Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Introductory texts for functionals and calculus of variation I am going to learn some math about functionALs (like functional derivative, functional integration, functional Fourier transform) and calculus of variation. Just looking forward to any good introductory text for this topic. Any idea will be appreciated.
| Have a look first at several chapters in Stone and Goldbart, "Mathematics for Physics" (the free preprint is here) before entering into more specific books. I think you may want to see chapters 1, and parts of 2 and 9.
You may find some parts of what you want in classic books of the "Comprehensive Mathematical Methods for Physics" type, but they don't usually cover that questions in detail. Stone&Goldbart, without being a dedicated book, is somewhere in between.
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Why is a classical formalism necessary for quantum mechanics? This is not a question pertaining to interpretations, after the last one I realized I should not open Pandora's Box ;)
For theories to be consistent, they must reduced to known laws in the classical domains. The classical domain can be summed up as:
$$\hbar=0 ; c=\infty$$
Which is OK. I need to know, however, is that if QM is an independent and fundamental theory, why does it rely so heavily on the classical formalism. Is it necessary for a classical formalism to exist in order to have a quantum formalism? From as far as I have read, it does not seem so, and I find this puzzling. Suppose you have a dissipative system, or an open system when you cannot write an autonomous Hamiltonian in the classical case, how then do we approach these quantum mechanically, when we cannot even write down the corresponding Hamiltonian.
| This is an excellent question and the answer is it follows from the crucial postulates of quantum mechanics. Namely, the operator corresponding to a dynamical variable is obtained by replacing the classical canonical variables by corresponding quantum mechanical operators and any pair of canonically conjugate operators will satisfy the Heisenberg commutation rules. Physically speaking, it has definite relevance to these remarks of the great Neils Bohr, "However far the phenomena transcend the scope of classical physical explanation, the account of all the evidence must be expressed in classical terms". This is because "by the word 'experiment' we refer to a situation where we can tell others what we have done and what we have learned" so that "the account of the experimental arrangement and of the results of the observations must be expressed in unambiguous language with suitable application of the terminology of the classical physics". (Note that for some observables like spin there is no classical analog and we take different approach.)
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Why does Venus spin in the opposite direction? Given: Law of Conservation of Angular Momentum.
*
*Reverse spinning with dense atmosphere (92 times > Earth & CO2 dominant sulphur based).
*Surface same degree of aging all over.
*Hypothetical large impact is not a sufficient answer.
Assuming any object large enough to alter a planets rotation or even orbit would likely destroy most of its shape, yet Venus has retained a spherical property with a seemingly flat, even terrain indicating no volcanoes,and few if any visible meteor impacts. It would be fragmented and dispersed for billions of years. Even the question of what meteor, comet, asteroid composition could survive traveling that close to the sun's temperature, radiation, electromagnetic energy, solar flares, or gravity to equal a mass reactionary change as to alter it's spin.
| IMO there is no solid explanation, as anna said. Only clues (WP).
In this simulation (2002) Long term evolution of the spin of Venus- II, Numerical simulations we find a mix of: 'chaotic zone', instability, large impact, close encounter, tidal effects, planetary perturbations,...
There is room for speculation:
I think that the heavy atmosphere is not a significant factor of slowing the rotation.
I discarded any violent event because it can easily change more than one parameter; in this case we have only one (the direction).
The planet's minute axial tilt (less than three degrees, compared with 23 degrees for Earth) make me think to keep only the tidal perturbation, although the present configuration Sun-Venus-Earth is not able to justify it.
As the Venusian surface rotates as low as 6.5 km/h (on Earth is about 1,670 km/h) we can think that Venus may have changed the direction of rotation not long time ago.
The solar system angular momentum problem is not solved (the planets strangely have almost all of it) and I think that the solution of this does one is not related to the present question.
The equation 35 of this paper (a new model, undiscussed) allows the slow evolution of the configuration Sun-Venus-Earth.
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Massless charged particles Are there any massless (zero invariant mass) particles carrying electric charge?
If not, why not? Do we expect to see any or are they a theoretical impossibility?
| Since a particle only can be said to exist if it can express its existence, its properties in interactions if it has energy and localized energy is a source of gravity and we define mass as something which exerts and feels gravity, then there cannot exist massless particles.
(That is, if we define a particle as an entity which at all times has a well-defined position -which a massless particle like a photon hasn't as from its own point of view, its transmission takes no time at all.)
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How do electrons "know" to share their voltage between two resistors? My physics teacher explained the difference between voltage and current using sandwiches. Each person gets a bag full of sandwiches when they pass through the battery. Current = the number of people passing through a particular point per unit time. Voltage = the (change in) number of sandwiches per person. In a parallel circuit the number of people (current) is divided between the two paths, but the number of sandwiches per person (voltage) remains the same. In a series circuit the number of people passing through a particular point remain the same, but they drop off a certain percentage of their sandwiches at every resistor. Therefore, there is a voltage drop that occurs between the points before and after every resistor.
This analogy naturally leads to the question: how do the electrons "know" that they are going to have to share their voltage between two resistors before they reach the second one? (In other words, not drop off all their sandwiches at the first resistor they find)
| Continuing Marek's reply, I will add that in electrical circuits the value of the current depends on the elements of the circuit: the voltage , the resistors summed when in series, summed inversely if in parallel. The whole defines the value of the current, and currents in the parallel parts, according to the electricity laws. The individual electrons participating in the current follow the flow, it is the current that "knows" where the resistances are .
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Are these two quantum systems distinguishable? Suppose Stanford Research Systems starts selling a two-level atom factory. Your grad student pushes a button, and bang, he gets a two level atom. Half the time the atom is produced in the ground state, and half the time the atom is produced in the excited state, but other than that you get the exact same atom every time.
National Instruments sells a cheap knockoff two-level atom factory that looks the same, but doesn't have the same output. In the NI machine, if your grad student pushes a button, he gets the same two-level atom the SRS machine makes, but the atom is always in a 50/50 superposition of ground and excited states with a random relative phase between the two states.
The "random relative phase between the two states" of the NI knockoff varies from atom to atom, and is unknown to the device's user.
Are these two machines distinguishable? What experiment would you do to distinguish their outputs?
| This is such a nice philosophical question with such a neat resolution that I can't resist dropping a comment. The reduced density matrices of the atom are the same for Stanford and National, but quantum mechanics is irreducibly holistic. The wave function describes the entire universe. If the atom was prepared by Stanford, it will be entangled with traces of the environmental record in Stanford in a particular way, but if it was prepared at National, it will be entangled with traces at National in a different but still specific way. Holistically, there is indeed a difference. To suppose the atom can be considered in isolation from the rest of the world is a major fallacy in quantum mechanics.
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Mathematical justification of Hartle-Hawking "no boundary" proposal In Hartle-Hawking "no boundary" proposal it is proposed that Riemannian spacetimes rather than Lorentzian dominated the path integral near the big bang.
