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Why does the homogeneity of the universe require inflation? They say inflation must have occured because the universe is very homogeneous. Otherwise, how could one part of the universe reach the same temperature as another when the distance between the parts is more than light could have traveled in the given time?
Why can't this problem be solved without inflation? If each part started with the same temperature to begin with, then they can have the same temperature irrespective of the distance between them. Am I missing something here?
| The short answer is that physicists/astronomers want to avoid fine tuning wherever possible. Creating a universe where the temperature everywhere was essentially the same requires exceptional fine tuning. Creating a universe where the temperatures were random in different parts of space and had an opportunity to come in thermal equilibrium before going out of causal contact (as a result of inflation) is more natural. If this was all that inflation solved, it maybe would not be considered as likely to have occurred as it is. It also solves another fine tuning problem though, in that the universe is very nearly flat (if not exactly so), and inflation naturally would produce such a universe as well.
Edit: I should also mention that inflation naturally explains the absence of observations of magnetic monopoles as well. This problem was actually the primary motivation of Alan Guth, who first developed the idea of inflation. This issue is not so much one of fine tuning though, unlike the flatness problem and the homogeneity problem.
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Why does Jupiter emit more energy than it receives? I hear that Jupiter and Saturn emit more energy than they receive from the Sun.
This excess energy is supposedly due to contraction.
*
*Is this accepted as fact (or is it controversial)?
*Does this mean that Jupiter is shrinking a little bit (its diameter decreases), or are there just changes in the internal density distribution?
| The book Jupiter: The Planet, Satellites and Magnetosphere, edited by Fran Bagenal, Timothy Dowling, and William McKinnon has, in its third chapter, the text: "[Jupiter] is still contracting at a rate of ~3 cm per year while its interior cools by ~1 K per million year."
The chapter does not give a specific source for that, but the chapter has an extensive list of references. I believe that the subject first came up in 1966 in a paper by Low who measured the infrared heat flux to be 1.9 times the incident solar. I was under the impression that this has been measured, but I was unable to find a direct reference to this; at this point it is likely to be a model and that's the amount required for the observed level of heating. One early paper that modeled this is by W.B. Hubbard in the journal Icarus, published in 1977 under the title "The Jovian surface condition and cooling rate."
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Why did the WMAP mission last so much longer than Planck? NASA endorsed 9 years of data taken with the Wilkinson Microwave Anisotropy Probe (WMAP). The High Frequency Instrument aboard the Planck satellite ran out of coolant at the start of 2012, after about two and a half years of operation. Even if the Low Frequency instrument runs for another two years, its operating life will be much shorter than WMAP's.
Why is there such a big difference between the operational life-spans of the two missions, given that they have basically the same objective: to measure the cosmic microwave background?
| From Wikipedia:
Passive thermal radiators cool the WMAP to ca. 90 degrees K; they are
connected to the low-noise amplifiers. The telescope consumes 419 W of
power. The available telescope heaters are emergency-survival heaters,
and there is a transmitter heater, used to warm them when off. The
WMAP spacecraft's temperature is monitored with platinum resistance
thermometers.
Because no coolant was used, there was no hard deadline on its operational lifetime. Its design lifetime was for two years of data collection. However the cost of continuing to collect data from an operational satellite is low so there was no pressing need to turn it off; and until Planck had returned its first dataset, adding more WMAP data was the only option available for scientists studying the CMB that wanted more data. However, diminishing returns come into play; each additional scan of data from the same instruments gives a smaller return. Planck's more sensitive instruments rapidly surpassed the quality of its data; at which point there was no reason to keep running it.
Cryogenic cooling reduces thermal noise. Using it was one way the Planck team could increase the sensitivity of their instruments. Planck completed five complete scans of the sky before exhausting the coolant for the HFI. The LFI is expected to have another 6-9 months of operation before its coolant is exhausted.
Finally the amount of coolant used isn't a finely tunable parameter. Rockets come in a relatively small number of discrete sizes; you can't just spend a few tens of thousands of dollars to add a few extra kg of payload. The next larger rocket size can easily cost millions of dollars more. As a result, most missions are designed around the payload capacity of the rocket they're budgeted for and have a fixed mass as a result. An extra kg of coolant means removing a kg somewhere else.
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Where can I find information for how to create amateur radio telescope? Where can I find good source of information for how to create amateur radio telescope.
Particularly interested in creating Fresnel Zone Plate Antennas (aka flat dish).
*
*design, diagrams, electronics.
*discussions about probes in production etc.
| Taking a step back, what do you want to study with your telescope? The telescope is just one component in your DIY astronomy lab. You are also going to need a plan for your study, a way to acquire signals from your telescope, and a way to process those signals.
Some telescope / antenna stuff -
*
*A guy who built a telescope for examining Hydrogen emissions
*Some thoughts on designing Fresnel Antennas
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Is it possible to make a hydrogen-alpha solar scope? Is the construction of an etalon / Fabry-Pérot interferometer within the reach of amateur telescope makers? Are there any resources pointing to such projects?
| You can buy etalons on eBay cheap, they even have some 1" 1047nm for $35 now. I am a holographer, and I know how to use these in a laser. What confuses me is how they are used in a telescope with converging, rather than parallel rays.
Can you just put mylar and a long pass filter (maybe a couple lighting gels) in front, place the etalon on a gimbal mount in the path, and put a camera on the end?
| {
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What is behind the cosmological horizon barrier? I'm wondering what is behind the cosmological horizon barrier?
| The optical horizon is calculated to be about 46 Gyrs away if we presume a flat Universe, from the simple relationship R_h = 3ct, where R_h is the optical horizon and t time. This relationship can be found in P.J.E.Peebles book, Principles of Physical Cosmology, and in the book by J.B. Hartle, Gravity. So the 46 Gyrs mentioned above is consistent with a t = 15 Gyrs, and our visible Universe is about 90 Gyrs across.
Now if the Universe is not flat, then things really go wild quickly and that value grows very quickly with increased curvature. Same if the accelerating universe turns out to be true.
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How to measure the diameter of a star? I am thinking about something I read somewhere (if only I could find it again) in a textbook. It is about the size of a star and its ER peaks. It has to do with the waves coming off the edge (maybe) and arriving later than those from "head on" and therefore you can know something about its diameter.
It has been puzzling me but I can't quite remember it. Just yesterday I was reading about a black hole that pulses at a minimum of 10 minutes and so it is at least 10 light-minutes across. (I probably am not getting it just right, but help me out!) Is this the same principle?
Anyone know what I'm talking about? I would love to have this explained and/or would like to know what it is called, so I can look it up.
| In the first case, in regards to star measurement, I believe you're thinking of how the diameter of very large stars are measured using interferometry. Because light waves from the edges of these stars arrive at us in parallel, and they are waves, we can determine the diameter of the star by measuring the interference pattern between these light waves. (That's probably what you were thinking of — the "peaks" and "troughs" in the interference pattern.)
Stars like Betelgeux, Antares and Aldebaran have been measured in this way and the size agrees with the Stefan–Boltzmann law which can be used to calculate the radius of a spherical body if the luminosity and temperature are known.
I found this 1921 Popular Science article which describes it in detail.
The black hole pulsing thing is a completely different concept. I'm not sure what the logic of that is, perhaps it was talking of the doppler shifts of light rotating around the black hole or its interaction with a binary companion star?
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If neutrinos travel faster than light, how much lead time would we have over detecting supernovas? In light of the recent story that neutrinos travel faster than photons, I realize the news about this is sensationalistic and many tests still remain, but let's ASSUME neutrinos are eventually proven to travel "60 ns faster than light". If so, how much lead time would they have over light from local supernovas (e.g. SN 1987A) and distant (e.g. SN 2011fe)?
What does the math look like to calculate this?
| If light is interacting with ions/atoms and neutrinos do not, that would mean that light has a variable speed no? Therefore neutrinos are more constant at the "speed of light". If this is true, then can we devise an experiment that slows light? It is energy and has mass, why not? If this is correct, then wouldn't nuetrinos be affected too and we should be able to slow them down? Do they change state? Interesting questions.
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Where are the Voyagers going? Given enough time, where are the Voyager spacecrafts heading? (Assuming some alien civilization doesn't pick them up.)
Will they pass by any interesting stars on the way to the black hole at the center of our galaxy or will it perhaps leave the galaxy?
What are the highlights on their journey that we can reasonably predict?
| So basically voyager 2 is not really moving at all. Our solar system is traveling away from the center of the galaxy at approximately 35,000 miles per hour and voyager is traveling toward the center of the galaxy or away from the our sun at the same speed.
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How do we know the masses of single stars? I have recently read that we can only know the masses of stars in binary systems, because we use Kepler's third law to indirectly measure the mass. However, it is not hard to find measurements for the mass of stars not in binary systems. So how is the mass of these stars determined?
| The mass of these single stars is sometimes determined by the effect of Gravitational Lensing. In General Relativity, it is stated that light is bent when it is influenced by a Gravitational Field, so actually the angle of bending of any light coming from near a single star, would give us the magnitude of its Gravitational Field and ultimately its mass value.
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How bright are auroras (aurorae)? Digital cameras are making the recent auroras look magnificent, but what are they like to the naked eye? Are they comparable in surface brightness to the Milky Way?
| They are much brighter than the Milky Way - I have once seen one bright enough to read by, think it was back in '76 or '77 and it was so powerful it was also audible in Orkney (59 degrees north.) Unfortunately haven't seen this week's one as it has been overcast here.
A full moon will drown out a lot of the detail to the naked eye, but if you are used to photographing night scenes, you should find it remarkably easy. Either go for a relatively high ISO to fix the movement, or allow for the blurring and get some of the longer timeframe effects along with a lot of bright colours.
Whether or not you see one will depend mostly on your latitude, but in any case, get out of the city and head for dark countryside - take a deck chair, blankets and a flask of soup and enjoy the show.
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Can the Hubble telescope bring any star into focus? Lets say I am talking about a view like this supernova - 13 billions light year away. In short can Hubble bring any star into focus in the entire universe? And if so, to what definition?
I also wonder, how much time would time would it need to focus on a distant star or planet, in comparison to a closer one?
| Telescopes do not focus. The objects they view (even the Moon) are so far away compared to the aperture and focal length of a telescope that they are all at infinite focus, so telescopes use a fixed focus.
Whether a telescope can resolve a distant object or not is dependent on aperture and brightness. Brightness in turn is dependent on intrinsic brightness and distance. Hubble can resolve even incredibly distant objects if they are bright enough, but cannot see any object at any distance.
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Do days and months on the Moon have names? On Earth we have various calendars, for example,
Days: Monday, Tuesday, Wednesday, etc., etc.
Months: January, February, March
Does the Moon have names for its "daily" rotations, etc.?
It sounds like a silly question, and I am not sure if I've asked it using the correct terminology. I suppose what I'm trying to ask is; from a viewpoint of someone living on the Moon - does it have "day" names?
| Since no one actually lives on the Moon, there is no call to have special names.
A day and a month on the Moon are the same length, 29.5 Earth days. Each new lunar month is called a "lunation" and numbered. Lunation 1094 started on 2011 Jun 01 and Lunation 1095 starts on 2011 Jul 01.
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Why is a new moon not the same as a solar eclipse? Forgive the elementary nature of this question:
Because a new moon occurs when the moon is positioned between the earth and sun, doesn't this also mean that somewhere on the Earth, a solar eclipse (or partial eclipse) is happening?
What, then, is the difference between a solar eclipse and a new moon?
| - This is because of the of moon’s tilted orbit around Earth with respect to the earth’s orbital plane (ecliptic).
*
*Solar and lunar eclipses happen only during an eclipse season when the plane of the Earth's orbit
around the Sun crosses with the plane of the Moon's orbit around the Earth.
*If the orbit of the Earth around the Sun and the Moon's orbit around the Earth were both in the same plane,
then there would be a lunar eclipse at every full moon, and a solar eclipse at every new moon.
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How could I translate a field of view value into a magnification value? When I zoom in with Stellarium, it indicates a field of view (FOV) value in degrees, but most binoculars and telescopes are advertised with value like "nX magnification power."
How could I translate this value so I get an idea of what I will see with a telescope or binocular?
For example, I if got a 30X telescope, how much should I zoom to get similar view?
| The magnification on a telescope can be calculated using its focal length divided by the focal length of the eyepiece you are using, this means, using a 150mm focal length scope with a 5mm eyepiece would give you roughly 30x but also a 1200mm FL telescope with a 40mm eyepiece will.
If you can get the values of aperture, focal length and the eyepiece focal length, you can use this values in the plugins section of stellarium to configure it to the approximate FOV you will experience, but the image in stellarium will never be exact to what you see in a telescope or binocular, as the view depends on many factors, as light gathering, optical aberrations, design of the telescope, collimation and atmospheric conditions.
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Mass of a galaxy via luminosity Is there a general formula for calculating the mass of a galaxy, or even a nebula from the luminosity? Or, is there a way of calculating the total mass of a galaxy from its energy output?
Is there a Hertzsprung–Russell diagram equivalent for galaxies?
I know about gravitational lensing or velocity dispersion via the virial equation, and the Schechter function, and using doppler spread to calculate a mass.
| There is not a straightforward relation between a galaxy's luminosity and its mass. The luminosity depends on how much present and recent star formation there has been. Some very massive elliptical galaxies have little star formation going on, so they are not particularly luminous for their mass.
