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# Why do photons continue in same direction after interaction with transparent media

1. Dec 11, 2005

### Phalanx

Ok, i have recently been trying to learn a lot about light, and I have found that the reason the speed of light is slower in dense materials is because of the photon-atom interactions.

However, I have also been taught that when a photon is absorbed by an atom that an electron in that atom jumps from its current energy shell to a higher one, then jumps back down emitting a pulse of light that is the same energy as the original. My question is how does that pulse of light "know" to go in the same direction as the original as opposed to scattering randomly. I feel like my understanding of the photon/particle interaction needs to be expanded upon.

2. Dec 11, 2005

### Physics Monkey

This is a very good question that is only properly answered in a nice quantum mechanical treatment of the atom-light interaction. Here is the basic idea: imagine you have a beam of light incident upon an atom sitting somewhere. The atom can absorb a photon and change its internal energy and the rate of this absorption is proportional to the intensity of the light beam. Now, this excited atom likes to decay back down to the ground state and re-emit a photon. Where does this photon go? Well, the atom can decay in two ways: spontaneous emission or stimulated emission. Spontaneous emission is caused by vacuum fluctuations of the electromagnetic field and every excited state of the atom has a lifetime due to spontaneous emission. When the atom emits by spontaneous emission, the photon can go pretty much anywhere i.e. the light is scattered. Now, the more interesting process for your question is stimulated emission. If the incident light is intense enough, the atom really likes to emit that photon back into the beam. This is basically because photons are bosons and they like to be near their friends. In precise language, the rate of stimulated emission is proportional to incident intensity, so the more intense the beam the more likely that the photon will be emitted right back into the beam. Now, if you put a bunch of these atoms together in a solid lots of things can happen and the story will change some. However, the very basic idea remains the same, if the intensity is high enough then the photon almost always decays by stimulated emission and continues on its way. Of course, some photons will always be emitted spontaneously, this is the scattered light you can see from any transparent material.

Hope this helps.

Last edited: Dec 11, 2005
3. Dec 11, 2005

### Danger

It sure helped me. That's something that had me wondering too. Thanks.

4. Dec 12, 2005

### pervect

Staff Emeritus
This explanation (stimulated emission vs spontaneous emission) would seem to imply that very weak light (for instance, sometimes beams are created that are expected to have only one photon at a time) would always be scatterered. I don't think this happens.

5. Dec 12, 2005

### Spin_Network

Take the explanation of PM to be correct, one can ask of photons inside the universe: At any one moment, are there more photons "inside" or external to atoms?

Is there a scattering amplitude to the Whole Universe, and is this related to the CC?

6. Dec 12, 2005

### DaTario

photons also have a cross section for the interaction process, so photons may pass through the material without suffering any disturbance. Perhaps in this case, neither spontaneous nor stimulated process has happened.

DaTario

7. Dec 12, 2005

### Tide

The classical argument is that the medium is essentially a continuum of electromagnetic dipoles that radiate due to interaction with the incident wave. Interference between the incident wave and the radiated field results in a combined field propagating in the forward direction.

8. Dec 12, 2005

### ZapperZ

Staff Emeritus
Again, as Tide as alluded to, optical transmission in solids, as in this case, does not involve any "atomic transition". The band structure of the solid does not look anything close to the atomic energy levels of atoms - the atoms have, for most practical purposes, lost their individuality due to the overlapping of their valence shell.

What is more important in the optical properties of such solids is the phonon structure, and if the optical wavelength being sent can be sustained by the phonons. These "dipoles" that Tide referred to can produce optical phonons that can dictate the tranparency of a material at a particular wavelength.

Zz.

9. Dec 12, 2005

### zoobyshoe

This is how Claude Bile explained it a few months ago, in more detail, though. Someone had asked if the photon that emerges from a piece of transparent material is the same one that went in. He explained how the energy and momentum of a photon propagates through the material by turning the first atom it hits into a dipole, which causes the next to respond by becoming a dipole, and so on through the thickness of the glass. This explanation made it clear to me why it continues in the same direction.

10. Dec 12, 2005

### ZapperZ

Staff Emeritus
Ah, but there is a subtle but crucial difference between the two explanation. An isolated atom turning into a "dipole" due to such interaction does not explain why there are opaque and transparent solids. Tide stated of a continuum of dipoles. I have, in another thread, tried to explain this as a chain of ions separated by negative charges and how optical modes from such a chain can be obtain.

I think there is a very crucial, moral-of-the-story point to be brought across here, and that is that in a solid, "atoms" or the properties of individual atoms are no longer that significant in a number of common physical characteristics of a solid. You can have the identical atom content, such as carbon, but arrange them in different ways and you get two very different material (graphite and diamond). The individuality of the atoms (i.e. the atomic carbon energy levels) alone doesn't tell you the if you're going to get graphite or diamond characteristics.

Zz.

11. Dec 12, 2005

### Physics Monkey

I see I must respond to some critics of my answer, but before I do, thank you all for your valid points.

First, let me point out that I indicated that the story would change in solids. I am perfectly aware of the fact that atoms tend to loose their identities in a solid. That being said, I don't think the OP has any idea what phonons are, let alone optical phonon modes. I was trying to give a simple yet realistic explanation in keeping with the level of the OP, and in doing so I ended up giving something like an Einstein model for the transition structure. Now, it is true that in insulators (and almost all transparent materials are insulators) much of the action happens with the phonon modes, but all you have to do is replace "atomic transition" with "excitation of phonon mode" to make the simple picture fit, at least roughly. Either way, a phonon mode that emits light will do so with a rate proportional to a "phase space" factor of $$n+1$$ (the number of photons already present plus one), this is just Fermi's Golden Rule. Now, there are plenty of complications, details, even other factors that can be just as important, and so on and so forth. For practical purposes, one often doesn't even need to quantize the field, but I think this was the flavor and level of explanation that the OP was looking for.

As an aside, I should note that phonons are also not the most important player for a number of very interesting processes. In particular, many solid state lasers use the discrete levels of color centers in their operation (like the very first laser built by Maiman). These same color centers, defects in the crystal lattice, are also largely responsible for the coloration of ionic crystals. So the individuality of defects can sometimes be important too.

Last edited: Dec 12, 2005
12. Dec 12, 2005

### ZapperZ

Staff Emeritus
I was more concerned with the OP's question in reference to the thread topic. The inference to "transparent media" indicated to me that this was a question of optical transmission in something like glass, for instance, and not a "pumped" solid to produce or amplify light. This is a very common question here on PF - it appears every couple of weeks, it seems, and have been repeatedly addressed. At the simplest level, I tried to emphasize the idea that most of the optical characteristics of solids do not resemble the atomic structure of the the atoms that make up the solid. If such take-away message gets across, I consider it a success.

As you have said, beyond the simple "transparent" issue, the phenomena and explanation can become involved and complicated (example: UV absorption in glass, but not for quartz, would require a very involved explanation).

Zz.

13. Dec 12, 2005

### zoobyshoe

Yes, I see there is a whole other level to go into as to why some solids are transparent and others not.

14. Dec 12, 2005

### Staff: Mentor

Really? I didn't know that. If a photon does pass through without suffering any disturbance does it pass through at c or at the "c" of the material? I would think that it would go at the lower "c", but then it would seem like it was disturbed, so I am probably missing something.

-Dale

15. Dec 12, 2005

### Hans de Vries

Good question indeed.

This is related to issues like the "Single Photon Interference".

A picture taken with an 8 meter telescope of an extremely faint
astronomical object is typically a "one-photon-a-time" process,
nevertheless, the single photon's trajectory after the lens is
extremely well defined.

In this case, even with a single photon there is still a spread wave-
function associated to it. According to Huygens principle (cq. Feynman's
path integral principle) the sum of all paths defines a well defined wave-
front in the direction of motion of the single photon.

All the paths of the wave function are slowed down in the transparent
material. The question then of how this works leads you directly to the
problematic issue of the Interpretation of Quantum Mechanics.

There is still a lot of active research going on and photon recoil
measurements have been made only recently:

Photon recoil momentum in dispersive media:
Gretchen K. Campbell, and al. PRL 94, 170403 (2005)

http://www.aip.org/pnu/2005/split/732-1.html [Broken]
http://cua.mit.edu/ketterle_group/Projects_2005/Pubs_05/camp05.pdf

A photon inside a dielectric media, however dilute, like the BEC in the
paper above has a momentum modified by the refraction index n of the
medium. With $\lambda$ as the wavelength in vacuum we get:

$$p\ =\ \frac{nh}{\lambda}$$

Thus when a photon hits an atom, even in a very diluted gas, the
interaction of the light with all the atoms has to be taken into account,
even if the specific interaction being measured, in effect, is that of
a single atom.

Regards, Hans

Last edited by a moderator: May 2, 2017
16. Dec 12, 2005

### Spin_Network

Which then leads to the puzzling Luminocity Function associated with ..say far away faint QSO's ?

The amount of outpouring light, has a specific "pressure", defined as Luminocity. Take a single far away faint object, this is sitting within a field of Radiative energy of its own making (outpouring photons), but it is also recieving a VAST amount of Photon energy that is "opposite"..it is infalling from every Luminous object from the rest of the Universe?

As mentioned elsewhere in thread, the infalling faint light "weak-field" from the rest of the Universe, will still impose an effect that is, or can be equated to that of a 'material-like', transparent field, or Dark Energy.

This will in effect impose a 'Grip' and restrain the outpouring light source, as percieved to have an extra "restraining" halo?

Just as the Electron is WRT an Atom (electron=shroud) that interacts with incoming photons, coming from the source atoms, and from a specific direction, and via a connecting medium as explained by QM or QCD, then on a Universal scale, Galaxies as source's, if are isolated enough, will have a greater number of infalling photons, across a great light spectrum , by fact that all other Galaxies that are providing light that is infalling upon an isolated single observed object?

In a simplistic way (and my interpretation may be challenged), a far away QSO may be thought as emmiting a single source signal, whilst recieving a vast, multi-number of incedent photons?

Last edited by a moderator: May 2, 2017
17. Dec 12, 2005

### Hans de Vries

This all has very little to do with the original question.

Are you suggesting Light Pressure as Dark Energy?

Can't be, because it would accelerate differently sized objects in different
ways (acceleration = exposed-surface/mass)

Also, why should in-falling light restrain the emission of light?

Regards, Hans

18. Dec 12, 2005

### Spin_Network

For some reason, I thought the original question:"Why do Photons continue in the same direction, after interaction with transparent media"
could be applied to this:

http://en.wikipedia.org/wiki/Gunn-Peterson_trough

and thus:
http://en.wikipedia.org/wiki/Lyman_series

Seems to be that the Original Question was not meant to be Universally applied!

19. Dec 15, 2005

### piercas

I have been reading all of the above.
What Happens with a photon hitting a mirror under a certain angle ?
And by the way : can any explanation be found for speed of light being constant

20. Aug 9, 2008

### shiam

Re: Why do photons continue in same direction after interaction with transparent medi

Hi physics monkey your answer was very usefull but nat complete please explain more about interaction between photon and the matter thx

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text:
# Homework Help: Quadratics homework help

1. Jul 14, 2011

### benjaminxx12

1. The problem statement, all variables and given/known data

I am meant to look for both points when thiobject is 6m above the ground

the equation is

-5x^2+10x+2=6

my question is how do i solve it for specific points in time,
Do i just make the y term 0 and solve from there?

2. Relevant equations

3. The attempt at a solution

-5x^2 +10x -4

-5(x-1)^2-5=0

2. Jul 14, 2011

### Isak BM

I'm guessing $y$ is the vertical displacement and $x$ is some other parameter (the point or points you want to find at which $y = 6$m). Thus you need to solve for $x$ given $y = 6$m.

