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# Photoelectric effect

(Redirected from Photoelectron)

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons. The phenomenon is studied in condensed matter physics, and solid state and quantum chemistry to draw inferences about the properties of atoms, molecules and solids. The effect has found use in electronic devices specialized for light detection and precisely timed electron emission.

The emission of electrons from a metal plate caused by light quanta – photons.

The experimental results disagree with classical electromagnetism, which predicts that continuous light waves transfer energy to electrons, which would then be emitted when they accumulate enough energy. An alteration in the intensity of light would theoretically change the kinetic energy of the emitted electrons, with sufficiently dim light resulting in a delayed emission. The experimental results instead show that electrons are dislodged only when the light exceeds a certain frequency—regardless of the light's intensity or duration of exposure. Because a low-frequency beam at a high intensity could not build up the energy required to produce photoelectrons, as it would have if light's energy were coming from a continuous wave, Albert Einstein proposed that a beam of light is not a wave propagating through space, but a swarm of discrete energy packets, known as photons.

Emission of conduction electrons from typical metals requires a few electron-volt (eV) light quanta, corresponding to short-wavelength visible or ultraviolet light. In extreme cases, emissions are induced with photons approaching zero energy, like in systems with negative electron affinity and the emission from excited states, or a few hundred keV photons for core electrons in elements with a high atomic number.[1] Study of the photoelectric effect led to important steps in understanding the quantum nature of light and electrons and influenced the formation of the concept of wave–particle duality.[2] Other phenomena where light affects the movement of electric charges include the photoconductive effect, the photovoltaic effect, and the photoelectrochemical effect.

## Emission mechanism

The photons of a light beam have a characteristic energy, called photon energy, which is proportional to the frequency of the light. In the photoemission process, when an electron within some material absorbs the energy of a photon and acquires more energy than its binding energy, it is likely to be ejected. If the photon energy is too low, the electron is unable to escape the material. Since an increase in the intensity of low-frequency light will only increase the number of low-energy photons, this change in intensity will not create any single photon with enough energy to dislodge an electron. Moreover, the energy of the emitted electrons will not depend on the intensity of the incoming light of a given frequency, but only on the energy of the individual photons.

While free electrons can absorb any energy when irradiated as long as this is followed by an immediate re-emission, like in the Compton effect, in quantum systems all of the energy from one photon is absorbed—if the process is allowed by quantum mechanics—or none at all. Part of the acquired energy is used to liberate the electron from its atomic binding, and the rest contributes to the electron's kinetic energy as a free particle.[3][4][5] Because electrons in a material occupy many different quantum states with different binding energies, and because they can sustain energy losses on their way out of the material, the emitted electrons will have a range of kinetic energies. The electrons from the highest occupied states will have the highest kinetic energy. In metals, those electrons will be emitted from the Fermi level.

When the photoelectron is emitted into a solid rather than into a vacuum, the term internal photoemission is often used, and emission into a vacuum is distinguished as external photoemission.

### Experimental observation of photoelectric emission

Even though photoemission can occur from any material, it is most readily observed from metals and other conductors. This is because the process produces a charge imbalance which, if not neutralized by current flow, results in the increasing potential barrier until the emission completely ceases. The energy barrier to photoemission is usually increased by nonconductive oxide layers on metal surfaces, so most practical experiments and devices based on the photoelectric effect use clean metal surfaces in evacuated tubes. Vacuum also helps observing the electrons since it prevents gases from impeding their flow between the electrodes.

As sunlight, due to atmosphere's absorption, does not provide much ultraviolet light, the light rich in ultraviolet rays used to be obtained by burning magnesium or from an arc lamp. At the present time, mercury-vapor lamps, noble-gas discharge UV lamps and radio-frequency plasma sources,[6][7][8] ultraviolet lasers,[9] and synchrotron insertion device[10] light sources prevail.

Schematic of the experiment to demonstrate the photoelectric effect. Filtered, monochromatic light of a certain wavelength strikes the emitting electrode (E) inside a vacuum tube. The collector electrode (C) is biased to a voltage VC that can be set to attract the emitted electrons, when positive, or prevent any of them from reaching the collector when negative.

The classical setup to observe the photoelectric effect includes a light source, a set of filters to monochromatize the light, a vacuum tube transparent to ultraviolet light, an emitting electrode (E) exposed to the light, and a collector (C) whose voltage VC can be externally controlled.

A positive external voltage is used to direct the photoemitted electrons onto the collector. If the frequency and the intensity of the incident radiation are fixed, the photoelectric current I increases with an increase in the positive voltage, as more and more electrons are directed onto the electrode. When no additional photoelectrons can be collected, the photoelectric current attains a saturation value. This current can only increase with the increase of the intensity of light.

An increasing negative voltage prevents all but the highest-energy electrons from reaching the collector. When no current is observed through the tube, the negative voltage has reached the value that is high enough to slow down and stop the most energetic photoelectrons of kinetic energy Kmax. This value of the retarding voltage is called the stopping potential or cut off potential Vo.[11] Since the work done by the retarding potential in stopping the electron of charge e is eVo, the following must hold eVo = Kmax.

The current-voltage curve is sigmoidal, but its exact shape depends on the experimental geometry and the electrode material properties.

For a given metal surface, there exists a certain minimum frequency of incident radiation below which no photoelectrons are emitted. This frequency is called the threshold frequency. Increasing the frequency of the incident beam increases the maximum kinetic energy of the emitted photoelectrons, and the stopping voltage has to increase. The number of emitted electrons may also change because the probability that each photon results in an emitted electron is a function of photon energy.

An increase in the intensity of the same monochromatic light (so long as the intensity is not too high[12]), which is proportional to the number of photons impinging on the surface in a given time, increases the rate at which electrons are ejected—the photoelectric current I—but the kinetic energy of the photoelectrons and the stopping voltage remain the same. For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.

The time lag between the incidence of radiation and the emission of a photoelectron is very small, less than 10−9 second. Angular distribution of the photoelectrons is highly dependent on polarization (the direction of the electric field) of the incident light, as well as the emitting material's quantum properties such as atomic and molecular orbital symmetries and the electronic band structure of crystalline solids. In materials without macroscopic order, the distribution of electrons tends to peak in the direction of polarization of linearly polarized light.[13] The experimental technique that can measure these distributions to infer the material's properties is angle-resolved photoemission spectroscopy.

### Theoretical explanation

Diagram of the maximum kinetic energy as a function of the frequency of light on zinc.

In 1905, Einstein proposed a theory of the photoelectric effect using a concept first put forward by Max Planck that light consists of tiny packets of energy known as photons or light quanta. Each packet carries energy ${\displaystyle h\nu }$  that is proportional to the frequency ${\displaystyle \nu }$  of the corresponding electromagnetic wave. The proportionality constant ${\displaystyle h}$  has become known as the Planck constant. The maximum kinetic energy ${\displaystyle K_{\max }}$  of the electrons that were delivered this much energy before being removed from their atomic binding is

${\displaystyle K_{\max }=h\,\nu -W,}$

where ${\displaystyle W}$  is the minimum energy required to remove an electron from the surface of the material. It is called the work function of the surface and is sometimes denoted ${\displaystyle \Phi }$  or ${\displaystyle \varphi }$ .[14] If the work function is written as
${\displaystyle W=h\,\nu _{o},}$

the formula for the maximum kinetic energy of the ejected electrons becomes
${\displaystyle K_{\max }=h\left(\nu -\nu _{o}\right).}$

Kinetic energy is positive, and ${\displaystyle \nu >\nu _{o}}$  is required for the photoelectric effect to occur.[15] The frequency ${\displaystyle \nu _{o}}$  is the threshold frequency for the given material. Above that frequency, the maximum kinetic energy of the photoelectrons as well as the stopping voltage in the experiment ${\textstyle V_{o}={\frac {h}{e}}\left(\nu -\nu _{o}\right)}$  rise linearly with the frequency, and have no dependence on the number of photons and the intensity of the impinging monochromatic light. Einstein's formula, however simple, explained all the phenomenology of the photoelectric effect, and had far-reaching consequences in the development of quantum mechanics.

### Photoemission from atoms, molecules and solids

Electrons that are bound in atoms, molecules and solids each occupy distinct states of well-defined binding energies. When light quanta deliver more than this amount of energy to an individual electron, the electron may be emitted into free space with excess (kinetic) energy that is ${\displaystyle h\nu }$  higher than the electron's binding energy. The distribution of kinetic energies thus reflects the distribution of the binding energies of the electrons in the atomic, molecular or crystalline system: an electron emitted from the state at binding energy ${\displaystyle E_{B}}$  is found at kinetic energy ${\displaystyle E_{k}=h\nu -E_{B}}$ . This distribution is one of the main characteristics of the quantum system, and can be used for further studies in quantum chemistry and quantum physics.