To understand why this is so, consider stars converting hydrogen to helium, and lying on the "Main Sequence" (see Wikipedia) in the Luminosity-Color (Hertzsprung-Russell) diagram. On the Main Sequence, stars burn with a luminosity which is proportional to their mass to about the 3.5 power. The Sun is a Main Sequence star, and a blue B-type star on the Main Sequence might have 30 times the Sun's mass and 100,000 times the luminosity, and a red M-type star on the Main Sequence might only one-tenth the Sun's mass and less than a thousandth of the luminosity. Since the lifetime of a star depends on the mass divided by the luminosity, stars much more massive than the Sun will have short lifetimes (millions of years) and stars much less massive than the Sun will have very long lifetimes (trillions of years).
After they exhaust the hydrogen, the stars will go through a relatively brief giant phase where they are even more luminous for a short while, but then gradually fade into a white dwarf that still has a lot of mass but very little luminosity. The upshot is that soon after a burst of star formation,a galaxy will glow brightly with massive main sequence stars and stars in the giant phase. But a long time later, the luminosity will be dominated by white dwarfs and red main sequence stars, both of which give off very little luminosity for their mass.
| {
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In astronomy what phenomena have theory predicted before observations? As far as I know, astronomy is generally an observational science. We see something and then try to explain why it is happening. The one exception that I know of is black holes: first it was thought of, then it was found.
Einstein's relativity is middle ground to me, he thought of light beams at the speed of light but obviously could observe gravity's effects.
Anyway, I guess my question is, what are the biggest discoveries that were thought of before they were seen in the sky?
| Aristotle (4th century BCE) first hypothesized that the Earth is geographically divided into three types of climatic zones based on their distance from the equator (circles of latitude). In my opinion, it's the basis of one of the greatest science experiments of antiquity that never was. If the ancient Greek astronomers had been modern scientists in their methods, confirmation of the existence of geographical zones would have been the foundational experiment justifying spherical astronomy.
Another basic theoretical prediction that preceded observation is Earth's rotation, which I believe was first firmly predicted by Newton's celestial mechanics, and experimentally observed of course by Foucault's pendulum experiment in 1851.
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Is there such a thing as "North" in outerspace? On Earth, North is determined by the magnetic poles of our planet. Is there such a thing as "North" in outerspace? To put it another way, is there any other way for astronauts to navigate besides starcharts? For instance, if an astronauts spaceship were to be placed somewhere (outside of our solar system) in the milkyway galaxy, would there be a way for them to orient themselves?
| You could orient yourself using the CMB fluctuations as your compass--- given a detailed WMAP picture of the CMB flutuations, you could tell which way is which anywhere in the local galactic region, and how fast you are going relative to the CMB.
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Why are stars, planets and larger moons (approximately) spherical in shape (like, the Sun, the Moon, the Earth, and other planets)? Why are stars, planets and larger moons (approximately) spherical in shape (like, the Sun, the Moon, the Earth, and other planets)?
| I'm not sure if this is the "done" thing, but the question is cross-posted from Physics.SE, so I'm cross-posting my answer...
In short, it's because gravity is "round". That is, it only depends on the distance between objects. All objects that are at a particular distance are attracted with the same acceleration, so we'd say it's constant on a sphere and thus, in a way, it's "round". This isn't the whole story, of course. Things aren't perfectly round because of effects like rotation. But if gravity were left to itself, they'd tend towards perfect spheres.
In physics, we tend to say these objects are in hydrostatic equilibrium. In fact, this is part of the new IAU definition of a planet. What it means is that the pressure of a star/planet balances gravity at each point, or each distance from the centre of gravity. Because gravity is round, the pressure gradient must also be round. This only applies when gravity is strong enough to force things into shape. A brick has its own self-gravity, but obviously it isn't nearly strong enough to turn the brick into a near-sphere. This is also true of smaller solar system bodies like some asteroids. They aren't quite big enough for the gravity to force them to match the shape of gravity.
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Why does Omega Centauri have a distinct chemical signature from the rest of the Milky Way? In answering a question about the orbital path of Omega Centauri, I learned that it has a distinct chemical signature from the rest of the Milky Way. Basically, it is very rich in s-process elements, which I think are primarily produced in Asymptotic Giant Branch stars. It is not totally clear to me why that would be the case. Are AGB stars dominating metallicity in Omega Centauri, and if so, why? If it is not what, what is the cause?
| After looking through a few papers, in particular Chemical Abundances and Kinematics in Globular Clusters and Local Group Dwarf Galaxies and Their Implications for Formation Theories of the Galactic Halo and references therein, I think I have a reasonable answer. Omega Centauri's chemical abundance seems to be most easily explained by it being an accreted dwarf spheroidal galaxy. The metallicities of its stars match up quite well with those of almost all of the Milky Way's dwarf spheroidal satellites.
The reason for the dwarf spheroidal galaxies having different metallicities is thought to be due to two factors: they form stars at lower efficiency than the Milky Way, and the material for star formation (that is, gas) is more easily driven out of dwarf spheroidals by galactic winds. Chemical Abundances for 855 Giants in the Globular Cluster Omega Centauri (NGC 5139) says that in this respect, Omega Centauri differs even from dwarf spheroidals, in that Type Ia supernovae played a minimal role in enriching the stars in the Omega Centauri system. Instead, its metallicity is dominated by the injection of elements produced by Type II supernovae early in the system's history, and "pollution" caused by intermediate mass stars in the asymptotic giant branch (AGB) phase, where certain elements found preferentially in Omega Centauri can form, and then are ejected because AGB stars are unstable.
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Determining cloud cover from observer to near the horizon Does there exist a Clear Sky Chart with the following enhancements?:
1 - Actual Cloud Cover (Offered Visually and not just Colors with a Legend, Over Time/Past & Predictive)
2 - Simulate/Predict Cloud Cover taking into account the direction from Observer to Observed Object and Angle of view - Close to the horizon (May be helpful to know when you can reasonably start/end tracking something you want to catch that night that is close to the horizon)
The reason I'm curious is:
A) I wonder if it's just not feasible for any/many reasons.
B) It would be great help to know this information.
In general, does anyone know of other Earth Weather, Clear Sky Clocks and Charts or anything else that gives more information?...anything related will be helpful.
EDIT: I would love to find this lecture "You can do better than Clear Sky Chart" mentioned: http://stjornuskodun.blog.is/blog/stjornuskodun/entry/966941/
| Rain Today provide a rain cloud image down to 15 minutes prior. But rain cloud doesn't always equate to total cloud cover.
I think the major drawback is that clouds can arise from clear air as the moisture condenses, and predicting where this is going to happen becomes too complicated to resolve at fine enough detail to be of much use...
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Can the Hanbury-Brown and Twiss effect be used to measure the size of composite objects like galaxies? I know that the Hanbury-Brown and Twiss effect can be used to measure the size of stars. Can it also be used to measure the size of galaxies?
| I assume you mean measure the apparent angular diameter of a galaxy too distant to be resolved, in which case I would say yes. (If you can resolve it, the HBT effect is unnecessary, of course!) First of all, stars are extended objects, just like galaxies, so there should be no fundamental difference. Secondly, the Wiki article discusses summing over different photon sources within an overall object, with the net result being that your detectors have to be sufficiently close together... "for a given angular diameter desired to be detected," I assume.
I don't know off the top of my head how stellar apparent angular diameters compare with very distant galaxies, but if anything, I would guess stars would be smaller.
The only wrinkle I can think of is that a star's image ought to be very nearly perfectly circular while a galaxy's need not be, but I think you should probably still get an average or effective diameter, even when your source is non-circular.
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Why don't we have a better telescope than the Hubble Space Telescope? The Hubble Space Telescope (HST) was launched in 1990, more than 20 years ago, but I know that it was supposed to be launched in 1986, 24 years ago. Since it only took 66 years from the fist plane to the first man on the Moon why don't we have a better telescope in space after 24 years?
| The key reason why there will never be a next-generation optical telescope in space is: adaptive optics.
In 1978 when funding for the HST started, adaptive optics was in its infancy. Only in the 1990's when computer technology had sufficiently advanced, did adaptive optics really take off. Modern earth-bound telescopes with adaptive optics are far more capable than anything we can bring into orbit at comparable costs.
To give you an idea: the European Extremely Large Telescope (E-ELT) sports a 1000 square meter light gathering area, and will be build and operated for a very tiny fraction of the 10+ billion USD total price tag of the HST (with a mere 4.5 square meter light gathering area).
It simply makes no sense anymore to send optical telescopes in space. This is different for telescopes operating at non-optical wavelengths for which the atmosphere has a low transmissibility. COBE, WMAP and Planck all operate at microwave wavelengths, Chandra at X-ray wavelengths, and the JWST is designed for infrared wavelengths,
| {
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Can black holes actually merge? If time stops at the event horizon, can we ever detect two black holes merging? In other words, if you are a short distance away, would you encounter a spherically symmetric gravitational field, or a dipole field?
| As a distant observer we can watch the shadow of the black holes forming in front of the background stars. According to a nice little paper by Daisuke Nitta, Takeshi Chiba, and Naoshi Sugiyama ("Shadows of Colliding Black Holes, 2011") the answer is yes. To a distant observer in a finite span of time two black holes form a shadow that is indistinguishable from a single black hole's. I would "conjecture" that in a similar time frame the gravitational field around the pair would be same as that of a single black hole too. The question of black holes "actually" merging is not answerable in the sense that we can never know what is actually going on in our universe; we have to contend with limited observables...
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What is exactly the density of a black hole and how can it be calculated? How do scientists calculate that density? What data do they have to calculate that?
| There might be no full-fledged theory of quantum gravity, but we can speculate a little on results from whatever the true theory is. Quantizing gravity usually implies quantizing spacetime- in other words, the entire universe is grainy. It is likely that you can pack no more than about one Planck mass into each Planck volume, i.e. cubic Planck length. This works out to 5.1555e96 kg/m^3. The implication of this calculation is that all black holes will have roughly the same density, and will simply increase in real volume with increasing mass.
I know I've mentioned this on another question, but I can't find it right now.
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What if our Sun were located in the middle of a globular cluster? Say you took our current solar system and relocated it deep in the heart of a globular cluster such as Omega Centauri. What would the night sky look like? Would the starshine of nearby stars be enough to turn the sky blue or cast shadows or are they still too far away to do that? What would the day sky look like? How would nearby starts affect the Earth's ecosystem? Is it even possible for a solar system such as ours to exist in a stable form in such surroundings?
| I just have to say that our planets would probably never get life because other suns would no doubt destroy the planets. however if we did live in a place like that night would be as bright as day.
| {
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What is Hawking radiation and how does it cause a black hole to evaporate? My understanding is that Hawking radiation isn't really radiated from a black hole, but rather occurs when a particle anti-particle pair spontaneously pop into existence, and before they can annihilate each other, the antiparticle gets sucked into the black hole while the particle escapes. In this way it appears that matter has escaped the black hole because it has lost some mass and that amount of mass is now zipping away from it.
Is this correct? If so, wouldn't it be equally likely that the particle be trapped in the black hole and the antiparticle go zipping away, appearing as if the black hole is spontaneously growing and emitting antimatter?
How is it that this process can become unbalanced and cause a black hole to eventually emerge from its event horizon and evaporate into cosmic soup over eons?
| The virtual particle/antiparticle explanation is common, but (from what I understand) not very accurate; see e.g. this explanation by John Baez. To summarize it in less technical terms, spacetime near the black hole's event horizon is so strongly curved that what a nearby observer would call "absolute zero" (i.e. zero emission of radiation) looks like a greater-than-zero temperature to someone far away. That means the black hole is emitting energy, and by conservation of mass/energy the hole must be getting smaller as a result.
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What is this shadow of the Sun on the Moon? I was reading the article Moon Phases on HowStuffWorks. In the picture, each moon has a dark green area which represents the shadow of the Sun. How is this shadow formed and why is this important?
| That is a very odd diagram - the area of the moon shown as dark is not necessarily dark: it just represents the side we cannot see.
The green area is meant to represent the side unlit by the sun, leaving the lit section as that which we can see from Earth and lit by the sun.
It is obviously done this way to help you understand why the phases look like they do, but using green for shadow, and dark for not visible is confusing.
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Counting complete sets of mutually unbiased bases composed of stabilizer states Consider $N$ qubits. There are many complete sets of $2^N+1$ mutually unbiased bases formed exclusively of stabilizer states. How many?
Each complete set can be constructed as follows: partition the set of $4^N-1$ Pauli operators (excluding the identity) into $(2^N+1)$ sets of $(2^N-1)$ mutually commuting operators. Each set of commuting Paulis forms a group (if you also include the identity and "copies" of the Paulis with added phases $\pm 1$, $\pm i$). The common eigenstates of the operators in each such group form a basis for the Hilbert space, and the bases are mutually unbiased. So the question is how many different such partitions there exist for $N$ qubits. For $N=2$ there are six partitions, for $N=3$ there are 960 (as I found computationally).
The construction above (due to Lawrence et al., see below) may be an example of a structure common in other discrete groups - a partition of the group elements into (almost) disjoint abelian subgroups having only the identity in common. Does anyone know about this?
Reference:
Mutually unbiased binary observable sets on N qubits - Jay Lawrence, Caslav Brukner, Anton Zeilinger, http://arxiv.org/abs/quant-ph/0104012
| For finite dimensional systems, R. Buniy and T Kephart in 1012.2630 quant-ph provide a tool for defining a set of equivalence classes for entanglement states based on their algebraic properties. Your answer should be in there.
