Saturday, January 11, 2014

Black-body radiation quantization

Radiation quantization

Black-body radiation, the emission of electromagnetic energy due to an object's heat, could not be explained from classical arguments alone. The equipartition theorem of classical mechanics, the basis of all classical thermodynamic theories, stated that an object's energy is partitioned equally among the object's vibrational modes. This worked well when describing thermal objects, whose vibrational modes were defined as the speeds of their constituent atoms, and the speed distribution derived from egalitarian partitioning of these vibrational modes closely matched experimental results. Speeds much higher than the average speed were suppressed by the fact that kinetic energy is quadratic—doubling the speed requires four times the energy—thus the number of atoms occupying high energy modes (high speeds) quickly drops off because the constant, equal partition can excite successively fewer atoms. Low speed modes would ostensibly dominate the distribution, since low speed modes would require ever less energy, and prima facie a zero-speed mode would require zero energy and its energy partition would contain an infinite number of atoms. But this would only occur in the absence of atomic interaction; when collisions are allowed, the low speed modes are immediately suppressed by jostling from the higher energy atoms, exciting them to higher energy modes. An equilibrium is swiftly reached where most atoms occupy a speed proportional to the temperature of the object (thus defining temperature as the average kinetic energy of the object).
But applying the same reasoning to the electromagnetic emission of such a thermal object was not so successful. It had been long known that thermal objects emit light. Hot metal glows red, and upon further heating, white (this is the underlying principle of the incandescent bulb). Since light was known to be waves of electromagnetism, physicists hoped to describe this emission via classical laws. This became known as the black body problem. Since the equipartition theorem worked so well in describing the vibrational modes of the thermal object itself, it was trivial to assume that it would perform equally well in describing the radiative emission of such objects. But a problem quickly arose when determining the vibrational modes of light. To simplify the problem (by limiting the vibrational modes) a longest allowable wavelength was defined by placing the thermal object in a cavity. Any electromagnetic mode at equilibrium (i.e. any standing wave) could only exist if it used the walls of the cavities as nodes. Thus there were no waves/modes with a wavelength larger than twice the length (L) of the cavity.
Standing waves in a cavity
The first few allowable modes would therefore have wavelengths of : 2L, L, 2L/3, L/2, etc. (each successive wavelength adding one node to the wave). However, while the wavelength could never exceed 2L, there was no such limit on decreasing the wavelength, and adding nodes to reduce the wavelength could proceed ad infinitum. Suddenly it became apparent that the short wavelength modes completely dominated the distribution, since ever shorter wavelength modes could be crammed into the cavity. If each mode received an equal partition of energy, the short wavelength modes would consume all the energy. This became clear when plotting the Rayleigh–Jeans law which, while correctly predicting the intensity of long wavelength emissions, predicted infinite total energy as the intensity diverges to infinity for short wavelengths. This became known as the ultraviolet catastrophe.
The solution arrived in 1900 when Max Planck hypothesized that the frequency of light emitted by the black body depended on the frequency of the oscillator that emitted it, and the energy of these oscillators increased linearly with frequency (according to his constant h, where E = hν). This was not an unsound proposal considering that macroscopic oscillators operate similarly: when studying five simple harmonic oscillators of equal amplitude but different frequency, the oscillator with the highest frequency possesses the highest energy (though this relationship is not linear like Planck's). By demanding that high-frequency light must be emitted by an oscillator of equal frequency, and further requiring that this oscillator occupy higher energy than one of a lesser frequency, Planck avoided any catastrophe; giving an equal partition to high-frequency oscillators produced successively fewer oscillators and less emitted light. And as in the Maxwell–Boltzmann distribution, the low-frequency, low-energy oscillators were suppressed by the onslaught of thermal jiggling from higher energy oscillators, which necessarily increased their energy and frequency.
The most revolutionary aspect of Planck's treatment of the black body is that it inherently relies on an integer number of oscillators in thermal equilibrium with the electromagnetic field. These oscillators give their entire energy to the electromagnetic field, creating a quantum of light, as often as they are excited by the electromagnetic field, absorbing a quantum of light and beginning to oscillate at the corresponding frequency. Planck had intentionally created an atomic theory of the black body, but had unintentionally generated an atomic theory of light, where the black body never generates quanta of light at a given frequency with an energy less than . However, once realizing that he had quantized the electromagnetic field, he denounced particles of light as a limitation of his approximation, not a property of reality.

