atomic bomd to achieve proper amount of heat and pressure, thus creating
an uncontrlled fusion.

Controlled nuclear reaction of fusion is what scientists are working on right now. So far, it's still in the experimental stage.

Monday, August 12, 2019

Unizor - Physics4Teens - Energy - Energy of a Nucleus - Fission



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

Nucleus Fission

Fission, first of all, is a nuclear reaction, when heavier nuclei are split into lighter ones.
The reason for this reaction to release the energy is the difference between amount of energy needed to break strong forces that hold the nucleus together (this energy is consumed by fission) and amount of potential energy in positively charged and repelling protons inside nucleus (this energy is released by fission).
The former is less than the latter.

When the heavy nucleus is broken into parts, the excess of potential energy of squeezed together protons against their repelling force over the energy of strong forces that keep nucleus together is converted into thermal and electromagnetic field energy.

Analogy of this is a spring squeezed tightly and held in this position by a thread. A thread plays the role of strong forces, while a potential energy of a squeezed spring plays the role of protons kept close to each other by a this force. When you cut a thread, the spring will release the potential energy, similarly to protons repelling from each other.

Electrically positively charged protons repel each other and, at the same time, are bonded together by strong forces inside a nucleus. At the same time neutrons are also bonded by strong forces among themselves and with protons without any repulsion.
So, the more neutrons the nucleus has - the stronger it is. Neutrons only add "bonding material" to a nucleus without adding any repelling forces that work against the nucleus' stability.

Uranium-238 with 92 protons and 146 neutrons (92U238) naturally occurs on Earth and is relatively stable.
Uranium-235 with the same 92 protons and 143 neutrons (92U235) has less "bonding material" (less neutrons) and is more susceptible to fission.

All it takes to break the nucleus of 92U235 is a little "push" from outside, which can be accomplished by bombarding it with neutrons. In the process of fission, caused by hitting a nucleus of 92U235 with a neutron, it can transforms into Barium-141 with 56 protons and 85 neutrons 56Ba141, Krypton-92 with 36 protons and 56 neutrons 36Kr92 and 3 free neutrons.
As we see, the numbers of protons is balanced (input: 92, output: 56 and 36), as well as a number of neutrons (input: 1 free hitting neutron and 143 in a nucleus of 92U235 total 144, output: 85 in a nucleus 56Ba141, 56 in a nucleus of 36Kr92 and 3 new free neutrons total 144).

Let's express this reaction in a formula (letter n denotes a neutron):
0n1 + 92U235 =
56Ba141 + 36Kr92 + 3·0n1


What's interesting in this reaction is that it not only produces energy because we break a heavy nucleus into lighter ones, but also that it produces 2 new neutrons that can bombard other atoms, causing a chain reaction and, potentially, an explosion (atomic bomb). However, if we absorb extra neutrons, it will allow to slowly release of nuclear energy (nuclear power stations).

Monday, August 5, 2019

Unizor - Physics4Teens - Energy - Energy of a Nucleus



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



Energy of Nucleus



In this lecture we will analyze the energy aspect of nucleus - the central part of an atom.



By now we have built a pyramid of energy types, related to the depth of our view inside the matter.



First, we analyzed the mechanical energy - the energy of moving macro-objects.



Our next view deep into the world of macro-objects uncovered the molecules - the smallest parts of macro-objects that retain their characteristics. The movement of these molecules was the source of thermal energy, which we often call the heat.



Next step inside the molecules uncovered atoms, as the molecules'
components. There are about 100 types of atoms and their composition
inside the molecules creates all the thousands of different molecules. Chemical reactions
change the composition of atoms in molecules, thereby creating new
molecules from the atoms of old molecules. This process broke some
inter-atomic bonds and created the new ones and is the source of chemical energy.



Now we look deep inside the atoms and find there 3 major elementary particles - electrically positively charged protons and electrically neutral neutrons inside a small but heavy nucleus and electrically negatively charged electrons,
circulating around nucleus on different orbits. For electrically
neutral atoms the numbers of protons and electrons are equal. Nuclear energy is hidden inside the nucleus and is the subject of this lecture.



The first question we would like to answer is "What holds nucleus, its
protons and neutrons, together, considering protons, as electrically
positively charged particles must repel each other?"