Moments after the big bang however spacetimes with Lorentzian metrics started to dominate over the Riemannian. The dominance of the Riemannian spacetimes is characterized by positive definite metrics obtained by applying Wick rotation.
Now to compute the path integral some approximations are made and the process of analytic continuation is applied. What bothers me is analytic continuation of an approximate holomorphic function is not guaranteed to remain analytic in another region.
Why then this unreliable process applied—can one rely on this scheme which seems to be mathematically dubious?
| What is bothering you here is not Hartle-Hawking per-se, but the process of analytic continuation inside a path integral. This can be mystifying, because the actual functions which dominate the integral are never analytic, they are hardly ever continuous or even bounded. Usually they are distributional, so that they fluctuate in terrible ways at short distances. How can you analytically continue them?
But you don't ever continue the functions, you continue the path integral. The functions which you integrate over in imaginary time are of no relation to the functions which you integrate over for real times. The reason you can continue the path integral is because of positive energy--- the contribution of high-energy intermediate states decay quickly in positive imaginary time, and there are no negative energy contributions.
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What are the conditions to be satisfied by a theory in order to be a quantum theory? This is in continuation to my previous question. It is not a duplicate of the previous one. This question arises because of the answers and discussions in that question.
Can we call a theory, quantum theory, if it is consistent with HUP? For example, suppose there is a finite and self consistent theory of gravity which incorporates the uncertainty principle. Can we at once call this theory a quantum theory of gravity or does it have to satisfy other conditions too?
This question may be too basic but it is intriguing my mind.
| In mathematical reasoning, and not only, there are what are defined as necessary conditions, and others that are defined as sufficient conditions.
Necessary means that without this condition the theory cannot be proven/ be consistent. Sufficient means that just from this condition the theory is consistent.
The QM postulates are necessary and sufficient for a quantum mechanical theory to be fully developed. The HUP is a theorem arising from the postulates of the QM theory, so it is a necessary attribute. But it is not sufficient to define the postulates, they cannot be derived from it, and thus not sufficient to define a theory as a quantum theory.
Look at the answer by @Roy Simpson in your previous question. I remember this "uncertainty principle" from my classical electrodynamics course. That it exists, it does not make classical electrodynamics a quantum theory.
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Are there examples in classical mechanics where D'Alembert's principle fails? D'Alembert's principle suggests that the work done by the internal forces for a virtual displacement of a mechanical system in harmony with the constraints is zero.
This is obviously true for the constraint of a rigid body where all the particles maintain a constant distance from one another. It's also true for constraining force where the virtual displacement is normal to it.
Can anyone think of a case where the virtual displacements are in harmony with the constraints of a mechanical system, yet the total work done by the internal forces is non-zero, making D'Alembert's principle false?
| You can have instances where there is no local extremum of the action--for instance, take the lagrangian $L=m\left(\dot x ^{2}+\dot y^{2}\right)$ over the space defined by a crescent embedded in $\mathbb{R}^2$--then, even though the tips of the crescent are both perfectly good starting and ending points in your domain, there is no extremal path connecting them--it would have to be the straight line that leaves the domain of your configuration space.
But this is admittedly a contrived example.
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Why do all the planets of the solar system orbit in roughly the same 2D plane?
*
*Most images you see of the solar system are 2D and all planets orbit in the same plane. In a 3D view, are really all planets orbiting in similar planes? Is there a reason for this? I'd expect that the orbits distributed all around the sun, in 3D.
*Has an object made by man (a probe) ever left the Solar System?
| *
*The orbital planes of different planets has small inclinations to the ecliptic plane. The corresponding wikipedia diagram should give a better view than the 2D images you've seen.
*Both Voyager 1 and 2 are beyond the Solar System
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Chern-Simons Theory in 3D For the CS theory on a three manifold $M$ with a gauge group $G$, it is said that the gauge field $A$ is a connection on the trivial bundle over $M$. Why the bundle should be trivial? I know that space of classical solutions is made of flat connections but why should we assume the bundle is trivial from the beginning?
Another question: Can anyone explain the meaning of framing of a three manifold in simple terms?
| Answering the first question. The Chern-Simons functional expressed in terms of the vector potential $A$:
$$S(A) = \frac{k}{8 \pi^2} \int_M tr(A \wedge dA + \frac{2}{3} A \wedge A \wedge A )$$
is well defined only for a trivial principal bundle, because in this case the gauge connection is
a Lie algebra one form over $M$. When $M$ is an oriented three manifold and the gauge group $G$ is compact connected and
simply connected, then every $G-$ bundle over $M$ is trivializable (see, for example the following article by D. Freed)
The case when the principal bundle is not trivial (for example when the above restrictions on the
gauge group are not satisfied) was treated by Dijgraaf and Witten
, using the standard construction by Witten: One can choose a 4-manifold $B$ such that $M$ is its boundary.
The CS action is defined:
$$S(A) = \frac{k}{8 \pi^2} \int_B tr(F \wedge F)$$
This action is the proper definition of the CS action for the following two reasons
*
*It reduces to the usual CS action when the bundle is trivial (Using Stoke's theorem).
*If k is an integer, the general form is independent modulo 1 of the extension $B$ of $M$.
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Can a disk like object (like UFO's) really fly? UFOs as shown in movies are shown as disk like objects with raised centers that emit some sort of light from bottom. Can such a thing fly?
My very limited knowledge in physics tell me that a disk like object may not be able to maneuver unless it has thrusters on sides and simple light can not be enough to make any object go up in the air.
Is it possible?
| Have a look at the flying car
There is even a video.
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Voltage drop along an idealized resistance-free wire in a circuit? If you connected the positive terminal of a battery to the negative terminal to a battery with a wire with (hypothetically) no resistance, and are asked to give the voltage drop of a segment of wire in the circuit, how would you determine this?
| Let $V$ be the e.m.f of the battery/ voltage source
Let $i$ be the current through the wire of resistance $R$.
Actually $i= \frac{V}{R}$, as per the question, $R$ tends to ZERO, hence $i$ tends to a very very large number.
Let the drop across the wire be $V_1$ then $V_1=$(current through the wire)*(resistance of the wire) {as per Ohm's law}
so: $V_1= iR$
hence: $V_1=\frac{V}{R}R$
so: V_1=V.
Thus, voltage drop across ANY segment of the wire connected will be equal to the drop $V$ {e.m.f of the battery}
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Have CMB photons "cooled" or been "stretched"? Introductory texts and popular accounts of why we see the "once hot" CMB as microwaves nearly always say something about the photons "cooling" since the Big Bang. But isn't that misleading? Don't those photons have long ("cool") wavelengths because space expanded since they were emitted. There's no separate "cooling" process, is there?
| Cooling and stretching essentially mean the same thing here. The temperature of any blackbody radiation is related to the peak wavelength by Wien's Law $$\lambda_{\mathrm{max}} = \frac{b}{T} $$ Therefore as the universe expands, all of the photon wavelengths get stretched out and so does the peak wavelength. $$\lambda \propto a(t)$$ This decreases the temperature of the radiation by the same factor that expands the wavelength of the photons. $$ T \propto \frac{1}{a(t)} $$ This factor $a(t)$ is the scale factor of the universe and increases with time for an expanding universe.
