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Strange modulation of radioactive decay rates with solar activity Recently I found out this strange article about nuclear decay rates somehow showing seasonal variations with a high correlation with sun activity.
*
*Has this been experimentally confirmed/disproved? An experiment using neutrinos from a fission reactor would be awesome, although probably a couple orders of magnitude below the required luminosity (at least to be comparable with solar sources)
*Could the standard model possibly allow neutrinos to modulate decay rates in this way? Or do we need new physics?
Link to the public version of the paper.
Another paper about this, but now the SuperKamiokande data is compared with data from Brookhaven: http://arxiv.org/abs/1301.3754 They also propose a model of the effect called neutrino "resonant spin-flavor precession".
| A quite detailed article (in german though) discussing this "effect" can be found here:
*
*Radioaktivität: Wer hat an der Uhr gedreht?
At the end of the article there is a list of interesting related papers and literature. According to this article first experiments trying to use artificially produced neutrinos to influence the decay rates of radioactive materials came out with negative results. So if the sun should influence the decay rate, it doesn't seem to be via neutrinos, but must be by some mysterious unknown force or particle. However most physicist seem to believe more in a systematic error produced by the used measurement equipment as explanation for this "effect".
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In what order should the subjects be studied in order to get to String Theory I know:
*
*Quantum Mechanics (Griffiths Level, currently doing Sakurai Level)
*Mechanics (Newtonian+ Lagrangian/Hamiltonian but at level lower than Goldstein/Landau)
*Classical Electrodynamics (Griffiths + electro/magnetostatics from Jackson)
*Statistical Physics (Pathria)
I know very little relativity from undergrad mechanics class. Nothing about General Relativity, nothing about QFT, etc. I wish to reach string theory in a proper way without leaving any gaping holes in my education. What subjects should be studied in what order?
| I would say Special Relativity, then General Relativity, and finally QFT.
Books:
Special Relativity -- I would recommend Wheeler & Taylor and Woodhouse
General Relativity -- Woodhouse
QFT -- Zee and Aitchison & Hey
There's also an undergrad physics book [junior level] on String Theory by Zwieback that's highly recommended by many.
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Laplace's equation: Why is $\ell$ to be a non-negative integer? I have got some mathematical difficulties in the following exercise:
Calculate the potential of the polarized sphere along the $z$-axis. There are no free charges.
For this, we need to solve Laplace's equation, by using the method of separation of variables.
$$\phi (r, \theta, \phi) = R(r) \Theta(\theta) $$
We obtain a partial differential equation on $r$ and $\theta$:
$$\frac{1}{R} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r} R = \ell(\ell+1) $$
$$\frac{1}{\Theta} \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \sin \theta \frac{\partial}{\partial \theta} \Theta = - \ell(\ell+1). $$
My question is : "Why is $\ell$ to be a non-negative integer?" The solution I got in some course takes it for granted, but I do not know why.
| It need not be positive but can also be zero. So let's suppose you're asking why it has to be non-negative.
The answer to this question is that Laplace's operator $\Delta$ splits into radial and angular parts (these are precisely the operators that can be found in your equations, if you discard the $\phi$ dependence). The angular part is then simply $L^2$ where $\mathbf L$ is an angular momentum operator. Now because $L^2$ is a non-negative operator (convince yourself that a square of an operator must be non-negative) it must have only non-negative eigenvalues.
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Is this a weather phenomenon or an instrumental artifact? The radar image of the midwest provided by Weatherunderground at 10:30 PM Central time, May 8 2011 has odd patterns.
Are these patterns real? Perhaps caused by large scale convection over cities? Or are they artifacts of radar placement?
Here is the image that I am referring to, where green indicates light and yellow moderate rain:
| I have a master's degree in meteorology so I think I can clear this up for you!
This is simply ground clutter. You will see this sort of thing happening on evenings where the relative humidity is very high, more so when the mixing ratio is high also. The radar beam can actually start to interact with water droplets in the air when your humidity values are very high. You are more likely to max out your humidity values during the night as the air temperature falls and approaches the dewpoint. Indeed, if you look at the image I posted below you will see that in the area your radar image depicts, the relative humidities are near 100% in most of these areas. You can see that in many areas the temperature/dewpoint ratio is near 100%. For example, there are values on the map such 44/40, 55/54, 57/54, 50/49...very humid.
(Surface Analysis on May 8, 2011 0300Z(10PM CDT). Image is taken from http://www.hpc.ncep.noaa.gov/html/sfc_archive.shtml where you can then retrieve the surface analysis for any day you wish through March 30, 2006.
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$cm^3/g$ as a unit of adsorption I recently saw $cm^3/g$ as a unit for amount adsorbed. Usually, you see either $\mathrm{kg_{adsorbate}/kg_{adsorbent}}$ or $\mathrm{mole_{adsorbate}/kg_{adsorbent}}$. Does anyone know the meaning of this unit?
| In the case of $cm^3/g$, the quantity of gas adsorbed (the adsorbate) is expressed as its volume in gaseous form at STP. These units are still commonally used although the International Union of Pure and Applied Chemistry (IUPAC) recommends that the quantity adsorbed be described in units of $mol/g$. The quantity of adsorbate in $cm^3$ units occurs from the manner by which the absolute adsorbed quantity is calculated in the volumetric adsorption analytical technique.
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Why is $\frac{dx}{dt}=0$ in this average momentum calculation? In the following excerpt from S. Gasiorowicz's Quantum Physics, he derives an expression for the average momentum of a free particle. $\psi(x,t)$ is the wave function of a free particle, $\psi^*$ denotes its complex conjugate.
We try the following: since classically,
$$ p = mv = m\frac{dx}{dt} $$
we shall write
$$ <p> = m\frac{d}{dt}<x> = m\frac{d}{dt}\int{dx \psi^*(x,t) x \psi(x,t)} $$
This yields
$$ <p> = m\int_{-\infty}^\infty{dx\left( \frac{\partial\psi^*}{\partial t} x \psi + \psi^* x \frac{\partial\psi}{\partial t} \right)} $$
Note that there is no $dx/dt$ under the integral sign. The only quantity that varies with time is $\psi(x,t)$, and it is this variation that gives rise to a change in $x$ with time.
I seem to have trouble understanding the difference between the position $x$ and the average position $<x>$. Why can it be assumed that $\frac{dx}{dt}=0$? What is x?
| This $x$ is a position in the reference frame's coordinate system, which is just static by design. You can imagine it as a ruler, with a probability cloud in a foreground; the ruler stays on its place while the cloud moves and deforms changing its mean position.
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What is "pure energy" in matter-antimatter annihilation made of? I used to read the term "pure energy" in the context of matter-antimatter annihilation. Is the "pure energy" spoken of photons? Is it some form of heat? Some kind of particles with mass?
Basically, what does "pure energy" in the context of matter-antimatter annihilation refer to?
| All the three above answers basically are correct and saying almost the same thing, i.e. the end product of matter-antimatter collision is photon at low energies and massive particles such as mesons... at high energies. However,by thinking of the reverse action namely production of matter from a form of energy ( radiation/light) and the source of that unimaginable huge energy,simply using E=mc*c,supports the above mentioned answers firmly. AA 7 June 20011 at10pm
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Feedback on the paper, 'CCC-predicted low-variance circles in CMB sky and LCDM' by V. G. Gurzadyan and R. Penrose Ref: CCC-predicted low-variance circles in CMB sky and LCDM
To all cosmology / theoretical physics / related or similar researchers and academics,
Are there some updates concerning the issue of these concentric circles observed in the CBM in view of the new Planck data? What do the new Planck data mean for the particular issues discussed in the paper?
Warmest gratitude in advance.
| It's very easy to go wrong in analyzing microwave background maps for this sort of signal. In particular, unless you're quite used to doing this sort of analysis, it's easy to make mistakes in calculating the probability of getting a false positive -- that is, of seeing patterns like the ones you're looking for, even if the standard model is correct.
I haven't reproduced the calculations myself (although I've done a number of very similar analyses in the past), but I have read papers by various other people arguing that the original result was incorrect. All of these papers present very clear and cogent explanations of the correct way to do this analysis and point out errors in the original analysis, which would have precisely the effect of inflating the statistical significance (by reducing the computed false-positive probability). Moreover, the rebuttals are written by people with extensive experience doing this sort of analysis, while the original paper was written by people with no such experience.
I conclude that there's no convincing evidence in support of the Gurzadyan-Penrose model. As far as I can tell, there's essentially complete consensus among experts in the field that this is the correct conclusion.
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If dark matter is a new type of particle, what does that imply? My understanding is that dark matter cannot be (or is at least highly unlikely to be) an exotic form of any known particle. On the other hand, articles about particle accelerators seem to say that the Higgs is the last piece missing in the Standard Model jigsaw puzzle.
If dark matter is determined to be some form of new particle, what are the certain implications? Might such a discovery "stand to the side" of the Standard Model or would it certainly change the foundations?
(Forgive my extremely lay understanding and vocabulary -- please feel free to correct my mistakes.)
| Let me address this part of the question, as the physics part is covered by Ted Bunn.
If dark matter is determined to be some form of new particle, what are the certain implications? Might such a discovery "stand to the side" of the Standard Model or would it certainly change the foundations?
Up to now, progress in understanding particle physics has been happening not by completely discarding previous theories, but by assimilating them. The reason is that previous theories, as the standard model will become too in the future, were based on solid data. Being based on data means that they are just a shorthand of describing them. The old data still exists when new data appear, as the example of dark matter , so what will happen is that the standard model will be incorporated in any new theory describing matter in the microcosm.
Already string theories include the standard model within their structure so no "stand aside" will happen if they are found sufficient to describe all new data. String theories have a plethora of matter forms that could well explain/describe dark matter.
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Online QFT video lectures I'm aware of Sidney Coleman's 1975/76 sequence of 54 lectures on Quantum Field Theory. Are there any other high-quality QFT lecture series available online?
| The Perimeter Institute has recorded all the courses of their one-year Master's in Theoretical Physics. The courses are given by professors and lecturers from different universities.
For a full list of the courses they recorded in 2010/w011 go to Pirsa.org, and under "4. Collections" select "Course" and "2010". The courses are labeled by the tag "10/11 PSI". (I couldnt find a decent link which neatly lists all the courses)
You'll find multiple courses on QM, QFT, string theory, Beyond the standard Model, General Relativity, condensed matter physics, Quantum Information and many more. The format is also very neat. The Flash presentation is a recording of the class in low-quality, combined with decent photos of the blackboard.
Here are some direct links:
QFT part I -- by K. Zarembo
QFT part II -- by F. David
Also check out the courses given in 09/11 (search for "2009" instead), where the curriculum was a bit different (also different lecturers). You have for instance David Tong from Cambridge who gives an introductary course on QFT.
QFT part I -- by D. Tong
(part II was also given by F. David)
The site also has a large collection of recorded seminars.
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Is there any simple proof of the no-ghost theorem? Is there any simple proof of the no-ghost theorem in string theory?
| The proof using DDF formalism involves constructing a set of operators that commute with the Virasoro operators, and when applied to the ground state, they give all possible physical states. These operators $A^{i}_{n}$, where $i$ runs over $d-2$ transverse dimensions of spacetime and $n$ is an arbitrary integer generate among themselves what is called the spectrum generating algebra. These operators are in one-to-one correspondence with the transverse components of $\alpha^{\mu}_{n}$, which arise as coefficients in the mode expansion of the string with appropriate boundary conditions and are promoted to operators upon quantization.
The proof is broadly (and nicely) sketched out for the open bosonic string case using the DDF formalism, given in [1]. Using this formalism, it is shown that there exists no ghosts in the Hilbert space after implementing the old covariant quantization scheme in $d = 26$. The way this is shown is by making contact with states resulting from implementing light cone quantization scheme, which is manifestly ghost-free. The same formalism can be used to prove the no-ghost theorem in $d=10$ for a string with world-sheet supersymmetry, as given in [2].
The two references do a very nice job in presenting the same, so there is no use replicating the proof.
[1]: Section 2.3.2, Superstring Theory, Volume I; Green, Schwarz, Witten
[2]: Section 4.3.2, Superstring Theory, Volume I; Green, Schwarz, Witten
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Gravitational time dilation at the earth's center I would like to know what happens with time dilation (relative to surface) at earth's center .
There is a way to calculate it?
Is time going faster at center of earth?
I've made other questions about this matter and the answers refers to:
$\Delta\Phi$ (difference in Newtonian gravitational potential between the locations) as directly related, but I think those equation can't be applied to this because were derived for the vecinity of a mass but not inside it.
Any clues? Thanks
| The CENTER of the earth will not have more gravity, but less. This is because half the mass will be "above" half "below" ( Regardless of orientation)...Sort of less g and in different direction/vectors. The thing is mass is not concentrated at a spot in the center with more and more g as one moves closer to the center. As you tunneled down, some of the mass, more and more would be behind you). More time dilation on surface... Where g is stronger. Time would not be slower at center of earth.
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What are the conditions for decoherence to be irreversible? Spin echo experiments have been able to reverse the motions of all the molecules in a gas in statistical mechanics in the manner of Loschmidt. The Fermi-Ulam-Pasta model has solutions with a single mode dispersing, only to recohere after quite some time has elapsed. Can the same thing happen for decoherence? What are the conditions fyor decoherence to be irreversible?
| Decoherence in a quantum mechanical system appears because the phase information is lost. If it were recorded/known then, as in your first example, one could reverse it. The answer for me seems to be complexity.
In some sense there exists a single quantum mechanical state function of the universe. To record it one would need oogleplexes of numbers and functions.
In my opinion, decoherence serves the same function that the transition from quantum statistical mechanics to thermodynamics does: it describes measurable experimentally quantities for the system under observation. If we knew all the parameters of the statistical ensemble we would get the same answers, but thermodynamics reduces the complexity.
Similarly working with a density matrix, the complexity of the quantum mechanical system is reduced to measurable quantities. So it is the number of parameters that have to be reversed that will decide whether decoherence is reversible or irreversible, and this will be time dependent as tools develop. Who would have thought that thousands of Feynman diagrams could be calculated as they are now doing for the needs of LHC experiments.
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Is the wave function objective or subjective? Here is a question I am curious about.
Is the wave function objective or subjective, or is such a question meaningless?