So all you need to do is solve the equation for $x$ like you have already started doing (either using the quadratic equation or by factorizing like you have done). You have however made a slight mistake in your factoring.

It should be

$-5{(x-1)}^2 + 1=0$

Multiply this out and you should see that this gives you the original expression.

From this it is clear that (subtract 1 and divide by -5 on both sides).

${(x-1)}^2 = \frac{1}{5}$

$x-1 = \pm \sqrt{\frac{1}{5}}$

$x = 1 \pm \sqrt{\frac{1}{5}}$

You could also have used the quadratic equation:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

To solve the general quadratic: $ax^2 + bx + c = 0$.
Thus in your case we solve for x like so...

$x = \frac{-10 \pm \sqrt{10^2-4(-5)(-4)}}{2(-5)}$

$= \frac{-10 \pm \sqrt{100-80}}{-10}$

$= \frac{-10 \pm \sqrt{20}}{-10}$

$= 1 \mp \sqrt{\frac{20}{100}}$

$= 1 \mp \sqrt{\frac{1}{5}}$

Last edited: Jul 14, 2011
3. Jul 14, 2011

### benjaminxx12

Thank you for the help, I am just curious as to where you get the +1 constant from.

You have my gratitude. :D

Last edited: Jul 14, 2011
4. Jul 14, 2011

### Isak BM

In case the +1 is still bothering you, consider first the expression that you ended up with.

$-5{(x-1)}^2-5=0$

This expression should be consistent with the original, when one multiplies out to see if it does agree with the initial equation one gets.

$-5x^2 -5(-2x) -5(1) -5= 0$

$-5x^2 +10x -5 -5= 0$

We see now that in order to get -4 and not -10 we need to replace the -5 with a +1.

format: plain


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# Math Help - MATLAB Hyperbolic Paraboloid

1. ## MATLAB Hyperbolic Paraboloid

Hi,

I am trying to create a hyperbolic paraboloid much like the one I have below, but instead of having a square region as shown, I would like to have an ellipse (to give an effect similar to the roof shape on the Calgary Olympic stadium).

My problem is that I have only used the "meshgrid" command in matlab to generate square/rectangular regions so I am unsure how to do this. I think I need to find an appropriate parametrization of an ellipse in order to produce the result I am after. I can see that if I wanted a circle I would use a parametrization of radius and the angle theta but for an ellipse where the radius is not constant I am pretty confused on how to achieve this.

Any help would be greatly appreciated,

Regards Elbarto

Code for images below

Code:
%Hyperbolic paraboloid
x = [-10:.5:10];
y=[-10:.5:10];
[X, Y] = meshgrid(x,y);
Z = X.^2-Y.^2;
surf(X,Y,Z)
xlabel x
ylabel y
zlabel z

2. Originally Posted by elbarto
Hi,

I am trying to create a hyperbolic paraboloid much like the one I have below, but instead of having a square region as shown, I would like to have an ellipse (to give an effect similar to the roof shape on the Calgary Olympic stadium).

My problem is that I have only used the "meshgrid" command in matlab to generate square/rectangular regions so I am unsure how to do this. I think I need to find an appropriate parametrization of an ellipse in order to produce the result I am after. I can see that if I wanted a circle I would use a parametrization of radius and the angle theta but for an ellipse where the radius is not constant I am pretty confused on how to achieve this.

Any help would be greatly appreciated,

Regards Elbarto

Code for images below

Code:
%Hyperbolic paraboloid
x = [-10:.5:10];
y=[-10:.5:10];
[X, Y] = meshgrid(x,y);
Z = X.^2-Y.^2;
surf(X,Y,Z)
xlabel x
ylabel y
zlabel z
I'm not sitting at a machine with Matlab at the moment so I don't know if M. works this way but:

Code:
dr=0.2;
r=dr/2:dr:10;
dtheta=2*pi/20;
theta=0:dtheta:dtheta*20;

x=kron(r,cos(theta));
y=kron(r,sin(theta));

[X, Y] = meshgrid(x,y);
Z = X.^2-Y.^2;
surf(X,Y,Z)
.

3. Thanks for the reply Constatine11.

Unfortunately the code you provided also produced a plot over a square domain. Does anyone know if MATLAB is even capable of plotting surfaces over a non square / rectangular domain?

4. ## Non rectangular domain in Matlab

%================================================= =========================
%"I am trying to create a hyperbolic paraboloid much like the one I have
% below, but instead of having a square region as shown, I would like to
% have an ellipse (to give an effect similar to the roof shape on the
% My problem is that I have only used the "meshgrid" command in matlab
% to generate square/rectangular regions so I am unsure how to do this.
% I think I need to find an appropriate parametrization of an ellipse
% in order to produce the result I am after. I can see that if I wanted
% a circle I would use a parametrization of radius and the angle theta
% but for an ellipse where the radius is not constant I am pretty confused
% on how to achieve this."
% Any help would be greatly appreciated,
%
% Regards Elbarto
%
% Prof. Diomar Cesar Lobao PhD.
% www.professores.uff.br/diomar_cesar_lobao
%================================================= =========================
%
clear all; clc; close all;
%
%Hyperbolic paraboloid
figure(1)
%
a = 2.0; % major axis
b = 1.0; % minor axis
x = [-10:0.015:10];
y=[-10:0.015:10];
[X, Y] = meshgrid(x,y);
Z = X.^2-Y.^2;
surf(X,Y,Z)
xlabel x
ylabel y
zlabel z
%
figure(2)
%
%
[I,J] = find((X.^2/a^2 + Y.^2/b^2) <= 1); % Trick here!!!
%
for k=1:size(I,1)
tmy(I(k),J(k))=X(I(k),J(k));
tmz(I(k),J(k))=Y(I(k),J(k));
tmu(I(k),J(k))=(tmy(I(k),J(k))^2 - tmz(I(k),J(k))^2);
end
end
%
%
surf(tmy,tmz,tmu);
%


text:
# 8th Class Science Force and Pressure NCERT Summary - Pressure

NCERT Summary - Pressure

Category : 8th Class

Pressure

• Pressure is defined as the force acting normally per unit area.
• Pressure on an area, A is the area of the surface on contact and F is the normal force applied.
• Pressure is a scalar quantity.

$P\,=\,\left( F/A \right)$

• If the force is due to the weight (W) of the object, the equation is then:

$P\,=\,W\,/\,A$

• Hydrostatic pressure is the pressure that is executed by a fluid at equilibrium at a given point within the fluid, due to the force of gravity.
• Hydrostatic pressure equation is P = hdg where P = pressure $(N/{{M}^{2}}\,or\,Pa)$, h =height (m), d = density$(kg/{{m}^{3}})$, g = acceleration due to gravity $(9.81\text{ }m/{{s}^{2}})$
• The SI unit for pressure is the pascal. Pa, where $1\,Pa=1\,N/{{m}^{2}}.$
• The standard atmospheric pressure is measured in various units.

1 Atmosphere = 760 mmHg = 101.3 kpa

• The mercury barometer and other manometer devices are reliable pressure measurement devices. A "rising" barometer indicates increasing air pressure i.e. calm weather, a "falling" barometer indicates decreasing air pressure, i.e., inclement weather or harsh weather.
• Factors Affecting the Magnitude of Pressure
• Magnitude of the force: The larger the force, the higher the pressure.
• Contact area: The larger the contact area, the lower the pressure.
• The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation.

Pascal's Principle

• Pascal's Principle is used to quantitatively relate the pressure at two points in an incompressible static fluid.
• It states that pressure is transmitted, undiminished, in a closed static fluid.
• The total pressure at any point within an incompressible, static fluid is equal to the sum of the applied pressure at any point in that fluid and the hydrostatic pressure changes due to a difference in height within that fluid.

Applications

• The underlying principle of the hydraulic pressure.
• Used for amplifying the force of the driver's foot in the braking system of most cars and trucks.
• Used in artesian wells, water towers, and dams.
• Scuba divers must understand this principle. At a depth of 10 metres under water, pressure is twice the atmospheric pressure at sea level, and increases by about 105 kPa for each increase of 10 m depth.

Surface Tension (T)

• Surface Tension (T): It is the property of a liquid by virtue of which it behaves like an elastically stretched membrane with a tendency to contract so as to occupy a minimum surface area.
• Mathematically $T=\frac{F}{L}$

F = Force per unit length

L = Length over which force acts

• I. Unit is : $N{{m}^{-\,1}}$ and Dimensional formula is ${{M}^{1}}{{L}^{0}}{{T}^{-\,2}}$

Surface Energy

• The energy responsible for the phenomenon of surface tension may be thought of as approximately equivalent to the work or energy required to remove the surface layer of molecules in a unit area. The potential energy per unit area of the surface film is called the surface energy.
• Surface tension is numerically equal to surface energy. Its unit is N/m.
• An increase in temperature lowers the net force of attraction among molecules and hence decreases surface tension.
• The Surface of Water: Surface tension creates a thin elastic film on the surface of water.
• Surface Tension at the Molecular Level: A water molecule has one oxygen atom (0) and two hydrogen atoms (H). The negative end of the water molecule is attracted to the positive end of other water molecules. Under the surface of water these attractions balance out, but on the surface they are unbalanced. This is because the molecules on the surface have one surface exposed to the air and no other water molecules are there to balance out the charges. So, the surface molecules are pulled inward and surface tension is created.
• Breaking up Surface Tension: Surface tension can be broken by adding soap to water. Soap disrupts the surface tension causing the thin film to split and fly apart.
• Adhesive Forces: Forces of attraction between a liquid and a solid surface are called adhesive forces. Force of attraction between like molecules of surface are called the cohesive forces. The difference in strength between cohesive forces and adhesive forces determines the behaviour of a liquid in contact with a solid surface.
• Water does not wet waxed surfaces because the cohesive forces within the drops are stronger than the adhesive forces between the drops and the wax.
• Water wets glass and spreads out on it because the adhesive forces between the liquid and the glass are stronger than the cohesive forces within the water.

Capillary Action

• Capillary Action can be defined as the ability of liquid to flow through slim tube, cylinder or permeable substance due to adhesive and cohesive forces interacting between the liquid and surface.
• The liquid creeps up the inside of the tube (as a result of adhesive forces between the liquid and the inner walls of the tube) until the adhesive and cohesive forces of the liquid are balanced by the weight of the liquid.
• The smaller the diameter of the tube, the higher the liquid rises.

Buoyancy

• It is an upward force exerted by a fluid that opposes the weight of an immersed object.

Archimedes' Principle

• It states that when a body is immersed completely or partly in a fluid, it loses weight equal to the weight of the fluid displaced by it.
• Apparent weight of body = Actual Weight of Body – Upthrust
• Buoyant force depends on the density of the fluid and not on the density of body acting on centre of gravity of fluid.

Archimedes' Law of Floatation

• When a body is immersed in a fluid then:
• The weight W of body is more than the upthrust W’ (W > W’), the body will sink.
• The weight W of the body is equal to upthrust W’ (W = W), the body floats with whole of its volume inside the liquid.
• The weight (W < W’), the body will float with some of its part outside the liquid.
• When a block of ice floats in a liquid having density greater than that of water, the liquid level rises when all the ice melts into water.
• When an ice block floats in a liquid of density less than that of water, the liquid level falls when all the ice melts into water.
• When an ice block floats in water, the water level will remain the same when all the ice melts into water.

Equilibrium of Floating Body

1. Stable equilibrium
2. Unstable equilibrium
3. Neutral equilibrium

Stable equilibrium: When a body is given a small angular displacement, i.e. it is tilted slightly by some external force and then it returns back to original position due to internal forces. Such equilibrium is called stable equilibrium.