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Is it possible to know the exact values of momentum and velocity of a particle simultaneously? I know that by Heisenberg's Uncertainty Principle that it is not possible to know the exact values of position and momentum of a particle simultaneously, but can we know the exact values of momentum and velocity of a particle simultaneously? I would think the answer would be no because even if we were 100% certain of the particle's position, we would be completely unsure of the particle's momentum, thus making us also completely unsure of the particle's velocity. Does anyone have any insight into this?
| If in your theory the momentum operator and velocity operator are proportional to each other, then yes. Knowing one's eigenvalue means knowing the other's. It is always the case with any function of a "known" operator.
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Supergravity calculation using computer algebra system in early days I was having a look at the original paper on supergravity by Ferrara, Freedman and van Nieuwenhuizen available here. The abstract has an interesting line saying that
Added note: This term has now been shown to vanish by a computer calculation, so that the action presented here does possess full local supersymmetry.
But the paper was written in 1976! Do you have any info what kind of computer and computer algebra system did they use? Is it documented anywhere?
| It sounds like it may have been an early version of SHEEP, or some extension thereof. SHEEP was 'officially' released in 1977, but its predecessor, ALAM, was developed by d'Inverno in 1969. It was used to automate some of the complicated algebra in early calculations of the Bondi mass. You can read a bit about the history here: notes on SHEEP.
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Is there a direct physical interpretation for the complex wavefunction? The Schrödinger equation in non-relativistic quantum mechanics yields the time-evolution of the so-called wavefunction corresponding to the system concerned under the action of the associated Hamiltonian. And this wavefunction is, in general, complex, and its modulus squared yields the probabilities observed experimentally. Though, perhaps, this question has been asked many times, I am wondering if there is a direct physical interpretation - something that physically corresponds to - the wavefunction. Or is it just an intermediate calculational tool to arrive at the appropriate predictions for experimental outcomes, and nothing more? Of course, things like superposition and interference effects follow from the complex nature of the probability amplitude. So there must be something physical about it. What is it? Or are we not supposed to ask that question?
Is it because the probability amplitude is complex that we have difficulty in relating it to something physical? Can we do quantum mechanics without complex numbers?
| I'll leave an answer for your last question on whether complex numbers are necessary for QM.
Scott Aaronson has a nice lecture here http://www.scottaaronson.com/democritus/lec9.html , scroll down to the section on Real vs. Complex numbers.
My favorite argument there is the first one -- that if you have a linear operator $U$, then you would want to have operators like $V$ where $V^2=U$, simply because you expect continuity; i.e. if you're allowed to do one full transformation, you should be able to do "half" of it too. (If waiting for one second is allowed, then waiting for half a second should also be allowed). In order to have square roots of operators in general, you'll need to allow operator matrices with complex elements. And once you allow that, the state vectors that they act on will also need to be complex in general. And so your wavefunction will also need to be complex.
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More on matter and anti-matter
*
*Does every particle that has rest mass also have an anti-particle with which it would annihilate?
*Does annihilation only occur between like particles? For example what happens if a antineutron (anti u, anti d, anti d) collides with a proton (uud)? What happens if a positron collides with a proton?
*Since the Tevatron accelerates antiprotons is this more difficult to handle and dump?
*I've read about WIMP annihilation detection. Why would one assume there is any different proportion of WIMPS to anti-WIMPS than as is for non-WIMP matter (where there is far more matter than anti-matter).
|
1.) Does every particle that has rest mass also have an anti-particle with which it would annihilate?
For fermions in the standard model scheme , yes.
The three columns on the left are fermions. Bosons are their own antiparticle and the term "annihilation" has no meaning other than "interaction" and whether it is probable.
2.) Does annihilation only occur between like particles? For example what happens if a antineutron (anti u, anti d, anti d) collides with a proton (uud)? What happens if a positron collides with a proton?
You have to figure it out from the SM scheme, linked above and the energies involved. An antiquark striking a quark will annihilate, the rest will interact.
3.) Since the Tevatron accelerates antiprotons is this more difficult to handle and dump?
It is more difficult to create and handle an anti proton beam upt to the energy needed for the experiment, as well as to build up a sufficient number of antiprotons in the beam. When entering the ring the handling difficulty is the same. Dumping is the same for both beams.
4.) I've read about WIMP annihilation detection. Why would one assume there is any different proportion of WIMPS to anti-WIMPS than as is for non-WIMP matter (where there is far more matter than anti-matter).
It depends on the particular model, and how one assumes the WIMPs appear during the cooling after the big bang. It is reasonable to assume that the same CP violation that created the observed asymmetry between matter and antimatter holds for WIMPS too.
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What is the Physical Meaning of Commutation of Two Operators? I understand the mathematics of commutation relations and anti-commutation relations, but what does it physically mean for an observable (self-adjoint operator) to commute with another observable (self-adjoint operator) in quantum mechanics?
E.g. an operator $A$ with the Hamiltonian $H$?
| You may consider commutativity of different variables as physical independence, something like separated independent variables:
$\frac{\partial}{\partial x} \frac{\partial}{\partial y} = \frac{\partial}{\partial y} \frac{\partial}{\partial x}$.
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Has quantum entanglement been demonstrated to be able to take place over infinite distances? In my poor understanding of quantum physics, quantum entanglement means that certain properties of one of two 'entangled' quantum particles can lead to change over infinitely large distances when the other particles' properties are changed.
Disregarding this already mind-boggling event taking place over say 10 meters distance; how have physicists been able to demonstrate, beyond reasonable doubt, that this can take place over infinitely large distances?
For instance: have they done some of these tests between ISS and Earth perhaps?
How can they be so sure?
| Since my Alma Mater was involved I can point to this:
A team of European scientists has
proved within an ESA study that the
weird quantum effect called
'entanglement' remains intact over a
distance of 144 kilometres.
(Source)
In September 2005, the European team
aimed ESA's one-metre telescope on the
Canary Island of Tenerife toward the
Roque de los Muchachos Observatory on
the neighbouring island of La Palma,
144 kilometres away. On La Palma, a
specially built quantum optical
terminal generated entangled photon
pairs, using the SPDC process, and
then sent one photon towards Tenerife,
whilst keeping the other for
comparison.
There is an effort for Quantum Entanglement in Space Experiments.
Experimentally is has not been proven to work over an infinite distance, since that would be impossible.
To better understand Quantum Entanglement (and why distance is not a problem) I suggest starting with the EPR Paradox.