Conventionally, subjectivity is as follows: if a quantity is subjective then it is possible for two different people to legitimately give it different values. For example, in Bayesian probability theory, probabilities are considered subjective, because two agents with access to different data will have different posteriors.
So suppose two scientists, A and B, have access to different information about the same quantum system. If A believes it has one wavefunction and B believes it has another, is one of them necessarily "right" and the other "wrong"? If so then the wavefunction is objective, but otherwise it must contain some subjective element.
| A wave function is as objective as a photo picture of somebody. It encodes an objective information about something. It consist, as any information, of many bits of information.
At the same time is is subjective because it depends on our perception. For one some information tells nothing, for somebody else it may be very speaking.
Concerning our minds, yes, everybody believes in the wrong wave function due to lack of knowledge.
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What happens if you connect a hot resistor to a cold resistor? Kind of an extension to this question:
If you heat up an object, and put it in contact with a colder object, in an ideal insulated box, the heat from one will transfer to the other through thermal conduction and they will eventually reach an equilibrium temperature at the midpoint, correct?
Now if you have a hot resistor (electrical component) and a cold resistor, and connect them by their leads, so that they make a circuit:
there will be the same conduction and radiation heat transfers. But also, the hotter resistor will have a larger noise current, right? So will there additionally be a transfer of electrical energy from one resistor to the other? Would completing the circuit allow them to reach equilibrium temperature faster than if they were just touching through an insulator with the same thermal conductivity?
| Actually, that's exactly what happens: heat is transported by freely moving electrons. - but you wouldn't call it noise current, you would call it heat transfer, regardless of whether it's phonons or electrons which carry the heat.
The whole thing really has nothing to do with the resistors, it would work equally well with any passive component (proof: otherwise you could use that to build a 2nd order perpetuum mobile).
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Graduate Physics Problems Books Need to brush up on my late-undergrad and early-grad physics and was wondering if anyone can recommend books or lecture notes (hard copy, or on-line) that also have solutions.
Two that I have come across are:
Princeton Problems in Physics with Solutions - Nathan Newbury
University of Chicago Graduate Problems in Physics with Solutions - Jeremiah A. Cronin
Spacetime Physics - Taylor & Wheeler (favorite book on special relativity; has a lot of problems with solutions at the back; a lot of the problems really enforce the material and discuss paradoxes)
If possible, please also provide a reason why you like the books as opposed to just listing them.
| Some review/problem set books that I like are:
Solid State Physics: problems and solutions - Laszalo Mihaly and Michael C. Martin
Problems in Quantum Mechanics with Solutions - Gordon Leslie Squires (The reviews are bare bones but I find the questions to be very good at making you think)
Hope this helps. I am interested to see what other people come up with.
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Do quantum states contain exponentially more information than classical states? Do quantum states contain exponentially more information than classical states? It might seem so at first sight, but what about in light of this talk?
| Experts agree that one qubit stores no more information than one bit.
But a zillion identical qubits (which you can prepare or manufacture to your specifications, even though you can not clone an unknown qubit) contains about a zillion times more information than a zillion identical bits which still contains only a single bit of information.
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Why don't waves erase out each other when looking onto a wall? If I stand exactly in front of a colorful wall, I imagine the light waves they emit, and they receive should randomly double or erase out each other.
So as a result, I imagine I should see a weird combination of colors, or a full-black/full-white/very lightly perception of the wall, when all the light waves that the wall receives and emits cancel out each other or double each other.
Why doesn't that actually happen? Any time I look into a wall, I never see the wall "cancel out" of my perception. Same for radio waves. Shouldn't radio waves not work at all? There are so many sources where they could reflect and cancel out or annoy each other...
| I will like to point out that the walls do not emit any light but instead they reflect what shines on them. You may also be interested in reading more about diffuse reflection which roughly and non mathematically attempts at explaining some of the physics at work here. If you are looking for a more mathematical treatment, this buffalo lecture on interference may help you --esp. the coherence bit.
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Why does the road look like it's wet on hot days? Often, I'll be driving down the road on a summer day, and as I look ahead toward the horizon, I notice that the road looks like there's a puddle of water on it, or that it was somehow wet. Of course, as I get closer, the effect disappears.
I know that it is some kind of atmospheric effect. What is it called, and how does it work?
| It is a mirage: in particular it is caused by hot air near the road and less hot air above it creates a gradient in the refractive index of the air and so making a virtual image of the sky appear to be on or below the road. Air currents make this shimmer, similar to a reflection of the sky on water, hence causing the illusion of wetness.
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Why doesn't a phone charge faster, rather than slower when it is in use In Physics class, we were building parallel circuits, and as more lights were attached in parallel, they got brighter (as more power was being provided to the lights, and the resistance decreases). So, when I charge a phone, why doesn't the battery charge faster when it is in use (eg. more devices powered on within the phone), than when it is not in use?
Clarification: All lights at once became brighter
| Assuming your phone is typical, the charger can only generate a limited current. When the phone is off, all the current from the charger goes to the battery. When the phone is on, the current from the charger goes partly to charging the battery and partly to running the phone. Hence the battery takes longer to charge.
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How do laser rangefinders work when the object surface is not perpendicular to the laser beam? I find the functioning of a laser rangefinder confusing.
The explanation usually goes like this: "you shine a laser beam onto the object, the laser beam gets reflected and gets back to the device and time required for that is used to calculate the distance".
Okay. But the object surface can be uneven and not perpendicular to the laser beam so only a tiny fraction of beam energy is reflected back to the device. And there's plenty of other radiation around, sunlight included.
How does a rangefinder manage to "see" that very weak reflected signal in a reliable manner?
| Some laser rangefinding uses a retroreflector, which will bounce the laser light back in the direction it came regardless of orientation.
Otherwise, lasers operate at a very specific frequency, so the signal/noise ratio only needs to be strong enough to be detectable at that frequency.
If you shine a normal laser pointer on a wall, even if the wall is pretty far away, you can see the spot it makes. That means your eye can pick out the reflected laser light. The electronics can be made better than your eye, so it's not too hard to see reflected laser light..
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Can heat be transfered via magnetic field in a vacuum? Say you want to store hot coffee in a container surrounded by a vacuum. To remove all sources of conductive energy loss the container is suspended in the vacuum by a magnetic field and does not have a physical connection to the sides of the vacuum chamber,
My question is would the magnetic field be a path for energy to be conducted out of the suspended container?
Another way to look at this question would be two magnets are suspended in a vacuum with their poles aligned. A heat source is attached to one of the magnets. Would the second magnet show a corresponding increase in temperature, excluding radiated heat transfer?
| No, a static magnetic field does not conduct heat.
The only way for thermal energy to cross a vacuum is by radiation. Of course, radiative heat transfer actually does consist of electric and magnetic fields: hot objects give off electromagnetic waves, which travel through the vacuum and are absorbed by, say, the walls of the vacuum chamber. But those are waves, not static fields, and besides, they are actually emitted by the hot object - it's not like it takes advantage of preexisting fields to move the energy.
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Why photons transfer to electrons perpendicular momentum? Linear antenna directed along z, photons (EM waves) propagate along x. Momentum of photons have only x component. Why electrons in antenna have z component of momentum?
| There are 2 different effects of the EM waves:
*
*Acceleration of charges due to EM field
*Radiation pressure of the EM field
If the EM wave propagates along x direction and your antenna is along the z direction, then electrons will be driven along the antenna by the electric field. On the other hand, the antenna will also receive an overall impact (momentum) along the x direction because of the radiation pressure (which does not affect the antenna practically since its effect is tiny and the antenna is fastened).
It is also worth pointing out that there are 2 descriptions of radiation:
*
*the classical description (EM waves) and
*the quantum description (photons)
The quantum description is more general. Both pictures are identical only in the limit of so many photons. Electromagnetic waves can be described in terms of photons, but individual photons cannot be described in terms of electromagnetic waves. Photons description is more fundamental.
| {
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Can the effects of gravity be broken by jumping? I was having a debate the other day with a work colleague where I explained that gravity is a weak force because it is easily broken. Then I remembered a lecture by someone, I forget who, that explained gravity is very weak because you can break its influence just by jumping or lifting a pencil, etc.
He countered that with something along the lines of 'that even though the pencil or your body is being moved away from the source of gravity it is still affected by gravity and thus it has weight'.
Is jumping a good example of gravity being a weak force?
P.S. You can probably tell, my colleague and I are not physicists but we enjoy our little debates, we just need to get our facts straight.
| Well, keep in mind gravity isn't being "broken" when you jump, it is still exerting a force of approximately F_g=G * m_1 * m_2 / r^2, where G is the gravitational constant (6.677 * 10^(-11)), m_1 and m_2 are the two masses, and r is the distance between them.
The idea of gravity being a weak force, relative the electromagnetic force is shown by jumping as the electromagnetic forces moving your muscles are able to overcome (a better word than "broken") the gravity of the entire earth for a period of time, so F_g < F_jump. Since you cannot jump again midair very effectively, it is temporary, as F_g is present still in the air. But assuming you had lots of energy and a space suit you could climb a very long ladder out to where gravities effects would be minimal.
Another good example that I like of electromagnetic force > gravity is the fact that when you jump out of a window (first story please!), land on the pavement. There the electromagnetic interactions between your electrons and the pavements electrons are stopping you from going all the way to the center of the earth.
| {
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Will the sun cool and produce a disrupting EMP?
Scientists say rare drop in sunspot activity could cause global cooling
http://www.globalpost.com/dispatch/news/business-tech/science/110615/science-news-solar-flares-sunspots-global-warming
Last week's event is almost certain to be just the start of a cycle that is expected to peak in 2013. It is during this rising activity that we can expect the Earth to be buffeted by some devastating solar storms.
http://timesofindia.indiatimes.com/home/science/Solar-storm-rising-Earth-may-be-hit-by-cosmic-hurricane/articleshow/8858351.cms
Solar Flares Could Cripple Earth's Tech Infrastructure in 2013
http://www.pcworld.com/article/229876/solar_flares_could_cripple_earths_tech_infrastructure_in_2013.html
Isn't the first article contradictory with the other two? Which one is closer to the truth?
| The first is front line research announced at a conference, and is news, and closer to the current truth as far as the way cycle 24 is manifesting.
The probability of a large solar eruption is another story, one large eruption was manifested even though the cycle is not yet at its maximum; there is no direct contradiction
Now as far as cooling goes, there is no solid model nor long range data to be able to predict whether low intensity sun cycles lead to Maunder minimum type cooling. The energy depletion is of the order of 1.4 watts/meter^2, not enough to make a large direct temperature effect. In 30 years we may have the data to know whether it is a coincidence that the world cooled during the Maunder minimum or somehow the various hypotheses of the sun's indirect influence lead to a causal connection. ( magnetic fields, cosmic rays, UV and planckton, etc)
| {
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Analyzing the motion of a ball rolling without slipping inside a hemispherical bowl Consider a solid ball of radius $r$ and mass $m$ rolling without slipping in a hemispherical bowl of radius $R$ (simple back and forth motion). Now, I assume the oscillations are small and so the small angle approximation holds. I wish to find the period of oscillation and I analyze the motion in two ways, first using conservation of energy and secondly using dynamics. However, I receive two inconsistent answers. One or both of the solutions must be wrong, but I cannot figure out which one and more importantly, I cannot figure out why.
Method 1: We write the energy conservation equation for the ball
$mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = Constant$
from the center of mass, we take the height as $h = R-(R-r)cos\theta$ where $\theta$ is the angle from the vertical. Applying the no slip condition $v = r\omega$ and taking the moment of inertia for a solid sphere $I = \frac{2}{5}mr^2$ we can write the energy equation as
$mg(R-(R-r)cos\theta) + \frac{7}{10}mr^2\omega^2 = Constant$
Differentiating with respect to time:
$mg(R-r)sin\theta\cdot\omega + \frac{7}{5}mr^2\omega\cdot\alpha = 0$
taking the small angle approximation $sin\theta = \theta$ we get
$g(R-r)\theta + \frac{7}{5}r^2\alpha=0$
$-\frac{5g(R-r)}{7r^2}\theta = \alpha$
from which we can get $T = 2\pi\sqrt{\frac{7r^2}{5g(R-r)}}$
Method 2: The only torque acting on the ball at any point in its motion is the friction force $f$. So we can write
$\tau = I\alpha = fr$
again using the rolling condition $a = r\alpha$ and the moment of inertia for a solid sphere,
$\frac{2}{5}ma = f$
The net force acting on the system is the tangential component of gravity and the force of friction, so
$F = ma = mgsin\theta - f$
$\frac{7}{5}a = gsin\theta$
taking the small angle approximation and converting $a$ to $\alpha$ we get
$\alpha = \frac{5g}{7r}\theta$
and a corresponding period of $T = 2\pi\sqrt{\frac{7r}{5g}}$
Now the solutions are very different and I would appreciate it if someone would point out where I went wrong.
| The answer you got in first method was wrong.
The answer in second is nearly reached, we have to take $R-r$ instead of $r$ because you have to take $r$ from centre of the sphere.
The correct answer was $2\pi\sqrt{7(R-r)/5g}$
| {
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what cools bottle of water faster: ice or snow Imagine you have a pile of snow and a pile of ice shards. You put a soda bottle which has a room temperature into both piles. Which bottle is going to cool down faster?
| For one you must define room temperature. because if the bottle is worm enough to to slightly melt the show first and then refreeze then this one will cool 1st of not the air in the snow can act as an insulator and this one will take longer. Now the shape and sise of the ice will make a large contribution/small to the surface contact on the bottle this will also make a difference on the air flow around the pieces of ice.
You need to make the experiment MUCH more specific to get a correct answer.
| {
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Non-Dimentionalization of equations I am trying to understand a paper in which they mention;
" We non-dimentionalize the problem hereafter by selecting mass, length and time units such that the fluid density
$\rho\equiv1$
the gravitational acceleration g
$\equiv1$
and initial tank depth
$h_0\equiv1$
".
The reson for doing this is to calculate a pressure from a plot of y vs
$P/\rho g h_0$
.
So if the value in the plot is 20, what is the value in Pascal for a
$\rho = 1025 kg/m^3$
,
$g = 9.81 m/s^2$
and
$h = 50 m$
.