Photoelectric effect illuminated[edit]

Yet while Planck had solved the ultraviolet catastrophe by using atoms and a quantized electromagnetic field, most physicists immediately agreed that Planck's "light quanta" were unavoidable flaws in his model. A more complete derivation of black body radiation would produce a fully continuous, fully wave-like electromagnetic field with no quantization. However, in 1905 Albert Einstein took Planck's black body model in itself and saw a wonderful solution to another outstanding problem of the day: the photoelectric effect. Ever since the discovery of electrons eight years previously, electrons had been the thing to study in physics laboratories worldwide.
In 1902 Philipp Lenard discovered that (within the range of the experimental parameters he was using) the energy of these ejected electrons did not depend on the intensity of the incoming light, but on its frequency. So if one shines a little low-frequency light upon a metal, a few low energy electrons are ejected. If one now shines a very intense beam of low-frequency light upon the same metal, a whole slew of electrons are ejected; however they possess the same low energy, there are merely more of them. In order to get high energy electrons, one must illuminate the metal with high-frequency light. The more light there is, the more electrons are ejected. Like blackbody radiation, this was at odds with a theory invoking continuous transfer of energy between radiation and matter. However, it can still be explained using a fully classical description of light, as long as matter is quantum mechanical in nature.[6]
If one used Planck's energy quanta, and demanded that electromagnetic radiation at a given frequency could only transfer energy to matter in integer multiples of an energy quantum , then the photoelectric effect could be explained very simply. Low-frequency light only ejects low-energy electrons because each electron is excited by the absorption of a single photon. Increasing the intensity of the low-frequency light (increasing the number of photons) only increases the number of excited electrons, not their energy, because the energy of each photon remains low. Only by increasing the frequency of the light, and thus increasing the energy of the photons, can one eject electrons with higher energy. Thus, using Planck's constant h to determine the energy of the photons based upon their frequency, the energy of ejected electrons should also increase linearly with frequency; the gradient of the line being Planck's constant. These results were not confirmed until 1915, when Robert Andrews Millikan, who had previously determined the charge of the electron, produced experimental results in perfect accord with Einstein's predictions. While the energy of ejected electrons reflected Planck's constant, the existence of photons was not explicitly proven until the discovery of the photon antibunching effect, of which a modern experiment can be performed in undergraduate-level labs.[7] This phenomenon could only be explained via photons, and not through any semi-classical theory (which could alternatively explain the photoelectric effect). When Einstein received his Nobel Prize in 1921, it was not for his more difficult and mathematically laborious special and general relativity, but for the simple, yet totally revolutionary, suggestion of quantized light. Einstein's "light quanta" would not be called photons until 1925, but even in 1905 they represented the quintessential example of wave–particle duality. Electromagnetic radiation propagates following linear wave equations, but can only be emitted or absorbed as discrete elements, thus acting as a wave and a particle simultaneously.

http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality

Monday, December 23, 2013

Gravitational Field Strength


 Gravitational Field Strength


In the last lesson we were able to combine our two formulas for force due to gravity to get a new

formula.


F
g

=Fg


m
test g

=


Gm
testMe

r


2


g


=


GM
e

r


2




The great news is that the mass (shown above as the mass of the Earth “Me”) can actually be


any mass. It could be the mass of the moon, Mars, an asteroid, whatever!




In fact, we should replace mass of the Earth in the formula with just mass, since it can be any


mass.


g


=Gm


r


2


g = gravitational field strength (m/s

2)


G = Universal Gravitational Constant

m = mass of object producing the field (kg)

r = distance from centre of mass (m)




This formula lets you calculate the the gravitational field strength (the acceleration due to


gravity) caused by that mass ata specific distance from its centre.


Example 1



: The planet Mars has a mass of 6.42e23 kg and a radius (from its centre to the surface) of


3.38e6 m.

a)

Determine how much a 60.0 kg person would weigh on Mars.


b)

Compare it to his weight on Earth.


c)

Determine how heavy the 60.0 kg person would “feel” as an apparent mass in kilograms on


Mars.

a) To determine someone's weight, we first need to know the acceleration due to gravity

on Mars.


g



=Gm


r



2


g



=6.67e-11(6.42e23)


(


3.38e6)2


g



=3.748240608=3.75m/s2


7/12/2012 © studyphysics.ca Page 1 of 4 / Section 4.3

So the person's weight will be...


F

g

=mg


F

g

=60.0(3.748240608)


F

g

=224.8944365=225N


b) On Earth the person has a weight of…


F

g

=mg


F

g

=60.0(9.81)


F

g

=588.6=589N


Probably the easiest way to compare the person's weight on Mars and the Earth is to find

the ratio of the two.