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Why is it hard to solve the Ising-model in 3D? The Ising model is a well-known and well-studied model of magnetism. Ising solved the model in one dimension in 1925. In 1944, Onsager obtained the exact free energy of the two-dimensional (2D) model in zero field and, in 1952, Yang presented a computation of the spontaneous magnetization. But, the three-dimensional (3D) model has withstood challenges and remains, to this date, an outstanding unsolved problem.
| Exaxt solvability has nothing to do with NP-completeness.
For equations on a lattice or a continuum, exact solvability happens to be equivalent with having enough symmetries to allow the solution to be determined by exploiting these. (To a large extent, this even holds for ordinary differential equations in more than a few variables.)
The reason that a few (classical or quantum) systems are integrable therefore comes from the fact that they have a much larger (infinite-dimensional) symmetry group, and hence infinitely many conservation laws, while a typical system has only a small, low-dimensional symmetry group. This is the (modern) explanation why Onsager's solution works, while there is no analogous solution in the next dimension.
If one looks at lists of integrable systems (e.g., the one at
http://en.wikipedia.org/wiki/Integrable_system#Exactly_solvable_models
which for the classical case seems fairly complete) one sees that they get very scarce in higher dimensions. There are just not enough possible large symmetry groups around....
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When water is about to boil Have ever noticed? When water is about to boil, no matters the kettle, there is some sound I have no idea where it comes from, sometimes long before it boils.
Is there any explanation for this phenomena?
| The water near the heating element turns into water vapor. This vapor then rises up to the surface but as it meets colder water upwards it turns back into water.
As the water/steam transition is not smooth (e.g. the volume changes rather rapidly during phase change), the constant transition between vapor and liquid states produces noise. As the water gets to +100C (boiling point), the vapor bubble will not turn back into liquid before reaching the surface thus making less noise.
For more on this click this.
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Anti-matter repelled by gravity - is it a serious hypothesis?
Possible Duplicate:
Why would Antimatter behave differently via Gravity?
Regarding the following statement in this article:
Most important of these is whether ordinary gravity attracts or repels antimatter. In other words, does antihydrogen fall up or down?
Is this a seriously considered hypothesis? What would be the consequences on general relativity?
If this is seriously studied, can you point to some not-too-cryptic studies on the (anti ;-)matter?
| EDIT: fixed thought experiment.
If antimatter anti-gravitates, it would be possible to build a perpetual motion machine:
1) Start with a zero-net momentum pair of photons. Have them collide, and make a particle/antiparticle pair. These can be moved around arbitrarily with zero net work, since any gravitational force on the particle will be equal and opposite to the force on the antiparticle.
2) raise them to an arbitrary height. Let them collide, producing a massive particle that is its own antiparticle, say a Higgs.
3) let this particle fall. It gains energy
4) let the particle decay upon falling. it now has the rest energy of the particle/antiparticle pair (equal to the energy of the two photons), plus any gravitational potential from falling.
5) reflect the two photons that are the decay product on antipodal mirrors attached by a wire and suspended from the ceiling. No net momentum or energy is transferred to this mirror system, and the photons can be freely merged again to repeat the process.
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Glycerol: refractive index & absorption spectra in 0.2-0.4um range Could anyone suggest where can I find absorption spectra & refractive index of Glycerol?
I am specifically interested in UV range, 200-400nm, everything I was able to find out was for standard conditions only...
Is there any software which can get these parameters through simulation?
| 1st, UV-spectrum.
There is none! To have some absorption above 200 nm You need some
conjugated double bonds. The standard solvent for UV-spectroscopy is
ethanol for that reasons, if one needed a much more polar solvent
one would use water or in extreme cases (of polarity) glycerol.
2nd, Refractive index.
This is a real problem, I do not have this values. Maybe
You can use this:
http://en.wikipedia.org/wiki/Cauchy%27s_equation
to extrapolate values from visible domain to the UV.
Here:
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UV-Vis/spectrum.htm
is some link which might help You. Just google for:
" UV/Vis spectroscopy " or " UV/Vis Spectroskopie"
the latter gives a lot of scripts from German universities,
the tables included and the formulas You should understand
without speaking German.
| {
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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?
| Indeed, when you just want to measure how much material has been adsorbed, it's more natural to use the same units for both materials.
I think that 1 cubic centimeter per gram is not a unit of "adsorption" per se - adsorption is a process, not a quantity, after all - but it's a unit of "specific pore volume". You take one gram of a material and measure the volume of the pores inside this material in cubic centimeters - and you get "specific pore volume" in cubic centimeters per gram. Note that pores - holes - are empty so they may only be measured by a unit of volume, and not a unit of mass or moles. ;-)
Gas adsorption is a major method to measure "specific pore volume". In some sense, the "specific pore volume" also does quantify adsorption because it tells you at how many microscopic places the gas may get adsorbed inside a material. See some papers about the two concepts to get a flavor:
http://scholar.google.com/scholar?q=%22specific+pore+volume%22+adsorption
| {
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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?
| In particle antiparticle annihilation the end products come from the table of elementary particles created so that quantum numbers are conserved.
photons, two to conserve momentum in the cmsystem, pions and other mesons etc.
| {
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Phase Accumulation of Hankel-waves upon propagation Hankel functions are solutions to the scalar Helmholtz-equation $$\Delta\psi + k_e^2\psi = 0$$ in cylindrical and spherical geometry (with respect to a separated angular dependence). Thus, they are very important describing spherical and cylindrical waves. Here is an example of such a propagation in the spherical case taken from Franz Zotter:
I am searching for a reference that states the phase accumulation of Hankel waves of the form $$F_H^{\mathrm{out/in}}(\mathbf{r}) = H_m^{1/2}(k\rho)\ .$$
Assumed is stationarity with an $e^{-\mathrm{i}\omega t}$ time dependence fixing the meaning of the two different Hankel-waves as outgoing/incoming.
For plane waves one finds that the accumulated phase of a wave in $x$-direction, $$F_p=e^{\mathrm{i}kx}$$ is simply related to its argument, $$\phi_{\mathrm{acc}}(x_1,x_2)=\mathrm{Arg}(F_p(x_2))-\mathrm{Arg}(F_p(x_1)) = k(x_2 - x_1)$$ and it is natural to just use this formula in the Hankel-case, e.g. $$\phi_{\mathrm{acc}}(\rho_1,\rho_2)=\mathrm{Arg}(F_H^{\mathrm{out/in}}(\rho_2))-\mathrm{Arg}(F_H^{\mathrm{out/in}}(\rho_1))$$
However, I was not able to find a suitable reference. Hence my question:
Is there a reference defining the phase accumulation of Hankel waves?