Thanks in advance.
| As you note, the quantity on the vertical axis of the plot is $P/\rho g h_0$. So if this quantity equals 20, then
$$
P/\rho g h_0=20,
$$
or
$$
P=20\rho g h_0.
$$
The units on both sides of this expression match, so this expression is correct in any system of units. If you want to know the pressure in SI units (pascals), just use SI units for the stuff on the right. If I'm not mistaken, $\rho gh_0=5.03 \times 10^5$ Pa, so $20\rho g h_0=1.01\times 10^7$ Pa.
Another way to think about it: If you want, you can actually work out the sizes of the various units required to make the various given values equal 1. That is, you can assume that lengths, masses, and times are measured in units called L, M, T. Then you want to know how many meters there are in an L, how many kg in an M, and how many seconds in a T. You know that
$$
1025 {\rm kg/m^3}=1 {\rm M/L^3},
$$
and similarly for the other two quantities to be set equal to 1. Solve these three equations for M,L,T, and you'll know the size of the various "new" units in SI units (kg,m,s). The pressure is 20 of the new units, and now that you know what the new units are you can convert back to SI.
(Note: I'd originally, incorrectly written M/T$^3$ instead of M/L$^3$ in the above equation. Someone edited it to Tonnes/m$^3$, which is not what I meant at all. In that expression, the right-hand side is supposed to be in the "made-up" units M,L,T, which don't correspond to any standard units. M/L$^3$ is correct.)
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Ascent rate and size of balloon I am part of a school project, Project Stratos to send a balloon to the edge of space (the closer side :P) and was wondering how you would work out the accent rate of a large balloon (roughly 1m^3 of helium with 100g of mass) and the size of it as it increases its Altitude. I am creating a live map (that will be based on predictions rather than its actual location) and want to know the speed it will float up into the atmosphere. Currently we are assuming the ascent rate will be about 5m/s but I doubt that is very accurate and would this speed increase as it gets higher?
Edit: I would also quite like to know the burst height of the balloon.
| Ascent rate of a balloon (assuming spherical symmetry) depends on the following forces:
(1) The upward buoyant force $F_B=\frac{4}{3}\pi r^3\rho_{air}g$
(2) The gravitational pull downwards: $F_G=\frac{4}{3}\pi r^3\rho g$
(3) The drag force acting: $F_D\, =\, \tfrac12\, \rho_{air}\, v^2\, C_D\, A$
On a first glance it might appear that the balloon soon reaches a terminal velocity. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air changes with altitude. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting,
$$F_B=F_G+F_D$$
leading to,
$$v = \sqrt{\frac{8 g r}{3 C_D} \left( \frac{\rho_{air} - \rho}{\rho_{air}} \right)}$$
| {
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Direct exposure to the vacuum of space I was watching a few sci-fi movies and was wondering the real science explaining what would happen if you were to be subject to the conditions of outerspace.
I read the wikipedia article on space exposure, but was still confused. If a person was about the same distance from the sun as earth is, would they still freeze to death? (as shown in the movie Sunshine)
I'm reading from all sorts of sites with conflicting information about what would actually happen when a person is exposed to the vacuum of space...
| They would freeze.
They wouldn't freeze to death - since they would die of something else (lack of oxygen) first
Although a small part of you is facing a hot sun at 6000K most of your surface is facing cold dark space at 3K. You can work out what temperature you will reach - it depends only on how reflective you are.
Assuming you have the same reflectivity as the earth (35%) and you aren't close enough to the earth to receive any significant heat from it then you would end up at about the same temperature the Earth would have without the greenhouse effect of it's atmosphere = which is about -20C.
If you were made of much darker material like the moon - you would get much colder. Parts of the moon not facing the sun or earth get down to around -150C
| {
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Why is the gravitational force always attractive? Why is the gravitational force always attractive? Is there another way to explain this without the curvature of space time?
PS: If the simple answer to this question is that mass makes space-time curve in a concave fashion, I can rephrase the question as why does mass make space-time always curve concavely?
| The inverse square is apparently a consequence of conservation of momentum. For two particles in orbit, Newton showed that the orbit is planar, and Bertrand https://en.wikipedia.org/wiki/Bertrand%27s_theorem showed that the forces between the two have to be either of the inverse square k/r^2 or space spring/Hook's law. So Newton's law of gravity and that of Coulomb have conservation of momentum as the origin.
It is also worth noting that Hook's law can be shown to come out of/limiting case of the inverse square in the case of 'crowdedness', where there are too many interactants and very small space to move. This can be shown easily by taking three particles along a line interacting under the inverse square and give the middle a nudge keeping the end particles fixed. If we take it that there is no charge with zero mass, Coulomb's law follows too.
Even without the help of Bertrand theorem, it is possible to derive the Maxwell equation from just charge conservation and its continuity equation. See this reg and quote: https://pdfs.semanticscholar.org/3251/31eadb62c8fdfdaaad7b21a308992ff3a4d2.pdf '' We show how the covariant form of Maxwell’s equations
can be obtained from the continuity equation for the electric charge ''. Clearly the same can be done using mass conservation and we get the gravitomagnetic equations out of it.
In general, Maxwell/gravitomagnetic equations work for both attractive and repulsive forces. But if two masses are locked in an orbit, they must be under attraction. The facts that similar masses attract whereas similar electric charges repel has to rely on experimental knowledge.
New research from MIT on a new long-range attraction force connected to spin. https://www.youtube.com/watch?time_continue=10&v=1ZZcgBmS5W4
| {
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How to find the principal point in an image? I need to find the principal point in an image. Its a point where the principal axis intersects the image sensor.
Due to misalignment this point is not at the center of image always(or image sensor). I need to precisely determine its location using any of the optical methods available(if any). Any suggestions are greatly appreciated.
More information: The camera is giving me a live feed over a screen and I am able to store, analyze any part of it in real time or later.
| I suggest shining a narrow laser beam along the optical axis. Just don't fry your sensor! Hopefully, the antireflective coating of the lens is not perfect, and you will be able to see on the object side two reflections of the beam: one from each side of the lens. Align the laser in such a way that both reflections come back exactly at the output port of the laser: the beam is then perpendicular to both faces of the lens, and therefore it is exactly along it's optical axis.
Now look at the image: you have one bright spot exactly at the focus.
| {
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Clebsch-Gordan Identity I'm trying to take advantage of a particular identity for the sum of the product of three Clebsch-Gordan coefficients, however, the present form of my equation is slightly different. Is there a symmetry relation that will allow me to change:
$\sum_{\alpha\beta\delta}C_{a\alpha b\beta}^{c\gamma}C_{d\delta b\beta}^{e\epsilon}C_{d\delta f\phi}^{a\alpha}$
Into:
$\sum_{\alpha\beta\delta}C_{a\alpha b\beta}^{c\gamma}C_{d\delta b\beta}^{e\epsilon}C_{a\alpha f\phi}^{d\delta}$
Notice I need to swap $j_2m_2$ with $jm$ in the last Clebsh-Gordan coefficient. Does anyone know a way to do this?
Note: My notation follows that of Varshalovich, $C_{j_1 m_1 j_2 m_2}^{jm}$
| In general you cannot make the change you suggest because of the condition on projections. In your first equation, the projections in your last CG must satisfy
$\delta +\phi=\alpha$
whereas in your second equation, the projections in your last CG must satisfy
$\alpha+\delta=\phi$.
Thus, unless there is further symmetry that you have not mentioned in your problem, for instance $\alpha=\delta$, there is no way to transform the first into the second.
| {
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On a bicycle, why does my back tyre wear so much more quickly than the front? This question is cross-posted from Bicycles.SE, but it is really one for those that know a bit about physics.
Why does the back tyre of a bicycle wear out quicker than the front tyre?
I have my uneducated suspicions but I would appreciate an educated answer.
| When braking, most of your weight is going to be on the front tire (due to your forward momentum). This is why the front brake is so much more effective than the rear brake. That also means that, if you use the rear brake, you are pretty likely to skid (which only rarely happens on the front tire), because the weight on it is reduced. I imagine that rear-wheel skidding, which should not occur very often with proper (front-wheel) braking, likely contributes to rear tires wearing out faster. (In my opinion, the rear brake is only really useful when the roads are slick and as an emergency back-up brake.)
| {
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How to calculate concentration of vapor at the surface of a water drop I'm reading a paper that examines the evaporation rates of water. In the final formula, it has the following constant:
$c_s - c_\infty $ where $c_s$ is the concentration of the vapor at the sphere surface and $c_\infty$ is the concentration of the vapor at infinity.
I'm fairly confident in how to derive $c_\infty$:
1) Calculate water vapor pressure $p_s = 610.78 e^{\frac{17.2694 T}{T+238.3}}$
2) Actual vapor pressure is then: $ p=(Relative Humidity)p_s$
3) Using ideal gas law gives concentration at infinity: $c_\infty = \frac{(Molar Mass of Water)p}{RT}$
This generally looks like it gives me answers consistent with the paper's values. But, if I screwed up, please let me know.
My problem is with $c_s$. It's weird, because I recognize how easy it should be to get this, but just can't. I've asked some of the other grad students here and they basically all are befuddled. I think this is the physicist's version of a "tip of my tongue" experience. So, can anyone give me a link or method by which I can get this?
Thanks.
| Very close to the liquid surface, you assume that the vapor and liquid are in diffusive equilibrium (equal chemical potentials).
If the droplet is very very large, you can start with the ordinary saturation vapor pressure and figure it out as in pballjew's answer.
If the droplet is small enough, you need to use the Kelvin equation which takes into account the way that the water's surface tension alters the diffusive equilibrium between vapor and solid.
| {
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If you view the Earth from far enough away can you observe its past? From my understanding of light, you are always looking into the past based on how much time it takes the light to reach you from what you are observing.
For example when you see a star burn out, if the star was 5 light years away then the star actually burnt out 5 years ago.
So I am 27 years old, if I was 27 light years away from Earth and had a telescope strong enough to view Earth, could I theoretically view myself being born?
| No, I don't think the concept of time works quite like that. Yes, if a star is 5 light years away and it burned out - yes it burned out 5 years ago for that point in time. Space time is relevant for that time period.
In your example:
You are 27 Years old (Present Day Earth Time)
You are teleported 27 Light Years Away to Planet X (Your Present Age is 27 on Planet X).
You look back at Earth - You are non-existent on Earth.
Because Earth's time does not bend back. It took 27 years for Earth's image to reach you at the time you left. So unless you were at both planets at the same time you would be 27 on Planet X while on Earth you would be 54 Years old. Earth's time does not bend back, it only moves forward.
| {
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Is spacetime simply connected? As I've stated in a prior question of mine, I am a mathematician with very little knowledge of Physics, and I ask here things I'm curious about/things that will help me learn.
This falls into the category of things I'm curious about. Have people considered whether spacetime is simply connected? Similarly, one can ask if it contractible, what its Betti numbers are, its Euler characteristic and so forth. What would be the physical significance of it being non-simply-connected?
| Benjamin Horowitz's answer covered a lot of the key points, but it's worth adding that the question of the topology of the universe has been investigated by astrophysical observations. If the universe is multiply connected, and if the length scale is shorter than the horizon scale, then we should be able to see evidence of it.
To take a simple example, imagine that the universe is geometrically flat but has the geometry of a 3-torus. Specifically, take a cubical volume, and identify opposite faces, so that if you "leave" the cube through one face you reenter through the opposite face. If the length of an edge of the cube is sufficiently small, then you could see multiple copies of any given object. Of course, if the length is much larger than the horizon, then there's no way to tell the difference between this model and one in which space is infinite.
The best way to test these models is the "circles in the sky" technique, in which you look for correlated circles in different directions in maps of the microwave background radiation. The result is negative: we don't live in a multiply-connected universe with a sufficiently short length scale to be observable.
| {
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Calculating time for a fully charged UPS I have a UPS of 1000 Volts connected with 2 batteries each of 150 Amp. How much time it will take to consume the whole UPS (after fully charged) when a device of 1Amp is getting electricity form that UPS.
Please also explain me the calculation.
| Battery capacity is normally rated in ampere-hours (Ah), not ampere. This means what it sounds like - if you have 2 * 150 Ah of batteries, you can (ideally) pull 1A for 300 hours like in your question (*), or 300 amps for 1 hour. This is an intuitive value of battery capacity but is not strictly speaking a measurement of the energy stored since this depends on the voltage which you pull the current at (and both of these vary during the discharge of a real battery).
You wrote that your batteries are rated at 150A, but you might have misread the 150 Ah figure, in which case your answer is above. You could also have quoted the maximum current capacity of the battery of 150A (typical of a car battery for example), which isn't related to the capacity at all and I agree with the other answer here that you cannot calculate it without further information.
To use the old water analogy - the voltage is, roughly speaking, the water pressure of the battery and the ampere is the rate of water in the stream coming from it. You would have no indication of the size of the reservoir so you cannot calculate for how long your device can run.
(*)
This is of course assuming that you can connect your device at all, i.e. an UPS with an output of 1000 volts will not work very well if you connect it to a domestic 110V or 220V piece of equipment :)
| {
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whats the rate of energy increase required for constant acceleration between 0.0c and 0.99c? I was wondering how much energy would be required to accelerate 1000kg to 0.99c at 1G.
What I don't understand is what the rate of increase of energy is required as velocity increases. I was looking at the Lorentz factor curve but I'm not sure if this affects the energy required or just time dilation. eg. 0.99c = Lorentz factor of 7. So travel only takes 1/7th the time from the point of view of the traveler but does that mean you need 7 times as much energy to get to 0.99c than from say 0.0c to 0.1c.
Does any body have a link to a graph that shows this rate of increase of energy required or is it the same as the Lorentz factor curve?
I'm a programmer not a physics or math person so go easy on me.
Thankyou.
| The energy of the mass at rest is
$$
E_0 = m c^2
$$
At speed $v = 0.99 c$ it becomes
$$
E = \frac{m c^2}{\sqrt{1-v^2/c^2}}
$$
So, the increase in energy is
$$
E - E_0 = m c^2\left(\frac{1}{\sqrt{1-v^2/c^2}} - 1\right) \approx 6.09 m c^2.
$$
irrespective of the acceleration.
| {
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What happens to light in a perfect reflective sphere? Let's say you have the ability to shine some light into a perfectly round sphere and the sphere's interior surface was perfectly smooth and reflective and there was no way for the light to escape.