Unstable equilibrium: If a body does not return to its original position from the slightly displaced angular position and moves farther away when given a small angular displacement, such equilibrium is called an unstable equilibrium.

Neutral equilibrium: The body remains at rest in any position to which it may be displaced, no net force tends to return the body to its original state or to drive it further away from the original position, is called neutral equilibrium.

Metacentre: When a small angular displacement is given to a body floating in a liquid, it starts oscillation about some point M. This point about which the body starts oscillating is called the metacentre.

Metacentric height: The distance between the centre of gravity of a floating body and the metacentre, i.e. distance GM is called meta-centric height. Relation between centre of gravity and metacentre in three different types of equilibrium:

• Stable equilibrium: In this equilibrium, the position of metacenter (M) remains higher than the centre of gravity of body.
• Unstable equilibrium: In this equilibrium, the position of metacenter (M) remains lower than the centre of gravity of body.
• Neutral equilibrium: In this equilibrium, the position of metacenter (M) coincides with the centre of gravity of body.

#### Other Topics

##### Notes - Pressure

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Children of PostProcessing in Documentation Web

#### Results from Documentation web retrieved at 00:52 (GMT)

Examples * distance between origin and node coordinates label 'nodeDistance' formula 'np.linalg.norm( #node# )'
Example Based on the example file at $DAMASK_ROOT/examples/SpectralMethod/Polycrystal/20grains64x64x64.geom the following command creates all three features using... AddIndexed Examples * add squared values Suppose we have a reference ASCIItable containing data to which we want to add the squared values of column y . 2 head # r... AddLinked Examples * add squared values Suppose we have a reference ASCIItable containing data to which we want to add the squared values of column y . 2 head # r... AddStrainTensors Note the »material stretch tensor«$ \tnsr U $following from the »right Cauchy–Green deformation tensor«:$ \tnsr C = \tnsr F^\text T\tnsr F = \tnsr U^\text ...
AverageDown
Details %BEGINFIGURE{ label="fig:packingStencil" caption="Packing stencil concept exemplified in two dimensions." }% For a periodic and regular grid stored in an ...
FilterTable
Examples * get every second row of the ASCII table listing only $F_{11}$, $F_{12}$, and $F_{13}$ components of deformation gradient and all (nine) $P_{ij}$ compo...
GroupTable
Example 1 head id 1_vector 2_vector 3_vector 12 1.0 2.0 3.0 2 4.0 4.0 4.0 12 4.0 6.0 1.0 label id function np.sum 1 head id ...
ImageData
Example Single data column Suppose we have an ASCIItable that contains a column of two dimensional data as in type below. The additional columns x and ...
MentatColorMap
mentat_colorMap #TableOfContents Purpose Changes the color map in MSC.Mentat. Interpolates colors between "lower_hls" and "upper_hls". Usage mentat_colorMap...
PerceptualUniformColorMap
References \$ 1 : K. Moreland Diverging Color Maps for Scientific Visualization Proc. 5th Int. Symp. Visual Computing (2009)
PermuteData
Example Suppose we have an ASCIItable 1 head x type 1 1 2 1 3 1 4 2 5 3 6 2 Independent shuffling With permuteData label type we may get 1 head x type 1 3 2 ...
PostResults
postResults #TableOfContents Purpose Extract data from a simulation result file into an ASCII table. Background Supported file formats * MSC.Marc (*.t16) ...
ReLabel
Examples * add suffix green to divergence of first PiolaKirchhoff stress (vector) with label 'div(p)' substitute 'div(p)_green' This results in '1_d...
ShowTable
showTable #TableOfContents Purpose Show structural components of an ASCII table. Operates on given file(s) or STDIN STDOUT. Usage showTable options file(s)...


text:
# Complex: Argument and log

M. next
Hello!

So I have questions on this equivalence:

Imlog[(1+x)/(1-x)] = arg [(1+x)/(1-x)] where x: complex number

How is this true? Is it always applicable no matter what form of complex function is under calculation?

Thank you.

M. next
Yes, indeed. Excuse me for the late reply.

Homework Helper
Note, a complex number z can always be written in the form z=a+ib: a,b are real.

M. next
Yes?

Staff Emeritus
Gold Member
If I ask you for the imaginary part of log(z) can you tell me what it is?

From there it should be fairly obvious what the imaginary part of log(f(z)) is in terms of f(z).

Homework Helper
I don't see that the "(1+ x)/(1- x)" is really relevant. If z is any complex number, $z= re^{i\theta}$ where "$\theta$" is the "argument" of z. Then $log(z)= log(re^{ix\theta})= log(r)+ i \theta$. That is, $Im(log(z))= \theta$, the argument of z.

Homework Helper
Yes?


text:
lilypond-user
[Top][All Lists]

Re: numbering collision

 From: Heikki Johannes Junes Subject: Re: numbering collision Date: Sat, 23 Feb 2002 14:00:33 +0200 (EET)

> \times 2/3 {a,8^5 c^""^""^1 b'8^2}

With 1.5.32, at least, I have to use ^#'(finger "1") instead

\times 2/3 {a8^1 c^""^""^""^#'(finger "1") b,8^2}

It looks now much better!

> BTW, there is not now nor has there ever been anything wrong with
> fingering of any sort being on staff lines or before notes.  It is
> unfortunate that the only "right way" of fingering in lilypond is
> usually not to do it at all.

My case was an exception. If there are a lot of tuplets, for clarity, one
may want to turn off the tuplet numbering, as shown in the manual:
http://lilypond.org/development/Documentation/user/out-www/lilypond/Tuplets.html
Then there are no number collisions. But if there are only a few tuplets,
then tuplet numbering is necessary and the collisions can be avoided
the way described above.

This was resolved quickly. Thank You!

--
Heikki Junes


text:
# Find the ∫sec^3 xdx?

Find the integral of ∫sec^3 xdx

I need the steps used and answer please. I'll give 10 points to the best answer!

### 1 Answer

Relevance
• J D
Lv 5
1 decade ago
Favorite Answer

∫sec^3 x dx

∫sec x sec^2 x dx

let u = sec x; dv = sec^2 x dx

du = sec x tan x; v = tan x

sec x tan x - ∫tan x sec x tan x dx

sec x tan x - ∫sec x tan^2 x dx

sec x tan x - ∫sec x (sec^2 x - 1) dx

so at this point we have...

∫sec^3 x dx = sec x tan x - ∫sec^3 x dx - ∫sec x dx

2∫sec^3 x dx = sec x tan x - ∫sec x dx

2∫sec^3 x dx = sec x tan x - ln|sec x + tan x| + c

∫sec^3 x dx = 1/2 sec x tan x - 1/2 ln|sec x + tan x| + c

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text:
In the visual arts, color theory is the body of practical guidance for color mixing and the visual effects of a specific color combination. Color terminology based on the color wheel and its geometry separates colors into primary color, secondary color, and tertiary color. The understanding of color theory dates to antiquity. Aristotle (d. 322 BCE) and Claudius Ptolemy (d. 168 CE) already discussed which and how colors can be produced by mixing other colors. The influence of light on color was investigated and revealed further by al-Kindi (d. 873) and Ibn al-Haytham (d.1039). Ibn Sina (d. 1037), Nasir al-Din al-Tusi (d. 1274), and Robert Grosseteste (d. 1253) discovered that contrary to the teachings of Aristotle, there are multiple color paths to get from black to white.[1][2] More modern approaches to color theory principles can be found in the writings of Leone Battista Alberti (c. 1435) and the notebooks of Leonardo da Vinci (c. 1490). A formalization of "color theory" began in the 18th century, initially within a partisan controversy over Isaac Newton's theory of color (Opticks, 1704) and the nature of primary colors. From there it developed as an independent artistic tradition with only superficial reference to colorimetry and vision science.[citation needed]

Classifications

This section needs expansion. You can help by adding to it. (September 2022)

Colors can be classified as:

1. Warm and cold
3. Positive and negative

Color abstractions

Additive color mixing (such as in a computer)
Subtractive color mixing (such as in a printer)

The foundations of pre-20th-century color theory were built around "pure" or ideal colors, characterized by different sensory experiences rather than attributes of the physical world. This has led to several inaccuracies in traditional color theory principles that are not always remedied in modern formulations.[3]

Another issue has been the tendency to describe color effects holistically or categorically, for example as a contrast between "yellow" and "blue" conceived as generic colors, when most color effects are due to contrasts on three relative attributes which define all colors:

1. Value (light vs. dark, or white vs. black),
2. Chroma [saturation, purity, strength, intensity] (intense vs. dull), and
3. Hue (e.g. the name of the color family: red, yellow, green, cyan, blue, magenta).

The visual impact of "yellow" vs. "blue" hues in visual design depends on the relative lightness and saturation of the hues.

These confusions are partly historical and arose in scientific uncertainty about the color perception that was not resolved until the late 19th-century when the artistic notions were already entrenched. They also arise from the attempt to describe the highly contextual and flexible behavior of color perception in terms of abstract color sensations that can be generated equivalently by any visual media.[citation needed]

Many historical "color theorists" have assumed that three "pure" primary colors can mix into all possible colors, and any failure of specific paints or inks to match this ideal performance is due to the impurity or imperfection of the colorants. In reality, only imaginary "primary colors" used in colorimetry can "mix" or quantify all visible (perceptually possible) colors; but to do this, these imaginary primaries are defined as lying outside the range of visible colors; i.e., they cannot be seen. Any three real "primary" colors of light, paint or ink can mix only a limited range of colors, called a gamut, which is always smaller (contains fewer colors) than the full range of colors humans can perceive.[4]

Historical background

Color theory was originally formulated in terms of three "primary" or "primitive" colors—red, yellow and blue (RYB)—because these colors were believed capable of mixing all other colors.[5]

Goethe's color wheel from his 1810 Theory of Colours

The RYB primary colors became the foundation of 18th-century theories of color vision,[citation needed] as the fundamental sensory qualities that are blended in the perception of all physical colors, and conversely, in the physical mixture of pigments or dyes. These theories were enhanced by 18th-century investigations of a variety of purely psychological color effects, in particular the contrast between "complementary" or opposing hues that are produced by color afterimages and in the contrasting shadows in colored light. These ideas and many personal color observations were summarized in two founding documents in color theory: the Theory of Colours (1810) by the German poet Johann Wolfgang von Goethe, and The Law of Simultaneous Color Contrast (1839) by the French industrial chemist Michel Eugène Chevreul. Charles Hayter published A New Practical Treatise on the Three Primitive Colours Assumed as a Perfect System of Rudimentary Information (London 1826), in which he described how all colors could be obtained from just three.

Page from 1826 A New Practical Treatise on the Three Primitive Colours Assumed as a Perfect System of Rudimentary Information by Charles Hayter

Subsequently, German and English scientists established in the late 19th century that color perception is best described in terms of a different set of primary colors—red, green and blue-violet (RGB)—modeled through the additive mixture of three monochromatic lights. Subsequent research anchored these primary colors in the differing responses to light by three types of color receptors or cones in the retina (trichromacy). On this basis the quantitative description of the color mixture or colorimetry developed in the early 20th century, along with a series of increasingly sophisticated models of color space and color perception, such as the opponent process theory.