Thank you in advance.
| I am not sure how you get outwardly travelling waves from Hanken functions alone without also using Bessel functions. My understanding is that by analogy with plane waves, you have the stading wave solutions eg.
sin(kx)*cos(wt) and
cos(kx)*sin(wt)
and the travelling wave comes from the sum of these two. Similarly in cylindrical geometry you have the standing waves
Bess(kr)*cos(wt) and
Hank(kr)*sin(wt)
and the travelling waves shown in the applets are really the sum of these. I'm not sure if this point is relative to the question being asked, but this is how I understand the pictures.
| {
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Assuming everything else is equal, which will stop first: a heavier car or a lighter car? If the friction from brakes, wind resistance and all such factors remain constant, which will stop first? A heavier car or a lighter car? How will the momentum of the car and gravitational pull on a heavier object influence the stopping of the car?
| dmckee has answered what I believe to be the intent of the question; but Joe has correctly answered the absolute letter of the question.
Level ground seems to be implicitly assumed.
If all other things are EQUAL (e.g. coefficients of friction), except where they need to be PROPORTIONAL to be equal (e.g. wind resistance), then the cars will stop in the same distance.
If the cars both ride the the same coefficients of friction and have the same wind resistance (assumed to be a retarding force, not a tail wind) the lighter car will stop first.
| {
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Are all superalgebras Clifford algebras? I believe the answer to be yes, but I realize that sometimes physicists place additional constraints that might not be obvious. If superalgebras are Clifford algebras, why make a literary distinction?
| The answer is No, as Lubos Motl has already pointed out. Here I would like to make a couple of general remarks.
*
*On one hand, the notion of superalgebras is a huge topic, which includes, e.g., associative superalgebras and Lie superalgebras. Important examples of Lie superalgebras are super-Poincaré algebras.
*On the other hand, a Clifford algebra $\mathrm{Cl}(V,g)$ over a vector space $V$ satisfies
$$vw+wv=2g(v,w){\bf 1}, \qquad v,w\in V.$$
In physics applications, the vector space $V$ is often a vector space spanned by a basis of gamma matrices,
$$\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}{\bf 1}. $$
In more mathematical applications, the vector space $V$ sometimes comes with an odd grading, so that the anticommutator is a supercommutator
$$[v,w]=2g(v,w){\bf 1},$$
which can be viewed as a super Heisenberg algebra with odd grading, and which is hence an example of a superalgebra.
More generally, the vector space $V$ could be a super vector space $V=V_0\oplus V_1$ with both an even and an odd sector, leading to a notion of super Clifford algebras.
| {
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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?
| CERN Lectures
--
Lecturer: Herman Nicolai
Title: Quantum field theory
Year: 1997
Abstract: The lectures are intended to provide an introduction to Quantum Field Theory at an elementary level. In particular, the following topics will be treated: 1, Basic principle of QFT; 2, Representation theory of the Poincaré group; 3, The free scalar field; 4, The free Dirac field 5, Interacting field theories; 6, Supersymmetry; 7, S-Matrix. The emphasis will be more on general concepts than on specific applications. Therefore, for most of the lectures, only a knowledge of quantum mechanics and special relativity will be assumed.
Link: http://cds.cern.ch/record/338008/
--
Lecturer: Erik Verlinde
Title: Introduction to quantum field theory
Year: 1998
Abstract: In these lectures the basic concepts of quantum field theory are introduced. Topics that will be discussed are: the quantization of fields; Scattering amplitudes and cross sections; Feynman diagrams; Lagrangian field theory; Quantum electrodynamics; Gauge invariance and renormalization.
Link: http://cds.cern.ch/record/387599?ln=en
| {
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Simple Quantum Mechanics question about the Free particle, (part2) Continuing from my first question titled, Simple Quantum Mechanics question about the Free particle, (part1)
Griffiths goes on and says,
"A fixed point on the waveform corresponds to a fixed value of the argument and hence to x and t such that,"
x $\pm$ vt = constant, or x = $\mp$ vt + constant
What does this mean? I am so confused.
He goes on and says that psi(x,t) might as well be the following:
$$\psi(x,t) = A e^{i(kx-(hk^2/2m)t}$$
because the original equation,
$$\psi(x,t) = A e^{ik(x-(\hbar k/2m)t)} + B e^{-ik(x + (\hbar k/2m)t)}$$
only differs by the sign in front of k. He then says let k run negative to cover the case of waves traveling to the left, $k = \pm \sqrt{(2mE)}/\hbar$.
Then after trying to normalize psi(x,t), you find out you can't do it! He then says the following,
"A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy."
How did he come to this logical deduction. I don't follow. Can someone please explain Griffith's statement to me?
| Quantum mechanics states that the wave function must be square integrable. So, even although the plane wave is a solution of the stationnary Schrödinger equation, its not a physically accepted wave function. Now the wave packets or linear combinations of wave planes are physically acceptable solution but are NOT a solution of the stationary time independent Schrödinger's equation.So, your professor is right No free particle can exist in a stationnary state. Neverless, the time dependent Schrödinger equation can describe the evolution of the wave packet starting from an initial wave packet.
| {
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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
| Dear HDE, it's not hard to estimate the gravitational potential at the Earth's center. Of course, it's smooth. Let me assume that the Earth's mass density is uniform which is an OK estimate - up to factors of two or so.
The gravitational acceleration at distance $R$ from the center is $GM/R^2$ if $R$ is greater than the Earth's radius $R_E$. However, for smaller values of $R$, you have to use Gauss' law
$$\int d\vec S\cdot \vec g \sim GM_{inside}$$
and determine the total mass inside a smaller sphere. Because $M_{inside}$ goes like $R^3$ for $R<R_E$, and this $R^3$ is still divided by $R^2$ from $\int d\vec S$, it follows that the gravitational acceleration inside the Earth is pretty much proportional to $R$:
$$ g(R) = g(R_E)\cdot \frac{R}{R_E} $$
In particular, the gravitational acceleration at the Earth's center is zero and near the center, a particle would oscillate like in a harmonic oscillator, $\vec F\sim -k\vec x$.
It's also trivial to calculate the extra decrease of the gravitational potential you get if you go from the surface to the center. On the surface, the gravitational potential is $-GM/R_E$, as you know, because the derivative of $-GM/R$ over $R$ gives the right acceleration. However, the potential is getting even more negative. If you integrate $g(R_E)\cdot R/R_E$ over $R$ from $0$ to $R_E$, you will get $g(R_E) R_E/2$. This has to be taken with the negative sign.
So the potential at the center, assuming uniformity, is
$$ \Phi = -\frac{GM}{R_E} - g(R_E) \frac{R_E}{2} = -\frac 32 \frac{GM}{R_E} = -\frac 32 g(R_E) R_E $$
This gravitational potential determines the slowing of time, too. In SI units, $g(R_E)=10$ Newtons per meter and $R_E=6,378,000$. The product, with the $3/2$ factor added, is almost exactly $10^{8}$. Divide it by $c^2=10^{17}$ to get about $10^{-9}$ - the relative red shift from the center of the Earth to infinity.