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Values of SM parameters at one certain scale The general question is:
What are the values of Standard Model parameters (in the $\bar{MS}$ renormalization scheme) at some scale e.g. $m_{Z}$? As its parametrization in Yukawa matrices is not unique - what are the values of gauge couplings, fermion masses and CKM matrix?
The background:
I want to solve renormalization group equations of MSSM and in order to have initial conditions for them I need to know SM parameters at one scale - not at few different, which one can find at Particle Data Group webpage.
| See Appendix A of my PhD thesis: http://arxiv.org/abs/0905.1425
I worked out the values of the parameters to use in RGE at $\mu=M_Z$.
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Physical interpretation to the category of CFTs This question comes from reading Andre's question where I wandered whether that question even makes sense physically. In mathematics, VOAs form a category, does this category as a whole have a physical interpretation?
| The category of CFT's (and related 3D TQFT) has been studied by
Kapustin and Saulinas in their recent paper Topological boundary conditions in abelian Chern-Simons theory.
The quote "we obtain a classification of such theories up to isomorphism" from their abstract refers to a notion of isomorphism that is clearly higher-categorical (more precisely, it appears to be 3-categorical).
See also Surface operators in 3d Topological Field Theory and 2d Rational Conformal Field Theory by the same authors.
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Unitarity of S-matrix in QFT I am a beginner in QFT, and my question is probably very basic.
As far as I understand, usually in QFT, in particular in QED, one postulates existence of IN and OUT states. Unitarity of the S-matrix is also essentially postulated. On the other hand, in more classical and better understood non-relativistic scattering theory unitarity of S-matrix is a non-trivial theorem which is proved under some assumptions on the
scattering potential, which are not satisfied automatically in general.
For example, unitarity of the S-matrix may be violated if the potential is too strongly attractive at small distances:
in that case a particle (or two interacting with each other particles) may approach each other from infinity and form a bound state.
(However the Coulomb potential is not enough attractive for this phenomenon.)
The first question is why this cannot happen in the relativistic situation, say in QED.
Why electron and positron (or better anti-muon) cannot approach each other from infinity and form a bound state?
As far as I understand, this would contradict the unitarity of S-matrix.
On the other hand, in principle S-matrix can be computed, using Feynmann rules, to any order of approximation in the coupling constants. Thus in principle unitarity of S-matrix could be probably checked in this sense to any order.
The second question is whether such a proof, for QED or any other theory, was done anywhere? Is it written somewhere?
| Unitarity of the S-matrix can be checked perturbatively. Bound states tend to be non-perturbative effects, so may not show up naive perturbative calculations. Unfortunately, the datailed proof is not discussed in many places. One book that has it is Scharf's book on QED. When looking through other books you should look for keywords like optical theorem and Cutkosky rules. Bound states are usefully discussed in the last chapter of vol.1 of Weinberg's tretease on QFT.
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Uniqueness of supersymmetric heterotic string theory Usually we say there are two types of heterotic strings, namely $E_8\times E_8$ and $Spin(32)/\mathbb{Z}_2$. (Let's forget about non-supersymmetric heterotic strings for now.)
The standard argument goes as follows.
*
*To have a supersymmetric heterotic string theory in 10d, you need to use a chiral CFT with central charge 16, such that its character $Z$ satisfies two conditions:
*
*$Z(-1/\tau)=Z(\tau)$
*$Z(\tau+1)=\exp(2\pi i/3) Z(\tau)$
*Such a chiral CFT, if we use the lattice construction, needs an even self-dual lattice of rank 16.
*There are only two such lattices, corresponding to the two already mentioned above.
We can replace the lattice construction with free fermion construction, and we still get the same result. But mathematically speaking, there might still be a chiral CFT of central charge 16, with the correct property, right? Is it studied anywhere?
| There are plenty of chiral CFTs with central charge 16 and nice properties studied in the mathematics literature. A nice example in this context would be chiral differential operators on a 8-manifold. If you want modularity of the character so that you want a holomorphic vertex algebra then the reference is
"Holomorphic vertex operator algebras of small central charge" Dong and Mason. Pacific Journal of Mathematics. Vol 213 (2) 2004.
as discussed in the comments and in Lubos' answer.
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Examples of heterotic CFTs I'm trying to get a global idea of the world of conformal field theories.
Many authors restrict attention to CFTs where the algebras of left and right movers agree. I'd like to increase my intuition for the cases where that fails (i.e. heterotic CFTs).
What are the simplest models of heterotic CFTs?
There exist beautiful classification results (due to Fuchs-Runkel-Schweigert) in the non-heterotic case that say that rational CFTs with a prescribed chiral algebras are classified by Morita equivalence classes of Frobenius algebras (a.k.a. Q-systems) in the corresponding modular category.
Is anything similar available in the heterotic case?
| I just found by incidence a simple example in some proceedings of Böckenhauer and Evans below. Namely for $\mathrm{Spin}(8\ell)_1$ (so $D_{4\ell}$ lattice) with $\ell=1,2,\ldots$ there exist modular invariants, which should give rise to heterotic models (by Rehrens paper).
see section 7 in
http://books.google.de/books?id=yV_RlDznAu8C&lpg=PA120&ots=HwZm5KlDCW&pg=PA119#v=onepage
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Bogomol'nyi-Prasad-Sommerfield (BPS) states: Mathematical definition What is the proper mathematical definition of BPS states?
In string theory the BPS states correspond either to coherent sheaves or special Lagrangians of Calabi-Yau manifold depending upon the type of string theory considered. but in SUSY quantum field theories in 4d, there are no CYs as far I know (which is very little) and in gravity theories, these corresponds to some Black Holes. So what is the general mathematical definition of BPS states which is independent of the theory in consideration, say a general SUSY Quantum field theories, be it QFT, string theory, Gravity and in any dimension.
| In any supersymmetric theory, a BPS state is a state which preserves some of the supersymmetry.
If we take as a definition of a supersymmetric theory, some theory (classical or quantum) which admits a Lie superalgebra of symmetries, then a BPS state (or configuration) of such a theory is one which is annihilated by some nonzero odd element in the superalgebra.
Of course, the original meaning comes from the study of magnetic monopoles. Solutions of the Bogomol'nyi equation are precisely those which saturate a bound, the so-called Bogomol'nyi-Prasad-Sommerfield (BPS) bound.
The relation between the two notions is related to the fact that monopole configurations can be thought of as configurations in a four-dimensional $N=2$ supersymmetric gauge theory which preserve half of the supersymmetry.
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Constructing a CP map with some decaying property Given some observable $\mathcal O \in \mathcal H$ it is simple to construct a CP (completely positive) map $\Phi:\mathcal{H}\mapsto \mathcal{H}$ that conserves this quantity. All one has to observe is that
$$ \text{Tr}(\mathcal O \, \Phi[\rho]) = \text{Tr}(\Phi^*[\mathcal O] \rho).$$
Therefore, if we impose $\Phi^*[\mathcal O] = \mathcal O$, then $\text{Tr}(\mathcal O \, \Phi[\rho])=\text{Tr}(\mathcal O \rho), \; \forall \rho\in \mathcal H$. That amounts to impose that the Kraus operators of $\Phi^*$ should commute with $\mathcal O$.
I'd like, however, to construct a trace-preserving CP map for which the expectation value of $\mathcal O$ does not increase for any $\rho \in \mathcal H$. More explicitly, given $\mathcal O\in \mathcal H$, I want to construct $\Gamma:\mathcal H \mapsto \mathcal H$ such that
$$ \text{Tr}(\mathcal O\, \Gamma[\rho]) \le \text{Tr}(\mathcal O \rho), \; \forall \rho \in \mathcal H .$$
How would you go about that? Any ideas?
| This is probably not exactly what you had in mind, but how about the channel that discards its input and always outputs the state corresponding the the minimum eigenvalue of $\mathcal{O}$?
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Rigorous proof of Bohr-Sommerfeld quantization Bohr-Sommerfeld quantization provides an approximate recipe for recovering the spectrum of a quantum integrable system. Is there a mathematically rigorous explanation why this recipe works? In particular, I suppose it gives an exact description of the large quantum number asymptotics, which should be a theorem.
Also, is there a way to make the recipe more precise by adding corrections of some sort?
| Yes, it can be made precise and corresponds to the leading order of the semiclassical expansion (WKB approximation) in $\hbar$. See Faddeev-Yakubovsky's "Lectures on quantum mechanics for mathematics students" (§20, formula (13)). An approach inspired by geometric quantization is explained in chapter 4 in Bates-Weinstein's Lectures on the Geometry of Quantization.
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Question from Schutz's In q. 22 in page 141, I am asked to show that if
$$U^{\alpha}\nabla_{\alpha} V^{\beta} = W^{\beta},$$
then
$$U^{\alpha}\nabla_{\alpha}V_{\beta}=W_{\beta}.$$
Here's what I have done:
$$V_{\beta}=g_{\beta \gamma} V^{\gamma},$$
so
$$U^{\alpha} \nabla_{\alpha} (g_{\beta \gamma} V^{\gamma})=U^{\alpha}(\nabla_{\alpha} g_{\beta \gamma}) V^{\gamma} + g_{\beta \gamma} (U^{\alpha} \nabla_{\alpha} V^{\gamma}).$$
Now, I understand that the second term is $W_{\beta}$, but how come the first term vanishes?
| I don't think you need metric compatibility to prove this although you can use it. There is a much simpler way with repeated use of (any) metric to lower the index.
$
U^\alpha\nabla_\alpha V^\beta=W^\beta
$
$
\Rightarrow U^\alpha g^{\beta\gamma}\nabla_\alpha V_\gamma=g^{\beta\gamma}V_\gamma
$
$
\Rightarrow U^\alpha g_{\mu\beta}g^{\beta\gamma}\nabla_\alpha V_\gamma=g_{\mu\beta}g^{\beta\gamma}V_\gamma
$
$
\Rightarrow U^\alpha\delta_\mu^\gamma\nabla_\alpha V_\gamma=\delta_\mu^\gamma V_\gamma
$
$
\Rightarrow U^\alpha\nabla_\alpha V_\mu=V_\mu
$
there are three ways to do the first step, only one uses metric compatibility. More at https://www.general-relativity.net/2019/10/symmetries-and-killing-vectors.html
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Hilbert-Schmidt basis for many qubits - reference Every density matrix of $n$ qubits can be written in the following way
$$\hat{\rho}=\frac{1}{2^n}\sum_{i_1,i_2,\ldots,i_n=0}^3 t_{i_1i_2\ldots i_n} \hat{\sigma}_{i_1}\otimes\hat{\sigma}_{i_2}\otimes\ldots\otimes\hat{\sigma}_{i_n},$$
where $-1 \leq t_{i_1i_2\ldots i_n} \leq 1$ are real numbers and $\{\hat{\sigma}_0,\hat{\sigma}_1,\hat{\sigma}_2,\hat{\sigma}_3\}$ are the Pauli matrices. In particular for one particle ($n=1$) it is the Bloch representation.
Such representation is used e.g. in a work by Horodecki arXiv:quant-ph/9607007 (they apply $n=2$ to investigate the entanglement of two qubit systems). It is called decomposition in the Hilbert-Schmidt basis.
The question is if there is any good reference for such representation for qubits - either introducing it for quantum applications or a review paper?
I am especially interested in the constrains on $t_{i_1i_2\ldots i_n}$.
| A good starting point, I have checked just chapter 4 but there is more, is
R. R. Puri, Mathematical Methods of Quantum Optics, Springer (2001) (see here).
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How to deterministically distinguish the following quantum states? (1) How to deterministically distinguish the following quantum states:
$$\frac{1}{\sqrt{2}}[|+0\rangle|0\rangle+|-1\rangle|1\rangle$$,
$$\frac{1}{\sqrt{2}}|-0\rangle|0\rangle+|+1\rangle|1\rangle$$,
$$\frac{1}{\sqrt{2}}|-0\rangle|0\rangle-|+1\rangle|1\rangle$$,
$$\frac{1}{\sqrt{2}}|+0\rangle|0\rangle-|-1\rangle|1\rangle$$
where $$|\pm\rangle=\frac{1}{\sqrt{2}}[|0\rangle\pm|1\rangle$$
Here the first two particles are with Alice and the third with Bob.
| If you want a deterministic local-operations-and-classical-communication protocol that will unambiguously determine which of the four states is present, it isn't possible. Alice can perform a degenerate measurement that collapses onto the spans of $|+0\rangle,|-1\rangle$ and $|-0\rangle,+1\rangle$, but after that the question is equivalent to determining under LOCC which of the Bell states $$|\phi^\pm\rangle_{AB}=\frac{1}{\sqrt{2}} \left[ |0\rangle_A |0\rangle_B \pm |1\rangle_A |1\rangle_B \right]$$ is present, and this is impossible - the best you can hope for using only classical communication is beating the CHSH Bell inequality.
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Given entanglement, why is it permissible to consider the quantum state of subsystems? Quantum entanglement is the norm, is it not? All that exists in reality is the wave function of the whole universe, true? So how come we can blithely talk about the quantum state of subsystems if everything is entangled? How is it even possible to consider subsystems in isolation? Anything less than the quantum state of the whole universe at once. Enlighten me.
| If you have two subsystems that are entangled and forming an (ideal) pure state, it is still meaningful to consider a subsystem experimentally. If I entangle two atoms and give one to you, nothing prevents you from making measurements on "your" atom only.