If you could observe the inside of the sphere, what would you observe? A glow? And would temperature affect the outcome?
Seems silly, it's just something I've always thought about but never spent enough time (until now) to actually find an answer.
| The interaction of the light with any particles present (there can be no transfer of energy even light without matter) would lead to an increase of particle density and a vacuum until the sphere collapses in on itself. The idea that the light would just stop if the source was "switched off is nonsense. How we see the light from a long dead star?, Light continues it course to it's conclusion (absorbsion).
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What is sector decomposition What is sector decomposition and how can it be used to 'disentangle' UV and IR divergences?
I have read about it in the paper SecDec: A general program for sector decomposition,
but I have no idea if, with a suitable change of variable $x=1/q$ on each coordinate, we can change the IR divergences to UV divergences or change the UV divergences to IR ones and then apply sector method to disentangle the divergences.
| UV divergence is due to self-action. It exists even in statics. No self-action is introduced, no UV divergence appears in calculation. Renormalizations discard the self-action contributions perturbatively rather than exactly, at the beginning.
IR divergence is due to too distant initial approximation, distant from the exact solution, of course. It is like using a Taylor expansion $f(x) = f(0) + f'(0)\cdot x + ...$ for calculations where $x$ is infinite (distant from zero).
So these divergences are not interchangeable.
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Simplest interferometer I want to build simplest interferometer which should be able to measure movements down to fraction of wavelength.
What is the simplest scheme for that, and what are the requirements for a laser?
I have a bunch of laser diode-based ones, and I guess they might be not coherent enough...
Are green DPSS ones any better?
| Lasers aside, the mirror translation stages are likely to be your biggest hassle. If you want to measure sub-optical-wavelength distances, then you should expect to need translation stages that are stable to this sub-micron range or your measurement will wash out. That is, you can have static interference patterns, but if you expect to move mirrors and count fringes to do a length measurement, then your mirror-mover should be stably controllable to the length scales you plan to measure.
On a practical level, this means you will really need a quite solid optical surface like, say, this one, and your mirrors should also be stably mounted with precision screws. If you have a moderate budget then this is nothing to balk at and I expect$^1$ you should really be able to build a quite reasonable one for something like $$100 or $200.
This is not to say, however, that you can't have something really good for really cheap - it just says that you won't be able to finely control the interference pattern. If seeing the pattern is all you really need, then resources like this instructable seem to show that getting a Michelson-Morley interference pattern is relatively easy if you have the patience.
$^1$Note, though, that I'm a theorist.
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Software for simulating 3D Newtonian dynamics of simple geometric objects (with force fields) I'm looking for something short of a molecular dynamics package, where I can build up simple geometric shapes with flexible linkages/etc and simulate the consequences of electrostatic repulsion between surfaces. Something, say, that would let me simulate billiards in a magnetic field.
Does anyone have good recommendations for a 3D Newtonian dynamics software package (free or not) with these sort of capabilities?
| I think that the 'agent based' approach is good for your purpose. I've already used MASON to play a little with 'a sort of' gravity.
(you will have to know java - easier if you work on top of the examples- and the documentation is very good)
MASON
is a fast discrete-event multiagent simulation library core in Java, designed to be the foundation for large custom-purpose Java simulations, and also to provide more than enough functionality for many lightweight simulation needs. MASON contains both a model library and an optional suite of visualization tools in 2D and 3D.
see the examples:
Bouncing Particles - is a tutorial demo of simple particles bouncing and interacting with one another.
Balls and Bands (Tutorial 5) (mass springs between objects)
simulates Hooke's Law with balls connected by rubber bands of different strengths.
HeatBugs -- shown in wireframe 3D.
you can also explore the BREVE package.
Breve A 3d Simulation Environment for Multi-Agent Simulations and Artificial Life (IA)
objects, joints, gravity.
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Do eyeballs exhibit chromatic aberration? Fairly straightforward question. If not, why not?
I suspect that if they do, it is not perceived due to the regions of highest dispersion being in one's region of lowest visual acuity.
| I have a 405nm laser pointer, which emits light right on the edge of the visible spectrum. It is easy to see that this wavelength of light appears out of focus through the eye likely due to chromatic aberration. I would imagine this is because our eyes are not made to see that kind of light. I've also noticed that mercury vapor streetlights have a faint violet halo around them due to what I believe is chromatic aberration, since the violet light on the edge of the spectrum is refracted the most through our eyes.
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What are the mathematical problems in introducing Spin 3/2 fermions? Can the physics complications of introducing spin 3/2 Rarita-Schwinger matter be put in geometric (or other) terms readily accessible to a mathematician?
| Physics complications of "introducing" spin 3/2 matter are the same as for spin 1/2 and spin 0 - the initial approximation in the corresponding interaction theory is physically wrong and calculations give too big (= just wrong) perturbative corrections. It is a complete failure of physics description and it cannot be casted in "geometric terms". Most people, however, does not see it.
Edit for downvoters: While in case of Rarita-Schwinger equation the solution violates even causality and it cannot be repaired with the constant renormalizations, this feature is still not considered as a failure of coupling. Indeed, we cannot be wrong. It is nature who is wrong, especially at short distances.
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'Getting in' to research physics? I'm going to be choosing a university course soon, and I want to go into a branch of physics. A dream job for me would be to work in research, however, I do realise that this isn't for everyone and is difficult to reach. So what is the best way to go about achieving this aim? What things can I do which will help me?
| First of all, if you consider science as an ordinary career choice, then the effort and the risk is not worth it.
But if you are genuinely interested in science, and really want to do research for your living, here they are:
*
*stay curious,
*learn hard, especially the things you are interested the most,
*participate in workshops, student conferences and summer internships,
*ask professors (or even grad students) if they can give you a research task, or know some opportunities.
However, most of time there is no secret. Try to be a good student and you will see if you are or are not (it does not require to be 'the best student'). The only two things which may not bre obvious (from the student's perspective) are:
*
*courses and grades are not everything and
*information and personal contacts are crucial.
While doing something extra requires extra effort, it's stimulating, allows to learn skill that are really needed and gives insight in how the real research work. Also, it is crucial to learn what are the possibilities, where is good to go for your PhD, etc.
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How beam focusing looks like in electron microscope? I mean I know there are electrostatic/electromagnetic lenses which does focus the beam, but I am not sure how it is possible to foсus beam down to a few 10nm while emitter might be 1mm thick while having large focusing distance (especially when looking at chromatic aberrations..) - optics approach says you would not be able even get 1mm spot at such NA...
Why doesn't electrons widen the beam by repelling each other en-route?
Is there any good books on the subject explaining all these tricky stuff of focusing electron beam?
| The answer is a combination of two things:
*
*The "source size" (the first crossover size, in fact) is about 50 μm even for thermionic cathodes. They are able to do that by using hairpin cathodes and a negative electrode known as the Wehnhelt cylinder (see this image for a diagram of a thermionic electron gun).
*The electron probe can be made much smaller than the source by accepting a reduction in beam current. See page 30 and 31 of Reimer's Scanning Electron Microscopy for more details.
The reduction in current associated with a large demagnification is problematic, so field emission guns are preferred when small probe sizes are desired, because they have apparent source sizes in the nanometer range.
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Why is 55-60 MPH optimal for gas mileage of a passenger car? My driver's education teacher back in high school said 55 MPH is optimal for gas mileage of a passenger car. Just last week, I read an article in a magazine saying 60 MPH is optimal. These numbers are pretty close, so there's some validity in the statement. What's the physics explanation for this 55-60 MPH sweet spot?
| There are two things to consider.
*
*The amount of energy required to maintain constant speed. This starts at a low value and increases. Energy rate can be measured in Watts (power) and it is proportional to $v^3$.
$$ P = F_{drag}\,v $$
*The amount of fuel supply needed to make the above energy rate. This is called the break specific fuel consumption (BSFC) and it varies widely with the % throttle applied and the engine rpm (and thus gear). An engine is typically at it's best efficiency near full thottle at mid low rpms (like 80% thottle with 4500 rpm). Incidentally, peak torque rpm usually corresponds the best BSFC area. With low throttle the air intake is restricted and you have pumping losses (it takes more energy to suck the air into the pistons). Engine friction increases with speed and combustion efficiency varies with flame temperarure and volumetric efficiency. Also at low speeds the combustion suffers and thus you cannot maintain low speeds at high gears. Remember though that the lower speeds increase travel time and there is limit on the minimum fuel per hour used, as travel times increase so is the total fuel used for a trip.
Obviously engineers target their designs to perform well during the EPA mandated duty cycles. To get the best miles per gallon you need a car with a small engine (low friction) that can operate near full throttle at low rpms, with very high gearing. On the other hand if you have a Corvette what you want to do accelerate full throttle to like 120mph and then coast down (no throttle) to 30 mph and then repeat the cycle.
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Could a human run horizontally inside a Wall of Death? A popular circus stunt is for a motorcycle rider to ride inside a bowl shaped depression called a "Wall of Death." The rider goes higher and higher up the wall until they are actually horizontal. I wonder if a human could do the same.
| Short answer: No.
Longer answer: I found it easiest to approach this by asking the question in a different way ---> If a runner were to run around in circles on an inverted conical ramp, how fast would he have to run to remain perdendicular to the ramp's surface? The answer depends on three things: radius for the running path; slope of the ramp (B) and height/mass distribution of the runner (shorter runners do better, since more of their mass is at a larger $v^2/R$).
Assumptions: 1.6m (5'3") tall runner, with a center of gravity (CoG) 0.8m from feet, on a 2m radius (approx 13 foot diameter) path. Since he is tilted in from vertical, his CoG traces a path which has a diameter somewhat less than that: for now, estimate B = 40 degrees tilt, which give a path for the CoG of radius (2 - 0.8 sin(40))m = about 1.5m.
$a_d = 9.8 m/s^2$ is acceleration (due to gravity) straight down. $a_L = V^2/R$ is acceleration laterally outward, due to centrifugal force. $a_t$ is total acceleration, which will be perpendicular to ramp surface.
$a_d = a_tcos(B)$
$a_L = a_tsin(B)$
Therefore, we have $a_L = a_d tan(B)$ ----> $B = tan^{-1}(V^2/(R a_d)$
Remembering that the R in that equation is the R for CoG, and that the velocity of that CoG will be slower than speed of the runner's feet, we have...
$B = tan^{-1}((6.7 (m/s)(1.2/2))^2/(1.5(m) 9.8 (m/s^2))$ = about 47.8 degrees from horizontal. Compare to the original guess of 40 degrees. I'm not making a second iteration to correct path length: instead guess about 45 degrees. This assumes that the runner is running a 4 minute mile, which is quite a feat to pull off on a conical ramp with path of radius 2 m, with your head (experiencing less centripetal force than CoG) constantly trying to fall in towards the center while your feet (experiencing more) trying to slide out from under you in the opposite direction.
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The philosophy behind the mathematics of quantum mechanics My field of study is computer science, and I recently had some readings on quantum physics and computation.
This is surely a basic question for the physics researcher, but the answer helps me a lot to get a better understanding of the formulas, rather than regarding them "as is."
Whenever I read an introductory text on quantum mechanics, it says that the states are demonstrated by vectors, and the operators are Hermitian matrices. It then describes the algebra of vector and matrix spaces, and proceeds.
I don't have any problem with the mathematics of quantum mechanics, but I don't understand the philosophy behind this math. To be more clear, I have the following questions (and the like) in my mind (all related to quantum mechanics):
*
*Why vector/Hilbert spaces?
*Why Hermitian matrices?
*Why tensor products?
*Why complex numbers?
(and a different question):
*
*When we talk of an n-dimensional space, what is "n" in the nature? For instance, when measuring the spin of an electron, n is 2. Why 2 and not 3? What does it mean?
Is the answer just "because the nature behaves this way," or there's a more profound explanation?
| Scott Aaronson, himself a (quantum) computer scientist, thinks and writes about a number of these subjects in his paper Is Quantum Mechanics An Island In Theoryspace? - at least the "why complex numbers and not the reals or the quaternions?", and I'm pretty sure he mentions it in his 'Democritus' lectures as well.
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Is the earth expanding? I recently saw this video on youtube:
http://www.youtube.com/watch?v=oJfBSc6e7QQ
and I don't know what to make of it. It seems as if the theory has enough evidence to be correct but where would all the water have appeared from? Would that much water have appeared over 60 million years? Also what would cause it to expand. The video suggests that since the time of dinosaurs the earths size has doubled in volume, how much of this is and can be true?
[could someone please tag this, I don't know what category this should come under]
| I don't know as a fact if the earth is expanding or not. But way is it so hard to believe it is. On the same principal that ice expands when it mixes whit gases. Lava expands when it cools specialty when water is present. It fells whit gas bobbles and increases in volume 3 fold. On the same principal way is it not possible for the earth crust to expand. I'm not saying that the earth mass is increasing just the volume.
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Creation of matter in the Big Bang I appreciate your patience to my neophyte question. I am working on my dissertation in philosophy (which has nothing or little to do with physics) about the "problem of naming." Briefly what I am arguing is that when we name something, we stop it from being anything or everything else. It is a phenomenological question and has a lot to do with language as an object.
My question for you is that, is it true that all matter was somehow formed in the Big Bang or in those famous three minutes following? I think I understand that helium and hydrogen were formed and are they then to be considered the basis of all matter today? A friend said to me a long time ago that we are made of the same atoms that were present at the Big Bang; could this possibly be true? (And how wonderful if it is...)
| Hydrogen and helium (and lithium) were formed in the big bang, but other elements arose later through a variety of different routes, including stellar and supernova nucleosynthesis, as well as fission and spallation. So only some of your atoms are truly primordial.
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Creating the opposite of an optical lattice Is it possible to create periodic potentials that instead of creating a well for an atom to be trapped in, repulsed by that specific location?
If yes, can we use this as a means to make artificial objects that we can feel, with let's say our hands, by putting lots of these repulsive points together?
This is an idea that I have in order to make holograms that we can feel like the holodeck in Star Trek.
| Atoms in an optical lattice are attracted to high/low field states depending on the quantum state the atom is in. Atoms in states that would be shifted down in energy by the application of an electric field are drawn to low-field regions and repulsed from high-field regions, and vice versa for atoms in states that would be shifted up in energy by the application of an electric field.