If you spend 1 billion years at the center of the Earth, your twin brother outside the gravitational field will get 1 billion and one years older. If you wish, you may interpret it by saying that it's healthy to live at the center of the Earth. Good luck.
| {
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A question on a system of particles governed by laws of gravity and electromagnetic field Consider a system of many point particles each having a certain mass and electric charge and certain initial velocity. This system is completely governed by the laws of gravitation and electromagnetic field. If this system is left on its own without any external influence, any physical quantity like velocity of a particle or electric field at a point, of this system will vary smoothly with time and space. (keeping aside any singularities that could occur eventually). My question is whether any such system which is governed by such laws of physics (except ofcourse QM) would remain smooth until the occurrence of a singularity, in the absence of any external influence.
PS : please feel free to edit/suggest changes to the question in case if it could make it more meaningful.
| You can always create discontinuous waves (which are solutions that exists even in the total absence of sources (be them gravitational or electromagnetic)
You can always write a discontinuous function and use it to define, say, the electric field at some time $t_1$, and ask what do the fields look at time $t_0 < t_1$. Of course such solutions are a bit ad hoc in the sense that they are simply a wave solution with no actual sources
You certainly can generate fields that approach a discontinuous one by adding enough particles that you can approximate them by discontinuous functions $\rho$ for densities and currents, but of course the discontinuity is only an approximation
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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.
| The wave function is not an observable, so you can characterize it however you want.
However, any alternative explanation you want to put forth must agree with experimental reality, which in the end will mean it must be mathematically equivalent to the usual approach and will accordingly have some object that is isomorphic to the wave function.
That makes it real enough for my purposes.
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Is there a "Size" Cutoff to Quantum Behaviour? We all know that subatomic particles exhibit quantum behavior. I was wondering if there's a cutoff in size where we stop exhibiting such behavior.
From what I have read, it seems to me that we still see quantum effects up to the nanometer level.
| The classic experiment demonstrating quantum effects, the 2 slit experiment, has been preformed with subsequently larger and larger particles as our technology available to do it advanced. Originally, it was performed with electrons, which are just as much matter as any other matter, but are extremely small. The largest particle it has been demonstrated with are Buckminsterfullerene, which contain 60 Carbon atoms. For size comparison:
$$m_e = 5.485 x 10^{-4}u$$
$$m_{buckyball} = 720.642 u$$
There is a good reason that the experiment gets more difficult with increasing mass, and to be sure, the buckball experiment was quite an accomplishment. To the basics of quantum mechanics:
$$\Delta x\, \Delta p \ge \frac{\hbar}{2}$$
Alternatively, the de Broglie wavelength is:
$$\lambda = \frac{h}{p} = \frac{h}{m v}$$
I believe that in order to obtain the same wavelength with the same mass you have to decrease the velocity. The reason this could be problematic for such experiments is that it is hard to successfully create the conditions needed with a large and slower moving particle, such as needing a better vacuum.
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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.
| (I have a suggestion to make this question a CW.)
General Physics: (Early undergrad and advanced high school)
*
*Problems in Physics I.E Irodov - (Highly recommended)
*Problems in Physics S S Krotov - (Once again, highly recommended but out of print)
*Physics Olympiad Books - (Haven't read but saw some olympiad problems back in the day)
*Physics by example (like this book a lot, lower undergrad)
*Feynman's Tips on Physics (Exercises to accompany the famous lectures )
General Qualifying exam books:
The following books are a part of a series dedicated to the qualifying exams in American Universities and has a large compilation of problems of all levels. Others in the series include Mechanics, Electromagnetism, Quantum Mechanics, Thermodynamics, Optics and Solid State Physics. Unlike other compilation of exercises for qualifiers (such as Princeton or Chicago Problems, or the one mentioned below), they make no excuse for economy and include as many problems from all levels for each subtopic.
Another good book that I read recently for my exam is the two volume series: A Guide to Physics Problems (Part 2 has some relatively easy but interesting problems. I haven't gone through the first part, which is much much more challenging.)
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Understanding the cause of sidebands in Amplitude Modulation I've read it many places that Amplitude Modulation produces sidebands in the frequency domain. But as best as I can imagine it, modulating the amplitude of a fixed-frequency carrier wave just makes that "louder" or "quieter", not higher-frequency or lower-frequency. That is, I believe I could sketch, on graph paper, a path of a wave function that touches a peak or a trough exactly every 1/f increments, regardless of the "volume". Why do the sidebands appear?
| Let your carrier signal be $A_0 \cdot \cos(\omega_c t)$ with amplitude $A_0$ and carrier frequency $\omega_c$.
Let your signal be a simple wave, $\phi(t) = A_s \cdot \cos(\omega_s t)$.
Then the modulated signal becomes
$$A_0 A_s \cdot \cos(\omega_c t) \cdot \cos(\omega_s t)$$.
In addition, as pointed out by George in the comments, the carrier also gets transmitted.
Using the trigonometric identity $\cos(u) \cdot \cos(v) = \frac{1}{2} [\cos(u-v) + \cos(u+v)]$, you get the final signal:
$$\frac{1}{2} A_0 A_s \cdot ( \cos((\omega_c - \omega_s) t) + \cos((\omega_c + \omega_s) t)) + A_0 \cdot \cos(\omega_c t)$$
Hence, the frequency becomes changed, you get the carrier frequency in the middle (at $\omega$) and two side-bands at $\omega_c \pm \omega_s$.
Now, in reality your signal is not a simple cosine, but you could do a Fourier decomposition of the signal and treat each frequency independently. The two frequencies then get smeared out and you get the two sidebands.
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What are the forces of constraint if there are multiple equivalent constraints? Suppose a large (rigid) block is sitting on top of two smaller blocks of equal height $1$, both of which rest on the ground. We wish to find the position of the block (easy) and the forces of constraint on the block coming from the two smaller blocks.
If we write the Lagrangian along with the equations of constraint $z=1$, we can find the motion of the object using Lagrange multipliers, and we can also find the force of constraint. If there were constraints $\{G_n\}$, then the force of constraint in the $z$ direction would be $F_{n} = \lambda_n \frac{\partial G_n}{\partial z}$ on the constraint $G_n$. However, this tells us nothing about how much force each individual block is exerting.
We know from experience that the upward force from the smaller blocks must equal $mg$, where $m$ is the mass of the larger block, and furthermore, that the force is evenly distributed, so that each of the two smaller blocks exerts an equal amount of force.
It seems natural to indicate the existence of two blocks by writing down the constraint $z=1$ twice and then trying to solve the resulting system of equations. However, this doesn't seem to get us very far.