By repeating this same experiment many times, with atoms prepared in the same way, you would be able to characterise the density matrix describing your subsystem. You wouldn't be able to characterise the joint wave function of the two atoms though.
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Which main physics journals publish the main types of physics papers? I'm thinking of writing a paper on a new way of deriving the conservation of energy from symmetry principles and the Galilean transformations, but I'm not sure where to publish. Taking a look at AJP, it tends to be there for teachers at universities writing papers on how best to teach a certain topic, and to clear up misunderstandings. On the other hand, my paper certainly isn't to do with cutting edge physics, but merely another approach to something that is understood in other ways, and therefore not worthy of being published in Physics review, say.
So where should I publish?
More generally, how are the papers physicists publish mainly categorised, and which main journals publish papers in these areas?
| If your argument is nice pedagogically, and very original, it might be suitable for the "American Journal of Physics". But in general, if you just want to get it out, you can just put in on Arxiv under "General Physics", as various people have done with pedagogical innovations. Or else you can just write it in a blog, or elsewhere online.
In the case you mention, I doubt there is significantly room for improvement.
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Would you die if you put your hands on a powerline? You know how birds perch on powerlines without getting electrocuted? What if by some chance that I find myself falling and I grab on one of them? Let's say both of my hands are on the same line, would i get electrocuted?
I am thinking I won't because the current won't rush through me and I won't be part of the circuit - me - powerline.
How does the ground play a role in this? I've heard people say that the ground creates a potential difference, but how? There is only voltage across the powerlines, the pole connecting to the ground is wood, an insulator?
Thanks
| There actually is a current that moves through your body, albeit it not enough to hurt you. You don't need to actually touch the powerline, just standing near it will cause a current to flow in your body. If we model a single powercable hanging a height H above the ground as an electric charged cable of infinite length hanging a height $H$ above a perfect conductor, then the electric potential relative to the ground a distance R below the cable is:
$$V(R) = V_0 \frac{\log\left(\frac{2 H}{R}-1\right)}{\log\left(\frac{2 H}{R_{0}}-1\right)}$$
where $R_0$ is the radius if the cable and $V_0$ is the voltage relative to the ground. E.g. if you take $R_0 = 10$ cm , $V_0 = 10^6$ Volt then the optential difference between the head and the feet of someone 1.8 meters tall a meter below the powerline will be approximately 166,000 Volt. Now, this voltage has a frequency of 50 Hz and the human body has a finite resistance and electric capacity, so currents will flow in the body in response to the electric field. But these currents are not very strong. Even if the body had zero resistance so its potential were constant, then that would only require a small amount of charge to be moved around to neutralize the 166,000 Volt.
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Coulomb potential in 2D I know that the Coulomb potential is logarithmic is two dimensions, and that (see for instance this paper: http://pil.phys.uniroma1.it/~satlongrange/abstracts/samaj.pdf) a length scale naturally arises:
$$ V(\mathbf{x}) = - \ln \left( \frac{\left| \mathbf{x} \right|}{L} \right) $$
I can't see what's the physical meaning of this length scale, and, most of all, I can't see how this length scale can come up while deriving the 2D Coulomb potential by means of a Fourier transform:
$$ V(\mathbf{x}) = \int \frac{\mathrm{d^2 k}}{\left( 2 \pi \right)^2} \frac{e^{\mathrm{i} \mathbf{k} \cdot \mathbf{x}}}{\left| \mathbf{k} \right|^2} $$
I would appreciate some references where the two-dimensional Fourier transform is carried out explicitly and some insight about the physical meaning of L and how can it arise from the aforementioned integral.
|
I can't see what's the physical meaning of this length scale, and, most of all, I can't see how this length scale can come up while deriving the 2D Coulomb potential by means of a Fourier transform:
zakk, the constant L in the logarithm is the integration constant which is present for any differential equation and means that the Poisson equation used to derive V has infinity of solutions. It can be fixed by prescribing the value of potential V(r_0) at some distance r_0.
The same happen in 3D; there we have
V = 1/r + C
and we choose C = 0.
In your case we cannot choose L = 0, but L = 1 meter is good.
If V is supposed to be potential of electric force, the value of L is otherwise completely arbitrary.
The Fourier transform method can only find one particular solution of the equation. It cannot supply the integration constant L. You still have to add a constant - ln L to your Fourier integral to get the general solution.
Your integral for self-energy does not seem right. What is your starting point?
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How do we recognize hardware used in accelerator physics When I see a new accelerator in real life or on a picture, I always find it interesting to see how many thing I can recognize. In that way, I can also get a small first idea of how the accelerator is working.
Here is a picture, I have taken of LEIR at CERN
Help me to be able to recognize even more stuff, than I can now(I will post a few answers myself)
Suggested answer form:
*
*Title
*Images
*One line description
*Link
| Quadrupole magnet
Quadrupole magnets are mostly used for beam focusing.
http://en.wikipedia.org/wiki/Quadrupole_magnet
| {
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A Query on the Trapped Null Surface A a compact, orientable, spacelike surface always has 2 independent forward-in-time pointing, lightlike, normal directions. For example, a (spacelike) sphere in Minkowski space has lightlike vectors pointing inward and outward along the radial direction. The inward-pointing lightlike normal vectors converge, while the outward-pointing lightlike normal vectors diverge. It can, however, happen that both inward-pointing and outward-pointing lightlike normal vectors converge. In such a case the surface is called trapped.--------
from Wikipedia: http://en.wikipedia.org/wiki/Apparent_horizon
Now a null vector is parallel to and perpendicular to itself at the same time. So the tangent plane on the concerned point on the spacelike surface should be a null surface.
What is the formal explanation for this?
| The surface is codimension two in spacetime, it is two dimensional inside a four dimensional space. This means that the tangent space to the surface does not consist of all vectors perpendicular to the normal to the surface, just as in three dimensions, the one-dimensional vector tangent space is not the collection of vectors perpendicular to some perpendicular line.
The tangent space to the surface does not include any null vectors, it just is what it is, some 2-dimensional space-like vector space sitting inside the manifold tangent space. To give a simple example, if the surface is the x-y plane at t=0, the future pointing perpendicular null directions are along the z axis going to the right and towards the future (1,0,0,1) and to the left and to the future (1,0,0,-1). The collection of all vectors perpendicular to (1,0,0,1) is all vectors of the form (a,b,c,a) (where a,b,c are real numbers), and includes more vectors than the tangent space of (0,b,c,0). The collection of vectors perpendicular to the other null vector, (1,0,0,-1), are (a,b,c,-a). The collection of all vectors perpendicular to both the outgoing light rays is the tangent space of the surface.
The same holds in curved space, since this is the situation locally.
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If blue light has a higher energy than red light, why does it scatter more? As $E=hf=\frac{hc}{\lambda}$, blue light - with a smaller wavelength - should have a higher energy. However, it is the case that blue light scatters the most. Why is it that higher energy rays scatter more?
| I assume you are thinking along the lines of why is the sky blue?
Its because of the interaction of light with matter. In this case the interaction is called Rayleigh scattering.
The intensity of Rayleigh scattering is proportional to $$(Energy\;of\; the\; Photon)^4$$ or more completely
$$I = I_0\frac{8 \pi ^4 \alpha ^2}{\lambda ^4 R^2}\left(1+ cos^2\theta \right)$$
See the wikipedia article on Rayleigh Scattering for More info:
http://en.wikipedia.org/wiki/Rayleigh_scattering
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Time Reversal Invariance in Quantum Mechanics I thought of a thought experiment that had me questioning how time reversal works in quantum mechanics and the implications. The idea is this ... you are going forward in time when you decide to measure a particle. The particle then collapses to the observed state. Now if physics were to be the same in reverses time, then if we stop and reverse time then measure that very same particle again ... then I would imagine that since the wave function has collapsed we ought to measure the same thing. What this says to me is that given some time evolution in the + direction, if we measure a particle and it collapses the wave function, then if you reverse the arrow of time to go in the - direction we ought to get the same answer as before. The future/present effects the past. This means if we theoretically had a time machine and went back in time, we would have traveled into a different past.
Another implication of this thought experiment is that the future would be indistinguishable from the past and would hence forth be the same. I would imagine that this is consistent with the 2nd law of thermodynamics since physics dictates that entropy only increases ... going in the reverse direction of time to decrease entropy would violate the laws of physics. Has anyone else out there thought about this?
From my studies in quantum mechanics, I don't remember any postulates stating anything like this, but this all makes sense to me. Are there any theories out there that go along these lines?
| Just a few pointers for you to explore more on this. Check out Aharonov's paper the time symmetric formulation of quantum mechanics: http://arxiv.org/abs/quant-ph/9501011
Tony Leggett talks about this: http://www.youtube.com/watch?v=IGim9uzcumk
It's a nice video and quite simple to understand.
| {
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How can a Photon have a "frequency"? I picture light ray as a composition of photons with an energy equal to the frequency of the light ray according to $E=hf$. Is this the good way to picture this? Although I can solve elementary problems with the formulas, I've never really been comfortable with the idea of an object having or being related to a "frequency". Do I need to learn quantum field theory to really understand this?
| All you need is quantum mechanics, i.e. that nature in the microcosm is dual,sometimes it can manifest wave properties and sometimes particle properties.
It depends on the measurement/experiment if the wave or the particle nature will manifest itself. Electrons manifest this duality: in the two slit experiment their wave nature appears governed by the de Broglie wavelength. Photons do the same too, displaying the wavelength/frequency associated with the collective classical electromagnetic wave.
The classical electromagnetic wave is built out of photons in a consistent way, and you could study this link if you are interested in this more complicated problem.
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Has any permanent magnet motor been proven to run? I have read lots of articles about permanent magnet motors, some of which claim the possibility and other which refute it. Is it possible to have a permanent magnet motor that runs on the magnetic force of permanent magnets?
| Magnetic field, and interactions of atoms with magnetic field, conserve energy, i.e. no net energy gets created at any point. There is not a lot of energy in the field of even a strong permanent magnet. One could, in principle, construct a 'motor' that would demagnetise the magnets somehow, converting the field energy into motion, but it wouldn't be some effective super energy storage or the like, and would run for some time then stop.
I think the reason people turn to magnets when trying to build perpetual motion devices is that it is harder to understand energy conservation in this context, and furthermore it seems more magical. The fact is, the laws governing electromagnetic interactions are very well known, incorporate conservation of energy. There is the point of "what if the scientists are wrong?" Science, indeed can be wrong sometimes. But if this particular knowledge was off by more than utterly microscopic amount, the computer you use to read this message, the communication equipment, the power equipment, the generators, the hard drive in the computer storing this message, and so on, would not have been possible to engineer.
| {
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Origin of the Higgs field Are there any attempts in the literature at addressing the origin of the Higgs field? And, which lines of research that find it inevitable to address this question?
| In quantum field theory, the Higgs field is – much like the electromagnetic field, the W-boson field, the electron's Dirac field, and other fields – an elementary entity that can't be decomposed to anything simpler.
This claim of course assumes that the Higgs field isn't composite. The mass of the Higgs boson around 125 GeV, nearly discovered at the LHC, strongly indicates that the Higgs field isn't composite.
So one simply has a Higgs doublet of fields, $h_1(x,y,z,t)$ and $h_2(x,y,z,t)$, at each point of the spacetime. They're quantum fields (with hats) which means that the energy carried in the waves upon these fields is quantized (it is effectively composed of particles).
The now-disfavored models would construct the Higgs field out of more elementary fields. For example, in the technicolor models, one doesn't get objects that are "strictly identical" to the Higgs boson but one may produce similar particles out of more fundamental fields and particles such as the techniquark fields. The construction of the Higgs boson out of techniquarks is fully analogous to the construction of mesons (e.g. pions) out of quarks in Quantum Chromodynamics.
In string theory, all fields – electromagnetic field, Higgs fields, electron field, and others – may be derived from more fundamental building blocks, namely strings (and branes) and the string fields that create them (if one uses the string field theory description). In this picture, a Higgs boson is a vibrating microscopic string. The typical size of the string is so short, however, that one can't probe the internal structure by direct experiments.
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Is there a point in universe that is observable at present? We know that we can see distant galaxies only billions years before now. We can observe the nearest stars just several years before the present. Something on the Moon can be observed only some seconds in the past.
Continuing this scale, is there an object in the universe that can be observed just now, at present, or at least closer to the present than any other object?
I suppose such object should be located in the brain of the observer, but where in the brain exactly, given that brain has finite dementions.
The question can be formulated differently: where exactly is located the center of the sphere of the cosmological event horizon for a given observer?
| Perception of an event by a brain is a process distributed over a large part of brain's neural network (rather than being a process performed by a single special neuron cell). Thus, for perception to occur a part of the network must reach a state distributed across multiple neuron cells. Due to the limited speed at which neural impulses travel along axons and dendrites and across synapses, there is certain amount of time between the instant when sensory inputs reach the brain and the instant when the neural network settles in a respective distributed state. This means that even the perception of brain's own state isn't instantaneous.
This in turn implies that even "here" isn't perceived "now" and that no point in the universe is observable at precent.