#### Models of photoemission from solids

The electronic properties of ordered, crystalline solids are determined by the distribution of the electronic states with respect to energy and momentum—the electronic band structure of the solid. Theoretical models of photoemission from solids show that this distribution is, for the most part, preserved in the photoelectric effect. The phenomenological three-step model[16] for ultraviolet and soft X-ray excitation decomposes the effect into these steps:[17][18][19]

1. Inner photoelectric effect in the bulk of the material that is a direct optical transition between an occupied and an unoccupied electronic state. This effect is subject to quantum-mechanical selection rules for dipole transitions. The hole left behind the electron can give rise to secondary electron emission, or the so-called Auger effect, which may be visible even when the primary photoelectron does not leave the material. In molecular solids phonons are excited in this step and may be visible as satellite lines in the final electron energy.
2. Electron propagation to the surface in which some electrons may be scattered because of interactions with other constituents of the solid. Electrons that originate deeper in the solid are much more likely to suffer collisions and emerge with altered energy and momentum. Their mean-free path is a universal curve dependent on electron's energy.
3. Electron escape through the surface barrier into free-electron-like states of the vacuum. In this step the electron loses energy in the amount of the work function of the surface, and suffers from the momentum loss in the direction perpendicular to the surface. Because the binding energy of electrons in solids is conveniently expressed with respect to the highest occupied state at the Fermi energy ${\displaystyle E_{F}}$ , and the difference to the free-space (vacuum) energy is the work function of the surface, the kinetic energy of the electrons emitted from solids is usually written as ${\displaystyle E_{k}=h\nu -W-E_{B}}$ .

There are cases where the three-step model fails to explain peculiarities of the photoelectron intensity distributions. The more elaborate one-step model[20] treats the effect as a coherent process of photoexcitation into the final state of a finite crystal for which the wave function is free-electron-like outside of the crystal, but has a decaying envelope inside.[19]

## History

### 19th century

In 1839, Alexandre Edmond Becquerel discovered the photovoltaic effect while studying the effect of light on electrolytic cells.[21] Though not equivalent to the photoelectric effect, his work on photovoltaics was instrumental in showing a strong relationship between light and electronic properties of materials. In 1873, Willoughby Smith discovered photoconductivity in selenium while testing the metal for its high resistance properties in conjunction with his work involving submarine telegraph cables.[22]

Johann Elster (1854–1920) and Hans Geitel (1855–1923), students in Heidelberg, investigated the effects produced by light on electrified bodies and developed the first practical photoelectric cells that could be used to measure the intensity of light.[23][24]: 458  They arranged metals with respect to their power of discharging negative electricity: rubidium, potassium, alloy of potassium and sodium, sodium, lithium, magnesium, thallium and zinc; for copper, platinum, lead, iron, cadmium, carbon, and mercury the effects with ordinary light were too small to be measurable. The order of the metals for this effect was the same as in Volta's series for contact-electricity, the most electropositive metals giving the largest photo-electric effect.

The gold leaf electroscope to demonstrate the photoelectric effect. When the electroscope is negatively charged, there is an excess of electrons and the leaves are separated. If short wavelength, high-frequency light (such as ultraviolet light obtained from an arc lamp, or by burning magnesium, or by using an induction coil between zinc or cadmium terminals to produce sparking) shines on the cap, the electroscope discharges, and the leaves fall limp. If, however, the frequency of the light waves is below the threshold value for the cap, the leaves will not discharge, no matter how long one shines the light at the cap.

In 1887, Heinrich Hertz observed the photoelectric effect[25] and reported on the production and reception[26] of electromagnetic waves.[27] The receiver in his apparatus consisted of a coil with a spark gap, where a spark would be seen upon detection of electromagnetic waves. He placed the apparatus in a darkened box to see the spark better. However, he noticed that the maximum spark length was reduced when inside the box. A glass panel placed between the source of electromagnetic waves and the receiver absorbed ultraviolet radiation that assisted the electrons in jumping across the gap. When removed, the spark length would increase. He observed no decrease in spark length when he replaced the glass with quartz, as quartz does not absorb UV radiation.

The discoveries by Hertz led to a series of investigations by Hallwachs,[28][29] Hoor,[30] Righi[31] and Stoletov[32][33] on the effect of light, and especially of ultraviolet light, on charged bodies. Hallwachs connected a zinc plate to an electroscope. He allowed ultraviolet light to fall on a freshly cleaned zinc plate and observed that the zinc plate became uncharged if initially negatively charged, positively charged if initially uncharged, and more positively charged if initially positively charged. From these observations he concluded that some negatively charged particles were emitted by the zinc plate when exposed to ultraviolet light.

With regard to the Hertz effect, the researchers from the start showed the complexity of the phenomenon of photoelectric fatigue—the progressive diminution of the effect observed upon fresh metallic surfaces. According to Hallwachs, ozone played an important part in the phenomenon,[34] and the emission was influenced by oxidation, humidity, and the degree of polishing of the surface. It was at the time unclear whether fatigue is absent in a vacuum.

In the period from 1888 until 1891, a detailed analysis of the photoeffect was performed by Aleksandr Stoletov with results reported in six publications.[33] Stoletov invented a new experimental setup which was more suitable for a quantitative analysis of the photoeffect. He discovered a direct proportionality between the intensity of light and the induced photoelectric current (the first law of photoeffect or Stoletov's law). He measured the dependence of the intensity of the photo electric current on the gas pressure, where he found the existence of an optimal gas pressure corresponding to a maximum photocurrent; this property was used for the creation of solar cells.[citation needed]

Many substances besides metals discharge negative electricity under the action of ultraviolet light. G. C. Schmidt[35] and O. Knoblauch[36] compiled a list of these substances.

In 1899, J. J. Thomson investigated ultraviolet light in Crookes tubes.[37] Thomson deduced that the ejected particles, which he called corpuscles, were of the same nature as cathode rays. These particles later became known as the electrons. Thomson enclosed a metal plate (a cathode) in a vacuum tube, and exposed it to high-frequency radiation.[38] It was thought that the oscillating electromagnetic fields caused the atoms' field to resonate and, after reaching a certain amplitude, caused subatomic corpuscles to be emitted, and current to be detected. The amount of this current varied with the intensity and color of the radiation. Larger radiation intensity or frequency would produce more current.[citation needed]

During the years 1886–1902, Wilhelm Hallwachs and Philipp Lenard investigated the phenomenon of photoelectric emission in detail. Lenard observed that a current flows through an evacuated glass tube enclosing two electrodes when ultraviolet radiation falls on one of them. As soon as ultraviolet radiation is stopped, the current also stops. This initiated the concept of photoelectric emission. The discovery of the ionization of gases by ultraviolet light was made by Philipp Lenard in 1900. As the effect was produced across several centimeters of air and yielded a greater number of positive ions than negative, it was natural to interpret the phenomenon, as J. J. Thomson did, as a Hertz effect upon the particles present in the gas.[27]

### 20th century

In 1902, Lenard observed that the energy of individual emitted electrons increased with the frequency (which is related to the color) of the light.[3] This appeared to be at odds with Maxwell's wave theory of light, which predicted that the electron energy would be proportional to the intensity of the radiation.

Lenard observed the variation in electron energy with light frequency using a powerful electric arc lamp which enabled him to investigate large changes in intensity, and that had sufficient power to enable him to investigate the variation of the electrode's potential with light frequency. He found the electron energy by relating it to the maximum stopping potential (voltage) in a phototube. He found that the maximum electron kinetic energy is determined by the frequency of the light. For example, an increase in frequency results in an increase in the maximum kinetic energy calculated for an electron upon liberation – ultraviolet radiation would require a higher applied stopping potential to stop current in a phototube than blue light. However, Lenard's results were qualitative rather than quantitative because of the difficulty in performing the experiments: the experiments needed to be done on freshly cut metal so that the pure metal was observed, but it oxidized in a matter of minutes even in the partial vacuums he used. The current emitted by the surface was determined by the light's intensity, or brightness: doubling the intensity of the light doubled the number of electrons emitted from the surface.

The researches of Langevin and those of Eugene Bloch[39] have shown that the greater part of the Lenard effect is certainly due to the Hertz effect. The Lenard effect upon the gas[clarification needed] itself nevertheless does exist. Refound by J. J. Thomson[40] and then more decisively by Frederic Palmer, Jr.,[41][42] the gas photoemission was studied and showed very different characteristics than those at first attributed to it by Lenard.[27]

In 1900, while studying black-body radiation, the German physicist Max Planck suggested in his "On the Law of Distribution of Energy in the Normal Spectrum"[43] paper that the energy carried by electromagnetic waves could only be released in packets of energy. In 1905, Albert Einstein published a paper advancing the hypothesis that light energy is carried in discrete quantized packets to explain experimental data from the photoelectric effect. Einstein theorized that the energy in each quantum of light was equal to the frequency of light multiplied by a constant, later called Planck's constant. A photon above a threshold frequency has the required energy to eject a single electron, creating the observed effect. This was a key step in the development of quantum mechanics. In 1914, Millikan's experiment supported Einstein's model of the photoelectric effect. Einstein was awarded the 1921 Nobel Prize in Physics for "his discovery of the law of the photoelectric effect",[44] and Robert Millikan was awarded the Nobel Prize in 1923 for "his work on the elementary charge of electricity and on the photoelectric effect".[45] In quantum perturbation theory of atoms and solids acted upon by electromagnetic radiation, the photoelectric effect is still commonly analyzed in terms of waves; the two approaches are equivalent because photon or wave absorption can only happen between quantized energy levels whose energy difference is that of the energy of photon.[46][17]

Albert Einstein's mathematical description of how the photoelectric effect was caused by absorption of quanta of light was in one of his Annus Mirabilis papers, named "On a Heuristic Viewpoint Concerning the Production and Transformation of Light". The paper proposed a simple description of light quanta, or photons, and showed how they explained such phenomena as the photoelectric effect. His simple explanation in terms of absorption of discrete quanta of light agreed with experimental results. It explained why the energy of photoelectrons was dependent only on the frequency of the incident light and not on its intensity: at low-intensity, the high-frequency source could supply a few high energy photons, whereas at high-intensity, the low-frequency source would supply no photons of sufficient individual energy to dislodge any electrons. This was an enormous theoretical leap, but the concept was strongly resisted at first because it contradicted the wave theory of light that followed naturally from James Clerk Maxwell's equations of electromagnetism, and more generally, the assumption of infinite divisibility of energy in physical systems. Even after experiments showed that Einstein's equations for the photoelectric effect were accurate, resistance to the idea of photons continued.