Across the same period, industrial chemistry radically expanded the color range of lightfast synthetic pigments, allowing for substantially improved saturation in color mixtures of dyes, paints, and inks. It also created the dyes and chemical processes necessary for color photography. As a result, three-color printing became aesthetically and economically feasible in mass printed media, and the artists' color theory was adapted to primary colors most effective in inks or photographic dyes: cyan, magenta, and yellow (CMY). (In printing, dark colors are supplemented by black ink, known as the CMYK system; in both printing and photography, white is provided by the color of the paper.) These CMY primary colors were reconciled with the RGB primaries, and subtractive color mixing with additive color mixing, by defining the CMY primaries as substances that absorbed only one of the retinal primary colors: cyan absorbs only red (−R+G+B), magenta only green (+R−G+B), and yellow only blue-violet (+R+G−B). It is important to add that the CMYK, or process, color printing is meant as an economical way of producing a wide range of colors for printing, but is deficient in reproducing certain colors, notably orange and slightly deficient in reproducing purples. A wider range of colors can be obtained with the addition of other colors to the printing process, such as in Pantone's Hexachrome printing ink system (six colors), among others.

Munsell's 1905 color system represents colors using three color-making attributes, value (lightness), chroma, and hue.

For much of the 19th-century artistic color theory either lagged behind scientific understanding or was augmented by science books written for the lay public, in particular Modern Chromatics (1879) by the American physicist Ogden Rood, and early color atlases developed by Albert Munsell (Munsell Book of Color, 1915, see Munsell color system) and Wilhelm Ostwald (Color Atlas, 1919). Major advances were made in the early 20th century by artists teaching or associated with the German Bauhaus, in particular Wassily Kandinsky, Johannes Itten, Faber Birren and Josef Albers, whose writings mix speculation with an empirical or demonstration-based study of color design principles.

Complementary colors

Chevreul's 1855 "chromatic diagram" based on the RYB color model, showing complementary colors and other relationships
 Main article: Complementary colors

For the mixing of colored light, Isaac Newton's color wheel is often used to describe complementary colors, which are colors that cancel each other's hue to produce an achromatic (white, gray or black) light mixture. Newton offered as a conjecture that colors exactly opposite one another on the hue circle cancel out each other's hue; this concept was demonstrated more thoroughly in the 19th century. An example of complementary colors would be red and green[5]

A key assumption in Newton's hue circle was that the "fiery" or maximum saturated hues are located on the outer circumference of the circle, while achromatic white is at the center. Then the saturation of the mixture of two spectral hues was predicted by the straight line between them; the mixture of three colors was predicted by the "center of gravity" or centroid of three triangle points, and so on.

Primary, secondary, and tertiary colors of the RYB color model

According to traditional color theory based on subtractive primary colors and the RYB color model, yellow mixed with purple, orange mixed with blue, or red mixed with green produces an equivalent gray and are the painter's complementary colors. These contrasts form the basis of Chevreul's law of color contrast: colors that appear together will be altered as if mixed with the complementary color of the other color. A piece of yellow fabric placed on a blue background will appear tinted orange because orange is the complementary color to blue.

However, when complementary colors are chosen based on the definition by light mixture, they are not the same as the artists' primary colors. This discrepancy becomes important when color theory is applied across media. Digital color management uses a hue circle defined according to additive primary colors (the RGB color model), as the colors in a computer monitor are additive mixtures of light, not subtractive mixtures of paints.

One reason the artist's primary colors work at all is due to the imperfect pigments being used have sloped absorption curves and change color with concentration. A pigment that is pure red at high concentrations can behave more like magenta at low concentrations. This allows it to make purples that would otherwise be impossible. Likewise, a blue that is ultramarine at high concentrations appears cyan at low concentrations, allowing it to be used to mix green. Chromium red pigments can appear orange, and then yellow, as the concentration is reduced. It is even possible to mix very low concentrations of the blue mentioned and the chromium red to get a greenish color. This works much better with oil colors than it does with watercolors and dyes.

The old primaries depend on sloped absorption curves and pigment leakages to work, while newer scientifically derived ones depend solely on controlling the amount of absorption in certain parts of the spectrum.

Another reason the correct primary colors were not used by early artists is they were not available as durable pigments. Modern methods in chemistry were needed to produce them.

Warm vs. cool colors

The distinction between "warm" and "cool" colors has been important since at least the late 18th century.[6] The difference (as traced by etymologies in the Oxford English Dictionary), seems related to the observed contrast in landscape light, between the "warm" colors associated with daylight or sunset, and the "cool" colors associated with a gray or overcast day. Warm colors are often said to be hues from red through yellow, browns, and tans included; cool colors are often said to be the hues from blue-green through blue violet, most grays included. There is a historical disagreement about the colors that anchor the polarity, but 19th-century sources put the peak contrast between red-orange and greenish-blue.

Color theory has described perceptual and psychological effects to this contrast. Warm colors are said to advance or appear more active in a painting, while cool colors tend to recede; used in interior design or fashion, warm colors are said to arouse or stimulate the viewer, while cool colors calm and relax.[7] Most of these effects, to the extent they are real, can be attributed to the higher saturation and lighter value of warm pigments in contrast to cool pigments; brown is a dark, unsaturated warm color that few people think of as visually active or psychologically arousing.

The traditional warm/cool association of a color is reversed relative to the color temperature of a theoretical radiating black body; the hottest stars radiate blue (cool) light, and the coolest radiate red (warm) light.

The hottest radiating bodies (e.g. stars) have a "cool" color, while the less hot bodies radiate with a "warm" color. (image is in Kelvin scale)
Doppler redshift for receding and blueshift for advancing

This contrast is further seen in the psychological associations of colors with the Doppler effect seen in astronomical objects. Traditional psychological associations, where warm colors are associated with advancing objects and cool colors with receding objects, are directly opposite those seen in astrophysics, where stars or galaxies moving towards our viewpoint on Earth are blueshifted (advancing) and stars or galaxies moving away from Earth are redshifted (receding).

Achromatic colors

Any color that lacks strong chromatic content is said to be unsaturated, achromatic, near-neutral, or neutral. Near neutrals include browns, tans, pastels, and darker colors. Near neutrals can be of any hue or lightness. Pure achromatic, or neutral colors include black, white and all grays.

Near neutrals are obtained by mixing pure colors with white, black or grey, or by mixing two complementary colors. In color theory, neutral colors are easily modified by adjacent more saturated colors, and they appear to take on the hue complementary to the saturated color; e.g., next to a bright red couch, a gray wall will appear distinctly greenish, this is a property of human vision.

Black and white have long been known to combine "well" with almost any other colors; black decreases the apparent saturation or brightness of colors paired with it and white shows off all hues to equal effect.[citation needed]

When mixing colored light (additive color models), the achromatic mixture of spectrally balanced red, green, and blue (RGB) is always white, not gray or black. When we mix colorants, such as the pigments in paint mixtures, a color is produced which is always darker and lower in chroma, or saturation, than the parent colors. This moves the mixed color toward a neutral color—a gray or near-black. Lights are made brighter or dimmer by adjusting their brightness, or energy level; in painting, lightness is adjusted through mixture with white, black, or a color's complement.

It is common among some painters to darken a paint color by adding black paint—producing colors called shades—or lighten a color by adding white—producing colors called tints. However, it is not always the best way for representational painting, as an unfortunate result is for colors to also shift in hue. For instance, darkening a color by adding black can cause colors such as yellows, reds, and oranges, to shift toward the greenish or bluish part of the spectrum. Lightening a color by adding white can cause a shift towards blue when mixed with reds and oranges. Another practice when darkening a color is to use its opposite, or complementary, color (e.g. purplish-red added to yellowish-green) in order to neutralize it without a shift in hue and darken it if the additive color is darker than the parent color. When lightening a color this hue shift can be corrected with the addition of a small amount of an adjacent color to bring the hue of the mixture back in line with the parent color (e.g. adding a small amount of orange to a mixture of red and white will correct the tendency of this mixture to shift slightly towards the blue end of the spectrum).

Split primary system

In painting and other visual arts, two-dimensional color wheels or three-dimensional color solids are used as tools to teach beginners the essential relationships between colors. The organization of colors in a particular color model depends on the purpose of that model: some models show relationships based on human color perception, whereas others are based on the color mixing properties of a particular medium such as a computer display or set of paints.

The split-primary palette, also called color-bias theory, is a color-wheel model that attempts to explain, and to compensate for, the unsatisfactory results sometimes produced when mixing the traditional primary colors, red, yellow, and blue.

Painters have long considered red, yellow, and blue to be primary colors. In practice, however, many of the mixtures produced from these colors lack chromatic intensity. Rather than adopting a more satisfactory set of primary colors, proponents of split-primary theory explain this lack of chroma by the purported presence of chemical impurities, small amounts of other colors, in the paints, or biases away from the ideal primary toward one or the other of the adjacent colors. Every red paint, for example, is said to be tainted with, or biased toward, either blue or yellow, every blue paint toward either red or green, and every yellow toward either green or orange. These biases are thought to result in mixtures that contain sets of complementary colors, darkening the resulting color. In order to obtain vivid mixed colors, according to split-primary theory, it is necessary to employ the primary colors whose biases both fall in the direction, on the color wheel, of the color to be mixed, combining, for example, green-biased blue with green-biased yellow to make bright green.

Though widely accepted and taught, the split-primary system is based on false assumptions. The perceived bias of colors is not due to chemical impurities. Rather, the color of any given colorant is inherent to its chemical and physical properties, the purity of such a chemical being unrelated to whether it conforms to our arbitrary conception of an ideal hue. Moreover, the identity of gamut-optimizing primary colors is determined by human color vision. Although no set of three primaries can be mixed to obtain the complete color gamut perceived by humans, red, yellow, and blue are a poor choice if high chroma mixtures are desired. From their incorrect premises, proponents of split-primary theory conclude that extra colors are needed in order to mix a wide gamut of high-chroma colors, an idea belied by the longtime success of three-color photographic printing.

Although flawed in principle, the split-primary system can be successful in practice, because the recommended blue-biased red and green-biased blue positions are often filled by near approximations of magenta and cyan, while orange-biased red and violet-biased blue serve as secondary colors, which may widen the gamut even further.

This system is still popular among contemporary painters,[citation needed] as it is basically a simplified version of Newton's geometrical rule that colors closer together on the hue circle will produce more vibrant mixtures. However, with the range of contemporary paints available, many artists simply add more paints to their palette as desired for a variety of practical reasons. For example, they may add a scarlet, purple and/or green paint to expand the mixable gamut; and they include one or more dark colors (especially "earth" colors such as yellow ochre or burnt sienna) simply because they are convenient to have premixed.[citation needed] Printers commonly augment a CMYK palette with spot (trademark specific) ink colors.

Color harmony

Georg Christoph Lichtenberg. Göttingen, 1775, plate III.
Ignaz Schiffermüller, Versuch eines Farbensystems (Vienna, 1772), plate I.

It has been suggested that "Colors seen together to produce a pleasing affective response are said to be in harmony".[8] However, color harmony is a complex notion because human responses to color are both affective and cognitive, involving emotional response and judgment. Hence, our responses to color and the notion of color harmony is open to the influence of a range of different factors. These factors include individual differences (such as age, gender, personal preference, affective state, etc.) as well as cultural, sub-cultural, and socially-based differences which gives rise to conditioning and learned responses about color. In addition, context always has an influence on responses about color and the notion of color harmony, and this concept is also influenced by temporal factors (such as changing trends) and perceptual factors (such as simultaneous contrast) which may impinge on human response to color. The following conceptual model illustrates this 21st-century approach to color harmony:

${\displaystyle {\text{Color harmony))=f(\operatorname {Col} 1,2,3,\dots ,n)\cdot (ID+CE+CX+P+T)}$

wherein color harmony is a function (f) of the interaction between color/s (Col 1, 2, 3, …, n) and the factors that influence positive aesthetic response to color: individual differences (ID) such as age, gender, personality and affective state; cultural experiences (CE), the prevailing context (CX) which includes setting and ambient lighting; intervening perceptual effects (P) and the effects of time (T) in terms of prevailing social trends.[9]

In addition, given that humans can perceive over 2.8 million different colors,[10] it has been suggested that the number of possible color combinations is virtually infinite thereby implying that predictive color harmony formulae are fundamentally unsound.[11] Despite this, many color theorists have devised formulae, principles or guidelines for color combination with the aim being to predict or specify positive aesthetic response or "color harmony".