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When would the proposed black hole at the centre of Milky Way gulp in our solar system? I've heard and read that our solar system lies near to the peripheral region of the Galaxy. Then accordingly we would have a greater probability of sustaining to eventual gulping down by the super-massive black hole. But how long ?
| We don't have to worry about falling into the black hole because we have far too much angular momentum from our motion about the galactic center. This can be made quantitative by considering the sign of the effective potential of our orbit. For reference, the sun is about 27000 light years from the galactic center, and its orbital speed is about 220 km/s. You'll find that the centrifugal term overwhelms the gravitational term by a factor of about $10^5$.
Also, on a galactic scale, the mass of the central black hole is tiny. It is a few million times as massive of the sun. On the other hand, the total mass of all the stars in the galaxy is about $10^5$ times larger, and the mass of all the dark matter is another factor of 10 or so larger than that.
Careers in astrophysics have been built on the question opposite of the one posed here - how does material ever shed its angular momentum to feed the black hole and allow it to grow to the size we observe? The answer is quite detailed and complex, and is still an active research topic.
| {
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Rotational speed of a discus I was wondering whether the rotational speed of a discus has any influence on the flight of the discus. Would slowing the rotation or speeding it up change the trajectory in any way or would the flight simply become unstable when slowing down?
| Given the physical conditions, this seems like an appropriate explanation: The faster the discus rotates, the more violently and quickly it displaces the air around it. Now the absence or scarcity of air causes a reduction in air friction or viscosity around the discus and this allows it to move onward in the direction of propulsion; now that depends on what angle the athlete projects it. After a certain distance there begins a constant deceleration of rotational speed because at some point, the air friction starts overpowering the rotation and this results in the discus entering the second half of its trajectory, i.e., moves downward along a curved path. I hope that answers your question.
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Is it possible to have incommensurable but equally valid theories of nature which fits all experimental data? Is it possible to have mutually incommensurable but equally valid theories of nature which fits all experimental data? The philosopher of science Paul Feyerabend defended this seemingly outrageous thesis and made a very strong case for it. In such a case, is it impossible to decide which of the incommensurable competing theories is the "right" one? In that case, does "anything goes"?
| The philosopher Saul Kripke has come up with a solution using modal logic and the possible worlds semantics. For each model or theory which fits all experimental data, assign a possible world. If a proposition P is true in all possible worlds, $\square P$. If a proposition Q is only true in some possible worlds, $\Diamond Q$. Philosophers have analyzed modal logic in incredible detail and come up with many deep insights. Incommensurability means $\Diamond P \wedge \Diamond Q$ but not $\Diamond (P\wedge Q)$. The possible worlds where P is true are not the worlds where Q is true.
Positivists will tell you to only consider those propositions which hold in all possible worlds, $\square P$. The problem is there are too many examples where to prove, $\square P$, you have to make use of $\Diamond Q$ type lemmas as intermediate steps. That kind of undermines the case for positivism.
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Lenses (refractor) or mirrors (reflector) telescope? What differentiates, in terms of practical quality, not technical implementation, a refractor from a reflector telescope?
Why would one prefer a refractor over a reflector, when reflectors come with such large diameters at a smaller price?
| Lenses are expensive, and have an upper limit on their size (when they get too big they get really expensive, and you start seeing a lot more aberration effects). Really cheap telescopes are usually refractors (at least from what I've seen), but if you want a respectable telescope, refractors climb in price rather quickly, in my experience.
Reflector telescopes are more common, smaller in size, generally less expensive and offer higher quality pictures without some of the aberration that comes with a refractor.
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Electrostatic Pressure Concept There was a Question bothering me.
I tried solving it But couldn't So I finally went up to my teacher asked him for help . He told me that there was a formula for Electrostatic pressure $\rightarrow$
$$\mbox{Pressure}= \frac{\sigma^2}{2\epsilon_0}$$
And we had just to multiply it to the projected area = $\pi r^2$
When i asked him about the pressure thing he never replied.
So what is it actually.Can someone Derive it/Explain it please.
| Electrostatic pressure is the tension developed inside the sphere due to mutual repulsion among the charges of the same sphere. Its same like a rubber band is stretched from all the points outwards so a stress is developed in it
| {
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Rotationally invariant body and principal axis Suppose a rigid body is invariant under a rotation around an axis $\mathsf{A}$ by a given angle $0 \leq \alpha_0 < 2\pi$ (and also every multiple of $\alpha_0$).
Is it true that in this case the axis $\mathsf{A}$ is a principal axis of the rigid body?
If so, how to prove it? Do you have any references for a proof?
| Your statement is true.
Proof:
Let $\rho$ be the mass density of the rigid body.
Remember that the tensor of inertia $I$ is given by:
$$
\vec{v}^t I \vec{w} = \int d^3b\, \rho(\vec{b}) (\vec{v} \cdot \vec{w} - (\vec{v}\cdot \vec{b})(\vec{w}\cdot \vec{b}))
$$
for all $\vec{v},\vec{w} \in \mathbb{R}^3$.
Now take an orthogonal matrix $O$ which represents a rotation around an axis $\mathsf{A}$ with direction vector $\vec{n}$, i.e. $O\vec{n} = \vec{n}$.
Your invariance means that $\rho$ is invariant under $O$, i.e. $\rho(O \vec{x}) = \rho(\vec{x})$ for all $\vec{x}$.
Next show that the inertia tensor commutes with $O$: $IO = OI$:
$$
\begin{align*}
\vec{v}^t I O \vec{w} &= \int d^3b\, \rho(\vec{b}) (\vec{v} \cdot O \vec{w} - (\vec{v}\cdot \vec{b})(O \vec{w}\cdot \vec{b})) \\
&= \int d^3b\, \rho(\vec{b}) (O^t \vec{v} \cdot \vec{w} - (\vec{v}\cdot \vec{b})(\vec{w}\cdot O^t \vec{b})) \\
&= \int d^3b\, \rho(O^t \vec{b}) (O^t \vec{v}\cdot \vec{w} - (O^t \vec{v}\cdot O^t \vec{b})(\vec{w}\cdot O^t \vec{b})) \\
&= \int d^3b\, \rho(\vec{b}) (O^t \vec{v}\cdot \vec{w} - (O^t \vec{v}\cdot \vec{b})(\vec{w}\cdot \vec{b})) \\
&= \int d^3b\, \rho(\vec{b}) ((\vec{v}^t O)^t\cdot \vec{w} - ((\vec{v}^t O)^t \cdot \vec{b})(\vec{w}\cdot \vec{b})) \\
&= \vec{v}^t O I \vec{w}
\end{align*}
$$
for all $\vec{v},\vec{w} \in \mathbb{R}^3$
Then one sees that $I\vec{n}$ is again an eigenvector of $O$ because $O(I\vec{n}) = IO\vec{n} = I\vec{n}$.
Now $I$ has only one real eigenvalue $1$ with eigenspace $\mathbb{R}\vec{n}$. This implies that there is a unique $\lambda$ with $I\vec{n} = \lambda \vec{n}$. Thus $\vec{n}$ is an eigenvector of $I$ i.e. $\vec{n}$ points along a principal axis of $I$.
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Power dissipated in resistor Suppose one has a circuit consisting of an inductor $L$ and resistor $R$ in series where $L$ and $R$ are known, passes an alternating voltage of frequency $\omega$ through it and that one wishes to calculate the mean power dissipated in the resistor.
Let the RMS voltage across the series combination be $V_0$. Then the RMS current through the components will be $I=\frac{V_0}{i\omega L + R}$ and the mean power dissipated in $R_2$ will be $\overline{P}=I^2R$. However, at this point, $I$ involves a complex quantity. How do you calculate the mean power? Do you calculate the magnitude of the complex current?
With very many thanks,
Froskoy.
| okay, This was really cool and I got some help from my physics professors on this one (apparently I won't learn this until next semester) and to find the magnitude of the square of a complex number you take it times it's complex conjugate. So in this case $$\frac{V_0}{i \omega L +R}$$ is multiplied with $$\frac{V_0}{-i \omega L + R}$$ leaving you with simply $$\frac{V_0^2}{w^2L^2+R^2}$$, then just take it times your $$R_2$$ giving you $$\frac{V_0^2}{\omega^2 L^2+R^2} \times R_2$$ for the average power.
Sorry that my equations aren't very pretty, LaTex isn't working on my ubuntu install yet.....
Hope this helps!!
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Why/how does an electron emit a photon when decelerating? I've had two special relativity courses so far but none really gave me a clear description of the process.
| This is a hand waving answer.
You ask in the comments:
I see that that particle gains/loses energy, but aren't there other ways to do that?
Down in the particle world everything is quantum mechanical and the only things that exist are the standard model particles, which may sometimes act as waves according to strict QM rules.
What are the possible interactions of an electron? Weak and electromagnetic. Weak is orders of magnitude weaker than the electromagnetic ( hence the name) and can be ignored.
Thus any measurable interaction an electron can have has to be electromagnetic. In the microcosm dimension any gaining or losing energy has to go through photons .
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Would the rate of ascent of an indestructible balloon increase as function of it's altitude? Assume a balloon filled with Hydrogen, fitted with a perfect valve, and capable of enduring vacuum (that is to say, it would retain it's shape and so well insulated that the extremes of temperature at high altitudes and in space would have little effect) were to be launched.
As long as the balloon were in atmosphere it would ascend upwards (and also affected by various winds/currents, and gravity). As the balloon passed through increasingly rare atmosphere, would it rise faster?
| There are several things to note here. First, the atmosphere stops being buoyant around 100-150 km above the surface of Earth (I believe the mesopause is roughly where this occurs). Balloons simply cannot float beyond this point, no matter how fluffy (lower density per unit volume) they are. Second, past the tropopause (where the mixing of the lower atmosphere stops), one starts to see the atmosphere lose its heavier components such as argon and diatomic oxygen (as well as trace molecules like carbon dioxide) to lighter molecules like atomic oxygen and nitrogen. This will have some affect on the rise of the balloon.
Third, you can get a more consistent rise, if you replace the rigid balloon of your question with a zero-pressure balloon, filled with a less dense gas, which maintains the same pressure on both sides of the balloon. Your perfect balloon expands so that the displacement remains the same even though the atmosphere becomes less dense, crudely exponentially, as one rises.
This is the sort of balloon actually used in placing things in the upper atmosphere. For example, the traditional latex weather balloon, while not zero-pressure, comes close with a low pressure drop between inside and outside of the balloon. It can easily reach the upper stratosphere. Proper zero-pressure balloons (which are effectively bags of helium or hydrogen with a hole on the bottom) can get significantly higher. I think somewhere over 50km is the current altitude record which is in the lower mesosphere roughly.
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Is there any way to survive solarwinter like in Sunshine - movie? Is there any way to survive solarwinter like in Sunshine - movie?
Solar winter is where for some reason sun looses its capasity to produce radiation( heat etc.). It doesn't loose everything but some of its radiation energy( say 50 %) That causes earth to cool down causing next "ice age"
| Currently all life lives off of the sun's energy, at least indirectly. One possible exception is some types of bacteria that need only heat energy plus some inorganic chemicals that abound in the ocean - those would be fine as long as the Earth is geologically active.
For us, however, it would be curtains, within months or several years at best. You could watch the movie The Road for at least one potential scenario where this happens (in the form of nuclear winter, where cloud cover never lets the sun through) - not good.
Energy would remain available for a long time to come - fossil fuels, nuclear fission fuels, and then even more if we figure out fusion. But a big question would be where do we get our food. Currently, science does not know how to keep humans alive without at least some input from the biosphere in the form of calories and minerals. Food is chemically complex, and we haven't fully characterized it yet, and don't know how to synthesize it. Because of that reason mostly, we would be doomed if the sun took a several-years nap. Since there's a lot of food stored around the earth, a lot of people would find ways to survive for a while - but the inevitable would be that once those thermophilic bacteria are the only other living creatures, and we've eaten up all canned food - we would be finished.
Interesting question! Whenever the answer to this becomes "yes we could survive" it would mean we are capable of surviving in the dead of space - which for me would mean a new level of evolution for our species.
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At what g is terminal velocity not terminal? How weak would gravity need to be in order for a human to reliably survive the terminal velocity of falling through air?
(Context: watching scifi on a space station with a variety of artificial gravities, it occurred to me that medium-strength gravity would have some advantages; also I note that insects already seem to have this luxury on earth, with their small masses and high air resistances...)
| The smallest body that we know has an atmosphere is Titan, which has about 1/7 of Earth's surface gravity ($1.4\ m/s^2$) but an atmospheric pressure of 1.45 Earth's ($146.7\ kPa$).
Using $pV=nRT$ where $T=95K$ and a mean molar mass of Titan's atmosphere being $28.6\ g/mol$ allows us to calculate an atmospheric density of $5.87 kg/m^3$, greater than Earth's.
So since there is less gravity but more atmosphere, I decided to look at the specific question of whether a terminal velocity fall on Titan is survivable. The answer was enlightening.
Using the formula for terminal velocity $$ V_t = \sqrt{\frac{2mg}{\rho AC_d}}$$
where (using reasonable estimates for the human coefficients):
$$ m_{human} = 75\ kg\\
g_{Titan}=1.4\ m/s^2\\
C_{d\ human}= 1.0\\
\rho_{Titan}= 5.87\ kg/m^3 \\
A_{human}= 0.75\ m^2$$
we get the interestingly low figure of
$$6.9\ m/s$$
Such a fall would be survivable.