The answer to part 1 is that this is possible, and in fact atoms are already "repulsed" by parts of the lattice anyway. Which part of the lattice is attractive/repulsive depends on the state the atom is in.
Because optical lattices are by definition periodic, these "repulsive" regions are surrounded by attractive regions, and the atoms in you hand tend to migrate from the repulsive to the attractive regions. I don't think you'd be able to make a macroscopic force you could feel with your hand.
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At what speed does our universe expand? Conceivably it expands with the speed of light. I do not know, but curious, if there is an answer. At what velocity, does our universe expand?
| The rate of expansion of the universe (the fact that all objects are receding from each other and more so if they're farther away from each other) is given by the Hubble constant $H_0= 69.32 ± 0.80 (km/s)/Mpc$ 1
Check out this plot from Wikipedia
On the y axis you have the velocity with which the object is receding from us and on the x axis the current distance in a common astrophysical unit called Megaparsec (parallax of one arc-second $1\text{pc} \approx 3.26$ light yrs).
Edit:
the discrepancy circled in blue is due to the galaxies having additional internal motions on top of their receding due to expansion. The galaxies measured there are (as the label says) part of the virgo cluster. The internal motion will induce a Doppler shift that will influence the overall redshift of the galaxy
1: According 20th Dec 2012 the Hubble constant, as measured by NASA's Wilkinson Microwave Anisotropy Probe (WMAP)
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In dimensional analysis, why the dimensionless constant is usually of order 1? Usually in all discussions and arguments of scaling or solving problems using dimensional analysis, the dimensionless constant is indeterminate but it is usually assumed that it is of order 1.
*
*What does "of order 1" mean? 0.1-10?
*Is there any way, qualitative or quantitative, to see why the dimensionless constant is of order 1?
*Are there exceptions to that? I mean cases where the dimensionless constant is very far from 1? Could you give some examples? Can such exceptions be figured out from dimensional analysis alone?
| To add to the other answers, it should be noted that when formulating laws and relationships it is custom to use only unitary values of the relevant physical quantities.
$F=ma$, $V=IR$ etc could just as well have had a constant inserted, and if you shift from SI units you will need to include some. Therefore lack of numerical constants at the beginning leads to a lack of constants at the end of the analysis...
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Why does the weighing balance restore when tilted and released I'm talking about a Weighing Balance shown in the figure:
Press & Hold on onside of the horizontal beam and then release it. It makes some oscillations and comes back to equilibrium like shown in the figure.
Both the pans are of equal equal masses. When the horizontal beam is tilted by an angle using external force, the torque due to these pan weights are equal in magnitude & opposite in direction. Then why does it come back to it position? What's making it to come back?
| Nice question! if the following analogy applies : imagine a seesaw on a half-sphere fulcrum (top of the picture). if it inclines e.g. to left side (bottom) - the length from the right edge to fulcrum ($L_2$) increases, the lever rule kicks in ($F_2>F_1$) and the weight of the right side brings the seesaw back to equilibrium (top) (which is then broken again by inertia)
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Must the action be a Lorentz scalar? Page 580, Chapter 12 in Jackson's 3rd edition text carries the statement:
From the first postulate of special relativity the action integral must be a Lorentz scalar because the equations of motion are determined by the extemum condition, $\delta A = 0$
Certainly the extremeum condition must be an invariant for the equation of motion between $t_1$ and $t_2$, whereas I don't see how the action integral must be a Lorentz scalar. Using basic classical mechanics as a guide, the action for a free particle isn't a Galilean scalar but still gives the correct equations of motion.
| Yest, it must. It does not guarantee that the equations have physical exact solutions but at least everything looks relativistic.
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GUT that includes all 3 particle families into a large group? Explaining SU(5) GUTs (using the first particle family as an example) in the last SUSY lecture 10, Lenny Susskind mentioned that there are at present no ideas how to combine simultaneously all 3 particle families into a large GUT theory.
I somehow dont believe him, suspecting that he just didnt want to talk about this :-P...
So, are there any ideas around how to incorporate all 3 families into a larger structer?
If so, I would appreciate explanations about how it works at a "Demystified" level :-)
| Jacob Bourjaily derives three Standard Model generations from an E8 singularity here and here. (See comments by Lubos, 1 2 3.)
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Introducing emf of a chemical cell as a hint towards quantum mechanics Today I had a discussion with a colleague who teaches electricity and magnetism to 2nd year undergraduate physics students. He is seeking the best way to explain how is the emf generated inside a battery with a minimal appeal to physics beyond classical. We have lamented that some textbooks refer to "non-electromagnetic chemical forces" since all of chemistry is essentially electrostatics+quantum mechanics.
Our proposal is to draw the students' attention to the existence of atoms which can not be explained by classical mechanics + classical electromagnetism. In the same vein the forces on charge carriers in galvanic cells are electromagnetic but the response is not classical. Thus a battery "amplifies" non-classicality to the level macroscopic electricity. It is not that forces are non electrostatic but the systems response is not classical. Can you recommend textbooks or online sources that use/expand this idea?
(the main tag for this question should be "teaching" but I'm too much of a rookie here to create one)
| What you're describing is electrochemistry, which is summarised here as
Electrochemistry is a branch of chemistry that studies chemical reactions which take place in a solution at the interface of an electron conductor (a metal or a semiconductor) and an ionic conductor (the electrolyte), and which involve electron transfer between the electrode and the electrolyte or species in solution.
You don't need to delve into quantum mechanics to understand what's going on at the macro level since this branch of science has been around since Faraday, and so basic chemistry will suffice. On the other hand, the inner workings of chemistry are down to non-relativistic quantum mechanics and the solution of Schrodinger's equation which determines how electrons behave within atoms and with other electrons. Here's a link that develops a classical picture of the physical processes involved which is easily accessible to your students:
http://www.chem1.com/acad/webtext/elchem/
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Why did my liquid soda freeze once I pulled it out of the fridge and opened it? This isn't a duplicate to "Why did my liquid soda freeze once I pulled it out of the fridge?". My question is why soda froze after it was opened. Opening a can or bottle seems to have a larger effect than just jostling it.
Is it because of the disturbance noted in the previous question? Is it related to the pressure decrease? Is it because of the release of some CO2 when it was opened?
| this is very simple. when a gas loses pressure it cools. when the top is removed from a carbonated drink gas is released which expands cooling the liquid enough to freeze. some time open the valve on a scuba tank. the air coming out will be cold, it is expanding. the same tank will get very hot while being filled. that is why they are placed in water tanks during filling.
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How does the internuclear repulsion vary in Hydrogenic atom collision? Hydrogen fusion requires two hydrogen nuclei to get close enough (typically a few fm) to fuse. Much of the problem of creating a fusion reactor is overcoming the Coulomb repulsion between a pair of nuclei - the millions of degrees for Maxwellian distributions, the Bremstrahlung losses for inertial confinement.
If we could align the paths of two neutral Hydrogen atoms (of whichever isotopes), what would the repulsion look like between them as they approach collision? Obviously at long range there is negligable force as both are neutral. But as they approach each other, what happens to the electron distribution?
Intuitively, I expect a bonding cloud to form between the nuclei, and antibonding clouds beyond them. This would presumably attract at first until reaching the usual Hydrogen covalent bond length, after which the internuclear repulsion would increasingly dominate.
But how does that compare to bare ionic collision? How much lower is the potential barrier?
Obviously if it was significantly lower and we could somehow engineer the collision to achieve fusion, the cross section would be larger than ionic fusion, but how much?
Or would the barrier be just as high over the final few femtometers?
| Your last sentence is exactly right: the energy cost for fusion is almost all in those last few femtometers, at which electronic effects are negligible. Although there is in principle a difference between colliding neutral atoms and nuclei, at the energies required for fusion the effect is tiny.
The energy differences associated with the presence, absence, arrangement, etc., of electrons are of order a few to a few tens of electron volts. One way to see this is to note that the size scale of the electron's wavefunction is of order the Bohr radius $a_0$, so the energy levels are of order $e^2/(4\pi\epsilon_0 a_0)$. But to get fusion to happen, as you say, you have to get the two protons to get within a fermi or so of each other. This length scale is something like $10^4$ to $10^5$ times smaller, so it involves energies that are larger by about the same factor -- MeV rather than eV.
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Physical interpretation of describing mass in units of length I'm working in Taylor and Wheeler's "Exploring Black Holes" and on p.2-14 they use two honorary constants: Newton's constant divided by the speed of light squared e.g. $G/c^2$ as a term to convert mass measured in $kg$ to distance.
Without doing the arithmetic here, the "length" of the Earth is 0.444 cm; and of the sun is 1.477 km. To what do these distances correspond? What is their physical significance, generally?
|
Earth is 0.444 cm; and of the sun is 1.477 km
It corresponds to half of the respective Schwarzschild radius.
The $\frac{G}{c^2}$ is covered there and also in Adam’s answer.
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Can superconducting magnets fly (or repel the earth's core)? If a superconducting magnet and appropriate power supply had just enough $I\cdot s$ (current $\cdot$ length) so that when it was perpendicular to the earth's magnetic field, the force of the interaction was just enough to excede the force exerted on the object from gravity. And it was rotating so the angular momentum of the vehicle was just high enough so it wouldn't flip over, would the vehicle fly?
Assuming the vehicle is a 1000 kg (and the earth's magnetic field is $0.3$ gauss) I calculated that with $6.54\cdot10^8$ meter amperes you just about reverse the force on the vehicle.
Now assuming a $100$ meter diameter, that leaves $6.54 \cdot 10^6$ A, which is less then the current in a railgun, but still a lot.
The problem is that the force normal is no longer so normal. It will want to flip the vehicle so the magnet is the other way. Now we would need to spin the vehicle fast enough, so that it has rotated 180 degrees faster then it would take for the force of the magnet to flip the vehicle 180 degrees. How would you go about calculating this part?
| The entire craft wouldn't need to be rotating, only a hollow torus containing sufficient mass (of a heavy superfluid?), rotating at a sufficient speed.
It would also make an incredibly good flywheel, assuming you could easily put energy in and take it out again. (super-ferrofluid?)
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What is the physical significance of dot & cross product of vectors? Why is division not defined for vectors? I get the physical significance of vector addition & subtraction. But I don't understand what do dot & cross products mean?
More specifically,
*
*Why is it that dot product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\cos\theta$?
*Why is it that cross product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\sin\theta$, times a unit vector determined from the right-hand rule?
To me, both these formulae seem to be arbitrarily defined (although, I know that it definitely wouldn't be the case).
If the cross product could be defined arbitrarily, why can't we define division of vectors? What's wrong with that? Why can't vectors be divided?
| In addition to nibot's answer: division something is finding a part of something. In case of a vector, its part has the same direction but a smaller length. So it is natural to divide vectors by numbers, not by vectors.
Those dot and cross products are not simple products because they depend not only on lengths but also on orientations. They are called correspondences between a couple of vectors and numbers or vectors.
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Nomenclature of radial solutions to the Schrodinger Equation For the free particle with quantum number $l=0$, the regular solution to the radial Schrodinger equation is $R_0 (\rho)=\frac{\sin{\rho}}{\rho}$ while the irregular solution is $R_0 (\rho)=\frac{\cos{\rho}}{\rho}$. Is there a reason for this nomenclature -- (ir)regular? Thanks.
| The reason for this nomenclature is the behavior at $r=0$:
$$\lim_{r\to0^{+}}\frac{\sin r}{r} = 1,$$
$$\lim_{r\to0^{+}}\frac{\cos r}{r} = \infty .$$
$\frac{\sin r}{r}$ is regular at $r=0$ while $\frac{\cos r}{r}$ is irregular.
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Collision of Phobos Mars has two moons: Phobos and Deimos. Both are irregular and are believed to have been captured from the nearby asteroid belt.
Phobos always shows the same face to Mars because of tidal forces exerted by the planet on its satellite. These same forces causes Phobos to drift increasingly closer to Mars, a situation that will cause their collision in about 50 to 100 million years.
How I can calculate, given appropriate data, the estimated time at which Phobos will collide with Mars?
| Because Phobos is already within the Roche Limit, within which it should disintegrate due to tidal forces, "appropriate data" would have to include quite a bit of detail about Phobos's structure and composition (information which we currently lack), which would let you determine not so much the "time at which" it will collide, but the period over which bits of Phobos would.
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Why Do Hurricane Balls Spin So Fast? I was wondering if anyone could offer an explanation as to why the balls described in this video spin so fast.
Here's the setup: Two metal balls are wielded together. When spun with air, they acquire a massive amount of rpm.
| He says that the balls are spinning at 2000 Hz, and that they are two half inch balls welded together. That says that their outer surface is moving at most something like $2\pi 0.5$ inches in 0.0005 seconds, for a speed of 6300 inches per second or 523 feet per second or 356 miles per hour. This is a difficult to believe speed, but blowguns achieve 350 feet per second or 240 mph and it's tougher to accelerate down a blowgun because of the long length one must blow through.
Two half inch steel balls have a lot of weight and that means they have a lot of angular momentum even at low speeds. They have a low coefficient of friction and so they keep spinning for a long time. In terms of getting the object to spin at high speed, the primary advantage having two balls welded together is that it makes it easier to get them moving. If it were just a single sphere, blowing on it wouldn't make it spin much.
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Since the universe is expanding, does anything ever occupy the same point in space? Let's say we can observe expansion in a supercluster.
We define origin of our frame of reference at the center of the supercluster.
We observe an object/atom at point A at time T. The object is motionless relative to the origin.
We wait for expansion until T+ΔT and again observe the object.
Is the object at A or somewhere else?
| In the classical description of general relativity, the spacetime points constitute a smooth manifold with local coordinates $x^{\mu}$. In order to compute distances and intervals between points, an extra piece of information is needed, namely the metric tensor field $g_{\mu\nu}$(x). Space expansion can be thought of as not moving the points around, but rather as simply a change of the spatial part of the metric tensor with time.
For example, if we write the spacetime coordinates as $x^{\mu}$ = (t, $x^i$) where i=1,2,3 then we can write the metric of a spatially expanding universe as $$ds^2 = c^2dt^2-a(t)^2g_{ij}dx^idx^j$$ where the spatial metric components $g_{ij}$ do not depend on time.