So, how is it possible to detect the fact that the force on the two smaller blocks is distributed equally?
| You can still do this with Lagrangians, even if it's the hard way around. I define $z_1$ and $z_2$ as the height of the boxes above the table, and the constraints are $f_1 = z_1-\frac{h}{2}=0$ and $f_2 = z_2-z_1-\frac{h}{2}=0$, or "the bottom box sits on the table and the top box sits on the bottom box." I assumed that the masses are equal, but you don't have to. Write the Lagrangian without assuming anything about the motion:
$$L = T-V = \frac{m}{2}(\dot{z_1}^2 + \dot{z_2}^2) - mg(z_1+z_2)$$
From the Lagrange multiplier method:
$$m\ddot{z_1}+mg + \lambda_1 \cdot 1 + \lambda_2 \cdot-1 = 0$$
$$m\ddot{z_2}+mg + \lambda_2 \cdot 1 = 0$$
These lead (after applying the condition that $\ddot{z_1}=\ddot{z_1}=0$) to $F_1 = \lambda_1\frac{\partial f_1}{\partial z_1}=\lambda_1\cdot1=-2mg$ and $F_2 = \lambda_2\frac{\partial f_2}{\partial z_2}=\lambda_2\cdot1=-mg$
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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?
|
so only a tiny fraction of beam energy
is reflected back to the device.
This tiny fraction is enough. With respect to ambient light:
One can modulate the laser beam, and filter the the voltage
of the receiving photodiode for this modulation frequency and phase.
Another precaution is to have a light filter in front of the
receiving photodiode which only lets the wavelength of the laser pass.
I think both precautions are used.
And of course the receiving photodiode is focussed to a spot of some
centimeters diameter around the laser spot.
Try to point the range finder to a mirror, in that case the
range finder should fail, exept the mirror happens to reflect
precisely back to the range finder (which is rather unlikely).
Reason is that from a (clean) mirror You don't get a stray reflection.
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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?
| Heat consists of random vibrations in a material. If a magnetic field connects two objects, then it creates a mechanical coupling between the two objects. Such a coupling will couple vibrations, therefore heat will be conducted by the magnetic field.
In practice, the effect will be very small, but given enough time, heat will be conducted by the magnetic field. The same applies to suspension by a combination electric and magnetic field.
This is not entirely incompatible with some of the other (wrong) answers. For example, since the suspended object is not at absolute zero, it is not possible for the magnetic field suspending it to be entirely static.
As an example, let's consider a superconducting metal container of gas suspended in a magnetic field:
Now suppose that the gas is at some non-zero temperature. Then there is a non-zero probability that the gas will concentrate onto one side:
In order for this to happen, assuming that the center of mass of the container remained constant, the metal part of the container had to have shifted in the opposite direction:
Assuming that the levitation is stable, a motion of the superconducting container must be opposed by the magnetic field. Thus the above motion is opposed by a force from the magnetic field. However, for every action there is an opposite and equal reaction (Newton's 3rd law), therefore there will be a corresponding opposing force on the suspension:
Therefore thermal motion in the suspended container will induce thermal motion in the base, hence there will be conduction of heat.
| {
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Why are smaller animals stronger than larger ones, when considered relative to their body weight? I am interested in why many small animals such as ants can lift many times their own weight, yet we don't see any large animals capable of such a feat.
It has been suggested to me that this is due to physics, but I am not even sure what to search for. Could someone explain why indeed it is easy for smaller objects/lifeforms to support several times their own weight, but this is harder as objects/animals become larger?
| To answer this question you just need to know that the force scale like the transection surface of a muscle. In other words the bigger the muscles the stronger. Therefore you have
$F ∝ σ_s S$,
where $σ_s$ is the maximum developed muscle stress. It so happens that on earth the muscles work pretty much the same in all the animals (from the ants all the way up to the elephant) so it is pretty universal (it can still vary from one species to another but not by one order of magnitude). For mammals it is around 10 N/cm$^2$.
Then the general allometric law (or spherical cow argument) gives :
$l∝ M^{1/3}$,
$S ∝ l^2 ∝ M^{2/3}$,
where M is the animal weight.
Ok now let's consider the problem of lifting up a weight of mass $M_0$. To do so you have to lift up your own weight plus this mass using your muscles :
$F \sim (M+M_0)g $
$\dfrac{σ_s}{ρ^{2/3}}M^{2/3} \sim (M+M_0)g$
$\dfrac{M_0}{M} \sim a M^{-1/3} -1$
You thus see that the mass that an animal can lift up compare to its weight scale like $M^{-1/3}$. This leads to the impressive strength of ants.
Interestingly enough, you can also use this in a single species to compare the stronger and the heaviest :
$M_0 \sim a_h M^{2/3}(1-\dfrac{M{1/3}}{a_h})$
This function as a maximum ($\dfrac{dM_0}{dM}(M_{strongest}) = 0$) and 2 values for which M_0 = 0 (M=0, boring ; and $M_{heaviest} = a_h^3$. You will easily check that :
$M_{strongest} =\dfrac{8}{27}M_{heaviest}$.
Ok so what? well if you use it for human, we can estimate the value of $a_h$ using the weight of the heaviest man ($M_{heaviest} = 635 kg, Jon Brower Minnoch). Therefore the strongest man should be around 190kg. Well this is exactly the weight of the "strongest man" the Iranian Hossein Reza Zadeh...
Best,
Rémi
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Experimental evidence showing the kinetic energy of an electron changes in a static non-uniform magnetic field? In a previous question, Does a magnetic field do work on an intrinsic magnetic dipole?, one highly rated answer suggested that static magnetic fields do work on intrinsic magnetic dipoles in a non-uniform magnetic field. I can visualise the change in kinetic energy of the nucleus of an atom coming from a change in the configuration of the electrons around the nucleus. But for an electron, since it's truly fundamental, I'm scratching my head over where the energy comes from to change its kinetic energy. If it does, then it really must come from the static magnetic field.
So what is the experimental evidence that shows the kinetic energy of an electron changes in a static non-uniform magnetic field?
| The Stern Gerlach experiment is such an example (although we have an extra complication because the wave function splits)
1) There is a change in the EM Field energy $\tfrac12 B$ because the total magnetic field (static field + electron's own magnetic field) becomes less. A static field in isolation doesn't change per definition and its energy contents doesn't change either. To do work it needs to provide energy. The energy is provided by the total magnetic field and not the static field in isolation.
2) The equivalent view is that the work is provided by the internal energy -$\mu\cdot B^2$. This is also true, both views are equivalent, even if the energies actually come from different parts of the Lagrangian since ultimately -$\mu\cdot B$ comes from the interaction term $j_\mu A^\mu$.
This equivalence also hold classically even though the math for dipoles is somewhat involved.
Regards, Hans
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What causes the Pauli exclusion principle (and why does spin 1/2 = fermion)? It seems to be related to exchange interaction, but I can't penetrate the Wikipedia article. What has the Pauli exclusion principle to do with indistinguishability?
| Indistinguishableness of particles is formulated in QM in terms of the total wave function symmetry. Then the wave functions can be symmetric or antisymmetric on their arguments. In case of antisymmetry ($\psi(x_1,x_2) = -\psi(x_2,x_1))$ the particles are called fermions and $\psi(x_1,x_2=x_1) = 0$. They say "the particles cannot occupy the same state". It's a feature of nature, an experimental fact.