Of course, the extreme cold and unbreathable atmosphere would not be, but at least the fall wouldn't kill you.
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Show that Bell states cannot be decomposed as tensor products of single qubits' states I'm trying to learn about the Bell state $\frac{1}{\sqrt{2}}|00\rangle+\frac{1}{\sqrt{2}}|11\rangle$. Question 10.1 in Algorithms asks us to show that this cannot be decomposed into the tensor product of two single qubits' states.
It seems to me however that this can be decomposed while still obeying the basic rules. Wolfram Alpha lists some solutions. What am I doing wrong?
| This property of the tensor product has really nothing to do with its interpretation in quantum mechanics or your specific example. Given two vector spaces $U,V$ the tensorproduct $\phi: U \times V \to U \otimes V$ is characterized by the property (up to isomorphism), that for any bilinear map $b \colon U \times V \to W$ into some third vectorspace $W$, there exists a unique linear map $l \colon U \otimes V \to W$, such that $b = l\circ\phi$. Because the two projections from the direct product $U\times V \to U$ and $U \times V \to V$ are not bilinear, they do not yield a map $U\otimes V \to U$ or $U \otimes V \to V$.
In quantum mechanical parlance this is called entanglement. As you probably know in this case $U$ and $V$ are spaces of state and $U \otimes V$ is the space of state for the combined system. Notice that just because the projections do not yield maps $U \otimes V \to V$ this does not mean there are no such maps. If for instance $U = H_e$ is the state space of a relativistic electron and $V = H_\gamma$ the state space of a photon, there is a map $H_e \otimes H_\gamma \to H_e$ that describes the absorption of a photon by an electron. In this case one usually draws a feynman diagram with a wiggly line for the photon meeting the electron line.
| {
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Changing the Half-Life of Radioactive Substances Is there a way to extend or reduce the half-life of a radioactive object? Perhaps by subjecting it to more radiation or some other method.
| Have a look at the paragraph "radioactive decay" .
The half life is characteristic of each radioactive nucleus and depends on the basic interactions holding the nucleus together.
It depends on the quantum mechanical probabilities of transition from one energy level to another, sometimes changing element in the periodic table.
Thus, to affect the half life, one would have to affect the basic interactions of the decay mechanism. There have been speculations on what would happen if the QFT vacuum is different, as in the Casimir effect, (a simpler explanation here), but I have not been able to find an experiment.
The simple answer is, no, the half life cannot change.
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How do we prove the existence of a multiverse? How do we prove that a multiverse exists?
Scientists are talking about our universe not being the only universe, but even if that is true, how can we prove the existence of multiverse? We are being 'confined' in this universe and there is no way we can know what is happening outside, right?
| By definition we'll never be able to observe other universes directly, because if we could they'd be in our universe.
However suppose (for example) string/M theory does get developed into a theory that is easily testable, and suppose we find that all the predictions string theory makes about our universe are experimentally found to be true. That means we'd be pretty convinced that string/M theory was the right theory to describe physics. If this hypothetical development of string theory still predicted a multiverse then we'd be inclined to believe in on the grounds the theory gets everything else right.
It's not unprecedented for a theory to predict things we can never observe. For example we cannot see what is behind the event horizon of a black hole (at least, not without jumping in :-). However we believe what General Relativity tells us about the interior of a black hole because it gives the correct results for everything that we can observe. The multiverse idea is a lot more extreme than the interior of a black hole, but you can see how we could be persuaded that it does exist.
I suppose for completeness I should mention that some scientists have suggested different universes may collide, and we could see evidence of the collision. See this Science Daily article for a introduction to this idea. I'm not sure how seriously this idea is taken.
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Collision of a black hole & a white hole A black hole and white hole experience a direct collision.
What happens? What shall be the result of such a collision?
| White hole is an impossible object in universe.
Mathematically it is a black hole under inverted time. This can be interpreted as a black hole in an universe where second law of thermodynamics is inverted, that is the entropy always diminishes.
Since second law of thermodynamics has probablistic nature, one can see a white hole as a highly unprobable state of black hole: the state where it consumes high-entropy Hawking radiation and exhales low-entropy objects instead of doing the opposite.
In theories which consider collisions between objects which have opposite arrows of time it is usually derived that upon such collision the object with reverse time arrow will quickly switch its time direction for which only a microscopic perturbation is enough. This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.
It should be noted that in the universe which reached termodynamic equilibrium, there is no difference between a black and a white hole, the both behave the same: consume and radiate high-entropy radiation.
| {
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Number of bits needed to express physical laws? What is the minimum number of bits that would be needed to express a given physical law, like the law of universal gravitation? How many bits are needed to express each of the four fundamental forces? Is there a pattern here?
| How many bits would it take to define the standard model plus gravity plus massive neutrinos plus dark matter and dark energy? It's an excellent question.
Let's consider the "new minimal standard model", which is a little out of date but still provides a template for the study of this question. The elements of the lagrangian appear in equations 1 through 6, and are assembled together in equation 7. I estimate that it takes about 300 symbols to write down. I count about 40 parameters. Suppose that on average they are known to two significant figures. Assuming 10 ln 2 bits per digit and 128 ln 2 bits per symbol, and we have about (300 x 128 + 40 x 20) ln 2, i.e. about 27000 bits.
Next, how many bits of knowledge do you need to possess in order to be able to interpret those 27000 bits correctly? E.g. what is the shortest set of functions that one could define in Mathematica, such that they correspond to all the basic quantities that make up predictions by the new minimal standard model? I don't think it would be much more than that 27000 bits, if we're careful not to include all the heuristics for computing the function, in the definition of the function. That is, we want to be able to say e.g. what an expectation value is, without necessarily specifying how to compute it.
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What are electromagnetic fields made of? I am trying to understand electromagnetic fields so I have two question related to them.
*
*What is a electromagnetic field made of? Is it made of photons / virtual photons?
*How about a static electric or magnetic field?
| A magnetic field is a essentially a cloud of virtual photon "place-holders" in a state of flux; it's what the electrons that produce the field "owe" to other nearby electrons (which have gained real photons), for having their spin-charge moments aligned in the same direction.
A magnetic field is even more fundamental an entity than particles such as electrons, protons and neutrons.
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In what sense is SUSY a spacetime symmetry? Clearly the SUSY anti-commutation relations involve momentum, and thus the generator of translations in spacetime:
$$\{ Q_\alpha, \bar{Q}_{\dot{\beta}} \} = 2 (\sigma^\mu)_{\alpha \dot{\beta}} P_\mu . $$
So I would say that naively SUSY has 'something to do with spacetime' since the any idiot can simply see the $ P_\mu $ in the above as I just have. But at a deeper level is there a relation between, say, the $Q$ and $\bar{Q}$ and spacetime? Alternatively, without writing down the above relation, how can we see that SUSY has something to do with spacetime?
| SUSY has something to do with spacetime since its generators $Q$ carry spin angular momentum, so they change the spin of the state they act on, hence SUSY is a spacetime symmetry. And this kind of generators is called fermionic generators. (compare with the generators of gauge symmetries, which are unphysical symmetries. They don't change the spin of the state they act on, and they are called bosonic generators.)
| {
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Path traced out by a point While studying uniform circular motion at school, one of my friends asked a question:
"How do I prove that the path traced out by a particle such that an applied force of constant magnitude acts on it perpendicular to its velocity is a circle?" Our physics teacher said it was not exactly a very simple thing to prove.
I really wish to know how one can prove it.Thank you!
| Try looking for uniform circular motion in google. It is not hard to prove it if you know something about vectors and what taking a derivative of vector function means. Force is a vector it is proportional to acceleration. Acceleration is change in velocity(remember a vector) divided by time(really shot period of time). Try to draw a circle yourself and some velocity, acceleration and position vectors
| {
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Why isn't it allowed to use a flash when taking pictures in a certain place? When I go to, for example, a museum I try to take some pictures.
Sometimes the museum staffs forbid me to use a flash. Do you know the reason? I don't think it is related to photo-electric effect, right?
| It really does depend on the place.
Sometimes it will be to minimise the (real or imagined) damaging effect of the light and/or heat from the flash, sometimes because it annoys or disturbs other visitors.
In a very popular place it may be to avoid triggering a photosensitive epileptic fit. Theoretically that's possible with a large number of flashbulbs going off in rapid succession in an enclosed space!
Quite often, in museums in particular, it's because the curator knows a bit about photography and is aware that with many of the exhibits being behind glass then flash photography is more likely to result in a rubbish, glare-filled image than a worthwhile shot.
Take your pick from those - or, if you really want to know, politely ask one of the staff for the reasons behind the ban.
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What's the difference between Fermi Energy and Fermi Level? I'm a bit confused about the difference between these two concepts. According to Wikipedia the Fermi energy and Fermi level are closely related concepts. From my understanding, the Fermi energy is the highest occupied energy level of a system in absolute zero? Is that correct? Then what's the difference between Fermi energy and Fermi level?
| It depends on who you ask.
If you ask someone with solid-state physics background, they will probably answer along the lines of Colin McFaul or John Rennie: The fermi level is the same as chemical potential (or maybe one should say "electrochemical potential"), i.e. the energy at which a state has 50% chance of being occupied, while the fermi energy is the fermi level at absolute zero.
If you ask someone with semiconductor engineering background, they will probably give the same definition of "fermi level", but they will say that "fermi energy" means exactly the same thing as fermi level. (The obvious question is, "Then what term would a semiconductor engineer use to describe the fermi level at absolute zero? The answer is, they call it "the fermi level at absolute zero"!)
| {
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Non linear QM and wave function collapse I heard that there have been some propositions about describing the collapse of the wave-function by adding non-linear terms, but I couldn't anything in any any textbooks or even articles (probably those propositions never reached a good level of consistency). However, I'd like to read about it. Could someone send me a reference?
| The Ghirardi-Rimini-Weber Model is such a theory.
See for instance http://arxiv.org/abs/quant-ph/0406094.
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How to compute the pure extensions of a given mixed state? Let us consider any pure state $|\psi\rangle\in\mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n$. Its reduced bipartite density matrix represent a pure state or mixed state depending on whether $|\psi\rangle$ is entangled or not (exactly how it is entangled, on which system we take the partial trace, etc).
My question is given any arbitrary (mixed) state $\rho\in\mathcal{B}(\mathbb{C}^n\otimes \mathbb{C}^n)$, can we find out a pure state $|\psi\rangle\in\mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n$ (or in some suitable higher dimension which needs to be determined) such that $\rho$ is the reduced density matrix of $|\psi\rangle$. In particular, I do not want only an existential result, I also want to an algorithmic method to determine such $|\psi\rangle$. Obviously such state will not be unique. Advanced thanks for any help.
| *
*The fact that every mixed state $\rho$ acting on a finite dimensional Hilbert spaces can be viewed as the reduced state of some pure state $|\psi\rangle$ on a bigger Hilbert space is known as purification, see this Wikipedia page, where also the algorithm is given.
*In OP's case of $$\rho~\in~\mathcal{B}(\mathbb{C}^n\otimes \mathbb{C}^n),$$ one may choose a pure state $|\psi\rangle$ in the following Hilbert space
$$|\psi\rangle~\in~\mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n.$$
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Intrinsic Viscosity I'm presently undergoing an Experiment for the determination of the viscosity of Ficoll-70 using Ostwald viscometer to calculate the time and a digital weighing balance to determine the weight of a specified volume of the solution at different concentration. I have reached a result but unable to correlate with any data provided by the journals or thesis on this area. It's because they all have given the value of 'Intrinsic Viscosity'. How different is the Dynamic viscosity from the Intrinsic one? Can I calculate the dynamic viscosity of a solution if I know its Intrinsic viscosity ?
| Your viscometer measures the viscosity of your Ficoll-70 solution. This viscosity is made up partly from the viscosity of the water and partly from the viscosity of the Ficoll-70 dissolved within it, and obviously the viscosity you measure increases with the concentration of Ficoll-70.
Intrinsic viscosity is a rather different concept. It's really a measure of the shape of the solute at infinite dilution, or at least sufficiently dilute that you can ignore three or more body interactions. For these low dilutions it's the shape of the solute that determines the viscosity, and the intrinsic viscosity depends on this shape.
At very low concentrations the dynamic viscosity will be given by:
$$ \eta = ([\eta]\phi + 1)\eta_0 $$
where $[\eta]$ is the intrinsic viscosity, $\eta_0$ is the viscosity of the solvent (presumably water in this case) and $\phi$ is the concentration of the Ficoll-70. You'll probably find $[\eta]$ is given in units of ml/g, so you need to give $\phi$ in g/ml. You could try calculating the viscosity using this formula, but I'd be suprised if it worked for the sort of concentrations Ficoll-70 is normally used at i.e. a few percent by weight.
Later: I note that https://somapps.med.upenn.edu/pbr/portal/immune/Ficcoll_info.pdf gives the relative viscosity of Ficoll-70 as a function of concentration. You could use this to check your results.
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Could someone remind me of what we mean by zero electric field "inside" a conductor? If I have a spherical conductor (perhaps a shell) and "inside", as in the hollow area there is nothing. The electric field is 0. But what happens if there is a charge "inside" (not like inside the conductor, but in the hollow region) or some type of insulator with a nonzero net charge? Would there exist a field?