Einstein's work predicted that the energy of individual ejected electrons increases linearly with the frequency of the light. Perhaps surprisingly, the precise relationship had not at that time been tested. By 1905 it was known that the energy of photoelectrons increases with increasing frequency of incident light and is independent of the intensity of the light. However, the manner of the increase was not experimentally determined until 1914 when Robert Andrews Millikan showed that Einstein's prediction was correct.[4]

The photoelectric effect helped to propel the then-emerging concept of wave–particle duality in the nature of light. Light simultaneously possesses the characteristics of both waves and particles, each being manifested according to the circumstances. The effect was impossible to understand in terms of the classical wave description of light,[47][48][49] as the energy of the emitted electrons did not depend on the intensity of the incident radiation. Classical theory predicted that the electrons would 'gather up' energy over a period of time, and then be emitted.[48][50]

## Uses and effects

### Photomultipliers

Photomultiplier

These are extremely light-sensitive vacuum tubes with a coated photocathode inside the envelope. The photo cathode contains combinations of materials such as cesium, rubidium, and antimony specially selected to provide a low work function, so when illuminated even by very low levels of light, the photocathode readily releases electrons. By means of a series of electrodes (dynodes) at ever-higher potentials, these electrons are accelerated and substantially increased in number through secondary emission to provide a readily detectable output current. Photomultipliers are still commonly used wherever low levels of light must be detected.[51]

### Image sensors

Video camera tubes in the early days of television used the photoelectric effect, for example, Philo Farnsworth's "Image dissector" used a screen charged by the photoelectric effect to transform an optical image into a scanned electronic signal.[52]

### Photoelectron spectroscopy

Angle-resolved photoemission spectroscopy (ARPES) experiment. Helium discharge lamp shines ultraviolet light onto the sample in ultra-high vacuum. Hemispherical electron analyzer measures the distribution of ejected electrons with respect to energy and momentum.

Because the kinetic energy of the emitted electrons is exactly the energy of the incident photon minus the energy of the electron's binding within an atom, molecule or solid, the binding energy can be determined by shining a monochromatic X-ray or UV light of a known energy and measuring the kinetic energies of the photoelectrons.[17] The distribution of electron energies is valuable for studying quantum properties of these systems. It can also be used to determine the elemental composition of the samples. For solids, the kinetic energy and emission angle distribution of the photoelectrons is measured for the complete determination of the electronic band structure in terms of the allowed binding energies and momenta of the electrons. Modern instruments for angle-resolved photoemission spectroscopy are capable of measuring these quantities with a precision better than 1 meV and 0.1°.

Photoelectron spectroscopy measurements are usually performed in a high-vacuum environment, because the electrons would be scattered by gas molecules if they were present. However, some companies are now selling products that allow photoemission in air. The light source can be a laser, a discharge tube, or a synchrotron radiation source.[53]

The concentric hemispherical analyzer is a typical electron energy analyzer. It uses an electric field between two hemispheres to change (disperse) the trajectories of incident electrons depending on their kinetic energies.

### Night vision devices

Photons hitting a thin film of alkali metal or semiconductor material such as gallium arsenide in an image intensifier tube cause the ejection of photoelectrons due to the photoelectric effect. These are accelerated by an electrostatic field where they strike a phosphor coated screen, converting the electrons back into photons. Intensification of the signal is achieved either through acceleration of the electrons or by increasing the number of electrons through secondary emissions, such as with a micro-channel plate. Sometimes a combination of both methods is used. Additional kinetic energy is required to move an electron out of the conduction band and into the vacuum level. This is known as the electron affinity of the photocathode and is another barrier to photoemission other than the forbidden band, explained by the band gap model. Some materials such as gallium arsenide have an effective electron affinity that is below the level of the conduction band. In these materials, electrons that move to the conduction band all have sufficient energy to be emitted from the material, so the film that absorbs photons can be quite thick. These materials are known as negative electron affinity materials.

### Spacecraft

The photoelectric effect will cause spacecraft exposed to sunlight to develop a positive charge. This can be a major problem, as other parts of the spacecraft are in shadow which will result in the spacecraft developing a negative charge from nearby plasmas. The imbalance can discharge through delicate electrical components. The static charge created by the photoelectric effect is self-limiting, because a higher charged object doesn't give up its electrons as easily as a lower charged object does.[54][55]

### Moon dust

Light from the Sun hitting lunar dust causes it to become positively charged from the photoelectric effect. The charged dust then repels itself and lifts off the surface of the Moon by electrostatic levitation.[56][57] This manifests itself almost like an "atmosphere of dust", visible as a thin haze and blurring of distant features, and visible as a dim glow after the sun has set. This was first photographed by the Surveyor program probes in the 1960s,[58] and most recently the Chang'e 3 rover observed dust deposition on lunar rocks as high as about 28 cm.[59] It is thought that the smallest particles are repelled kilometers from the surface and that the particles move in "fountains" as they charge and discharge.[60]

## Competing processes and photoemission cross section

When photon energies are as high as the electron rest energy of 511 keV, yet another process, the Compton scattering, may take place. Above twice this energy, at 1.022 MeV pair production is also more likely.[61] Compton scattering and pair production are examples of two other competing mechanisms.

Even if the photoelectric effect is the favoured reaction for a particular interaction of a single photon with a bound electron, the result is also subject to quantum statistics and is not guaranteed. The probability of the photoelectric effect occurring is measured by the cross section of the interaction, σ. This has been found to be a function of the atomic number of the target atom and photon energy. In a crude approximation, for photon energies above the highest atomic binding energy, the cross section is given by:[62]

${\displaystyle \sigma =\mathrm {constant} \cdot {\frac {Z^{n}}{E^{3}}}}$

Here Z is the atomic number and n is a number which varies between 4 and 5. The photoelectric effect rapidly decreases in significance in the gamma-ray region of the spectrum, with increasing photon energy. It is also more likely from elements with high atomic number. Consequently, high-Z materials make good gamma-ray shields, which is the principal reason why lead (Z = 82) is preferred and most widely used.[63]

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34. ^ Hallwachs, W. (1907). "Über die lichtelektrische Ermüdung". Annalen der Physik. 328 (8): 459–516. Bibcode:1907AnP...328..459H. doi:10.1002/andp.19073280807.
35. ^ Schmidt, G. C. (1898) Wied. Ann. Uiv. p. 708.
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42. ^ Palmer, Frederic (1911). "Volume Ionization Produced by Light of Extremely Short Wave-Length". Physical Review. Series I. 32 (1): 1–22. Bibcode:1911PhRvI..32....1P. doi:10.1103/PhysRevSeriesI.32.1.
43. ^ Planck, Max (1901). "Ueber das Gesetz der Energieverteilung im Normalspectrum (On the Law of Distribution of Energy in the Normal Spectrum)". Annalen der Physik. 4 (3): 553. Bibcode:1901AnP...309..553P. doi:10.1002/andp.19013090310.
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46. ^ Lamb, Jr., Willis E.; Scully, Marlan O. (1968). "The photoelectric effect without photons" (PDF). Coral Gables, FL: Center for Theoretical Physics, University of Miami. we understand the photoeffect as being the result of a classical field falling on a quantized atomic electron
47. ^ Resnick, Robert (1972) Basic Concepts in Relativity and Early Quantum Theory, Wiley, p. 137, ISBN 0-471-71702-9.
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49. ^ Penrose, Roger (2005) The Road to Reality: A Complete Guide to the Laws of the Universe, Knopf, p. 502, ISBN 0-679-45443-8
50. ^ Resnick, Robert (1972) Basic Concepts in Relativity and Early Quantum Theory, Wiley, p. 138, ISBN 0-471-71702-9.
51. ^ Timothy, J. Gethyn (2010) in Huber, Martin C.E. (ed.) Observing Photons in Space, ISSI Scientific Report 009, ESA Communications, pp. 365–408, ISBN 978-92-9221-938-3
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53. ^ Weaver, J. H.; Margaritondo, G. (1979). "Solid-State Photoelectron Spectroscopy with Synchrotron Radiation". Science. 206 (4415): 151–156. Bibcode:1979Sci...206..151W. doi:10.1126/science.206.4415.151. PMID 17801770. S2CID 23594185.
54. ^ Lai, Shu T. (2011). Fundamentals of Spacecraft Charging: Spacecraft Interactions with Space Plasmas (illustrated ed.). Princeton University Press. pp. 1–6. ISBN 978-0-691-12947-1.
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# Integration along a keyhole