So spatial expansion is a change in the metric tensor field, rather than any "motion" of the spacetime points themselves.
An alternative way to think of this is that points of a spacetime manifold in isolation (i.e. with no other physical fields defined) do not have any physical significance. This is connected with Einstein's hole argument.
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The energy carried from one winding of a transformer to another, in quantum terms I have read in wikipedia this statement
"The energy carried from one winding of a transformer to another, in quantum terms is carried by virtual photons, not real photons" (wikipedia src: virtual particle)
Of course anything in wikipedia could be solved "changing it" =), but I wonder, someone put it for a reason, then in the case it have sense, I don't know what is the limit for that, what are the frequencies for virtual or real photon interactions. (if is the frequency what makes the difference, or what ?)
Perhaps it's related with the kind of interaction, I mean photoelectric effect and the induction are different mechanism , but I don't see clearly how "real photons" disappear from the picture, if they are the lesser energy that can be transfered anyway.
I have put that into a comment for this answer Low frequency electromagnetic waves but I think is better to open an specific question.
| In your case it is a near field that stands for "virtual" photons. The near field does not propagate like EMW in the sense it is "attached to the charges and currents. In fact, it is the Coulomb and quasi-static magnetic interactions of charges and currents. Despite being time-dependent, the near field decays with distance differently (faster) and does not carry away any energy-momentum. Real photons do.
There is no quantum terms for the near field. It is not quantized. It exists, as I said, as a potential interaction $\propto \frac{1}{|\vec{r}_1 - \vec{r}_2|}$.
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Particle antiparticle annihilation-do they have to be of the same type? I read that a particle will meet its antiparticle and annihilate to generate a photon. Is it important for the pairs to be of the same type? What will happen when for example a neutron meets an antiproton or a proton meets a positron? Are there any rules to determine what happens when such particles meet?
| Yes, there are rules that depend on the quantum numbers carried by the particles under question and the energy available for the interaction.
In general we label as annihilation when particle meets antiparticle because all the characterising quantum numbers are equal and opposite in sign and add and become 0, allowing for the decay into two photons, two because you need momentum conservation.
A positron meeting a proton will be repulsed by the electromagnetic interaction, unless it has very high energy and can interact with the quarks inside the proton, according to the rules of the standard model interactions.
When a neutron meets an antiproton the only quantum number that is not equal and opposite is the charge, so we cannot have annihilation to just photons, but the constituent antiquarks of the antiproton will annihilate with some of the quarks in the neutron there will no longer be any baryons, just mesons and photons, and all these interactions are given by the rules and crossections of the standard model.
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How long does a permanent magnet remain a magnet? I have a bunch of magnets (one of those game-board thingies) given to me when I was a school-going lad over 20 years ago, and the magnets feel just as strong as it was the day it was given.
As a corollary to this question Do magnets lose their magnetism?, is there a way to determine how long a permanent magnet will remain a magnet?
Addendum:
Would two magnets remain a magnet for a shorter duration if they were glued in close proximity with like poles facing each other?
| Some magnets, e.g. AlNiCo, require a keeper (essentially, a piece of iron placed between the poles) to keep magnetic flux lines concentrated inside the magnet, to keep from spontaneously demagnetizing (the material reaches an unstable point in the intrinsic B-H curve). They can also be demagnetized by mechanical means (e.g., by being dropped). I think this is because the shock provides energy necessary to change domain boundaries.
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Why are color values stored as Red, Green, Blue? I learned in elementary school that you could get green by mixing blue with yellow.
However with LEDs, TFTs, etc. you always have RGB (red, green, blue) values?
Why is that? From what you learned in elementary yellow would be the 'natural' choice instead of green.
| What @Michael said, plus your retina has sensors for roughly red, roughly green, and roughly blue, and not for any other colors.
(BTW, the green sensors are more sensitive, so it takes less green to make the same brightness.)
When you see something yellow, it's in between red and green, so it excites both of them, and your visual cortex happens to call that combination yellow.
If your TV set turns on a red and a green pixel so close together that you can't tell they are separate, what does your brain call it? Yellow, because there's no way it can tell the difference between real yellow and red plus green.
Paints work by subtracting colors, not by adding them.
Blue subtracts red and green, and yellow subtracts blue.
When you mix blue and yellow paint, the color that is least-subtracted is green.
Colors you get by mixing paints tend to be muddy, as opposed to brilliant.
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Wind turbine: Single large propeller or multiple small propellers? Assuming smaller propellers would spin even in very low wind, whereas a large propeller would stall;
Would it be efficient to use a bunch of propellers from axial fans geared to a single motor instead of a single large propeller?
| The power output of a wind turbine is proportional to the square of the blade length $L$ (specifically, the "swept area" enclosed by the tips, which will be $\pi L^2$). A single, large turbine is more efficient because it has fewer moving parts than several smaller ones. If you added up the friction from every bearing and gear in all of the smaller turbines, it would be greater than the equivalent from one large turbine.
Thus, for almost any given situation a single large turbine is preferable to multiple smaller ones. As an added bonus, a large turbine will have to be higher above the ground where wind speed is generally higher; capturing even more energy.
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An example of a quantum system for which Wigner function transitions to negative values I want to check my understanding of the Wigner transform and try to understand why and how exactly the probabilistic interpretation drops down as the function goes to zero and then to negative values
So, suppose we have a free quantum oscillator with integer eigenvalues of energy and (boson) occupation.
Questions:
*
*Is it possible to start the system in a state that will have a well defined Wigner phase space probability density (with an analog interpretation as a Liouville phase space density), and it will freely evolve into a state for which the Wigner density becomes negative-valued?. In other words, is the positive-definite property of a Wigner density invariant under free evolution?
*is there a simple system (hopefully a simple harmonic oscillator based state) for which the Wigner density transitions from being positive-definite to being negative at certain regions?
*what happens with the probability interpretation in the boundary of the transition between positive-definite and negative at certain regions in general?
| The times that the Wigner function is positive does not mean it should be interpreted as a probability distribution. (What events would it be the probability distribution of? Certainly not a fuzzy joint measurement of position and momentum, e.g, with coherent states; that's the Husimi Q function.)
Is there a simple system (hopefully a simple harmonic oscillator based state) for which the Wigner density transitions from being positive-definite to being negative at certain regions?
The only non-negative Wigner functions for pure states are mixtures of pure Gaussian wavepackets. Quadratic Hamiltonians (like the simple Harmonic oscillator) preserve the Gaussianity of pure states, so the Wigner function never becomes partially negative after starting out non-negative, or vice versa.
On the other hand, non-quadratic Hamiltonians do not have this property. So just about any non-quadratic Hamiltonian you pull out of a hat will in general destroy and create strict positivity. However, there really isn't anything profound going on, because you shouldn't be taking a probability interpretation of the Wigner function either way; it's called a quasiprobability distribution for a reason.
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How are neutrino beams emitted at CERN? As far I know they come from accelerator collisions, but I have read confusing things like magnetically focused. How could neutrinos be guided magnetically if they aren't affected by
the electromagnetic field?
I would like to have a better idea of how neutrinos are emitted.
| The basis of all neutrino beams is a less exotic (protons most of the time) beam smashing some mundane target and making scads of assorted particles---many of them charged. Those charged particles are focused (and possible subjected to a second filtering for energy by using collimators and more magnets, though this step is not being done at CERN), then they are allow to fly for a while during which time many of them decay, and the decay products include neutrinos which are well collimated in the beam direction by the Lorentz focussing. The un-decayed particles are stopped with a thick pile of something that the neutrinos go right through.
The particles that are most interesting for this purpose are those that decay only by weak processes. Both because it takes more time to focus them that strong-decays allow, and because weak interactions are necessary to make neutrinos.
So mostly we have
$$ k^- \to \mu^- + \bar{\nu}_\mu $$
$$ \pi^- \to \mu^- + \bar{\nu}_\mu $$
and several other channels (or the charge conjugates, of course (or not and because the horn selects for one sign)); end-state muons subsequently decaying as
$$ \mu^- \to e^- + \nu_\mu + \bar{\nu}_e ,$$
but we arrange the decay beam line so that few of them do this before reaching the beam stop (which means that few of the products end up in the final beam as decays from rest are isotropic).
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Graphene space elevator possible? I just read this story on MIT working on industrial scale, km^2 sheet production of graphene.
A quick check of Wikipedia on graphene and Wikipedia on space elevator tells me
Measurements have shown that graphene has a breaking strength 200 times greater than steel, with a tensile strength of 130 GPa (19,000,000 psi)
and
The largest holdup to Edwards' proposed design is the technological limit of the tether material. His calculations call for a fiber composed of epoxy-bonded carbon nanotubes with a minimal tensile strength of 130 GPa (19 million psi) (including a safety factor of 2)
Does this mean we may soon actually have the material for a space elevator?
| Though we may be a while away from the ability to make the all important material, we could in fact start working towards this now. A while ago I penned a small idea for building 2 elevators, the first of which would be a "Lunar Elevator". With greatly reduced gravity, the tensile strength of the cable would be reduced to well within the art of the possible as there are materials around now which could happily (and cheaply) do the job. M5 or even normal Kevlar or Spectra promise to be more than strong enough.
Once established then we have our system of moving large amounts of Mass from the Lunar surface to the GEO orbit to create our counterbalance. This in itself would take time (decades) but allow the team on earth to have developed the graphene tech to allow the second elevator to take place.
I have some workings on the idea if you're keen, though I'm aware I'm not the first.
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How do you find the uncertainty of a weighted average? The following is taken from a practice GRE question:
Two experimental techniques determine the
mass of an object to be $11\pm 1\, \mathrm{kg}$ and $10\pm 2\, \mathrm{kg}$.
These two measurements can be combined to
give a weighted average. What is the uncertainty of the weighted average?
What's the correct procedure to find the uncertainty of the average?
I know what the correct answer is (because of the answer key), but I do not know how to obtain this answer.
| Since other questions as well as Google point to this question, I would like to extend the already existing answer from luksen by a motivation from the stochastic for using the error propagation equation.
Let's assume we have $n$ random measurements $X_i$, typically denoted by $E\{X_i\} \pm \sigma_i$ (or $x_i \pm \Delta x_i$), whereas $E\{\cdot\}$ and $\sigma_i$ denote the expected value and the standard deviation, respectively. According to the question, we are interested in the weighted average of these measurements, calculated by $Y = \frac{\sum_i w_i X_i}{\sum_i w_i}$ ($=c$). Thanks to the linearity of the expected value, it is rather easy to get
$$E\{Y\} = E\left\{\frac{\sum_i w_i X_i}{\sum_i w_i} \right\} = \frac{\sum_i w_i E\{ X_i\}}{\sum_i w_i}.$$
For the variance $\sigma^2$, we need a few more lines but no hacky tricks.
$$\begin{align}
\sigma^2 &= E\left\{(Y-E\{Y\})^2\right\} = E\left\{\left(\frac{\sum_i w_i X_i}{\sum_i w_i} - E\left\{\frac{\sum_i w_i X_i}{\sum_i w_i} \right\} \right)^2\right\} \\
&= \frac{1}{(\sum_i w_i)^2} E\left\{ \left(\sum_i w_i X_i\right)^2 - 2 \left(\sum_i w_i X_i\right) \left(\sum_j w_j E\{X_j\}\right) + \left( \sum_i w_i E\{X_i\} \right)^2 \right\} \\
&= \frac{1}{(\sum_i w_i)^2} E\left\{ \sum_{i,j} w_i X_i w_j X_j - 2 \sum_{i,j} w_i X_i w_j E\{X_j\} + \sum_{i,j} w_i E\{X_i\} w_j E\{X_j\} \right\} \\
&= \frac{1}{(\sum_i w_i)^2} \sum_{i,j} w_i w_j \left( E\{X_i X_j\} - 2 \cdot E\{X_i\} E\{X_j\} + E\{X_i\} E\{X_j\} \right) \\
&= \frac{1}{(\sum_i w_i)^2} \sum_{i,j} w_i w_j \left( E\{X_i X_j\} - E\{X_i\} E\{X_j\} \right) \\
&= \frac{1}{(\sum_i w_i)^2} \sum_{i,j} w_i w_j \cdot Cov(X_i, X_j)
\end{align}$$
The covariance holds $Cov(X_i, X_i) = \sigma_i^2$ and, as long as the original measurements $X_i$ and $X_j$ are independent, $Cov(X_i, X_j) = 0$ otherwise. This leads to the answer of the original question, the variance (squared standard deviation) of the average $Y$ of $n$ measurements $X_i$ with random uncertainty $\sigma_i$:
$$\sigma^2 = \frac{1}{(\sum_i w_i)^2} \sum_i w_i^2 \sigma_i^2,$$
which yields
$$\Delta c = \sigma = \frac{1}{\sum_i w_i} \sqrt{\sum_i (w_i \sigma_i)^2}.$$
That's exactly what the error propagation equation yields after inserting the derivations of $c$. Why is it exact while the error propagation equation is an approximation? The error propagation equation approximates by a first-order Tayler expansion. Since the average is a linear function, it is exact and not only approximated, here.
Additional information: For the unweighted average ($w_i = 1~ \forall i$), we get
$$
E\{Y\} = \frac{1}{n} \sum_i E\{ X_i\} \quad\quad\quad
\sigma^2 = \frac{1}{n^2} \sum_i \sigma_i^2$$
If all original samples $X_i$ have the same variance $\tilde\sigma^2$, this leads also to the well know variance of the average:
$$\sigma^2 = \frac{\tilde\sigma^2}{n}.$$
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What does it take to understand Maxwell's equations? Assume I want to learn math and physics enough to reach a level where I understand Maxwell's equations (The terms and reasoning in the equations I.e. why they "work"). What would I have to learn in order to have the tools I need to make sense out of it?
I'm kindof looking for a road map which I can use to get started and know what to focus on. The fields are pretty big so some pruning would be very helpful (if possible)
| Most books have a detailed section on the level of the text, with the assumed background. Pick up a book on the subject, and read this section carefully to understand the prerequisites. This section is usually in the forward, before the actual text begins. There are frequently course maps in these sections as well, to guide instructors/self taught students on what the author/publisher feels would be an appropriate setup for teaching/learning the material.