F

gearth

F

gMars


=


589


225


=2.6172279=2.62


Since this is a ratio, it has no units (the Newtons canceled each other out). It simply

means that you weigh 2.62 times more on the Earth as compared to Mars.

c) The reason I asked for the person's

apparent mass is because I want to know how


heavy the person thinks he feels in kilograms. In reality,

the true mass of a person


never changes


. I am asking how heavy he feels based on the facts that he feels lighter


on Mars, and that they are used to the effects of gravity on Earth. We will take the

person's weight on Mars and the gravity of Earth to find out the apparent mass.


F

g

=mg


m



=


F

g

g

m



=224.8944365


9.81


m



=22.925019=22.9kg


Keep in mind that the person's mass is still really 60.0 kg. He just feels like he is only

22.9 kg because he is on Mars.


The Elevator Question


The concept of gravity becomes a bit more complicated when you examine a complex system like an

elevator going up and down.




It might sound strange to call an elevator complex, but it really does make a challenging


problem.




How do you think you would solve a question that asks you about your weight as an elevator


accelerates up or down?

7/12/2012 © studyphysics.ca Page 2 of 4 / Section 4.3


Going up...


When the elevator is accelerating up, what would happen to your weight?




Have you ever noticed that when an elevator first starts to move up, you feel as though you are being


pushed down a little?




This is because you can feel the elevator’s acceleration. The elevator is pushing you up, so


(according to Newton's Third Law) you push down against the floor.




It basically makes you feel a bit heavier for a moment.




A scale would show this as an increase in your weight (temporarily). If the scale shows


kilograms, it would show your


apparent mass as being bigger than your true mass.


Going down...


What would happen to your weight if the elevator started to accelerate down?




You would feel the elevator drop out underneath you.




If it really dropped out underneath you, it would feel just like being on the “Space Shot ” at


Galaxyland as it is falling … you’d feel weightless!




You feel this way because as the elevator accelerates down (away from you) it is not pushing up against


you as hard as it was before. Since the floor is not pushing as hard up against you, you are not pushing

as hard down against the floor (Newton's Third Law again).




A scale would show your weight as being less. If the scale shows kilograms, it would show your


apparent




mass as being smaller than your true mass.


Let’s look at how we would actually figure out some numbers for this type of question by looking at an example.




Keep the following in the back of your mind. A regular scale that you buy in a regular store is made to


measure things in kilograms, and it is built for Earth’s regular gravity of 9.81m/s


2. You’ll see why this is


important later.


Example 2




: You are standing on a scale in an elevator. You have a mass of 75kg. Determine what a scale would


show as your


apparent mass (in kilograms) if…


a) the elevator starts to accelerate upwards at 3.0m/s


2.


We have three parts in the formula that we will have to use:

F


NET = the overall force acting on the person causing acceleration upwards.


F


g = the force due to gravity pulling the person down.


F


N = the force of the scale on the floor pushing up against the person.


We should mostly be concerned with what the normal force is, since however hard the scale has to push

the person upwards will show up as a reading on the scale. We know the force due to gravity, since that's

just the person's weight. The net force is just overall what is happening to the person.


F

NET

=Fg+FN


ma



=mg+FN


ma



mg=FN


m



(ag)=FN


75


[3.0−(−9.81)]=FN


75


[12.81]=FN


F

N

=960.75


F

N

=9.6e2N


7/12/2012 © studyphysics.ca Page 3 of 4 / Section 4.3


We start off with a standard net force formula. We do

a quick substitution, since F


NET = ma and Fg = mg.


Next step is to move “mg” to the other side. Since

both formulas on the left side have “m” we can factor

it out. Now start putting in the numbers, watching

out for the directions of the accelerations. This lets us

calculate the normal force. As hard as the scale is

pushing the person up, the person must be pushing

down just as hard which tells us the weight the scale

will show in Newtons.


The question asked about the

apparent mass, not the weight of the person.




I can change this into a reading in kilograms by remembering that the scale we're using


has no idea what is going on... it was originally calibrated to be sitting in someone's

bathroom where gravity is a nice constant 9.81m/s

2.




This is not the true mass of the person, since mass never really changes.


F

N

=mg


m



=


F

N

g

m



=960.75


9.81


m



=97.93577982=98kg


So a regular scale shows an

apparent mass of 98 kg!


b) the elevator starts to accelerate downwards at 4.0m/s

2.


We’ll handle this part of the question the same way.


F

NET

=Fg+FN


ma



=mg+FN


ma



mg=FN


m



(ag)=FN


75


[−4.0−(−9.81)]=FN


75


[5.81]=FN


F

N

=435.75=4.4e2N


Since the scale is pushing up against the person with 4.4e2 N of force (the weight that shows on

the scale in Newtons), the person's apparent mass will be...


F

N

=mg


m



=


F

N

g

m



=435.75


9.81


m



=44.418960=44kg

So a regular scale shows your apparent mass as 44 kg.

http://www.studyphysics.ca/