(H. Priestley complex Analysis Chapter 7 Exercise 9)

Suppose $f$ is holomorphic inside and on $\gamma(0,1)$. By integration around the usual keyhole like this one : Integration of $\ln$ around a keyhole contour , prove that:

$$\int_0^1 f(x)\,dx=\frac{1}{2i \pi} \int_{\gamma(0,1)} f(z) (\ln(z) - i \pi)\, dz$$

where $\ln(z)$ denotes the branch of the logarithm with imaginary part between $[0,2\pi]$

I am not sure how to tackle this question :(

the branch of the $\ln$ will be $\ln(z) = \ln (|z|) + i \theta$

I thought that maybe if I integrate $f(z) (\ln(z) - i \pi)$ along the entire contour I will get $0$ by Cauchy's theorem as it is holomorphic inside by design:

$$\int_{\gamma(0,1)} f(z) (\ln(z) - i \pi)\, dz + \int_{\text{top edge}} f(z) (\ln(z) - i \pi)\, dz + \int_{\text{bottom edge}} f(z) (\ln(z) - i \pi)\, dz - \int_{\gamma(0,\epsilon)} f(z) (\ln(z) - i \pi)\, dz = 0,$$

i.e.

$$\int_{\gamma(0,1)} f(z) (\ln(z) - i \pi) i\, dz + \int_{0}^{1} f(x) (\ln(x) - i \pi)\, dx + \int_{1}^{0} f(x) (\ln(x)+2i \pi - i \pi)\, dx - \int_{\gamma(0,\epsilon)} f(\epsilon e^{i\theta}) (\ln(\epsilon) +i\theta - i \pi)\, dz = 0.$$

That is, with a fortunate cancellation:

$$\int_{\gamma(0,1)} f(z) (\ln(z) - i \pi)\, dz -2i \pi \int_{0}^{1} f(x)\, dx - \int_{\gamma(0,\epsilon)} f(\epsilon e^{i\theta}) (\ln(\epsilon) +i\theta - i \pi) \epsilon i e^{i\theta}\, d\theta = 0.$$

however I'm not sure about what happens with the very last integral, if the integrand were holomorphic apart from a simple pole then it would just be $2\pi \times\text{residue}$, i.e. $0$ since $f$ is holomorphic inside but with the $\ln()$ it makes it complicated...

You have tackled it correctly, all that is missing is that you see

$$\lim_{\epsilon \to 0} \int_{\gamma(0,\epsilon)} f(z)(\ln z - i\pi)\,dz = 0.$$

The standard estimate

$$\biggl\lvert \int_\Gamma g(z)\,dz\biggr\rvert \leqslant L(\Gamma)\cdot \sup \{ \lvert g(\zeta)\rvert : \zeta \in \operatorname{Tr}(\Gamma)\}$$

yields that, since $f$ is bounded, and

$$\lim_{\epsilon \to 0} \epsilon\cdot \lvert\ln\epsilon\rvert = 0.$$

The length of $\gamma(0,\epsilon)$ is bounded by $2\pi\epsilon$, and the integrand is bounded by $\lVert f\rVert_\infty(\pi + \lvert\ln\epsilon\rvert)$.

• Of course ! Many thanks – user3203476 Jul 5 '15 at 13:57

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# 生产者-消费者问题（The Producer-Consumer Problem）

In the producer–consumer problem, one process (the producer) generates data items and another process (the consumer) receives and uses them. They communicate using a queue of maximum size N and are subject to the following conditions:

• the consumer must wait for the producer to produce something if the queue is empty;
• the producer must wait for the consumer to consume something if the queue is full.

The semaphore solution to the producer–consumer problem tracks the state of the queue with two semaphores: emptyCount, the number of empty places in the queue, and fullCount, the number of elements in the queue. To maintain integrity, emptyCount may be lower (but never higher) than the actual number of empty places in the queue, and fullCount may be lower (but never higher) than the actual number of items in the queue. Empty places and items represent two kinds of resources, empty boxes and full boxes, and the semaphores emptyCount and fullCount maintain control over these resources.

The binary semaphore useQueue ensures that the integrity of the state of the queue itself is not compromised, for example by two producers attempting to add items to an empty queue simultaneously, thereby corrupting its internal state. Alternatively a mutex could be used in place of the binary semaphore.


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# Differential Amplifier not working properly.

1. May 18, 2012

### Xyius

I want to learn more about Op-Amps and amplification so I am making a simple differential amplifier. The circuit looks like this...

I set it up and it isn't working correctly. First I tried putting 3 volts for Vin+ and ground for Vin-, and the voltage accross Vout was around half a volt. I don't understand this result. Shouldn't it be an amplification of the difference in voltage? (Which is 3 volts correct?) How can I get half a volt? I didn't know what value of resistors to use so I used 100 ohm for RC1 and RC2 and 150 ohm for Re (Completely arbitrarily chosen).

Can anyone help me out?
Thanks!!

2. May 18, 2012

### Jony130

You need to tell as about V+ ? , -V and Re = ?.

3. May 18, 2012

### Xyius

In my original post I said...
V+ = 3V
V- = Ground
Re = 150 ohms
And the other two resistors are 100 ohms.

EDIT: Ohh and I am using a 6.5 volt power source, if that is what you meant. Sorry!

4. May 18, 2012

### the_emi_guy

Xyius,
You need to be more specific. Do you have v+ at 6.5V and v- at 0V, or do you have v+ at 0V and v- at -6.5V (or something else). Also, when you say the input is 3V where exactly are you measuring. In other words, what node are you assuming to be 0V (where are you putting the black lead of your meter when you are putting red lead on Vin and seeing 3V).

5. May 18, 2012

### Xyius

Sorry about the ambiguity. As you can probably tell I am new at this!
The way I have it set up is. I put the positive lead of the 6.5 volt battery (It WAS a 9V but I have been using it for awhile..) for V+ and I put the ground for V-. As for the inputs, I put the positive lead of the 3V battery on one of the inputs and the ground to the other input. When measuring V+ and V- my black lead is on the ground for that battery, and likewise for the 3V power source.

6. May 18, 2012

### Jony130

This is how you should set your differential amplifier.

You need to use two battery one for Vcc and second one for Vee.
Also kept in mind that the linear region for BJT differential amplifier is very small.
If +Vin - (-Vin) = is greater then 150mV the one of a bjt will be in cut-off and the second one will be in saturation.

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7. May 18, 2012

### Jony130

If I understand your post correctly you have build this circuit

If so. This circuit has nothing to do with real differential amplifier.
Q2 is in cut-off and Q1 is in active region.

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8. May 18, 2012

### Xyius

Thank you very much! I will study this later and try to re-create it. I just have a few questions..
1.) What roles to the values of the resistors play? What values should I use?
2.) Same question, but for the batteries.
3.) Why is it that one BJT will cut off and one will be in saturation after a 150 mV difference? Would that mean that one BJT is doing all the work while the other is doing nothing?

Thanks again!

9. May 18, 2012

### yungman

First, for a differential pair to work properly, you have to have a matched pair for Q1 and Q2. But this is not your problem. Your differential voltage is too large. You have +3 at base of Q1, and 0 at base of Q2, Q1 is totally on and Q2 is totally off. You don't have a differential pair anymore.

Differential voltage range between Q1 and Q2 can only be very small by equation:

$$V_{B1}-V_{B2}=V_T ln\left[ \frac{I_{C1}}{I_{C2}}\right]\;\Rightarrow\; \frac{I_{C1}}{I_{C2}}\;=\; e^{\left({\frac{V_{B1}-V_{B2}}{V_T}}\right)}\;\hbox { where }\; V_T≈\hbox { 25mV.}$$

As you can see, if the input differential voltage is much larger than 25mV ( 250mV) the ratio is going to be very large and consider cocked to one side. Q2 is turn off and disappeared.

This will hold true even in the more clearly defined biasing condition in the later posts.

Last edited: May 18, 2012
10. May 20, 2012

### Jony130

Well the resistor sets various of the amplifier parameters. The most important one is of course the voltage gain. But also the proper DC bias point is also important.
For example if you use two 9V batteries then. It is good to have:
Rc = (0.5*Vcc)/Ic and Re = (Vee - Vbe)/ (2Ic)

So for example

Rc = 4.5V/4.5mA = 1KΩ

And

Re = (9V - 0.65V)/9mA = 920Ω ≈ 1KΩ

I use Vcc = 9V and Vee = 9V

We use symmetric power supply because it is a natural way to power differential Amplifier.
Only in this case we can connect the BJT bases to gnd but we still have bout bjt to work in there linear region. So we made a true DC amplifier.

exactly

Last edited by a moderator: May 6, 2017
11. May 21, 2012

### Xyius

Thank you very much!!! The link you provided is proving to be very helpful! I just have one question. Why do they use an exponential relationship for the current in section 7.3.2? How would I derive or deduce that it is an exponential relationship?