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Why Is String Theory Called A Theory
Possible Duplicate:
Laws of Atomic Theory - how is this possible?
Generally in science, a theory is effectively a fact. The theory of evolution is not a guess, or a hypothesis. It's currently understood to be a fact, backed up by evidence from numerous scientific fields spanning many decades.
String theory on the other hand seems much more contentious, much less settled in the scientific community. Why isn't it called the string hypothesis?
| These terms are not used consistently, nor in a way consistent with the way people who talk about science interpret them. For example, the thing called the "standard model" is not really a model anymore (except for the Higgs sector), but an excellent theory, perhaps even a fact of nature, but it is still called the "standard model", not the "standard fact".
An open-ended program you can publish new fundamental papers about is always called a "theory". A "model" is something that was perfectly and precisely well defined in the original paper, like the Weinberg-Salam model. A "law" is a simple mathematical relation that comes either from experiment or theory. A "hypothesis" is a tentative guess, and turns into a theory when you can start writing papers about it. A "principle" is a hypothesis that you really believe in.
These terms are more publishing terms than philosophical terms, and don't give them too much respect.
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What is the difference between electric potential, electrostatic potential, potential difference (PD), voltage and electromotive force (EMF)? This is a confused part ever since I started learning electricity. What is the difference between electric potential, electrostatic potential, potential difference (PD), voltage and electromotive force (EMF)? All of them have the same SI unit of Volt, right? I would appreciate an answer.
| Anyway the simple answer is e.m.f. is not a force in the mechanical sense. It measures the amount of work to be done for a unit charge to travel in a closed loop of a conducting material.
Let's make it more clear. In static case (ignoring time variation of any magnetic field), electric field at a point can be derived solely from a scalar as the negative of the gradient of this scalar. This scalar at any point is called the "electric potential" at that point. If two points are at different potentials then we say there exists a potential difference. Obviously it is the difference in the potentials that matters and not their absolute values. One can therefore arbitrarily assign a value zero for some fixed point who's potential may be considered constant and compare the potentials of other points with respect to it. In this way one need not have to always speak of potential difference but simply potentials.
Now, often this "electric potential" at some point in a conductor or a dielectric is called "voltage" at that point assigning the value of the voltage to be zero for earth since the potential of earth is constant for all practical purposes.
If there is no variation of magnetic field then the work done by an unit charge in a closed loop will be $0$. But if the magnetic field varies then it will be nonzero. Recall the formula: $$\nabla \times {E} = -\frac {\partial {B}}{\partial {t}}.$$
What it really implies is, it is impossible for an electric field, derived solely from a scalar potential, to maintain an electric current in a closed circuit. So an e.m.f. implies presence of some source other then a source which can only produce a scalar potential.
The following equation tells the whole story:
$$E = -\nabla \phi - \frac{\partial A}{\partial t},$$
where $\phi$ is the scalar potential and $A$ is the vector potential.
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Numerical software to manipulate a light beam in its plane wave representation? Any light field can be expressed as a sum of plane waves. Such an ensemble of plane waves is called the plane wave spectrum of the light field. The plane wave spectrum is the Fourier transform of the light field in the real space representation.
Since, this is such a basic technique, I wonder whether there is a standard tool, preferably open-source, with a simple interface via Python, Matlab or similar, to do the following (numerically):
*
*Start with a light field in real space, say a monochromatic *-Gaussian beam.
*compute the Fourier transform
*apply some operation that changes the plane wave spectrum, say a boundary between two media
*compute the inverse Fourier transfrom
*produce some output, e.g. a plot
Computationally, all it takes is a library to perform the FFT. I'm looking for a framework that would wrap this along with some phyiscal concepts.
| You could try MEEP. It is an opensource tool for light manipulation. It automatically does all the FFTs and has a built-in Gaussian source and it also produces nice plots of 1D, 2D, 3D numerical simulations. It also has a python wrapper.
| {
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Open quantum systems and measuring devices The Copenhagen interpretation by Niels Bohr insists that quantum systems do not exist independently of the measuring apparatus but only comes into being by the process of measurement itself. It is only through the apparatus that anything can be said about the system. By necessity, the apparatus has to be outside the system. An open quantum system. Can quantum mechanics be applied to closed systems where the measuring apparatus is itself part of the system? Can a measuring apparatus measure itself and bring itself into existence?
| Definitely, yes. Sometimes it is even necessary, such as in the case when you place an atom ("system") in between two mirrors ("apparatus"). The resulting quantum mechanical model is that of cavity quantum electrodynamics. Now if you want to know what's going on there, you have to bring in a second measuring apparatus (a photon detector, say).
Q: Can a measuring apparatus measure itself and bring itself into existence?
By definition, the "apparatus" is the thing doing the measuring. The "system" is the phenomena under investigation. If you choose make the object of interest system+apparatus then you've just redefined what the "system" is and you are going to need a new apparatus to do the "measuring". So, if you believe the sentence "It is only through the apparatus that anything can be said about the system", then the answer is no.
| {
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How does one calculate the force applied on an object by a magnetic field? I've tried very hard to find an answer to this question, and every path leads me to an abstract discussion of fundamental forces. Therefore, I will propose two very specific scenarios and see if they yield the result that I am looking for.
Scenario One
Let's say that I have a vertical tube exactly 1 inch in diameter that is completely incapable of holding an electromagnetic charge and has a frictionless surface. Resting inside this tube is a steel ball also exactly 1 inch in diameter. If a cylindrical magnet, also exactly 1 inch in diameter is slowly lowered into the tube, how does one determine the exact point at which the force being applied to the steel ball by the magnet will cause the ball to overcome gravity and rise toward the magnet? Is there even any way to determine this? What further information would I need?
Scenario Two
I have the same tube from above with the cylindrical magnet resting on the bottom of the tube, north pole facing upward. Suspended by a weightless string in the tube is an identical magnet, north pole facing downward. If the bottom magnet is slowly raised, how does one calculate the exact point at which the suspended magnet will begin to move upward? Is this calculation possible? What further information would I need for this calculation?
Extra Question
How are weight capacities on magnets calculated? I.e. if a whitepaper says that a magnet is capable of lifting 25 pounds, how is the correct size magnet calculated?
| I would measure the force using a well calibrated weighing scale. You will see the weight decrease as a function of distance. The difference is the magnetic force. Note that the weighing scale gives kilogram force, 1 kgf being about 9.81 Newton.
| {
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Why do ships lean to the outside, but boats lean to the inside of a turn? Small vessels generally lean into a turn, whereas big vessels lean out.
Why do ships lean to the outside, but boats lean to the inside of a turn?
For example, a boat leaning into a turn:
Image Source.
And a ship leaning out:
Image source
| The answer is much simpler than explained above. Small boats turn with a change in the direction of the propeller. Large boats turn with a rudder. To turn left, the turn of the propeller applies a force in the back of the boat that makes it lean to the left. This does not happens with a rudder.
| {
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The topology of a "closed" universe - is it really closed? The spatial part of the positive curvature FRW metric has the form
\begin{equation}
ds^2=\frac{dr^2}{1-(r/R)^2}+r^2d\Omega^2
\end{equation}
or
\begin{equation}
ds^2=R^2(d\chi^2+\sin{\chi}^2d\Omega^2)
\end{equation}
This is described as "closed", as it has the metric of a three-sphere, but I want to know what this actually means. Is $\chi$ not limited to the interval $[0,\pi/2)$, since beyond that we are simply reproducing the same $r$? In which case, is the universe not really one "hemi-threesphere" and not closed in the same way as a full sphere is?
(Edit: I note that Carroll states that the "only possible global structure is the complete three-sphere", but he doesn't go into why.)
| The first metric form uses coordinates that are valid only in a patch of the whole space. This is a fairly common situation. Consider the example of a 2-sphere projected onto a tangent plane from the center. This is the gnomonic projection which maps all great circles into straight lines. The $(x,y)$ coordinate system on the tangent plane define a coordinate system on a patch of the sphere but not the whole sphere. The sphere is still all there though. Projecting from infinity onto the tangent plane maps the the sphere onto the unit disk, and $(r,\theta)$ on this disk are the equivalent to the first metric form. But each point on the disk represents two points on the sphere, so you restrict the mapping to a patch which is a hemisphere.
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Can an arbitrary RLC-circuit network have non-minimum phase zeros? I am working with certain input-output maps that can be thought of as large RLC-networks. I thought maybe this might be a place to get some thoughts/ideas/answers.
My basic question is, given some large connected RLC network (all linear and ideal elements) and two ports in the network, say 'a' and 'b', can the transfer-function between the ports have non-minimum phase zeros?
More generally, is it true that every transfer function representing an RLC-circuit network is minimum phase?
I suspect the answer is true, but I am having a hard time proving it.
Thanks!
|
More generally, is it true that every transfer function representing
an RLC-circuit network is minimum phase?
I suspect the answer is true, but I am having a hard time proving it.
It's not true because you can have an RLC all-pass filter. To see a more specific example, let's analyse a lattice phase equaliser topology:
Writing the node equations:
$$(V_A - 1)Z^{-1} + (V_A - 0)Z'^{-1} = 0\quad{\rm (node\ A)}$$
$$(V_B - 1)Z'^{-1} + (V_B - 0)Z^{-1} = 0\quad{\rm (node\ B)}$$
Reordering:
$$V_A(Z^{-1} + Z'^{-1}) - Z^{-1} = 0\quad{\rm (node\ A)}$$
$$V_B(Z^{-1} + Z'^{-1}) - Z'^{-1} = 0\quad{\rm (node\ B)}$$
Subtracting the equations and reordering:
$$(V_A - V_B)(Z^{-1} + Z'^{-1}) - (Z^{-1} - Z'^{-1}) = 0$$
$$(V_A - V_B)(Z^{-1} + Z'^{-1}) = Z^{-1} - Z'^{-1}$$
$$V_A - V_B = \frac{Z^{-1} - Z'^{-1}}{Z^{-1} + Z'^{-1}}$$
By linearity and definition of transfer function:
$$ H(s) = \frac{Z(s)^{-1} - Z'(s)^{-1}}{Z(s)^{-1} + Z'(s)^{-1}}$$
If we use an inductor $L$ as impedance $Z$ and a capacitor as impedance $Z'$ we get:
$$Z(s) = sL$$
$$Z'(s) = (sC)^{-1}$$
$$H(s) = \frac{\frac{1}{sL} - sC}{\frac{1}{sL} + sC}$$
$$H(s) = \frac{\frac{1 - s^2LC}{sL}}{\frac{1 + s^2LC}{sL}}$$
$$H(s) = \frac{1 - s^2LC}{1 + s^2LC}$$
$H(s)$ has zeroes at $s = \pm(LC)^{-\frac{1}{2}}$, so it cannot be minimum phase.
[Added 10/15]
Zeroes in the right half-plane can be obtained even when limited to RC circuits. To see that, consider the transfer function of this filter:
We can get the node voltages directly, because both branches are generalized voltage dividers:
$$\displaystyle V_A = \frac{(sC)^{-1}}{R + (sC)^{-1}}$$
$$\displaystyle V_B = \frac{R}{R + (sC)^{-1}}$$
$$\displaystyle H(s) = V_A - V_B = \frac{(sC)^{-1} - R}{R + (sC)^{-1}} = \frac{1 - sRC}{sRC + 1} = -\frac{s - (RC)^{-1}}{s + (RC)^{-1}}$$
The general restrictions in RC (and RL) transfer functions are:
*
*All poles are simple and on the negative real axis.
*All residues are real but can be positive or negative.
*Zeros can be anywhere in the s-plane, but complex zeros must be in conjugate pairs.
*Zero and infinite frequency cannot be poles.
(Extracted from p. 5 of The synthesis of voltage transfer functions, the best online reference I've been able to find.)
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How does a state vector be projected onto an eigenspace after measurement In http://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics#Degenerate_spectra, it is said that
If there are multiple eigenstates with the same eigenvalue (called degeneracies),..., The probability of measuring a particular eigenvalue is the squared component of the state vector in the corresponding eigenspace, and the new state after measurement is the projection of the original state vector into the appropriate eigenspace.
My question: Is the state vector after measurement when the eigenspace is degenerated a pure state or mixed state? And what is the mathematical formulation of the mentioned "projection" onto the eigenspace?
| Suppose you are in the state
$$|\Psi \rangle = a|\alpha_1\rangle + b|\alpha_2\rangle + c|\beta \rangle$$
$|\alpha_1\rangle$ and $|\alpha_2\rangle$ are eigenvectors of the observable $A$, both with eigenvalue $\alpha$. $|\beta\rangle$ is also an eigenvector of this observable but with the different eigenvalue $\beta$.
If you make a measurement of $A$ and the result is $\alpha$, the state of the system becomes the state
$$N\left(a|\alpha_1\rangle + b|\alpha_2\rangle\right)$$
with $N$ a factor chosen to keep the state normalized.
If you do not know the state of $|\Psi\rangle$ before making the measurement, then you make the measurement and get the value $\alpha$, you don't know the wavefunction after the measurement. Assuming the three states $|\alpha_1\rangle, |\alpha_2\rangle, |\beta\rangle$ form a basis, all you can say is that after the measurement
$$|\Psi\rangle = p|\alpha_1\rangle + q|\alpha_2\rangle$$
with $p^*p + q^*q = 1$, but you don't know what $p$ and $q$ are. If you want to determine them, you should find some other operator $B$ which commutes with $A$, so that $|\alpha_1\rangle$ and $|\alpha_2\rangle$ are eigenvectors of $B$, but have different eigenvalues. Then measuring $B$ will allow you to determine the state completely.
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How can a conductor be grounded yet there are induced charges on it? A classic example for the method of images is the following, quoted from Griffiths's Introduction to Electrodynamics, page 121:
Suppose a point charge $q$ is held a distance $d$ above an infinite grounded conducting plane. Question: What is the potential in the region above the plane?
Griffiths continued on solving the example using the method of images setting V=0 on the plane as one of the boundary conditions saying "since the conducting plane is grounded".
Now, of course there will be an induced surface charge density. My question is, how can this be since the plane is grounded?