Color wheel models have often been used as a basis for color combination principles or guidelines and for defining relationships between colors. Some theorists and artists believe juxtapositions of complementary color will produce strong contrast, a sense of visual tension as well as "color harmony"; while others believe juxtapositions of analogous colors will elicit a positive aesthetic response. Color combination guidelines (or formulas) suggest that colors next to each other on the color wheel model (analogous colors) tend to produce a single-hued or monochromatic color experience and some theorists also refer to these as "simple harmonies".[12]

In addition, split complementary color schemes usually depict a modified complementary pair, with instead of the "true" second color being chosen, a range of analogous hues around it are chosen, i.e. the split complements of red are blue-green and yellow-green. A triadic color scheme adopts any three colors approximately equidistant around a color wheel model. Feisner and Mahnke are among a number of authors who provide color combination guidelines in greater detail.[13][14]

Color combination formulae and principles may provide some guidance but have limited practical application. This is due to the influence of contextual, perceptual, and temporal factors which will influence how color/s are perceived in any given situation, setting, or context. Such formulae and principles may be useful in fashion, interior and graphic design, but much depends on the tastes, lifestyle, and cultural norms of the viewer or consumer.

As early as the ancient Greek philosophers, many theorists have devised color associations and linked particular connotative meanings to specific colors.[15] However, connotative color associations and color symbolism tends to be culture-bound and may also vary across different contexts and circumstances. For example, red has many different connotative and symbolic meanings from exciting, arousing, sensual, romantic, and feminine; to a symbol of good luck; and also acts as a signal of danger. Such color associations tend to be learned and do not necessarily hold irrespective of individual and cultural differences or contextual, temporal or perceptual factors.[16] It is important to note that while color symbolism and color associations exist, their existence does not provide evidential support for color psychology or claims that color has therapeutic properties.[17]

Monochromatic

The monochromatic formula chooses only one color (or hue). Variations of the color are created by changing the value and saturation of the color. Since only one hue is used, the color and its variations are guaranteed to work.

Current status

Color theory has not developed an explicit explanation of how specific media affect color appearance: colors have always been defined in the abstract, and whether the colors were inks or paints, oils or watercolors, transparencies or reflecting prints, computer displays or movie theaters, was not considered especially relevant.[18] Josef Albers investigated the effects of relative contrast and color saturation on the illusion of transparency, but this is an exception to the rule.[19]

References

1. ^ Smithson, H.E.; Dinkova-Bruun, G.; Gasper, G.E.M.; Huxtable, M.; McLeish, T.C.B.; Panti, C.P. (2012). "A three-dimensional color space from the 13th century". J. Opt. Soc. Am. A. 29 (2): A346–A352. Bibcode:2012JOSAA..29A.346S. doi:10.1364/josaa.29.00A346. PMC 3287286. PMID 22330399.
2. ^ Kirchner, E. (2013). "Color theory and color order in medieval Islam: A review". Color Research & Application. 40 (1): 5-16. doi:10.1002/col.21861.
3. ^ "handprint : colormaking attributes". www.handprint.com. Retrieved 2021-07-31.
4. ^ "Traditional and Modern Colour Theory Part 1: Modern Colour Theory". Retrieved 2021-10-15.
5. ^ a b "The History of Color Theory: Must-Know Facts for Creatives - Pigment Pool". 30 July 2021. Retrieved 2021-07-31.
6. ^ "color temperature". handprint. 2009-04-19. Retrieved 2011-06-09.
7. ^ Singh, Satyendra (2006-01-01). "Impact of color on marketing". Management Decision. 44 (6): 783–789. doi:10.1108/00251740610673332. ISSN 0025-1747.
8. ^ Burchett, K. E. (2002). "Color Harmony". Color Research and Application, 27 (1), pp. 28–31.
9. ^ O'Connor, Z. (2010). "Color harmony revisited". Color Research and Application, 35 (4), pp. 267–273.
10. ^ Pointer, M. R. & Attridge, G.G. (1998). "The number of discernible colors". Color Research and Application, 23 (1), pp. 52–54.
11. ^ Hard, A. & Sivik, L. (2001). "A theory of colors in combination – A descriptive model related to the NCS color-order system". Color Research and Application, 26 (1), pp. 4–28.
12. ^ Garau, Augusto (1993). Color Harmonies. University of Chicago press. p. 7. ISBN 0226281965.
13. ^ Feisner, E. A. (2000). Colour: How to use colour in art and design. London: Laurence King.
14. ^ Mahnke, F. (1996). Color, environment and human response. New York: John Wiley & Sons.
15. ^ Benson, J. L. (2000). Greek Color Theory and the Four Elements. Greek Color Theory and the Four Elements. Full text, not including figures.
16. ^ Bellantoni, Patti (2005). If it's Purple, Someone's Gonna Die. Elsevier, Focal Press. ISBN 0-240-80688-3.
17. ^ O'Connor, Z. (2010). "Colour psychology and color therapy: Caveat emptor". Color Research and Application
18. ^ "Pigments through the Ages - Renaissance and Baroque (1400-1600)". www.webexhibits.org.
19. ^ Albers, Josef (2006). Interaction of Color. Revised and Expanded Edition. Yale University Press. ISBN 0-300-11595-4.


text:
## The Annals of Probability

### Rounding of continuous random variables and oscillatory asymptotics

Svante Janson

#### Abstract

We study the characteristic function and moments of the integer-valued random variable ⌊X+α⌋, where X is a continuous random variables. The results can be regarded as exact versions of Sheppard’s correction. Rounded variables of this type often occur as subsequence limits of sequences of integer-valued random variables. This leads to oscillatory terms in asymptotics for these variables, something that has often been observed, for example in the analysis of several algorithms. We give some examples, including applications to tries, digital search trees and Patricia tries.

#### Article information

Source
Ann. Probab., Volume 34, Number 5 (2006), 1807-1826.

Dates
First available in Project Euclid: 14 November 2006

https://projecteuclid.org/euclid.aop/1163517225

Digital Object Identifier
doi:10.1214/009117906000000232

Mathematical Reviews number (MathSciNet)
MR2271483

Zentralblatt MATH identifier
1113.60017

#### Citation

Janson, Svante. Rounding of continuous random variables and oscillatory asymptotics. Ann. Probab. 34 (2006), no. 5, 1807--1826. doi:10.1214/009117906000000232. https://projecteuclid.org/euclid.aop/1163517225

#### References

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• Barbour, A. D., Holst, L. and Janson, S. (1992). Poisson Approximation. Oxford Univ. Press.
• Carver, H. C. (1936). The fundamental nature and proof of Sheppard's adjustments. Ann. Math. Statist. 7 154--163.
• Craig, C. C. (1941). A note on Sheppard's corrections. Ann. Math. Statist. 12 339--345.
• Cramér, H. (1945). Mathematical Methods of Statistics. Almqvist and Wiksell, Uppsala.
• Devroye, L. (1984). A probabilistic analysis of the height of tries and of the complexity of triesort. Acta Inform. 21 229--237.
• Flajolet, P. (1985). Approximate counting: A detailed analysis. BIT 25 113--134.
• Hitczenko, P. and Louchard, G. (2001). Distinctness of compositions of an integer: A probabilistic analysis. Random Structures Algorithms 19 407--437.
• Hwang, H.-K. and Yeh, Y.-N. (1997). Measures of distinctness for random partitions and compositions of an integer. Adv. in Appl. Math. 19 378--414.
• Janson, S. and Szpankowski, W. (1997). Analysis of an asymmetric leader election algorithm. Electronic J. Combin. 4 R17.
• Kirschenhofer, P., Prodinger, H. and Schoiß engeier, J. (1987). Zur Auswertung gewisser numerischer Reihen mit Hilfe modularer Funktionen. Zahlentheoretische Analysis II. Lecture Notes in Math. 1262 108--110. Springer, Berlin.
• Knuth, D. E. (1998). The Art of Computer Programming. 3. Sorting and Searching, 2nd ed. Addison--Wesley, Reading, MA.
• Langdon, W. H. and Ore, O. (1930). Semi-invariants and Sheppard's correction. Ann. Math. 31 230--232.
• Leadbetter, M. R., Lindgren, G. and Rootzén, H. (1983). Extremes and Related Properties of Random Sequences and Processes. Springer, New York.
• Louchard, G. (1987). Exact and asymptotic distributions in digital and binary search trees. RAIRO Inform. Théor. Appl. 21 479--495.
• Louchard, G. and Prodinger, H. (2006). The moments problem of extreme-value related distribution functions. Algorithmica. To appear.
• Mahmoud, H. (1992). Evolution of Random Search Trees. Wiley, New York.
• Parviainen, R. (2004). Random assignment with integer costs. Combin. Probab. Comput. 13 103--113.
• Pittel, B. (1986). Paths in a random digital tree: Limiting distributions. Adv. in Appl. Probab. 18 139--155.
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text:
# Correct way to write the eigenvector of a diagonalized hamiltonian in second quantization

I am studying diagonalization of a quadratic bosonic Hamiltonian of the type:

$$H = \displaystyle\sum_{<i,j>} A_{ij} a_i^\dagger a_j + \frac{1}{2}\displaystyle\sum_{<i,j>} [B_{ij} a_i^\dagger a_j^\dagger + B_{ij}^* a_j a_i ]$$

in second quantization.
Also, $A = A^\dagger$ and $B = B^T$, so that the Hamitonian is Hermitian.

This can be written in matrix form as $$H = \frac{1}{2} \alpha^\dagger M \alpha - \frac{1}{2} tr(A)$$ where $\alpha = \begin{pmatrix} a \\ a^\dagger \\ \end{pmatrix}$ , $\alpha^\dagger = \begin{pmatrix} a & a^\dagger \end{pmatrix}$ and M is $M = \begin{pmatrix} A & B\\ B^* & A^* \\ \end{pmatrix}$

Question: Can I numerically diagonalize the matrix M to get eigenvalues and eigenvectors of the Hamiltonian? If yes, then what would be the right way to write those eigenvectors in second quantization?

e.g. If for 2$\times$2 matrix M, one of the numerically calculated eigenvectors is $\begin{pmatrix} p\\ q \\ \end{pmatrix}$ , then, should it be written as $$p \,a|0\rangle + q \, a^\dagger|0\rangle$$ (where the column $\alpha$ has been used as the basis) or $$p \,a^\dagger|0\rangle + q \, a|0\rangle$$ (where $\alpha^\dagger$ has been used as the basis)?

Note: Here $|0\rangle$ is the vacuum state for 'a' type (bosonic) particles. End of Question.