Last edited by a moderator: May 6, 2017
12. May 22, 2012

### Jony130

When we star learn about BJT the told as that BJT is a current control (base current) current source.
Ic = β * Ib
But this simply model don't tell as the full story about BJT.

In reality the collector current is equal to

$$I_{C}≈I_{S}*\hbox e^{(\frac{V_{BE}}{V_T})}$$

And this e in the equation tell as that this is a exponential relationship.

http://www.wolframalpha.com/input/?i=e^x

13. May 22, 2012

### yungman

And if you use this equation for Q1 and Q2 and divide equation of Q1 by Q2, you get the equation I posted.

14. May 22, 2012

### Xyius

I was asking on how would I derive the exponential relationship. (I am very well versed in mathematics, I will have a bachelors in applied physics in a year) Is there a differential equation I can solve that gives it to me?

15. May 22, 2012

### Jony130

OK I understand what you want. But I don't know the answer to your question.
But if you interest in this you should buy some book about semiconductor physics.

16. May 28, 2012

### Bob S

Here is a simulation (first attachment) of a 2N5210 differential amplifier using LT Spice (free software). Vcc = + 9 volts, and Vee = -9 volts, with the base quiescent voltages at ground. The common emitter resistor Re is 420 ohms, to set the collector currents at about 10 mA. The collector resistors are both 420 ohms. The quiescent collector voltages are about 4.9 volts and the emitters about -0.7 volts. The input pulse is a 10 mV, 200 ns long pulse with a 2 ns rise and fall time. The full differential output is about 1.1 volts, representing a full differential gain of about 110.

The second attachment from Fairchild Semiconductor http://pdf1.alldatasheet.com/datasheet-pdf/view/101319/FAIRCHILD/2N5210.html shows the 2N5210 gain-bandwidth product contours vs. collector voltage and collector current. With a collector voltage of 5.4 volts, and a collector current of 10 milliamps, the gain-bandwidth product is 175 MHz. (I don't think the simulation is accurate).

Fairchild used to publish the gain-bandwidth plot with all its fast transistors.

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17. May 28, 2012

### rbj

your diff amp will have better common-mode rejection if you can make the Re as large as possible (which will require Vee to be even more negative). what you really want is a constant current source instead of the Vee/Re combination and you can get that with a transistor (also NPN) with a very low hoe parameter.

if you do that and your transistor pair is well matched (and two Rc are well matched), you should have a very good diff amp.


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# Equality between k and p forms

1. Nov 3, 2014

### davi2686

if i have $\int_{\partial S} \omega=0$ by stokes theorem $\int_{S} d \omega=0$, can i say $d \omega=0$? even 0 as a scalar is a 0-form?

2. Nov 3, 2014

### ShayanJ

Consider $d\omega=x^3 dx$ integrated over $S=(-a,a)$. The integral gives zero but the integrand is zero in only one point of the region of integration. So this is a counterexample to $\int_S d\omega=0 \Rightarrow d\omega=0$.

3. Nov 3, 2014

### davi2686

thanks, but have no problem with 0 is a 0-form and $d\omega$ a k-form? so can i work with something like $d\omega=4$?

4. Nov 3, 2014

### HallsofIvy

Staff Emeritus
I have no idea what "$d\omega= 4$", a differential form equal to a number, could even mean. Could you please explain that?

5. Nov 3, 2014

### davi2686

my initial motivation is in Gauss's Law, $\int_{\partial V} \vec{E}\cdot d\vec{S}$=$\int_V \frac{\rho}{\epsilon_0}dV$, i rewrite the left side with differential forms, $\int_{\partial V} \star\vec{E}^{\flat}=\int_V \frac{\rho}{\epsilon_0}dV$ which by the Stokes Theorem $\int_{V} d(\star\vec{E}^{\flat})=\int_V \frac{\rho}{\epsilon_0}dV\Rightarrow d(\star\vec{E}^{\flat})=\frac{\rho}{\epsilon_0}$, if i dont make something wrong in these steps, in left side we get a n-form and right side a 0-form, and that i dont know if i can do.

6. Nov 3, 2014

### ShayanJ

Its correct that 0 is a 0-form but by a zero 1-form we actually mean $\omega= 0 dx$ and write it as $\omega= 0$ when there is no chance of confusion.

You missed something. You should have written $d(\star \vec E ^\flat)=\frac{\rho}{\epsilon_0} dV$.(What's $\flat$ anyway?)

7. Nov 5, 2014

### davi2686

Thanks i did not know that.

That is musical isomorphism $\flat:M \mapsto M^*$, in fact i understand it works like a lower indice, $\vec{B}^{\flat}$ give me a co-variant B or it related 1-form.


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# Probability that the average is within range

I'm currently preparing for an exam, and I just can't seem to grasp how to go about the question given the following information:

Information: A brick factory produce bricks whose length $X_i$ can be assumed normally distributed with mean length of $228 mm$ and standard deviation $4 mm$ ..

Question: At the brick factory a random sample of $50$ bricks is taken and the average of the lengths calculated by $$\bar X=\frac{1}{50}\sum^n_{i=1}X_i$$ What is the probability that the average of the observed lengths $\bar X$, is within the range of $[227;229]$.

My thoughts: In the question before this, I was asked to calculate the 0.95 CI and found it to be in the range: $220 mm < X < 236 mm$, so I would assume when we averaged the random sample of 50 bricks, the probability would be over $90$ pct.

I hope someone can point me in the right direction.

• Do you know what the mean and standard deviation of $\bar{X}$ are? – Dilip Sarwate Dec 12 '15 at 15:03
• I don't know anything besides what I stated under "Information:" .. So to sum up I know the overall mean of 228 and the standard deviation 4. But I of course know nothing about the values of the random sample of 50 bricks :) – Akudo Dec 12 '15 at 15:07
• Then you really need to read your textbook instead of trying to solve this problem. The question I asked was not about the individual values of the random sample, but about the average of the 50 values, that $\bar{X}$ in your "question" Do you know what $\bar{X}$ might be expected to be? 228 mm? 220 mm? – Dilip Sarwate Dec 12 '15 at 15:13
• Alright, I will do that instead.. $\bar X$ is expected to be roughly 228 mm, roughly because we are estimating a proportion (50) of the original sample. – Akudo Dec 12 '15 at 15:19
• Think about what happens when you sum up normally distributed variables? – abaumann Dec 12 '15 at 15:40

You need to use some basic results about means and variances

• The expectation of a sum is the sum of the expectations

• The expectation of a constant times a random variable is the constant times the expectation of the variable

• The variance of a sum of independent random variables is the sum of the variances

• The variance of a constant times a random variable is the square of the constant times the variance of the variable

The two results about expectation mean that the distribution of a sample mean of identically distributed quantities has the same mean as the original variable. The two results about variance mean that the variance of a sample mean (of independent identically distributed variates) has variance that's the original variance divided by the sample size.

Hence the sample mean of a sample of 50 bricks has mean $228$ mm and standard deviation $\frac{4}{\sqrt{50}}$ mm.

• You also need that linear combinations of independent normal random variates will also be normally distributed.

As a result, the sample mean will be $\sim N(228,0.5657)$

From there you should be able to find the probability called for in the question.


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# All-sky search for short gravitational-wave bursts in the third Advanced LIGO and Advanced Virgo run

Document #:
LIGO-P2100045-v8
Document type:
P - Publications
Other Versions:
 LIGO-P2100045-v7 08 Jul 2021, 06:56
Abstract:
This paper presents the results of a search for generic short-duration gravitational-wave transients in data from the third observing run of Advanced LIGO and Advanced Virgo. Transients with durations of milliseconds to a few seconds in the 24-4096 Hz frequency band are targeted by the search, with no assumptions made regarding the incoming signal direction, polarization or morphology. Gravitational waves from compact binary coalescences that have been identified by other targeted analyses are detected, but no statistically significant evidence for other gravitational wave bursts is found. Sensitivities to a variety of signals are presented. These include updated upper limits on the source rate-density as a function of the characteristic frequency of the signal, which are roughly an order of magnitude better than previous upper limits. This search is sensitive to sources radiating as little as $$\sim$$10$$^{-10} M_{\odot} c^2$$ in gravitational waves at $$\sim$$70 Hz from a distance of 10 kpc, with 50% detection efficiency at a false alarm rate of one per century. The sensitivity of this search to two plausible astrophysical sources is estimated: neutron star f-modes, which may be excited by pulsar glitches, as well as selected core-collapse supernova models.
Files in Document:
Topics:
Author Groups:
Referenced by:

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text:
# Why use a laurent series in complex analysis?

ENgez
In complex analysis, what exactly is the purpuse of the luarent series, i mean, i know that it apporximates the function like a taylor series, an if the function is analytic in the whole domain it simplifies in to a taylor series. But i fail to see its purpose - what does it do that the taylor series can't?

spamiam
Hello!

