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.

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/R³
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 r² 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/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/R³
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/R³
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]x·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:

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 R·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·dy /(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.

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·dm /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·dx /√x²+H² =
= −G·M·2πx·dx /(πR²√x²+H²) =
= −G·M·2x·dx /(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·dx /√x²+H²
where k = G·M /R²

Substituting y=x²+H² and noticing the dy=2x·dx, we get
V = −k·∫_{[H²;H²+R²]} dy /√y
The derivative of √y is 1 /(2√y) Therefore, the indefinite integral of 1 /√y is
2√y + C

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