I drew a picture to illustrate. Not to scale by the way
| For a static situation (i.e. no charges moving or current flowing) the net electric field is always zero inside of a conductor. Where by 'inside' I mean actually inside the material itself, not a region of free space that is simply enclosed by the material. The reason for this is, say there is an electric field $\vec{E}_0$, applied to the conductor- by definition there is an essentially infinite well of free charge that can move. This charge will move in response to the applied electric field and set up its own induced field, $\vec{E}_{ind}$. This field will oppose the orignal field and charge will keeping moving until it reaches equilibrium such that $\vec{E}_0+\vec{E}_{ind}= \vec{E}_{net} = 0 $. In practice this happens almost instantaneously. Above I haven't mentioned anything about shapes or charges outside the conductor so this holds for all static electric field configurations, including the one you describe above.
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Dimension of vector resulting from tensorial product I'm quoting what I found in a book about quantum computation:
Individual state spaces of $n$ particles combine quantum mechanically through the tensor product. If $X$ and $Y$ are vectors, then their tensor product $X\otimes Y$ is also a vector, but its dimension is $\dim(X) \times \dim(Y)$, while the vector product $X\times Y$ has dimension $\dim(X)+\dim(Y)$. For example, if $\dim(X)= \dim(Y)=10$, then the tensor product of the two vectors has dimension $100$, while the vector product has dimension $20$.
I don't understand: how can he state that the result of a vector product has dimension $\dim(X) + \dim(Y)$? What does he intend for dim?
| The tensor product is the natural extension of the ordinary product
$(a+b)(c+d)=ac+ad+bc+bd$.
If you have two vector $x,y$ of dimension $n$ the tensor product become
$$
x_\mu \otimes y_\nu= x_\mu y_\nu=\Theta_{\mu\nu}
$$
where $\Theta_{\mu\nu}$ is a matrix of dimension $n\times n$.
The vector product of two vectors $x,y$ generate a third vector $z$ orthogonal to $x,y$. This means that if you want fully define the vector $z$ you must define $x$ and $y$ and then, in terms of degrees of freedom, you must add the degrees of freedom of $x$ and $y$.
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How long was a day at the creation of Earth? Since the earth is slowing its rotation, and as far as I know, each day is 1 second longer every about 1.5 years, how long was an earth day near the formation of earth (4.5 billion years ago)?
I wouldn't assume to just do 4.5b/1.5 and subtract, because you would think the rate of change is changing itself, as seen here from wikimedia. It is a graphical representation of data from INTERNATIONAL EARTH ROTATION AND REFERENCE SYSTEMS SERVICE. They decide when its time for a leap second (the last one being on Jun 30, 2012) The data can be found here.
| Actually we only gain 1.3 milliseconds every 96-100 years, not 1 second every 1.5 years! :) the shortest known Earth day was 6 hours and the longest is 24 hours & 2.5 milliseconds (today's current day), in 1820 the day was exactly 24 hours, but since it's been nearly 200 years we've gained 2.5 milliseconds to our day. So the days get longer just very shortly, I believe it'll be 15 minutes longer in 50 million years.
| {
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Can a huge gravitational force cause visible distortions on an object In space, would it be possible to have an object generating such a huge gravitational force so it would be possible for an observer (not affected directly by gravitational force and the space time distortion) to see some visual distortions (bending) on another small object placed near it ?
(eg : a building on a very huge planet would have his lower base having a different size than the roof).
We assume that object would not collapse on himself because of the important gravitational force.
| To elaborate a bit: when Einstein published the General Relativity papers, he included a calculation of the precession of the orbit of Mercury which showed GR made up for the rest of the corrections needed to account for the observed values, which was the first empirical evidence for GR. As a second piece of evidence, Einstein proposed using a solar eclipse to test precisely for gravitational lensing and included a calculation of the effect. (Two calculations, in fact - the first was off by a factor of 2 but he fixed it in time.)
This was verified by Eddington's expedition to Principe in 1919, which showed that stars near an eclipse are in fact shifted from where they ought to be in a Newtonian universe.
To highlight just how strange this is, here's the New York Times headlines reporting on the event:
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Friction at zero temperature? By the fluctuation-dissipation theorem (detailed-balance for Langevin equation), $$\sigma^2 = 2 \gamma k_B T$$ where $\sigma$ is the variance of noise, $\gamma$ is a friction coefficient, $k_B$ is Boltzmann's constant, and $T$ is temperature. So in principle, one can have $\gamma\neq 0$ while $T=0$ and $\sigma=0$.
Is it indeed possible to experimentally achieve a system whose temperature and noise approach zero, but whose friction coefficient $\gamma$ does not approach zero?
*
*If yes, what would be an example of such a system? What is the physical source of friction for such a system?
*If not, why not? Is there some sort of "quantum" correction to the fluctuation-dissipation theorem that rules out such zero-noise, non-zero friction systems?
| Mechanical friction is a perfectly fine example. The coefficient of friction between two materials does not approach zero at absolute zero.
Electrical resistance (as pointed out by Alexander) is another example. Some materials (superconductors) have zero resistance at absolute zero, but by no means all of them!
I would say that $\gamma \neq 0$ while $T=0$ and $\sigma=0$ is the "default" expectation that occurs most of the time. Things like superconductivity and superfluidity are interesting surprises that go against the normal expectation.
The physical sources of friction at absolute zero are generally the same as the physical sources of friction at other temperatures. For example, electrical resistance can come from electrons bumping into grain boundaries or impurities or defects etc. Mechanical friction comes from phonons (vibrations) that are created as the two materials rub against each other ... same as usual.
If a source of friction is temperature-dependent, it can either increase or decrease as you approach absolute zero.
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Magnetic field inside a charged stream Outside a narrow charged stream (say, a beam of ions or electrons) is the same as observing a current through a conducting wire - there is a circular magnetic field around it.
What would happen inside a charged stream (for example, inside a conducting wire or inside a solar flare)? I have a feeling that symmetry will rule that there is no magnetic field, but I am not sure.
| The mmf due to a current is determined by the current through the surface bounded by the closed path along which the magnetic field is integrated. A closed path within a cross-section of a conductor with, say, a uniform current density, will have a non-zero mmf associated with it and thus, a non-zero magnetic field exists within the conductor.
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Sound frequency of dropping bomb Everyone has seen cartoons of bombs being dropped, accompanied by a whistling sound as they drop. This sound gets lower in frequency as the bomb nears the ground.
I've been lucky enough to not be near falling bombs, but I assume this sound is based on reality.
Why does the frequency drop? Or does it only drop when it is fallling at an oblique angle away from you, and is produced by doppler shift?
I would have thought that most bombs would fall pretty much straight down (after decelerating horizontally), and therefore they would always be coming slightly closer to me (if I'm on the ground), and thus the frequency should increase..
| When a baby cries, the sound of his/her cry becomes lighter. But actually the frequency of the baby's vocal-cord arises. Due to the limitation of our ears (20-20000Hz) we failed to feel the loudness of the baby's cry.
Same way, when the bomb closer to ground the frequency of bomb reaches near 20000Hz (with respect to our ears). So we feel the frequency lighter.
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Will adding heat to a material increase or decrease entropy? Does adding heat to a material, thereby increasing electrical resistance in the material increase or decrease entropy?
Follow up questions:
Is there a situation were Heat flux ie. thermal flux, will change entropy?
Does increasing resistance to em transfer prevent work from being done?
| A higher temperature will cause the atoms in the material to vibrate more, increasing the number of microstates available to the material. Thus, the entropy also increases, since the (microscopic) definition of entropy is
$$ S = k \log \Omega $$
where $k$ is the Bolzmann constant, and $\Omega$ is the number of microstates.
See also the "Statistical Thermodynamics" section of the wikipedia entry on entropy.1
This does not directly have anything to do with the electrical resistance, although the electrons carry part of the entropy of the material.
| {
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Non-Newtonian Fluid Stop a Bullet? I just saw a YouTube video about Non-Newtonian fluids where people could actually walk on the surface of the fluid but if they stood still, they'd sink. Cool stuff.
Now, I'm wondering: Could a pool of Non-Newtonian fluid stop a bullet? Why or why not?
If so, if you put this stuff inside of a vest, it would make an effective bullet-proof vest, wouldn't it?
| Yes, here is an example where they made a bullet proof vest by soaking Kevlar fabric in a non-newtonian fluid:
http://www.youtube.com/watch?v=LlEo5MbcaX0
The video is from 2006, I don't know if this has been developed further.
| {
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Is there an intuitive description of vacuum entanglement? People often refer to the fact that the vacuum is an entangled state (It's even described as a maximally entangled state).
I was trying to get a feeling for what that really means. The problem is that most descriptions of this are done in the formalism of AQFT, which I'm not very familiar with. The entanglement definitions which I have some feeling for are those of the form
System S Hilbert space $\mathcal{H}$ factorizes as
$\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$ where A and B are
two subsystems of S. An entangled state can't be written in the form
$\phi_A \otimes \phi_B$
There are then various measures of this, such as entanglement entropy.
So my question is - is it possible to describe the entanglement of the QFT vacuum in these more familiar terms?
Can such a description be given for a simple QFT example, say a Klein Gordon field on Minkowski space?
| For a non-interacting quantum field, the whole mathematical structure of purely Gaussian VEVs that is the vacuum state is contained in the 2-point VEV, which for the KG field is the distribution
$$\left<0\right|\hat\phi(x+y)\hat\phi(y)\left|0\right>=\frac{m\theta(x^2)}{8\pi\sqrt{x^2}}\left[Y_1(m\sqrt{x^2})+\epsilon(x_0)iJ_1(m\sqrt{x^2})\right]-\frac{\epsilon(x_0)i}{4\pi}\delta(x^2)$$
$$\hspace{7em}+\frac{m\theta(-x^2)}{4\pi^2\sqrt{-x^2}}K_1(m\sqrt{-x^2}).$$
The second line gives the correlation function at space-like separation, where joint measurements are always possible, whereas at time-like or light-like separation the imaginary component of the first line causes measurements to be incompatible. Measurement incompatibility introduces issues that are not easily given an intuitive gloss, of course, but the above shows the nature of the correlations for the free field case.
The Bessel function term at space-like separation is $\frac{1}{4\pi^2(-x^2)}$ at small $x$, whereas it is asymptotically becomes $\sqrt{\frac{2m}{\pi^3\sqrt{-x^2}^3}}\,\frac{\exp{\left(-m\sqrt{-x^2}\right)}}{8}$ for large $x$.
For interacting fields, the 2-point function is always of a comparable form, smeared by a mass density, the Källén–Lehmann representation, but higher order VEVs are relatively nontrivial.
| {
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Phase shift of 180 degrees of transversal wave on reflection from denser medium Can anyone please provide an intuitive explanation of why phase shift of 180 degrees occurs in the Electric Field of a EM wave, when reflected from an optically denser medium?
I tried searching for it but everywhere the result is just used.The reason behind it is never specified.
| Mathematical explanation: The is because the boundary is rigid and the
disturbance must have zero displacement at all
times at the boundary. By the principle of
superposition, this is possible only if the reflected
and incident waves differ by a phase of π, so that
the resultant displacement is zero.
Using Newton's laws:We can arrive at the same conclusion
dynamically also. As the pulse arrives at the wall,
it exerts a force on the wall. By Newton’s Third
Law, the wall exerts an equal and opposite force
on the string generating a reflected pulse that
differs by a phase of π.
Source: page-374 of NCERT physics class 11,
| {
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Where do the terms microcanonical, canonical and grand canonical (ensemble) come from? Where do the terms microcanonical, canonical and grand canonical (ensemble) come from?
When were they coined and by whom? Is there any reason for the names or are they historical accidents?
| I'm not completely sure, but I think they are introduced by Gibbs, and that book (available for download) is of historic importance.
The word ensemble really just means "set" in French, you consider the space of canonical coordinates of the detailed mechanics = microstates and you impose statistics by the fundamental postulate.
| {
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About an electrostatics integral and a delta-function kernel I'm having trouble with an integral and I would like some pointers on how to "take" it:
$$
\int \limits_{-\infty}^{\infty}\frac{3\gamma a^{2}d^{3}\mathbf r}{4 \pi \left( r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf r \cdot \mathbf u)^{2} + a^{2}\right)^{\frac{5}{2}}}
$$
Here $\mathbf u$, $a$ and $\gamma$ are constants, and the integrand converges to a Dirac delta $\delta(\mathbf r)$ as $a\rightarrow 0$. The integral must be equal to 1.
| Set the $z$ axis along the direction of $\mathbf{u}$ and use spherical coordinates, which reduces your integral to something like
$$\int_0^\infty dr\int_0^\pi d\theta\int_0^{2\pi}d\phi\frac{r^2 \sin(\theta)}{\left(a^2 +r^2(1+\frac{\gamma^2}{c^2}\cos^2(\theta))\right)^{5/2}}.$$
Do the $\phi$ integral first and then the $\theta$ integral, transforming to $u=\cos(\theta)$. Be careful to use absolute values for the roots when necessary. After that the $r$ integral should be tough but doable.
EDIT to take some discussion off comments.