In complex analysis, we're often interested in functions that are not holomorphic/analytic, but are meromorphic: http://en.wikipedia.org/wiki/Meromorphic_function.

In this case, Laurent series are a generalization of Taylor series, since now we can approximate functions that have poles using a series. Take a look at this article: http://en.wikipedia.org/wiki/Residue_(complex_analysis).

Also, just like power series, the collection of Laurent series has a rich algebraic structure as well.


text:
# LXX International conference "NUCLEUS – 2020. Nuclear physics and elementary particle physics. Nuclear physics technologies"

Oct 11 – 17, 2020
Online
Europe/Moscow timezone

## Physics and astrophysics of ultra-high energy cosmic rays: recent results from the Pierre Auger Observatory

Oct 13, 2020, 1:20 PM
35m
Online

#### Online

Plenary report Section 5. Neutrino physics and astrophysics.

### Speaker

Prof. Joao de Mello Neto (Physics Institute - Federal University of Ro de Janeiro)

### Description

Ultra-high-energy cosmic rays (UHECRs) are the highest energy messengers in the universe, with energies up to $10^{20}$ eV. Studies of astrophysical particles (nuclei, electrons, neutrinos and photons) at their highest observed energies have implications for fundamental physics as well as astrophysics. The primary particles interact in the atmosphere (or in the Earth) and generate extensive air showers. Analysis of those showers enables one not only to estimate the energy, direction and most probable mass of the primary cosmic particles, but also to obtain information about the properties of their hadronic interactions at an energy more than one order of magnitude above that accessible with the current highest energy human-made accelerator.
The Pierre Auger Observatory, located in the province of Mendoza, Argentina, is the biggest cosmic ray experiment ever built. The Observatory was designed as a hybrid detector covering an area of 3000 km$^2$ and it has been taking data for more than twenty years.
In this talk a selection of the latest results is presented: the cosmic ray energy spectrum, searches for a directional anisotropy and studies of mass composition (including the photon and neutrino searches). Finally, the current upgrade ("AugerPrime“) of the Observatory, which is mostly aimed at improving the sensitivity to the particle type and mass of ultra-high energy cosmic rays, is described.

### Primary author

Prof. Joao de Mello Neto (Physics Institute - Federal University of Ro de Janeiro)


text:
# Battery Backup System

Joined Mar 22, 2020
2
I am building a 150 plant hydroponic system and want the piece of mind of a battery backup.

Brand new to an Arduino, but have played quite a bit with a PI. I purchased an Arduino and plan to use it as the smarts of my system/remote monitoring.

1) I have a 110v automation system that I want to switch over to battery/inverter power in the case of a power outage.
2) Ideas or suggestions? I found similar posts on battery/relays, but not for Arduino I am thinking...

-Arduino pulls power directly from battery backup system with fused 12v step down (from ~14.4v battery)
-solar keeps batteries topped off
-Scrounging the parts bin for any 110v transformer with an Arduino friendly input current... this will signal Arduino for power outage event.
-Arduino to cuts signal to NO relay A if power outage >5 minutes...
-Relay cuts 110V power to automation system
-Arduino sends signal to NO relay B after a couple seconds, this kicks on my inverter and sends 110v to my internet modem and automation system.
-Pumps, timers, etc run as usual or until batteries die.
-Arduino monitors for 110v restoration
-Arduino sees 110v is back online
-Arduino kills signal to relay B, Inverter cut off.
-Arduino waits a few seconds and sends signal to relay A, input now routed from grid, power restored to automation system
-Arduino sends me an update that system is back online/health check.
-This would also give me a nice piece of mind if someone killed my power while I wasn't home, and could kick on my security cameras, onto battery power.

Am I overthinking this? Is there an easier way!? What am I not thinking of?

A UPS is too expensive when I can get almost 300AH out of a couple deep cycle batteries for a couple hundred bucks.

#### nsaspook

Joined Aug 27, 2009
8,520
Are you sure your plants are NOT worth the cost of a good UPS?

Joined Mar 22, 2020
2
I live on the top of a mountain, I need days of backup power. Even a $500 ups will be out in hours. Im growing fruits and vegetables, and can build this for$20. I have everything but a dpdt.

#### nsaspook

Joined Aug 27, 2009
8,520
I live on the top of a mountain, I need days of backup power. Even a $500 ups will be out in hours. Im growing fruits and vegetables, and can build this for$20. I have everything but a dpdt.
What are your daily KWH requirements? A standby generator might be more practical depending on the power several days requirements unless you need silence at all times while growing your fruits and vegetables.


text:
# Iterate over all defined PGF keys

Is there a way to iterate over all PGF keys that have been defined previously? My idea is to capture the value of all keys in order to do Proper nesting of tikzpicture environments: Reset all PGF values to their defaults.

EDIT: I imagine something along these lines:

\documentclass{article}
\beginrecordallpgfkeys
\usepackage{tikz}
\endrecordallpgfkeys

\begin{document}
\restoreallrecordedpgfkeys
\end{document}

I wouldn't mind if recording and restoring would use a temporary file.

-

I think an option is to use a macro for setting the keys in the first place \def\setkeysdefault{\pgfkeysset{...}} then is easy to have them reset. Is this a good option? –  Yiannis Lazarides Apr 5 '12 at 13:04
Who would have to implement/maintain \setkeysdefault? –  krlmlr Apr 5 '12 at 15:40
Collecting/saving defined pgf keys should presumably be cheaper than in the case of ltxkeys. In ltxkeys there is a way to iterate over all defined ltxkeys that have been previously defined. Yes, it is expensive but not too expensive. All defined keys and their default/initial values are available at any time in the macro \<prefix>@<family>@intialkeyvalues. This is used to recover all default values of keys at once and to, if needed, reinitialize keys (ie, reset keys to their default values) at any time. –  Ahmed Musa Apr 5 '12 at 17:01


text:
# 'Thermal Analysis of CPU with different Heat Sinks' simulation project by pankajkumar979

#1

I created a new simulation project called 'Thermal Analysis of CPU with different Heat Sinks':

The cooling mechanism is very crucial to ensure the rated performance of CPU, at the same time it accounts for the major operating cost of the system. Therefore, an efficient electronics cooling is necessary to provide the continuous operations at low operating costs. Tools of computational Fluid Dynamics are widely being used to optimize the cooling mechanism of the chips. In this study the focus is on finding the optimum heat sink design for electronics cooling.

More of my public projects can be found here.

#2

# Description

Heat transfer can be defined as energy transfer between systems and occurs only because of temperature differences. Heat flows spontaneously from a hotter body. Electronic devices produce heat as a by-product of normal operation. When electronic current flows through a semiconductor or a passive device, a portion of the power is dissipated as heat energy.  In above mentioned example heat was transferred by both conduction and convection. In total, there are three modes of heat transfer, by conduction, by convection and by radiation. Heat transfer in solids occurs by conduction mainly because of temperature gradient in different parts of the material.

Conjugate heat transfer (CHT) refers to the coupling of conduction in solids with the convective and radiative het transfer in the surrounding fluids. The heat transfer between solid and fluid occurs by the process of conduction and then it gets carried away the current of the fluid. When the solid body is sufficiently hot then heat transfer also occurs via radiation, however radiation is not considered in this example. Accurately predicting CHT is an essential part of the design process in many industrial applications including electronics cooling, turbo-machinery and HVAC.

The heat transfer rate in a fluid flow problem depends upon velocity, temperature and Geometry. For better understanding, velocity profile and temperature profile inside the cavity at a steady state are drawn in the next sections and a parametric study at different Fin width is done.

## Problem statement

Heat transfer between a solid and fluid is investigated in this section. Heat is generated inside the solid at 80W. Flow under investigation is assumed to be incompressible and laminar.

Figure 1.1.: Domain of the case with heat source, inflow on left side and outflow on the right side.

Figure 1.2.: Detail view of the fins geometry(in mm) No. of Fins 26

Schematic diagram of the problem is shown above and the boundary conditions. Walls are isolated and heat flux across them is zero. Heat is generated inside Silicon chip. Zero Dirichlet boundary condition is used for velocity on all walls. The material properties of air at temp 300.0K are used for inlet flow and also for the initial condition.

## Mesh

Case is solved using a increasing grid density, highest at the fins for high resolution which enables in capturing the high variation in temperature gradient.

Figure1.3 Mesh for case-0.50mm Fin size, no. of cells = 513557.

## Results and discussion

All simulations were run till a stable temperature at the face of a silicon chip is reached signifying that a steady solution is achieved.

Figure 1.4 Temperature at different Fin Size.

As the Fin pitch remains constant at 2.5mm but fin size is increased, it can be seen that heat transfer because of convection is much more in case of a smaller fin size.

Figure 1.5 Pressure at different Fin Size.

It is observed from the pressure and temperature plots that stacking too many fins is not a solution for decreasing the Temp on the heat sink since this prevents the passage of air coming from the inlet to the hottest part of the heat sink.

Figure 1.6 Variation of Temperature with the vertical distance at the centre of geometry.