Note : Consider a simpler Hamiltonian $$H = \displaystyle\sum_{<i,j>} A_{ij} a_i^\dagger a_j$$ and note that its eigenvectors are of the form $$(a_1^\dagger a_2^\dagger ... ) |0\rangle$$

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I forgot to mention that the Hamiltonian is Hermitian. I will edit the question to add the conditions on A and B. –  user1906035 Mar 31 '14 at 13:56
Is your rewritten hamitonian correct? somthing seems missing –  hwlau Apr 6 '14 at 7:48
It seems correct. –  user1906035 Apr 6 '14 at 9:10
–  Qmechanic Apr 6 '14 at 11:48

Take the Hamiltonian you write at end of your question as the example(where your statement about the eigenvector is not right), the procedure to obtain the eigenvalue and eigenvectors of other Hamiltonian is nearly the same.

$$H=\sum_{ij}A_{ij}a_i^\dagger a_j$$

Writing in matrix form you have(suppose it is a $n\times n$ matrix):

$$H=\alpha^\dagger A \alpha$$

with:

$$\alpha^\dagger=(a_1^\dagger, a_2^\dagger,\dotsb,a_n^\dagger )$$

Now you are ready to diagonalize Matrix analytically or numerically:

$$A=X^\dagger D X$$

Where $D=[E_1,E_2,\dotsb,E_n]$ is the eigenvalue you want, and $X$ is just a Unitary matrix that diagonalize $A$. Indeed, it is the normalized mathematical eigenvectors of $A$ write in column way(which is same in Fortran if you do it numerically).

Now substitute this to the original Hamiltonian:

$$H=\alpha^\dagger A \alpha=\alpha^\dagger X^\dagger DX\alpha=\beta^\dagger D \beta$$

where $\beta$ is $\beta =X\alpha=(\beta_1,\beta_2,\dotsb,\beta_n)^T$. Because this is a unitary transformation, you can easily check that the commute or anti-commute relation still holds for $\beta_i,\beta_2,\dotsb$.

Now you can consider the system is consisting of non-interacting qusi-particles $\beta_i,\beta_2,\dotsb$, then the eigenstate of the system is easily obtained: $$|\psi \rangle=(\beta_1^\dagger)^{n_1}(\beta_2^\dagger)^{n_2}\dotsb(\beta_n^\dagger)^{n_n}|0\rangle$$

$n_i=0,1$ for fermions and $n_i=0,1,2,3\dotsb$ for bosons.

Also, see this thread with the very similar question.

-
Thanks for the reply. But, I don't understand how to get the exponents $n_1$, $n_2$, $n_3$,.... for a particular eigenvector from the matrix X. –  user1906035 Apr 14 '14 at 19:46

What you're talking about looks a lot like a Bogoliubov transformation. The idea would be that you would decompose your Hamiltonian into something along the lines of $$H = \sum_{i, j} \mathcal{C}_i ~c_i^\dagger c_i$$ so that the eigenvalues are the $\mathcal{C}$ and the eigenvectors are the eigenvectors of $c_i^\dagger c_i$.

Specifically, if you consider $U$ to be the diagonalizing unitary matrix that takes your original matrix $M \mapsto U \mathcal{C} U^\dagger$, then by directly inserting this into your definition for your Hamiltonian you would get $$H = \alpha^\dagger \left (U \mathcal{C} U^\dagger \right ) \alpha$$ and, hence, the individual elements of $U^\dagger \alpha$ are the new "eigen-operators" of the Hamiltonian. The spectral properties of those operators, combined with the diagonal $\mathcal{C}$, will give the new eigenvalues and eigenstates in terms of the old $a$-basis.

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This is unfortunately not correct, but this is a common mistake. The creation and annihilation operators associated with the eigenvectors of $M$ would not be bosonic. It would be correct for fermions, where the transformations preserve the euclidean norm, but for bosons you are working on a symplectic space (as in classical mechanics). The right way to do it is to look at the eigenvectors of $\Sigma M$, where $\Sigma=\mbox{diag}(1,-1)$ and you furthermore have to normalize them with respect to the norm induced by $\Sigma$. –  ulf Dec 4 '14 at 14:31
Please explain where the logical error is, and how it contrasts with the other answer given to this question. I've just made a change of basis, nowhere do I require that the $\alpha$ commute or anti-commute. –  webb Dec 4 '14 at 18:45

@webb's comment (since the answer is longer than a comment allows for).

Firstly, you have to distinguish between diagonalizing a matrix representation of an operator in a given basis and diagonalizing the coefficient matrix of an operator in a set of creation and annihilation operators.

We are dealing with the latter here. If (as you point out correctly) we need a Bogoliubov transformation, the transformation of the form $U^\dagger \alpha$ is not sufficient. You need to allow for mixing of creation and annihilation operators.

But then you can't just diagonalize the coefficient matrix. The logical error is that you imply that $c_i^\dagger c_i$ have the properties of a number operator, but why should this be true? The best you can do is bring the coefficient matrix into Jordan normal form, which leads to additional terms of different type.

For bosons it turns out that you cannot generally write the resulting Hamiltonian in the form $$\mathcal H= \sum_i \mathcal C_i \, c_i^\dagger c_i.$$ This would assume that the matrix is generally diagonalizable. This is for instance not true when Goldstone modes exist.

-


text:
# Problem: A gold nucleus has a radius of 7.3x10-15 m and a charge of +79e. Through what voltage must a particle with its charge of +2e be accelerated so that it has just enough energy to reach a distance of 2.0*10-14 m from the surface of a gold nucleus? (Assume the gold nucleus stays stationary and serves as a point charge).

###### FREE Expert Solution

We can write, from the law of conservation of energy:

$\overline{)\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}{\mathbf{=}}{\mathbf{V}}{\mathbf{q}}}$

92% (7 ratings)
###### Problem Details

A gold nucleus has a radius of 7.3x10-15 m and a charge of +79e. Through what voltage must a particle with its charge of +2e be accelerated so that it has just enough energy to reach a distance of 2.0*10-14 m from the surface of a gold nucleus? (Assume the gold nucleus stays stationary and serves as a point charge).


text:
# Properties

 Label 28T7 Order $$56$$ n $$28$$ Cyclic No Abelian No Solvable Yes Primitive No $p$-group No Group: $D_{14}:C_2$

## Group action invariants

 Degree $n$ : $28$ Transitive number $t$ : $7$ Group : $D_{14}:C_2$ Parity: $-1$ Primitive: No Nilpotency class: $-1$ (not nilpotent) Generators: (1,11,22,3,14,23,5,15,25,7,17,28,9,20)(2,12,21,4,13,24,6,16,26,8,18,27,10,19), (1,14)(2,13)(3,12)(4,11)(5,9)(6,10)(7,8)(15,27)(16,28)(17,25)(18,26)(19,23)(20,24) $|\Aut(F/K)|$: $2$

## Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$ x 3
4:  $C_2^2$
8:  $D_{4}$
14:  $D_{7}$
28:  $D_{14}$

Resolvents shown for degrees $\leq 47$

## Subfields

Degree 2: $C_2$

Degree 4: $D_{4}$

Degree 7: $D_{7}$

Degree 14: $D_{14}$

## Low degree siblings

28T6

Siblings are shown with degree $\leq 47$

A number field with this Galois group has no arithmetically equivalent fields.

## Conjugacy Classes

 Cycle Type Size Order Representative $1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$ $1$ $1$ $()$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1$ $14$ $2$ $( 3,27)( 4,28)( 5,25)( 6,26)( 7,24)( 8,23)( 9,22)(10,21)(11,19)(12,20)(13,18) (14,17)(15,16)$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2$ $1$ $2$ $( 1, 2)( 3, 4)( 5, 6)( 7, 8)( 9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22) (23,24)(25,26)(27,28)$ $4, 4, 4, 4, 4, 4, 4$ $14$ $4$ $( 1, 3, 2, 4)( 5,28, 6,27)( 7,26, 8,25)( 9,23,10,24)(11,21,12,22)(13,19,14,20) (15,18,16,17)$ $14, 14$ $2$ $14$ $( 1, 3, 5, 7, 9,11,14,15,17,20,22,23,25,28)( 2, 4, 6, 8,10,12,13,16,18,19,21, 24,26,27)$ $14, 14$ $2$ $14$ $( 1, 4, 5, 8, 9,12,14,16,17,19,22,24,25,27)( 2, 3, 6, 7,10,11,13,15,18,20,21, 23,26,28)$ $7, 7, 7, 7$ $2$ $7$ $( 1, 5, 9,14,17,22,25)( 2, 6,10,13,18,21,26)( 3, 7,11,15,20,23,28) ( 4, 8,12,16,19,24,27)$ $14, 14$ $2$ $14$ $( 1, 6, 9,13,17,21,25, 2, 5,10,14,18,22,26)( 3, 8,11,16,20,24,28, 4, 7,12,15, 19,23,27)$ $14, 14$ $2$ $14$ $( 1, 7,14,20,25, 3, 9,15,22,28, 5,11,17,23)( 2, 8,13,19,26, 4,10,16,21,27, 6, 12,18,24)$ $14, 14$ $2$ $14$ $( 1, 8,14,19,25, 4, 9,16,22,27, 5,12,17,24)( 2, 7,13,20,26, 3,10,15,21,28, 6, 11,18,23)$ $7, 7, 7, 7$ $2$ $7$ $( 1, 9,17,25, 5,14,22)( 2,10,18,26, 6,13,21)( 3,11,20,28, 7,15,23) ( 4,12,19,27, 8,16,24)$ $14, 14$ $2$ $14$ $( 1,10,17,26, 5,13,22, 2, 9,18,25, 6,14,21)( 3,12,20,27, 7,16,23, 4,11,19,28, 8,15,24)$ $14, 14$ $2$ $14$ $( 1,11,22, 3,14,23, 5,15,25, 7,17,28, 9,20)( 2,12,21, 4,13,24, 6,16,26, 8,18, 27,10,19)$ $14, 14$ $2$ $14$ $( 1,12,22, 4,14,24, 5,16,25, 8,17,27, 9,19)( 2,11,21, 3,13,23, 6,15,26, 7,18, 28,10,20)$ $14, 14$ $2$ $14$ $( 1,13,25,10,22, 6,17, 2,14,26, 9,21, 5,18)( 3,16,28,12,23, 8,20, 4,15,27,11, 24, 7,19)$ $7, 7, 7, 7$ $2$ $7$ $( 1,14,25, 9,22, 5,17)( 2,13,26,10,21, 6,18)( 3,15,28,11,23, 7,20) ( 4,16,27,12,24, 8,19)$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2$ $2$ $2$ $( 1,15)( 2,16)( 3,17)( 4,18)( 5,20)( 6,19)( 7,22)( 8,21)( 9,23)(10,24)(11,25) (12,26)(13,27)(14,28)$