Does the word grounded have different meanings? sometimes it means not charged and the others it means the potential there is 0?
| Here grounded means V=0; V=0 means, there is no work to be done to bring a positive charge near the plane of the grounded conductor. This means that there are no charges on the plane. Had there been any charges on the plane it would either push or pull the test charge and V would not be zero. I think this particular example is poorly worded. later it assumes a different situation of ignoring the plane and introducing a image charge. This is just a substitute to the original problem.Originally, due to +kq(look page 123 griffith, later happens to be +q) at distance d induces -kq((look page 123 griffith, later happens to be -q)) charge at the conductor and introducing the negative image charge essentially induces positive charges on the plane which adds up to 0 and the no charge hence no V.
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why making a surface "super" smooth increases the coefficient of friction? I read that:
If you take a rough surface and make it smooth, the coefficient of friction decreases. But if you make it super smooth, then the coefficient of friction increases. How come?
| If I think about friction geometrically, i.e. ignore for now chemical interactions, (although they are very important) you get an argument about the surface area in contact. If you take two flat smooth surfaces, they will contact along a plane. If you roughen up, one or more of the surfaces, contact will only take place at high spots, i.e. there will be a lot of gaps where there is no contact, so the area in contact is much lower. Conversely the forces/stresses on the much smaller area that is in contact will be higher.
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Does the Opera result hint to a discrete spacetime? Could the Opera result be interpreted as some kind of hint to a discrete spacetime that is only seen for high enough energy neutrinos?
I think I've read (some time ago) something like this in a popular article where among other things tests of quantum gravity theories, that assume a discrete spacetime, are explained.
Looking around in blogs and other places in the web, I notice that this is disussed seldom or not at all... ``
| Searching on Google there is nothing new . Considering the plethora of arxiv papers coming out with theoretical comments on the superluminal result I would think that if the LQG model had something to say, it would have said it, particularly if it were vindicated.
So the answer is "no" . For the nonce.
Because if one reads the wiki article there exists the cryptic:
led Lee Smolin and others to suggest that spin network states must break Lorentz invariance. Lee Smolin and Joao Magueijo then went on to study doubly special relativity, in which not only there is a constant velocity c but also a constant distance l. They showed that there are nonlinear representations of the Lorentz lie algebra with these properties (the usual Lorentz group being obtained from a linear representation). Doubly special relativity predicts deviations from the special relativity dispersion relation at large energies (corresponding to small wavelengths of the order of the constant length l in the doubly special theory)
It may be that LQG might be able to accommodate the OPERA result, though again, from not having jumped at the opportunity I would not hold my breath.
p.s. I am an experimentalist and am treating theories statistically :).
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Problem with an electricity / thermodynamics assignment I've been trying to figure this one out for a while on my own, so I'd like to ask for your help if you could offer some.
The task states:
A heater made out of a wire with a diameter $R = 0.2\text{ mm}$, length $4\pi\text{ m}$ and electrical resistivity of $0.5\times 10^{-6}\ \Omega\;\mathrm{m}$ is connected to a voltage source of $220\text{ V}$, sinked in the water.
Which mass of water will it heat up from $20^{\circ}\mathrm{C}$ to $50^{\circ}\mathrm{C}$ in the time of 10 minutes? (C of water = $4200\ \mathrm{J\;kg}/\mathrm{K}$)
I know I have the electrical properties of the wire and the thermodynamic properties of the water, but I don't know how to proceed from there. We've been studying electricity and I am not really aware how I can connect it with thermodynamics?
| You may consider this question from perspective of energy view: the electric energy is consumed by the resistor and convert this energy to the thermal energy (the source of heat that heat up the water). So from this point of view, if you can assume the 100% electric energy converting to thermal energy, and usually, this assumption is right for resistors since there is no other kind of energy that electric energy can convert to, since this is not a motor or a light bulb. Thus you will have the following equation:
$Heat = I^2Rt = \frac{U^2}{R}t$
So this amount of heat will be absorbed by water and heat the water up, so
$Heat = c_{water}\cdot m\cdot \Delta T$
By solving these two equations, you can get the amount of water ($m_{water}$) you need as
$m_{water} = \frac{U^2 t}{R\cdot c\cdot \Delta T}$
where $R$ can be easily calculated as $R=\frac{\rho L}{\pi r^2}$ for this cylindrical resistor.
So the key point to connect the electricity to thermal dynamics is the conservation of energy so that energy has to convert from one form (electric energy in your case) to another form (thermal energy or heat in your case), and little or no energy is converted into other form such as light.
| {
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The visibility of air For pilots of gliders or sailplanes, the 'thermal' is the most important phenomena of the air. A thermal is classically described as an upward flow of air caused by ground level heating of air that rises in bubbles or a connected stream of warmed air. Given sufficient velocity of the rising air, a gliding craft, bird or even trash and debris can be lifted thousands of feet. It can be also noted that 'dust devils' can result from especially vigorous flows and that other even stronger phenomena like tornadoes and cyclonic storms are related.
But in the absence of markers like dust, the air flow is generally invisible until possibly the flow reaches an altitude where the water vapor contained, condenses to form a cloud. And again if the flow is strong enough and contains enough water, a thunder storm is possible.
So, the question. Given all of the above is it possible to see the mass of air that makes up the thermal? Is there anything in the difference between the thermal and the surrounding air that could be detected and presented graphically?
| Mirages are a visible effect of the inhomogeneous, temperature-dependent refractive index of air. They're visible because of the relatively large heat gradient close to the earth's surface, and the excellent reference line provided by a uniform, flat horizon. I think those conditions would be difficult to replicate in-flight - perhaps you could have a drone could fly ahead of you with a uniform grid in tow behind it?
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Is the cooling rate of a (very) cold object, sitting next to an AC higher or lower? In more detail:
If i have two soda cans, both are cooled to exactly 4 degrees celsius,
And i put one in a 25 degrees room, and the other next to an AC vent set to 16 degrees.
After three minutes, which one should be colder than the other and why?
Edit: To clarify - if I have a cold soda can, should I place it near the AC vent or not (if I like my drink cold)? Which location will cause faster heating?
| AlanSE's answer is pretty good. I'll try to add to it by keeping it math free and strictly using everyday experiences.
Imagine you're in a pool of cool, still, water. If you keep still you're not too cold but as soon as you move a bit you get very cold. A blanket of warm water forms around you that protects you. As soon as you disturb this blanket by moving, you encourage heat to flow out of your body. The Cs in AlanSEs answer account for differences in fluid flow.
The same thing is happening to your soda cans. If the AC is blowing fast enough, the cooler air will warm the soda can faster - despite the smaller temperature gradient (difference in T). If the forced air current is slow enough, the can at a higher temperature will always warm quicker than the one at a lower temperature.
After a long time, the soda cans will reach equilibrium. The soda placed on the unmoving air has a higher equilibrium temperature, so it will eventually be warmer.
Note that the change in T (dT/dt) is also proportional to the difference in T. You can see this in AlanSE’s answer. If you haven’t taken a differential equations class yet, take it seriously no matter how hard you find the subject because they are beautiful and so powerful.
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Will the water added to an ice piece freeze? Water, at room temperature is poured into a hole made of a block of melting ice(kept at room temperature).I was wondering if the water will ever freeze?
Thank you.
| Ice coming from the freezer will typically be around -19 deg. celsius, and can only be stored for a limited time at room temperature. As soon as the ice is heated to 0 deg. or above, the ice will melt into liquid water. Liquid water coming into contact with ice will be cooled, and if cooled below 0 deg. it will also freeze.
The answer to your question is that it will depend on how much ice, how much water, and the starting temperatures of these(and much more if you really goes into small detail like the dynamic of energy transport). Everything is controlled by energy, to do the real calculations, you need constants like the heat capacity of water, and ice, and the melting energy.
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Dependence of Friction on Area Is friction really independent of area? The friction force, $f_s = \mu_s N$. The equation says that friction only depends on the normal force, which is $ N = W = mg$, and nature of sliding surface, due to $\mu_S$.
Now, less inflated tires experiences more friction compared to well inflated tire. Can someone give clear explanation, why friction does not depend on area, as the textbooks says?
| When you say underinflated tires experience more friction, do you mean static friction (i.e., resistance to slipping) or rolling resistance, which is something quite different?
Afaik the origin of the friction law is very much phenomenological, and has it's limits of applicability (especially at the static - dynamic transition). My understanding as to why drag racing vehicles have such enormous tires is spread out the shear forces and dissipate more heat. The force of friction is the same regardless, until the tires turn to liquid!
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Are specific heat and thermal conductivity related? Are there any logical relationship between specific heat capacity and thermal conductivity ?
I was wondering about this when I was reading an article on whether to choose cast iron or aluminium vessels for kitchen.
Aluminium has more thermal conductivity and specific heat than iron ( source ).
This must mean more energy is required to raise an unit of aluminium than iron yet aluminium conducts heat better than cast iron.
Does it mean that aluminium also retains heat better ?
How does mass of the vessel affect the heat retention?
| Imagine a substance in the size and form of an ice cube. If you could keep shooting it with a photon of say energy $1$ and you shot $10$ of these photons and noticed that the substance had gained a temperature difference say from $25$ to $26^\circ\mathrm{C}$, then its specific heat capacity would be $10$. (Specific heat capacity is more like a measure of the external energy given to produce the temperature change.) And it might even give off this temperature as fast as it got it.
Now for thermal conductivity (this guy is more like a range thing). If you could place a finger on one side of this substance and start your photon shooting on the other side, you may notice that even if the photon-receiving side has done the $25$ to $26^\circ\mathrm{C}$ climb, the side your finger is on might not have. (What you're doing now is obtaining the thermal conductivity of that substance.) $20$ photons might get the climb or not. Going on to $30$, $40$, ......
So basically to obtain this climb for the same cube of aluminium or iron, it might take $10$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/16255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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Inverted Harmonic oscillator what are the energies of the inverted Harmonic oscillator?
$$ H=p^{2}-\omega^{2}x^{2} $$
since the eigenfunctions of this operator do not belong to any $ L^{2}(R)$ space I believe that the spectrum will be continuous, anyway in spite of the inverted oscillator having a continuum spectrum are there discrete 'gaps' inside it?
Also if I use the notation of 'complex frequency' the energies of this operator should be
$$ E_{n}= \hbar (n+1/2)i\omega $$
by analytic continuation of the frequency to imaginary values.
| The QHO does not permit analytic continuation, because its energies and wavefunctions depend not on $\omega$, but on $|\omega|$. Thus, their dependence on $\omega$ is not analytic and $\omega$ cannot be simply replaced by $i\omega$.
Moreover, this Hamiltonian is not Hermitian. Still, just like few other interesting cases ($ix^3$, $-x^4$), it has real spectrum.
Here you can find a short but comprehensive explanation:
http://arxiv.org/abs/quant-ph/0703234
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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How do we perceive hotness or coldness of an object? Some objects, especially metallic ones, feel cold on touching and others like wood, etc. feel warm on touching. Both are exposed to same environment and are in their stable state, so some kind of equilibrium must be being reached. What is this equilibrium?
And how do we perceive hotness or coldness of an object? Does skin have some kind of heat sensors, etc., which transmit signals to brain? Like, how do the eye transmit/convey an image formed on the retina to brain?
| sensation of heat activates an electrical impulse which is passed through neurons and synapsis as electrical an chemical reactions and reaches our brain which interprets the message as heat some times if the heat is unbearable the message is interpreted by the spinal chord
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/16333",
"timestamp": "2023-03-29T00:00:00",
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The shape of the earth$\ldots$ ....is an oblate spheroid because centrifugal force stretches the tropical regions to a point farther from the center than they would be if the planet did not rotate. So we all learned in childhood, and it seems perfectly obvious. However...
I am at $45^\circ$ north latitude. Does that mean
*
*An angle with vertex at the center of the earth and one ray pointing toward the equator at the same longitude as mine, and one ray pointing toward me, is $45^\circ$ (that would mean I'm closer to the north pole than to the equator, measured along the surface, as becomes obvious if you think about really extreme oblateness); or
*The normal to the ground where I stand makes a $45^\circ$ angle with the normal to the ground at the equator at the same latitude (this puts me closer to the equator than to the north pole); or
*something else?
If for the sake of simplicity we assume the earth is a fluid of uniform density, it seems one's potential energy relative to the center of the earth would be the same at all points on the surface.
*
*Would the force of gravity at my location, assuming no rotation, be directly toward the center? Would it be just as strong as if the whole mass of the earth were at the center and my location is just as far from the center as it is now?
*Would the sum of the force of gravity (toward the center or in whichever direction it is) and the centrifugal force (away from the axis) be normal to the surface at my location?
*Given all this, how does one find the exact shape?
*How well does that shape in this idealized problem match that of the actual earth?
| 1, At 45 deg (N) latitude you are closer to the North pole, to picture this just draw the Earth as a much more extreme oblate spheroid.
2, The shape of the Earth is set by the outward rotational force exactly balancing the inward gravitational force at every point (except for local geography). So the overall potential is always down (except for local geology)
There is a useful intro to the difference between mean sea level and the Earth's surface at ESRI (makers of the popular GIS/mapping software)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Compound lenses and virtual images Ok so I have this problem where I have a system of two lenses. All I know is that the distance between the object and the first lens is 30cm, the distance between the object and the final image is 70cm, the focal distance of the first lens is 20 and the total magnification is -0.666. I need to find the focal distance of the second lens and the distance between lenses 1 and 2.
Now I've tried calculating the parameters of the first lens' image (it should be 60cm past the first lens and the magnification should be -2). My problem is that if I now use this image as an object for the second lens and solve the equations, I obtain absurd results (values should put the lens between the first lens and the final image but they don't).
How should I go about solving such a problem?
| Most likely, your problem is the sign of the distance from the image of the object through the first lens, to the second lens. When you use the formula for the ideal thin lens you must be really careful about the convention used.
For example, I usually use $\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$, where $f$ is the lens' focal distance, $o$ is the distance from the object to the lens, and $i$ is the distance from the lens to the image. The sign convention for this formula is
letter positive negative
o left of the lens right of the lens
i right of the lens left of the lens
f converging diverging
This means that the reference system is different for $o$ and $i$, which can be confusing.
Other common error for this kind of problems is that $o$ and $i$ is the distance to the lens, so you can't use the number you get for $i$ from the first lens as the $o$ for he second lens. You need to take into account the distance between the lenses.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/16434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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