I gave
Integrate[Sin[[Theta]]/(a^2+r^2 (1+[Gamma]^2 Cos[[Theta]]^2))^(5/2),[Theta]]/.[Theta]->[Pi]
to Mathematica to get
$$\frac{3 \left(a^2+r^2\right)+2 r^2 \gamma ^2}{3 \left(a^2+r^2\right)^2 \left(a^2+r^2+r^2 \gamma ^2\right)^{3/2}},$$
and then doing the radial integration by
Integrate[r^2 (3 (a^2+r^2)+2 r^2 [Gamma]^2)/(3 (a^2+r^2)^2 (a^2+r^2+r^2 [Gamma]^2)^(3/2)),r]
gives
$$\frac{r^3}{3 a^2 \left(a^2+r^2\right) \sqrt{a^2+r^2 \left(1+\gamma ^2\right)}}$$
for the antiderivative. I agree that the roots make one suspect nonelementary antiderivatives but it is only one root so that elliptic integrals are out. Once one has the antiderivative, of course, it is routine to check that it does differentiate to what it should.
| {
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Measurement and uncertainty principle in QM The Wikipedia says on the page for the uncertainty principle:
Mathematically, the uncertainty relation between position and momentum arises because the expressions of the wave function in the two corresponding bases are Fourier transforms of one another (i.e., position and momentum are conjugate variables).
Does that mean that position and momentum are just 2 different measurements of the same wave function? I.e., it is the same thing that is being measured, just in two different ways? Meaning, they are not really two different things, but two different views on the same thing?
| Any measurement in physics is in general described by a probability distribution of different outcomes. This distribution depends both on the state of the system being measured and on the measurement apparatus, which are two different things. In quantum mechanics states are described by vectors in Hilbert space $\left|\psi\right>$ (wavefunctions may be seen as their coordinates in some basis), and measurements by Hermitian operators $\hat{A}$ acting on this space (this is the simpliest case, actually the formalizm is a bit more complicated). Probability distribution of measurement outcomes is given by eigenvalues of these operators, and average values of measured quantities by $\hat{\left<A\right>}=\left<\psi\right|\hat{A}\left|\psi\right>$.
Position and momentum measurements are described by two different operators $\hat{x}$ and $\hat{p}$, such that $\hat{x}\hat{p}-\hat{p}\hat{x}=i\hbar$. Their noncommutativity leads to Heisenberg uncertainty relations for variances of corresponding measurements, as described in wikipedia. So the answer is no, they are different things, measured with different apparatus, but if their are done on a system in a given state, their variances turn out to be related.
| {
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How do mirrors work? My physics professor explained to me that electromagnetic waves are consisted of two components - electric and magnetic - which cause each other.
*
*Which part of the mirror actually reflects the wave?
*Which of those two wave components? Both?
*How come the wave doesn't get heavily distorted in the process?
I guess the actual electrons of atoms of silver play a role, but why isn't every material reflective, then? Because is isn't "perfectly" flat? If I lined up atoms of a non-metal element in a perfect plane (maybe several rows), would it reflect light just as mirrors do?
| Trying to separate electric and magnetic parts of a wave is not possible (Maxwell's equations couple them for propagation), so I will ignore your first two paragraphs.
The mirror conductivity is the key. The electric field from light reaches the mirror's
metal and thereby causes a current to flow (which actually generates an opposite-moving
electro-magnetic wave to nearly-cancel the surface electric field). The reflected image
you see really is generated by these induced currents.
"Flatness" does not matter. "Free electrons to move" matters.
| {
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What is the physical meaning of diffusion coefficient? In Fick's first law, the diffusion coefficient is velocity, but I do not understand the two-dimensional concept of this velocity. Imagine that solutes are diffusing from one side of a tube to another (this would be the same as persons running from one side of a street) to unify the concentration across the tube.
Here we have a one-dimensional flow in x direction. The diffusion coefficient should define the velocity of solutes or persons across the tube or street direction. How the two-dimensional velocity does this? I wish to understand the concept to imagine the actual meaning of the diffusion coefficient.
| Diffusion is a stochastic process where a single particle can move in each direction with the same probability.
Another description of the diffusion coefficient is the following equation:
$$D = x^2/(2t)$$
where $t$ is the time and $x^2$ is the mean squared displacement of the particles at this time.
The mean squared displacement, $x^2$, can be interpreted as the statistical variance of the particle positions, so the diffusion coefficient can be interpreted as the rate at which the variance changes.
This idea was first proposed by Einstein (1905. Ann. Phys., 17, 549--560. http://www.zbp.univie.ac.at/dokumente/einstein2.pdf)
| {
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Can light exist in $2+1$ or $1+1$ spacetime dimensions? Spacetime of special relativity is frequently illustrated with its spatial part reduced to one or two spatial dimension (with light sector or cone, respectively). Taken literally, is it possible for $2+1$ or $1+1$ (flat) spacetime dimensions to accommodate Maxwell's equations and their particular solution - electromagnetic radiation (light)?
| I agree. Light can not exist in 2D spacetime as well as a $\vec{B}$ component that has to be perpendicular to $\vec{E}$. Also, the Gauss law requires
$$E \propto \frac{q}{r^{D-1}}$$ where D is the number of spatial dimensions.
Therefore it is absent in 2D.
However, 2D or 3D spacetimes still can have the speed of light and
Lorentz invariance!
$$ ds^2= - c^2 dt^2 + dx^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What does symplecticity imply? Symplectic systems are a common object of studies in classical physics and nonlinearity sciences.
At first I assumed it was just another way of saying Hamiltonian, but I also heard it in the context of dissipative systems, so I am no longer confident in my assumption.
My question now is, why do authors emphasize symplecticity and what is the property they typically imply with that? Or in other more provocative terms: Why is it worth mentioning that something is symplectic?
| Symplectic geometry is may be the cornerstone of the geometrization of physics. In addition to the very known fact that classical mechanics can be described by symplectic geometry, given some other structures, symplectic spaces can be quantized to produce quantum mechanics as well. A subclass of symplectic geometries namely Kaehler geometry is especially important to quantization problems.
Many physical theories such as Yang-Mills and gravity have descriptions in the context of symplectic geometry, please see the review: THE SYMPLECTIZATION OF SCIENCE by Gotay and Isenberg.
Also many types of dissipative systems can be treated using symplectic geometry if we allow complex Hamiltonians please see S.G. Rajeev's article.
Finally, I want to remark that in the symplectic geometry terminology there is a distinction between symplectic and Hamiltonian vector fields, while a symplectic vector field is required to leave the symplectic structure invariant, a Hamiltonian vector field is required in addition to produce an exact form upon the contraction with the symplectic form.
For example the vector fields along the generators of the two-torus are symplectic but not hamiltonian. This distinction exists only if the symplectic manifold is nonsimply connected.
| {
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Motion in the body-fixed frame? This is really basic, I'm sure: For rigid body motion, Euler's equations refer to $L_i$ and $\omega_i$ as measured in the fixed-body frame. But that frame is just that: fixed in the body. So how could such an observer ever measure non-zero $L$ or $\omega$?
| I have yet to find a physics book that doesn't make this really confusing. If one has a vector fixed in inertial space, its components as viewed in a moving frame are obtained by the dot product of the vector with the moving unit triad fixed to the body but moving relative to inertial space. While the inertial frame would measure its components as constants with time, the moving system would measure components that vary with time because the unit triad to which the components refer are moving relative to the fixed vector under investigation.
The body rotates about an axis through an angle that can be described in the inertial frame, but all of its points are not moving if one is tied to and moving with the moving body frame. In fact, the only way one can deduce that he or she is tied to the body is via the coriolis force due to the rotational acceleration experienced.
The body rotates through an axis and angle, each of which, in general, varies with time, relative to a secondary frame (often inertial) within which the motion of every point on the body can be observed as moving.
| {
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Is it theoretically possible to reach $0$ Kelvin? I'm having a discussion with someone.
I said that it is -even theoretically- impossible to reach $0$ K, because that would imply that all molecules in the substance would stand perfectly still.
He said that this isn't true, because my theory violates energy-time uncertainty principle.
He also told me to look up the Schrödinger equation and solve it for an oscillator approximating a molecule. See that it's lowest energy state is still non-zero.
Is he right in saying this and if so, can you explain me a bit better what he is talking about.
| For a temperature to be definable and measurable the distribution of the kinetic energies of the molecules in the medium under discussion should be known.
The process of cooling involves removing thermal energy from a system. When no more energy can be removed, the system is at absolute zero, which cannot be achieved experimentally. Absolute zero is the null point of the thermodynamic temperature scale, also called absolute temperature. If it were possible to cool a system to absolute zero, all motion of the particles comprising matter would cease and they would be at complete rest in this classical sense. Microscopically in the description of quantum mechanics, however, matter still has zero-point energy even at absolute zero, because of the uncertainty principle.
The uncertainty principle assures that molecules cannot stay perfectly still and continue being in a certain position , i.e. in the material under study. Certainly not all molecules of the material, this would be necessary to define a 0K temperature.
The solution with the vibrational degrees of freedom that molecules may have is not conclusive , though sufficient as proof for that the specific material that displays these vibrational modes cannot go to 0K. It is the HUP that is general for all materials.
| {
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Can I study Quantum Computing or Quantum Mechanics with an Engineering background? I am currently studying Electrical & Electronic Engineering. I wish to pursue Quantum Mechanics or Quantum Computing as my research subject. Is it possible for me to do my M.Tech. and then pursue my research subject? What are the prerequisites for studying these subjects? I would be grateful if you could help me.
| Yes, it is possible. Your first step would be to learn Quantum mechanics from s standard textbook such as Modern Quantum mechanics: by sakurai, or Principles of Quantum Mechanics: by Shankar. There is nothing such as research in pure quantum mechanics, though there are a small handful of people working on the foundations of quantum mechanics. If you want go to the the physics side of quantum mechanics, you could study Quantum field theory, or condensed matter theory. but it would ve very difficult for you to study these, unless you go through the whole physics undergrad curriculum starting from classical mechanics, statistical mechanics, electromagnetism and a great deal of math. If you want to pursue quantum computing it is not that difficult if you have some knowledge of theoretical computer science. Refer to this question for further information. After reading QM, you may pick up a standard textbook on quantum computing. Many universities will accwept PhD applications in these areas as long as you have the knowledge in these areas, though you should have sufficicient knowledge in Physics to crack exams like the GRE, or JEST, for PhD admissions.
| {
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Explanation for $E~$ not falling off at $1/r^2$ for infinite line and sheet charges? For an infinite line charge, $E$ falls off with $1/r$; for an infinite sheet of charge it's independent of r! The infinitesimal contributions to $E$ fall off with $1/r^2$, so why doesn't the total $E$ fall off the same way for the infinite line and sheet charges?
| Loosely speaking, as we walk away from a sphere it looks smaller, as we walk away from a cylinder just the radius looks smaller, but not the infinite length, and finally as we walk away from an infinite sheet of charge it never looks any smaller (we can never 'get away' from an infinite sheet).
At more mathematical level I would say the best way to see this is with Gauss's Law. I am going ignore the details below and just assume you have a clue what Gauss's Law is, otherwise the rest of the post is pretty irrelevant for you. Below I will use the following 3 Gaussian surfaces I have borrowed from
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html :
For any case where you are interested in obtaining the electric field using Gauss's Law $\displaystyle \frac{ q_{enc}}{\epsilon_o} = \Phi= \int \bf{dA} \cdot {E} ~$ you will end up picking a gaussian surface such that the electric field is constant or zero over the Gaussian surface so that it just evaluates to $ |\bf{E}| \int |d\bf{A}_{||}|$ where the surviving integral is just over portions of the area parallel to the electric field. For example, for a point charge you will draw a concentric sphere. You can see from the above the further away you get from the point charge the bigger the area gets and so the electric field must get smaller in order to keep the flux constant. For a line charge we would draw a concentric cylinder and only the portion of the area that wraps around will survive. Again the flux must be constant but the area that wraps around a cylinder that grows much more slowly as the radius grows than that of the surface area of a sphere as the sphere's radius grows. Finally for a sheet we would draw a cylinder so that only the ends of the cylinder that are parallel with the sheet survive. Now as the length of the cylinder increases the surface area of the ends doesn't grow at all, so the electric field doesn't fall off at all.
| {
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Reflectance vs. Thin Metal film Thickness Graph Is there formula that gives reflectance of very thin film of given metal (tens of nanometers) to the visible light of given wavelength(808nm) ? Which properties of metals are needed for the formula ?
I would like to draw a plot of reflectance that is a function of titanium film thickness. Thanks
| Have a look at my answer to Make a semi transparent mirror with copper. To a good approximation the transmission falls exponentially with thickness. Just work out what tranmission you need, e.g. if you want 80% reflectance choose 20% transmission, and work out the film thickness accordingly.
You can find the optical constants for titanium at http://refractiveindex.info/?group=METALS&material=Titanium
| {
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What would happen if a hydrogen bomb were to explode in Saturn's atmosphere? Purely hypothetical since any kind of testing in atmosphere/space is banned by international legislation/agreement.
The humans have already bombed Luna so ... what could be expected to happen on Saturn if a hydrogen bomb were to explode in it's atmosphere? Would the explosion set the planet's atmosphere ablaze?
| Nothing devastating would happen. When the comet Shoemaker Levy hit Jupiter, with considerably more energy than an H-bomb, it made a big bang but Jupiter is still there.
Saturn's atmosphere can't burn because there is no free oxygen present. In fact there is regular lightning on Saturn, so if the atmosphere was going to catch fire it would have done so by now.
I wonder if you were thinking the H-bomb would start a hydrogen fusion reaction in Saturn's atmosphere. If so, no runaway fusion reaction would occur as the density and temperature is far too low.
| {
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