The answer is simple. There are other forces in the Universe, not only
electrostatic ones, that act in this case. These intra-nucleus forces
that hold the nucleus together are called strong forces. They are strong
because they are the source of attraction between the protons that is
stronger than electrostatic repelling. However, these strong forces act
only on a very small distance, comparable to the size of a nucleus
inside an atom. For example, at a distance 10−15m the strong force is more than 100 times stronger than electrostatic one.



If, by regrouping protons and neutrons, we will be able to create different atoms (inasmuch as regrouping atoms in chemical reaction we create new molecules), a new source of energy, based on strong forces, the nuclear energy, can be uncovered in the course of nuclear reaction.



There is another form of nuclear reaction related to
transformation of elementary particles. Under certain circumstance a
neutron inside a nucleus can transform into proton and, to keep the
total electrical charge in balance, it emits an electron. This reaction
is called beta-decay and it also produces energy in the form of electromagnetic waves of very high frequency (gamma-rays).



Nuclear reactions are a very powerful source of nuclear energy, which is
so much more powerful than other types of energy, that, if misused, it
might represent a danger for life on our planet.



There is a clear analogy between nuclear and chemical reactions.

What happens with atoms in the chemical reaction, happens with protons
and neutrons in nuclear reaction. Some atomic bonds break in a chemical
reaction, some are created. Some nuclear bonds between protons and
neutrons break in a nuclear reaction, some are created.



Sometimes the chemical reaction happens by itself, as long as
participating substances are close together, but sometimes we have to
initiate it, like lighting methane gas with a spark or a flame of a
match to initiate continuous burning.

Similar approach is valid for nuclear reaction. Sometimes it happens by
itself, but sometimes it should be started, like bombarding the nucleus
with neutrons, after which it continues by itself.



Here is an interesting fact.

Physicists have measured the masses of protons, neutrons and many
different nuclei that contain these protons and neutrons and have
discovered that the sum of masses of individual protons and neutrons is
greater than the mass of a nucleus that contain these exact particles.

For example,

mass of proton is 1.0072766 atomic mass units or 1.6726·10-27kg,

mass of neutron is 1.0086654 atomic mass units or 1.6749·10-27kg.

At the same time, mass of deuterium nucleus, that contains 1 proton and 1
neutron is 2.0135532 atomic mass units, which is smaller than the sum
of masses of proton and neutron (1.0072766 + 1.0086654 = 2.015942).

This so-called "mass defect" is directly related to nuclear energy - the energy of strong forces that hold the nucleus together.



A simplified explanation of this effect is based on the law of energy
conservation. Consider the force of gravity between a planet and an
object above its surface. The object has certain potential energy and,
if dropped to the ground, this potential energy transforms into other
forms, like kinetic, thermal etc.



Similarly, if we consider two independent neutrons (or neutron and
proton, or two protons) on a very small distance from each other, but
not forming a nucleus, there is a potential energy of the strong forces
acting between them. If we let these two particles to form a nucleus,
analogously to an object falling towards the surface of a planet, this
potential energy should be transformed into other forms, like thermal.



Now the Theory of Relativity comes to play, that has established the equivalence of mass and energy by a famous formula E=m·c².
According to this equivalence, if some energy is released during the
formation of a nucleus from individual protons and neutrons, there must
be certain amount of mass released associated with this energy. That is
the explanation of "mass defect".



It should be noted that to form a nucleus of deuterium from 1 proton and
1 neutron is easier than to form a nucleus that contains more than one
proton, because electrostatic repulsion between positively charged
protons prevents their bonding. So, to bring protons sufficiently close
to each other for strong forces to overcome the electrostatic
repulsion, we have to spend some energy. The net energy released by
forming a nucleus from protons and neutrons is the difference between
the energy released from strong forces taking hold of these particles inside a nucleus and the energy consumed to overcome repulsion of protons.



Actually, as we attempt to form bigger nuclei, the energy we have to
spend to overcome electrostatic repulsion forces become greater than
amount of energy released by forming a nucleus. This border line is
approximately around the nucleus of iron Fe. Forming iron
and heavier elements from protons and electrons is a process that
consumes more energy than releases. These heavier nuclei will produce
energy, if we reverse the procedure, breaking them into individual
protons and neutrons.



The mechanisms described above are used in nuclear reactors and atomic
bomb, where heavier elements are broken into lighter ones (fission),
releasing energy, and in hydrogen bomb, where lighter elements are
bonded together to release the energy (fusion).