## Group invariants

 Order: $56=2^{3} \cdot 7$ Cyclic: No Abelian: No Solvable: Yes GAP id: [56, 7]
 Character table:  2 3 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 7 1 . 1 . 1 1 1 1 1 1 1 1 1 1 1 1 1 1a 2a 2b 4a 14a 14b 7a 14c 14d 14e 7b 14f 14g 14h 14i 7c 2c 2P 1a 1a 1a 2b 7a 7a 7b 7b 7c 7c 7c 7c 7b 7b 7a 7a 1a 3P 1a 2a 2b 4a 14d 14e 7c 14i 14h 14g 7a 14c 14a 14b 14f 7b 2c 5P 1a 2a 2b 4a 14g 14h 7b 14f 14a 14b 7c 14i 14e 14d 14c 7a 2c 7P 1a 2a 2b 4a 2c 2c 1a 2b 2c 2c 1a 2b 2c 2c 2b 1a 2c 11P 1a 2a 2b 4a 14e 14d 7c 14i 14g 14h 7a 14c 14b 14a 14f 7b 2c 13P 1a 2a 2b 4a 14b 14a 7a 14c 14e 14d 7b 14f 14h 14g 14i 7c 2c X.1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 X.2 1 -1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 X.3 1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 X.4 1 1 1 -1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 X.5 2 . -2 . . . 2 -2 . . 2 -2 . . -2 2 . X.6 2 . 2 . A A -B -B C C -C -C B B -A -A -2 X.7 2 . 2 . B B -C -C A A -A -A C C -B -B -2 X.8 2 . 2 . C C -A -A B B -B -B A A -C -C -2 X.9 2 . 2 . -A -A -B -B -C -C -C -C -B -B -A -A 2 X.10 2 . 2 . -B -B -C -C -A -A -A -A -C -C -B -B 2 X.11 2 . 2 . -C -C -A -A -B -B -B -B -A -A -C -C 2 X.12 2 . -2 . D -D -C C F -F -A A -E E B -B . X.13 2 . -2 . E -E -A A -D D -B B F -F C -C . X.14 2 . -2 . F -F -B B E -E -C C D -D A -A . X.15 2 . -2 . -F F -B B -E E -C C -D D A -A . X.16 2 . -2 . -E E -A A D -D -B B -F F C -C . X.17 2 . -2 . -D D -C C -F F -A A E -E B -B . A = -E(7)^3-E(7)^4 B = -E(7)-E(7)^6 C = -E(7)^2-E(7)^5 D = -E(7)+E(7)^6 E = -E(7)^2+E(7)^5 F = -E(7)^3+E(7)^4


text: Integrable Systems Associated to a Hopf Surface A Hopf surface is the quotient of the complex surface $\mathbb{C}^2 \setminus \{0\}$ by an infinite cyclic group of dilations of $\mathbb{C}^2$. In this paper, we study the moduli spaces $\mathcal{M}^n$ of stable $\SL (2,\mathbb{C})$-bundles on a Hopf surface $\mathcal{H}$, from the point of view of symplectic geometry. An important point is that the surface $\mathcal{H}$ is an elliptic fibration, which implies that a vector bundle on $\mathcal{H}$ can be considered as a family of vector bundles over an elliptic curve. We define a map $G \colon \mathcal{M}^n \rightarrow \mathbb{P}^{2n+1}$ that associates to every bundle on $\mathcal{H}$ a divisor, called the graph of the bundle, which encodes the isomorphism class of the bundle over each elliptic curve. We then prove that the map $G$ is an algebraically completely integrable Hamiltonian system, with respect to a given Poisson structure on $\mathcal{M}^n$. We also give an explicit description of the fibres of the integrable system. This example is interesting for several reasons; in particular, since the Hopf surface is not K\"ahler, it is an elliptic fibration that does not admit a section. Categories:14J60, 14D21, 14H70, 14J27


text:
# Monte carlo error: Combining “experimental” and statistical errors

I'm doing a slightly involved Monte Carlo approximation of a quantity $E$ where I end up with the following formula:

$E=\frac{\sum_{i=1}^np_ie_iG_i}{\sum_{i=1}^np_iG_i}\ .\ \ \ \ \ \ \ \ \$ (1)

The quantities $e_i$ and $p_i$ can be calculated directly from the MC configurations but the quantity $G_i$ is computed with another, secondary, Monte Carlo - and so comes with an error $\delta_i$. Taking the denominator and numerator of (1) separately I tend to think of it like this: We have a bunch of measurements of the quantities $G_i$ which we use together with the parameters $e_i$ and $p_i$ to find the averages as estimates of the true values (although this shouldn't be taken too far, there is the statistical error associated with approximating the MC integral with a sum averaged over MC samples). How to estimate the error on the value $E$?

Normally the error estimate for $E$ in a MC is just the standard deviation of the mean $\sigma_M=\sigma/\sqrt{n-1}$ (possibly computed using bootstrap for a robust estimate). However in this case this does not give the right value as we'll see below.

A few possibilities for the error estimate come to mind, but so far all of them have some problem. I'll describe them in turn:

I) The standard deviation of the mean, as suggested above. The problem with this can be described with the following example: Say I compute (1) using 1000 MC samples, i.e. $n=1000$. With a bootstrap I calculate $\sigma_M=\sqrt{\overline{E^2}-\overline{E}^2}/\sqrt{n-1}$ as the error of the mean. I then do this whole calculation 100 times and so get 100 values $\{E_k\}$, and calculate the actual standard deviation from this - and get a different (much higher) value than $\sigma_M$. I think the reason could be related to systematic error on $E$ depending on the number of samples in the secondary MC computing $G_i$ (in other words $\delta_i$ inducing a systematic error in $E$).

II) Error propagation of the $\delta_i$'s. I could use the formula $\delta_E=\sqrt{\sum_{i=1}^n(\frac{\partial E}{\partial G_i}\delta_i)^2}$ (either with the numerator and denominator separately or the whole of $E$ as above). However this would give zero error $\delta_E$ if $\delta_i=0$, when this situation should correspond to a normal "single" MC where the error is given by the standard deviation, and so cannot be right.

III) Thinking of $G_i$ as measurements. This I alluded to earlier, where as an analogy I think of each $G_i$ as measured with errors $\delta_i$ and then want to find the total error on $E$. It can't just be the standard deviation of E, because imagine having $n$ a very low number, e.g. $n=3$ - it could then happen that the three values are so close as to give a better error than the individual $\delta_i$'s should permit. I assume there is some textbook answer to this problem but I cannot find it.

Does anyone have a suggestion for the error?

-


text:
# 5-cell

For the sequence of fifth element numbers of Pascal's triangle, see Pentatope number.
Regular 5-cell
(pentachoron)
(4-simplex)
Schlegel diagram
(vertices and edges)
Type Convex regular 4-polytope
Schläfli symbol {3,3,3}
Coxeter diagram
Cells 5 {3,3}
Faces 10 {3}
Edges 10
Vertices 5
Vertex figure
(tetrahedron)
Petrie polygon pentagon
Coxeter group A4, [3,3,3]
Dual Self-dual
Properties convex, isogonal, isotoxal, isohedral
Uniform index 1
Vertex figure: tetrahedron

In geometry, the 5-cell is a four-dimensional object bounded by 5 tetrahedral cells. It is also known as a C5, pentachoron, pentatope, pentahedroid,[1] or tetrahedral pyramid. It is a 4-simplex, the simplest possible convex regular 4-polytope (four-dimensional analogue of a Platonic solid), and is analogous to the tetrahedron in three dimensions and the triangle in two dimensions. The pentachoron is a four dimensional pyramid with a tetrahedral base.

The regular 5-cell is bounded by regular tetrahedra, and is one of the six regular convex 4-polytopes, represented by Schläfli symbol {3,3,3}.

## Alternative names

• Pentachoron
• 4-simplex
• Pentatope
• Pentahedroid (Henry Parker Manning)
• Pen (Jonathan Bowers: for pentachoron)[2]
• Hyperpyramid, tetrahedral pyramid

## Geometry

The 5-cell is self-dual, and its vertex figure is a tetrahedron. Its maximal intersection with 3-dimensional space is the triangular prism. Its dihedral angle is cos−1(1/4), or approximately 75.52°.

### Construction

The 5-cell can be constructed from a tetrahedron by adding a 5th vertex such that it is equidistant from all the other vertices of the tetrahedron. (The 5-cell is essentially a 4-dimensional pyramid with a tetrahedral base.)

The simplest set of coordinates is: (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (τ,τ,τ,τ), with edge length 2√2, where τ is the golden ratio.[3]

The Cartesian coordinates of the vertices of an origin-centered regular 5-cell having edge length 2 are:

${\displaystyle \left({\frac {1}{\sqrt {10}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {1}{\sqrt {3}}},\ \pm 1\right)}$
${\displaystyle \left({\frac {1}{\sqrt {10}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {-2}{\sqrt {3}}},\ 0\right)}$
${\displaystyle \left({\frac {1}{\sqrt {10}}},\ -{\sqrt {\frac {3}{2}}},\ 0,\ 0\right)}$
${\displaystyle \left(-2{\sqrt {\frac {2}{5}}},\ 0,\ 0,\ 0\right)}$

Another set of origin-centered coordinates in 4-space can be seen as a hyperpyramid with a regular tetrahedral base in 3-space, with edge length 2√2:

${\displaystyle \left(1,1,1,-1/{\sqrt {5}}\right)}$
${\displaystyle \left(1,-1,-1,-1/{\sqrt {5}}\right)}$
${\displaystyle \left(-1,1,-1,-1/{\sqrt {5}}\right)}$
${\displaystyle \left(-1,-1,1,-1/{\sqrt {5}}\right)}$
${\displaystyle \left(0,0,0,{\sqrt {5}}-1/{\sqrt {5}}\right)}$

The vertices of a 4-simplex (with edge √2) can be more simply constructed on a hyperplane in 5-space, as (distinct) permutations of (0,0,0,0,1) or (0,1,1,1,1); in these positions it is a facet of, respectively, the 5-orthoplex or the rectified penteract.

### Boerdijk–Coxeter helix

A 5-cell can be constructed as a Boerdijk–Coxeter helix of five chained tetrahedra, folded into a 4-dimensional ring. The 10 triangle faces can be seen in a 2D net within a triangular tiling, with 6 triangles around every vertex, although folding into 4-dimensions causes edges to coincide. The purple edges represent the Petrie polygon of the 5-cell.

### Projections

The A4 Coxeter plane projects the 5-cell into a regular pentagon and pentagram.

orthographic projections
Ak
Coxeter plane
A4 A3 A2
Graph
Dihedral symmetry [5] [4] [3]
Projections to 3 dimensions

Stereographic projection wireframe (edge projected onto a 3-sphere)

A 3D projection of a 5-cell performing a simple rotation

The vertex-first projection of the 5-cell into 3 dimensions has a tetrahedral projection envelope. The closest vertex of the 5-cell projects to the center of the tetrahedron, as shown here in red. The farthest cell projects onto the tetrahedral envelope itself, while the other 4 cells project onto the 4 flattened tetrahedral regions surrounding the central vertex.

The edge-first projection of the 5-cell into 3 dimensions has a triangular dipyramidal envelope. The closest edge (shown here in red) projects to the axis of the dipyramid, with the three cells surrounding it projecting to 3 tetrahedral volumes arranged around this axis at 120 degrees to each other. The remaining 2 cells project to the two halves of the dipyramid and are on the far side of the pentatope.

The face-first projection of the 5-cell into 3 dimensions also has a triangular dipyramidal envelope. The nearest face is shown here in red. The two cells that meet at this face projects to the two halves of the dipyramid. The remaining three cells are on the far side of the pentatope from the 4D viewpoint, and are culled from the image for clarity. They are arranged around the central axis of the dipyramid, just as in the edge-first projection.

The cell-first projection of the 5-cell into 3 dimensions has a tetrahedral envelope. The nearest cell projects onto the entire envelope, and, from the 4D viewpoint, obscures the other 4 cells; hence, they are not rendered here.

## Irregular 5-cell

There are many lower symmetry forms, including these found in uniform polytope vertex figures:

Symmetry [3,3,3]
Order 120
[3,3,1]
Order 24
[3,2,1]
Order 12
[3,1,1]
Order 6
[5,2]+
Order 10
Name Regular 5-cell Tetrahedral pyramid Triangular-pyramidal pyramid Pentagonal hyperdisphenoid
Schläfli symbol {3,3,3} {3,3} ∨ ( ) {3} ∨ { }
Example
Vertex
figure

5-simplex

Truncated 5-simplex

Bitruncated 5-simplex

Cantitruncated 5-simplex

Omnitruncated 4-simplex honeycomb

The tetrahedral pyramid is a special case of a 5-cell, a polyhedral pyramid, constructed as a regular tetrahedron base in a 3-space hyperplane, and an apex point above the hyperplane. The four sides of the pyramid are made of tetrahedron cells.

Many uniform 5-polytopes have tetrahedral pyramid vertex figures:

Other uniform 5-polytopes have irregular 5-cell vertex figures. The symmetry of a vertex figure of a uniform polytope is represented by removing the ringed nodes of the Coxeter diagram.