Above Figure presents the comparison between the Temperature at the centre of geometry for different fin pitch. We can see that there is not much variation in the temperature inside the solids and but there is a sharp decline near the contact between fins and fluid.

Figure 1.7 Average Temperature at the face of the chip. F 0.50 P2.50 refers that the Fin size is 0.5mm and fin pitch is 2.50mm.

Temperature centres around 342K when Fin pitch is increased more than 3mm.

# Comparison with Experimental Data

In the paper ( 1 ) experiment results are compared with the difference in maximum and the minimum temperature obtained in the simulations. Figure 1.8 belows shows the method used for finding the the maximum and the minimum temperature in the Plate Heat Sink.

Final Result obtained after comparison is shown below.

## Comparison with Analytical solution

Re_{d} = \frac{U_{\infty L}}{\nu} = \frac{2.562 \times 0.04}{1.8 \times 10^{-5}} = 5693

Nusselt number in case of heat transfer from a rectangular fin can be described from the equation below:

$$Nu_{d} = c Re^{m} Pr^{\frac{1}{3}}$$

Here \mathbf{c} is a constant 0.228
\mathbf{Re} is the Reynolds number
\mathbf{Pr} is the Prandtl number (0.71 in case of air
\mathbf{m} depends on the type of fin( 0.731 )
After calculation Nu_{d} is 113.112, with the help of Nu_{d}, heat transfer coefficient can be found easily

By using the formula h = Nu\times \frac{\kappa}{L}, it comes out to be 79.127

\vspace{100pt}

Heat gained by air is equal to the heat released by the Fins.

Heat content of air at inflow =

= \rho A * \textbf{inlet velocity} * \textbf{Specific Heat} *\textbf{inlet temperature(Avg 300K)} \\ = 8753.520182

Heat content of air at outflow =

= \rho* A * \textbf{outlet velocity} * \textbf{Specific Heat} * \textbf{outlet temperature(Avg 309.35) } \\ = 9026.417549

Net energy gained = 272.8973664

Heat transfer equation in solids is as follows:

\frac{Q}{A} = \kappa \frac{T_{0} - T_{1}}{L}

Q is the heat produced in W, A is the area of cross-section of solids. \kappa is the thermal conductivity.

In our case, Q = 80W
As total no. of fins = 22 and 2 sides
A = 0.054 \times 0.064 \times (44-1) = 0.09288m^{2}
L = 0.04m
\kappa = 240 W/m^{2}.K

Temperature difference comes out to be = 37.108K
Average Temperature at CPU face in Simulation = 343.0K
Average Temperature at CPU calculated analytically = 337.108K

Error = 15.94 \%

\textbf{Temperature variation comparison in fin between Simulation and Analytical Solution}

\frac{T-T_{\infty}}{T_{0} - T_{\infty}} = \frac{cosh[(1-\frac{x}{L})\sqrt{\frac{hP}{\kappa A}}L]}{cosh[\sqrt{\frac{hP}{\kappa A}}L]}

Error between the above curves comes out to be 7.16/43.0 = 16.65 %

## References

[1] R.Mohan and Dr.P.Govindarajan, Thermal Analysis of CPU with variable Heat Sink Base Plate
Thickness using CFD.
[2] Mohan, R. & Govindarajan, P. J Mech Sci Technol (2011) 25: 2003. https://doi.org/10.1007/s12206-011-0531-8.
[3] Ralph Remsburg, Thermal Design of Electronic Equipment.

#3

That’s amazing @pankajkumar979! Very good job

All the best!

Jousef

#4

Fantastic job @pankajkumar979! Congrats!

Cheers, Bruno

#5

Very nice @pankajkumar979 - thanks a lot for sharing!

#6

Thanks! This project took 5000 core hours.