Symmetry [3,2,1], order 12 [3,1,1], order 6 [2+,4,1], order 8 [2,1,1], order 4
Schlegel
diagram
Name
Coxeter
diagram
t12α5
t12γ5
t012α5
t012γ5
t123α5
t123γ5
Symmetry [2,1,1], order 2 [2+,1,1], order 2 [ ]+, order 1
Schlegel
diagram
Name
Coxeter
diagram
t0123α5
t0123γ5
t0123β5
t01234α5
t01234γ5

## Compound

The compound of two 5-cells in dual configurations can be seen in this A5 Coxeter plane projection, with a red and blue 5-cell vertices and edges. This compound has [[3,3,3]] symmetry, order 240. The intersection of these two 5-cells is a uniform birectified 5-cell. = .

## Related polytopes and honeycomb

The pentachoron (5-cell) is the simplest of 9 uniform polychora constructed from the [3,3,3] Coxeter group.

Schläfli {3,3,3} t{3,3,3} r{3,3,3} rr{3,3,3} 2t{3,3,3} tr{3,3,3} t0,3{3,3,3} t0,1,3{3,3,3} t0,1,2,3{3,3,3}
Coxeter
Schlegel

It is in the sequence of regular polychora: the tesseract {4,3,3}, 120-cell {5,3,3}, of Euclidean 4-space, and hexagonal tiling honeycomb {6,3,3} of hyperbolic space. All of these have a tetrahedral vertex figure.

It is similar to three regular polychora: the tesseract {4,3,3}, 600-cell {3,3,5} of Euclidean 4-space, and the order-6 tetrahedral honeycomb {3,3,6} of hyperbolic space. All of these have a tetrahedral cell.

## References

1. ^ Matila Ghyka, The geometry of Art and Life (1977), p.68
2. ^ Category 1: Regular Polychora
3. ^ Coxeter, Regular Complex Polytopes, 1991, p. 30. 4.2 The Crystalographic regular polytopes
• T. Gosset: On the Regular and Semi-Regular Figures in Space of n Dimensions, Messenger of Mathematics, Macmillan, 1900
• H.S.M. Coxeter:
• Coxeter, Regular Polytopes, (3rd edition, 1973), Dover edition, ISBN 0-486-61480-8, p. 296, Table I (iii): Regular Polytopes, three regular polytopes in n-dimensions (n≥5)
• H.S.M. Coxeter, Regular Polytopes, 3rd Edition, Dover New York, 1973, p. 296, Table I (iii): Regular Polytopes, three regular polytopes in n-dimensions (n≥5)
• Kaleidoscopes: Selected Writings of H.S.M. Coxeter, edited by F. Arthur Sherk, Peter McMullen, Anthony C. Thompson, Asia Ivic Weiss, Wiley-Interscience Publication, 1995, ISBN 978-0-471-01003-6 [1]
• (Paper 22) H.S.M. Coxeter, Regular and Semi Regular Polytopes I, [Math. Zeit. 46 (1940) 380-407, MR 2,10]
• (Paper 23) H.S.M. Coxeter, Regular and Semi-Regular Polytopes II, [Math. Zeit. 188 (1985) 559-591]
• (Paper 24) H.S.M. Coxeter, Regular and Semi-Regular Polytopes III, [Math. Zeit. 200 (1988) 3-45]
• John H. Conway, Heidi Burgiel, Chaim Goodman-Strass, The Symmetries of Things 2008, ISBN 978-1-56881-220-5 (Chapter 26. pp. 409: Hemicubes: 1n1)
• Norman Johnson Uniform Polytopes, Manuscript (1991)
• N.W. Johnson: The Theory of Uniform Polytopes and Honeycombs, Ph.D. (1966)


text:
# Surface area of a Pyramid 22 meters by 59 foot by 36 yards Calculator

Enter the Base(base 1)

Enter the Base (base 2)

Enter the height

The surface area of a pyramid is equal to the sum of the areas of each side of the pyramid. The base of the pyramid has area l x w , and sl and sw represent the slant height on the length and slant height on the width. If the Pyramid has a length 22 meters by width 59 foot by height 36 yards is 1739.8442543 meters2 or 2117.5254591 yards2 or 19057.7223689 foot2

Surface Area of a Pyramid 22 m by 59 ft by 36 yd in other units

Value unit
1.7398443 km2
1.0810918 mi2
1739.8442543 m2
5708.1504406 ft2
68497.8052874 in2
1902.7168135 yd2
173984.42543 cm2
1739844.2543 mm2

## Surface area of a Pyramid 22 meters by 59 foot by 36 yards

Steps:

The Surface area of Pyramid A = $lw+l$$\sqrt{(\frac{w}{2})^2+(h)^2}$$+w$$\sqrt{(\frac{l}{2})^2+(h)^2} Substitute the values of the length l = 22 m, the width w = 59 ft and the height h = 36 yd into the formula for surface area of a pyramid Unit Conversion of 36 yd = 32.9184 m To convert Yard to Meter we know that, 1 Yard = 0.9144 Meter To convert Yards to meters, multiply the yard value by 0.9144. Result in Meter: 36 yd × 0.9144 × m/yd Cancel The Comman factor of yd Result in Meters: (36 x 0.9144 m) Multiply 36 into 0.9144 Result in Meters: 32.9184 meters ∴ 36 Yard = 32.9184 meters Unit Conversion of 59 ft = 17.9832 m To convert Feet to Meter we know that, 1 Foot = 0.3048 Meter To convert Foot to meters, multiply the feet value by 0.3048. Result in Meter: 59 ft × 0.3048 × m/ft Cancel The Comman factor of ft Result in Meters: (59 x 0.3048 m) Multiply 59 into 0.3048 Result in Meters: 17.9832 meters ∴ 59 Foot = 17.9832 meters A=(22 \cdot17.9832+22$$\sqrt{(\frac{17.9832}{2})^2+(32.9184)^2}$$+17.9832$$\sqrt{(\frac{22}{2})^2+(32.9184)^2}$) m

Simplify each term.

Multiply 22 m by 17.9832 m

A = $395.6304+22$$\sqrt{(\frac{17.9832}{2})^2+(32.9184)^2}$$+17.9832$$\sqrt{(\frac{22}{2})^2+(32.9184)^2} Square root of \sqrt{(\frac{17.9832}{2})^2+(32.9184)^2} is 34.1243305 Put The values in Area Formula: A= 395.6304+ 22 \cdot 34.1243305+17.9832$$\sqrt{(\frac{22}{2})^2+(32.9184)^2}$

Square Root of $\sqrt{(\frac{22}{2})^2+(32.9184)^2}$ is 33.0018342

Put The values in Area Formula:

A= 395.6304 + (22 x 34.1243305) + (17.9832 x 33.0018342)

Multiply 22 and 34.1243305

A= 395.6304 + 750.73527 + (17.9832 x 33.0018342)

Multiply 17.9832 and 33.0018342

A= 395.6304 + 750.73527 + 593.4785842

A= 1146.36567 + 593.4785842

A= 1739.8442543 m2

∴ The Surface Area of Pyramid length 22 m , width 59 ft and height 36 yd is 1739.8442543 m2

or

Unit Conversion of 59 ft = 19.6666667 yd

To convert Feet to Yards

we know that, 1 Foot = 0.333333 or
1 Foot = 1/3 Yards

To convert Feet to Yards, divide the feet value by 3

Result in Yards: 59 × ft/yd × ft/yd

Cancel The Comman factor of ft

Result in Yards: 59/3 yd

Divide the 59 by 3

Result in Yards: 19.6666667 yards

∴ 59 Feet = 19.6666667 yards

Unit Conversion of 22 m = 24.0595058 yd

To convert Meter to Yards

we know that, 1 Meter = 1.0936139 Yards

To convert Meter to Yards, multiply the mile value by 1.0936139

Result in Yards: 22 m × 1.0936139 × yd/m

Cancel The Comman factor of m

Result in Yards: (22 x 1.0936139 yd)

Multiply 22 into 1.0936139

Result in Yards: 24.0595058 yards

∴ 22 Meter = 24.0595058 yards

A=($24.0595058 \cdot19.6666667+24.0595058$$\sqrt{(\frac{19.6666667}{2})^2+(36)^2}$$+19.6666667$$\sqrt{(\frac{24.0595058}{2})^2+(36)^2}) yd Simplify each term. Multiply 24.0595058 yd by 19.6666667 yd A = 473.17028154+24.0595058$$\sqrt{(\frac{19.6666667}{2})^2+(36)^2}$$+19.6666667$$\sqrt{(\frac{24.0595058}{2})^2+(36)^2}$

Square root of $\sqrt{(\frac{19.6666667}{2})^2+(36)^2}$ is 37.3188216

Put The values in Area Formula:

A= $473.17028154+ 24.0595058 \cdot 37.3188216+19.6666667$$\sqrt{(\frac{24.0595058}{2})^2+(36)^2} Square Root of \sqrt{(\frac{24.0595058}{2})^2+(36)^2} is 37.9567511 Put The values in Area Formula: A= 473.1702815 + (24.0595058 x 37.3188216) + (19.6666667 x 37.9567511) Multiply 24.0595058 and 37.3188216 A= 473.1702815 + 897.8724045 + (19.6666667 x 37.9567511) Multiply 19.6666667 and 37.9567511 A= 473.1702815 + 897.8724045 + 746.4827731 Add 473.1702815 and 897.8724045 A= 1371.042686 + 746.4827731 A=2117.5254591 yd2 ∴ The Surface Area of Pyramid length 22 m , width 59 ft and height 36 yd is 2117.5254591 yd2 or Unit Conversion of 22 m = 72.17848 ft To convert Meters to Foot we know that, 1 Meter = 3.28084 Foot To convert Meter to Foot, multiply the meter value by 3.28084. Result in Foot: 22 m × 3.28084 × ft/m Cancel The Comman factor of m Result in Foot: (22 x 3.28084 ft) Multiply 22 into 3.28084 Result in Foot: 72.17848 foot ∴ 22 Meters = 72.17848 foot Unit Conversion of 36 yd = 108.0 ft To convert Yards to Foot we know that, 1 Yards = 3 Feet To convert Yards to Foot, multiply the yard value by 3. Result in Feet: 36 yd x 3 × ft/yd Cancel The Comman factor of yd Result in Feet: (36 x 3 ft) Multiply 36 into 3 Result in Feet: 108.0 feets ∴ 36 Yards = 108.0 foots A=(72.17848 \cdot59+72.17848$$\sqrt{(\frac{59}{2})^2+(108.0)^2}$$+59$$\sqrt{(\frac{72.17848}{2})^2+(108.0)^2}$) ft

Simplify each term.

Multiply 72.17848 ft by 59 ft

A = $4258.53032+72.17848$$\sqrt{(\frac{59}{2})^2+(108.0)^2}$$+59$$\sqrt{(\frac{72.17848}{2})^2+(108.0)^2} Square root of \sqrt{(\frac{59}{2})^2+(36)^2} is 111.9564648 Put The values in Area Formula: A= 4258.53032+ 72.17848 \cdot 111.9564648+59$$\sqrt{(\frac{72.17848}{2})^2+(108.0)^2}$

Square Root of $\sqrt{(\frac{72.17848}{2})^2+(108.0)^2}$ is 113.8702474

Put The values in Area Formula:

A = 4258.53032 + (72.17848 x 111.9564648) + (59 x 113.8702474)

Multiply 72.17848 and 111.9564648

A = 4258.53032 + 8080.847452 +(59 x 113.8702474)

Multiply 59 and 113.8702474

A= 4258.53032 + 8080.847452 + 6718.3445968