text: How can I hit studs and avoid cables when installing a TV mount? What environmental conditions would result in Crude oil being far easier to access than coal? See pages that link to and include this page. The direction of the cross product of a and b is perpendicular to the plane which contains a and b. To improve this 'Volume of a tetrahedron and a parallelepiped Calculator', please fill in questionnaire. So we have-- … Then the area of the base is. It follows that is the volume of the parallelepiped defined by vectors , , and (see Fig. What should I do? , site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. The volume of any tetrahedron that shares three converging edges of a parallelepiped is equal to one sixth of the volume of that parallelepiped (see proof). \text{volume of parallelopiped} &= \text{area of base} \times \text{height}\\ Notify administrators if there is objectionable content in this page. Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Truesight and Darkvision, why does a monster have both? Theorem: Given an m-dimensional parallelepiped, P, the square of the m-volume of P is the determinant of the matrix obtained from multiplying A by its transpose, where A is the matrix whose rows are defined by the edges of P. &= \mathbf a\cdot(\mathbf b \times \mathbf c) General Wikidot.com documentation and help section. The height is the perpendicular distance between the base and the opposite face. Proof of the theorem Theorem The volume 푉 of the parallelepiped with? Track 11. Depending on how rigorous you want the proof to be, you need to say what you mean by volume first. \text{volume of parallelopiped} &= \text{area of base} \times \text{height}\\ The Volume of a Parallelepiped in 3-Space, \begin{align} h = \| \mathrm{proj}_{\vec{u} \times \vec{v}} \vec{w} \| = \frac{ \mid \vec{w} \cdot (\vec{u} \times \vec{v}) \mid}{\| \vec{u} \times \vec{v} \|} \end{align}, \begin{align} V = \| \vec{u} \times \vec{v} \| \frac{ \mid \vec{w} \cdot (\vec{u} \times \vec{v}) \mid}{\| \vec{u} \times \vec{v} \|} \\ V = \mid \vec{w} \cdot (\vec{u} \times \vec{v}) \mid \end{align}, \begin{align} V = \mathrm{abs} \begin{vmatrix} w_1 & w_2 & w_3 \\ v_1 & v_2 & v_3\\ u_1 & u_2 & u_3 \end{vmatrix} \end{align}, \begin{align} \begin{vmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ w_1 & 0 & 1 \end{vmatrix} = 0 \end{align}, Unless otherwise stated, the content of this page is licensed under. Then how to show that volume is = [a b c] The length and width of a rectangular parallelepiped are 20 m and 30 m. Knowing that the total area is 6200 m² calculates the height of the box and measure the volume. View wiki source for this page without editing. area of base of parallelepiped (parallelogram) = \mathbf b \times \mathbf c, the vector \mathbf b \times \mathbf c will be perpendicular to base, therefore: With Area and volume interpretation of the determinant: (1) ± a b1 1 a b2 = area of parallelogram with edges A = (a1,a2), B = (b1,b2). Notice that we SSH to multiple hosts in file and run command fails - only goes to the first host. Consider three vectors , , and .The scalar triple product is defined .Now, is the vector area of the parallelogram defined by and .So, is the scalar area of this parallelogram times the component of in the direction of its normal. The volume of one of these tetrahedra is one third of the parallelepiped that contains it. As soos as, scalar triple product of the vectors can be the negative number, and the volume of geometric body is not, one needs to take the magnitude of the result of the scalar triple product of the vectors when calculating the volume of the parallelepiped: View/set parent page (used for creating breadcrumbs and structured layout). It only takes a minute to sign up. Proof of (1). The volume of a parallelepiped based on another. How many dimensions does a neural network have? [duplicate], determination of the volume of a parallelepiped, Formula for $n$-dimensional parallelepiped. A parallelepiped can be considered as an oblique prism with a parallelogram as base. &= (\mathbf b \times \mathbf c) \times A \cos \theta\\ Parallelepiped is a 3-D shape whose faces are all parallelograms. The volume of this parallelepiped (is the product of area of the base and altitude) is equal to the scalar triple product. &= \mathbf a\cdot(\mathbf b \times \mathbf c) It displays vol(P) in such a way that we no longer need theassumption P ‰ R3.For if the ambient space is RN, we can simply regard x 1, x2, x3 as lying in a 3-dimensional subspace of RN and use the formula we have just derived. $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, $\mathrm{Volume} = \mathrm{abs} ( \vec{u} \cdot (\vec{v} \times \vec{w}) ) = \mathrm{abs} \begin{vmatrix}u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix}$, $V = (\mathrm{Area \: of \: base})(\mathrm{height})$, $h = \| \mathrm{proj}_{\vec{u} \times \vec{v}} \vec{w} \|$, $\begin{vmatrix}u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix} = 0$, $\mathrm{abs} ( \vec{u} \cdot (\vec{v} \times \vec{w})) = 0$, $w_1 \begin{vmatrix}0 & 1\\ 1 & 0\end{vmatrix} + \begin{vmatrix} 1 & 0\\ 1 & 1 \end{vmatrix} = 0$, Creative Commons Attribution-ShareAlike 3.0 License. $\begingroup$ Depending on how rigorous you want the proof to be, you need to say what you mean by volume first. Is cycling on this 35mph road too dangerous? The triple product indicates the volume of a parallelepiped. The volume of a parallelepiped is the product of the area of its base A and its height h.The base is any of the six faces of the parallelepiped. Volume of the parallelepiped equals to the scalar triple product of the vectors which it is build on: . Vector Triple Product Up: Vector Algebra and Vector Previous: Rotation Scalar Triple Product Consider three vectors , , and .The scalar triple product is defined .Now, is the vector area of the parallelogram defined by and .So, is the scalar area of this parallelogram multiplied by the component of in the direction of its normal. rev 2021.1.20.38359, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Why are two 555 timers in separate sub-circuits cross-talking? As we just learned, three vectors lie on the same plane if their scalar triple product is zero, and thus we must evaluate the following determinant to equal zero: Let's evaluate this determinant along the third row to get $w_1 \begin{vmatrix}0 & 1\\ 1 & 0\end{vmatrix} + \begin{vmatrix} 1 & 0\\ 1 & 1 \end{vmatrix} = 0$, which when simplified is $-w_1 + 1 = 0$. In particular, all six faces of a parallelepiped are parallelograms, with pairs of opposite ones equal. Check out how this page has evolved in the past. Surface area. $\endgroup$ – tomasz Feb 27 '17 at 15:02 add a comment | 2 Answers 2 If you want to discuss contents of this page - this is the easiest way to do it. Code to add this calci to your website . So the first thing that we need to do is we need to remember that computing volumes of parallelepipeds is the same thing as computing 3 by 3 determinants. Of course the interchanging of rows does in this determinant does not affect the determinant when we absolute value the result, and so our proof is complete. An alternative method defines the vectors a = (a 1, a 2, a 3), b = (b 1, b 2, b 3) and c = (c 1, c 2, c 3) to represent three edges that meet at one vertex. First, let's consult the following image: We note that the height of the parallelepiped is simply the norm of projection of the cross product. Therefore if $w_1 = 1$, then all three vectors lie on the same plane. The sum of two well-ordered subsets is well-ordered. volume of parallelepiped with undefined angles, Volume of parallelepiped given three parallel planes, tetrahedron volume given rectangular parallelepiped. Given that $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ and $\vec{u} = (1, 0, 1)$, $\vec{v} = (1, 1, 0)$, and $\vec{w} = (w_1, 0, 1)$, find a value of $w_1$ that makes all three vectors lie on the same plane. How would a theoretically perfect language work? For permissions beyond … Change the name (also URL address, possibly the category) of the page. , How to prove volume of parallelepiped? These three vectors form three edges of a parallelepiped. + x n e n of R n lies in one and only one set T z. \begin{align} Online calculator to find the volume of parallelepiped and tetrahedron when the values of all the four vertices are given. How does one defend against supply chain attacks? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Theorem 1: If \vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3, then the volume of the parallelepiped formed between these three vectors can be calculated with the following formula: \mathrm{Volume} = \mathrm{abs} ( \vec{u} \cdot (\vec{v} \times \vec{w}) ) = \mathrm{abs} \begin{vmatrix}u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix}. c 1 c2 c3 In each case, choose the sign which makes the left side non-negative. The volume of the parallelepiped is the area of the base times the height. a 1 a2 a3 (2) ± b 1 b2 b3 = volume of parallelepiped with edges row-vectors A,B,C. Corollary: If three vectors are complanar then the scalar triple product is equal to zero. The surface area of a parallelepiped is the sum of the areas of the bounding parallelograms: We can now deﬁne the volume of P by induction on k. The volume is the product of a certain “base” and “altitude” of P. The base of P is the area of the (k−1)-dimensional parallelepiped with edges x 2,...,x k. The Lemma gives x 1 = B + C so that B is orthogonal to all of the x i, i ≥ 2 and C is in the span of the x i,i ≥ 2. (Poltergeist in the Breadboard). We can build a tetrahedron using modular origami and a cardboard cubic box. Find out what you can do. The altitude is the length of B. After 20 years of AES, what are the retrospective changes that should have been made? Let's say that three consecutive edges of a parallelepiped be a , b , c . How to get the least number of flips to a plastic chips to get a certain figure? Volumes of parallelograms 3 This is our desired formula. Hence, the theorem. u=−3, 5,1 v= 0,2,−2 w= 3,1,1. For each i write the real number x i in the form x i = k i, + α i, where k i, is a rational integer and α i satisfies the condition 0 ≤ α i < 1. By the theorem of scalar product, , where the quantity equals the area of the parallelogram, and the product equals the height of the parallelepiped. Or = a. Proof: The volume of a parallelepiped is equal to the product of the area of the base and its height. \begin{align} The point is How do you calculate the volume of a $3D$ parallelepiped? Substituting this back into our formula for the volume of a parallelepiped we get that: We note that this formula gives up the absolute value of the scalar triple product between the vectors. Wikidot.com Terms of Service - what you can, what you should not etc. Something does not work as expected? Recall uv⋅×(w)= the volume of a parallelepiped have u, v& was adjacent edges. Page 57 of 80 Geometric Interpretation of triple scalar product Geometrically, one can use triple scalar product to obtain the volume of a parallelepiped. Prism is a $3D$ shape with two equal polygonal bases whose corresponding vertices can be (and are) joined by parallel segments.Parallelepiped is a prism with parallelogram bases. The triple scalar product can be found using: 12 12 12. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. The height of the parallelogram is orthogonal to the base, so it is the component of $\vec c$ onto $\vec a \times \vec b$ which is perpendicular to the base, \text{comp}_{\vec a \times \vec b}\vec c=\frac{|c. (b × c) ? Suppose three vectors and in three dimensional space are given so that they do not lie in the same plane. Proof: The proof is straightforward by induction over the number of dimensions. Checking if an array of dates are within a date range, I found stock certificates for Disney and Sony that were given to me in 2011. \end{align} The height is the perpendicular distance between the base and the opposite face. The cross product a × b is shown by the red vector; its magnitude is the area of the highlighted parallelogram, which is one face of the parallelepiped. An alternative method defines the vectors a = (a 1, a 2, a 3), b = (b 1, b 2, b 3) and c = (c 1, c 2, c 3) to represent three edges that meet at one vertex. Append content without editing the whole page source. So the volume is just equal to the determinant, which is built out of the vectors, the row vectors determining the edges. Tetrahedron in Parallelepiped. Calculate the volume and the diagonal of the rectangular parallelepiped that has … How were four wires replaced with two wires in early telephone? &= (\mathbf b \times \mathbf c) \times A \cos \theta\\ Is it possible to generate an exact 15kHz clock pulse using an Arduino? Male or Female ? It is obtained from a Greek word which means ‘an object having parallel plane’.Basically, it is formed by six parallelogram sides to result in a three-dimensional figure or a Prism, which has a parallelogram base. Hence the volume{\displaystyle V}$$of a parallelepiped is the product of the base area$${\displaystyle B}$$and the height$${\displaystyle h}(see diagram). ; Scalar or pseudoscalar. One nice application of vectors in \mathbb{R}^3 is in calculating the volumes of certain shapes. It is obviously true for m=1. Finally we have the volume of the parallelepiped given by Volume of parallelepiped = (Base)(height) = (jB Cj)(jAjjcos()j) = jAjjB Cjjcos()j = jA(B C)j aIt is also possible for B C to make an angle = 180 ˚which does not a ect the result since jcos(180 ˚)j= jcos(˚)j 9 The volume of the spanned parallelepiped (outlined) is the magnitude ∥ (a × b) ⋅ c ∥. My previous university email account got hacked and spam messages were sent to many people. Watch headings for an "edit" link when available. Let \vec a and \vec b form the base. How can I cut 4x4 posts that are already mounted? This is a … The volume of a parallelepiped is the product of the area of its base A and its height h.The base is any of the six faces of the parallelepiped. The volume of any tetrahedron that shares three converging edges of a parallelepiped has a volume equal to one sixth of the volume of that parallelepiped (see proof). Click here to toggle editing of individual sections of the page (if possible). One such shape that we can calculate the volume of with vectors are parallelepipeds. If it is zero, then such a case could only arise when any one of the three vectors is of zero magnitude. \end{align} (\vec a \times \vec b)|}{|\vec a \times \vec b|}. Multiplying the two together gives the desired result. Volume of parallelepiped by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. Click here to edit contents of this page. The three-dimensional perspective … View and manage file attachments for this page. What difference does it make changing the order of arguments to 'append'. As a special case, the square of a triple product is a Gram determinant. Can Pluto be seen with the naked eye from Neptune when Pluto and Neptune are closest. Row-Vectors a, b, c when Pluto and Neptune are closest to many.. ) of the base and its height, you need to say what you not. If possible ) zero, then such a case could only arise when any one of these tetrahedra one! Particular, all six faces of a and b is perpendicular to determinant. Number of flips to a plastic chips to get the least number of dimensions in each case, the. The determinants of two 3×3 matrices equals the determinant, which is built of! Truesight and Darkvision, why does a monster have both with volume of this.! ) = the volume of volume of parallelepiped proof $and$ \vec b ) | {... 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# Help with a problem of a Sphere!

1. Dec 1, 2007

We have a hollow Sphere. It weighs 240 newtons in the air, and 150 newtons when it's totally immersed in the water.

If the density of the metal is 7 g/cm^3. Find the metal thickness

thanks

Last edited: Dec 1, 2007
2. Dec 2, 2007

### Kurdt

Staff Emeritus
What have you tried so far?