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Calculating diffraction-limited resolution for a lens setup Supposed a lens arrangement is prepared where light from an object is collimated, focused and recollimated etc. before entering a CCD array. Given that we can calculate the diffraction-limited resolution for each lens in the system, how do we measure the diffraction limited resolution for the whole setup?
| (Edited based on your comments)
I want to briefly clarify what exactly is meant when we talk about being "diffraction limited." As light is focused, it will reach some minimum spot size before it begins to expand again. The size of this spot depends on how much the light beam is distorted.
A perfectly collimated beam (with perfectly planar wavefront) passing through a perfect lens would come out of the lens with perfectly spherical wavefront, and all of the rays in the beam would be converging to a single point. In this case, the spot size is determined solely by the angle occupied by the converging cone of light*. This is what we call "diffraction limited."
If the beam is abberated, for example by a poorly manufactured lens, then the beam will not have perfectly spherical converging wavefronts, and the resulting focal spot will be spread out over a larger area. The magnitude of these abberations is what determines the resolution of an optical system when it is not diffraction limited.
The size of the diffraction limited spot is a function of the f-number at the image plane. So, if you know the beam diameter after the last lens element, and the back focal distance, you can compute the diffraction limited spot size just like you would for any other lens.
*this is the case only assuming that the beam is always the same shape. In practice most beams are circular, so all we need to worry about is its diameter. If the beam is a different shape, then its diffraction limited spot size (and shape!) will change.
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Thought experiment that seems to involve something growing at twice the speed of light. Is anything wrong? Let foo be some unit of distance and bar be some unit of time which have been chosen so that the speed of light c = 1 foo/bar. Position several observers along a line each separated by one foo, and place light sources some distance apart amongst the observers, initially off. All observers and light sources are stationary with respect to one another.
So in my diagram the tick marks are the observers, and the scale is one foo. For the example, the light sources are ten foo apart at points L and R.
Now at time 0 bars, ignite both light sources.
In the diagram, at time 4 bars, A has observed source L but not R, C has observed R but not L, and B has not yet observed either light source. At time 5 bars, B observes both light sources, and at time 6 bars all three labeled observers have observed both light sources. At time 7 bar, a total of five observes (A, B, C, and two unlabeled) have observed both sources.
Let r be the length of the line segment containing all observers that have observed both light sources. In other words, r is the length of the intersection of the event horizons on the two ignition events (I think I am using that term right, if not, please let me know). It seems that dr/dt = 2 foo/bar; for instance, in the 1 bar interval from 6 to 7, the change in r was 2 foo. As c = 1 foo/ bar, r is growing at twice the speed of light.
Since r is not a particle and cannot be used to transmit information amongst the observers, is there any contradiction here? Have I made any logical falicies?
| Lagerbaer gives an excellent detailed explanation, but to put it more simply: there is no contradiction, no logical fallacy, the distance $r(t)$ as you have defined it does indeed grow at a rate of $2c$ as measured by any of the observers A, B, C, etc. But $r$ does not correspond to a real thing. It does not measure the position of a particle or an information signal relative to that of an observer, so there's no reason it can't grow faster than light.
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What do we consider "Perpetual Motion" I know this is a bad question to most serious Physics but I have a question about what is considered “Perpetual motion.” The Foucault pendulum in the UN consists of sphere that passes directly over a raised metal ring at the centre that contains an electromagnet, which induces a current in the copper inside the ball. This supplies the necessary energy to overcome friction and air resistance and keeps it swinging uniformly. Now the swing of the pendulum is induced but the 36h 45m clockwise shift generated bay the earths rotation is perpetual or as long as the earth rotates. Is this assumption correct?
Does a generator that works on tidal movements not fall in to the same assumption?
| From the webster.com dictionary , perpetual:
a : continuing forever : everlasting
In our universe as far as we have learned, nothing continues forever. So the definition of perpetual motion is a relative definition in time.
For a human lifetime the earth will be continuously turning on its axis,
as well as with the planets around the sun.
An atom will have its electrons in continuous "motion" around the nucleus,( until the expansion of the universe destroys everything,) etc
In that sense, a generator absorbing the energy of the tides is taking advantage of this, limited in definition, perpetual motion. In detail the energy turned into electricity is taken infinitesimally out of the angular momentum of the earth, so the perpetual motion of the earth is delayed by a tiny bit.
The perpetual motion machines on the other hand that crank inventors try to demonstrate, would get more energy out than the one coming in, thus breaking the law of conservation of energy.
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Why are WW gg ττ branching ratios so similar for a 115 GeV SM Higgs? In a previous question on Higgs branching ratios, I find this image
(originally from page 15 here).
I am VERY intrigued by the fact that decays to WW, gg, and ττ are almost equally probable, for a standard model Higgs with a mass in the vicinity of 115 GeV. I've noted previously that this is a special value: it lies in a narrow range of values for which the SM vacuum is metastable. Is there a connection?
| In this graph, you have clearly found the triangle composed of three curves which is the smallest one. The triangle made of $gg$, $ZZ$, $\tau\tau$ near $m_H=130$ GeV is comparably small but larger.
Still, none of the triangles is infinitesimal. Even though your triangle is the winner among the small ones, it doesn't show an intersection of three lines. So it's not exact. This property of "three lines intersecting at a single point" is fully analogous to the discussion of "gauge coupling unification". In the Standard Model, it's not exact although the triangle is similarly small and thin, just like yours.
In the MSSM, one modifies the spectrum so that the intersection of three lines, depicting the gauge couplings for $U(1)$, $SU(2)$, and $SU(3)$, is exact within the error margins. It also has a good reason: the groups become subgroups of a larger unified group such as $SU(5)$ which only has one coupling so the smaller groups' couplings have to agree at the GUT scale. In your case, no such improvement is known. At any rate, your lucky hunch is a coincidence, the triangle is not "anomalously small". Its size is not far from what you expect from the smallest triangle in a similarly large chaotic graph with intersecting curves.
So it's a coincidence. If you wanted to use this "near perfect intersection" to argue that 115 GeV is special, the LHC has proved that your intuition was wrong because the Higgs mass is known to be 125 GeV today.
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Length of a curve in D dimensional euclidean space In a book I am reading on special relativity, the infinitesimal line element is defined as $dl^2=\delta_{ij}dx^idx^j$ (Einstein summation convention) where $\delta_{ij}$ is the euclidean metric. Next, if we have some curve C between two points $P_1$ and $P_2$ in this space then the length of the curve is given as $\Delta L = \int_{P_1}^{P_2}dl$
I am having trouble deriving the next statement, which I quote:
A curve in D-dimensional Euclidean
space can be described as a subspace
of the D-dimensional spce where the D
co-ordinates $x^i$ are given by single
valued functions of some parameter
$t$, in which case the length of the
curve from $P_1=x(t_1)$ to
$P_2=x(t_2)$ can be written
$$\Delta L = \int_{t_1}^{t_2}\sqrt{\delta_{ij}
\dot{x}^i \dot{x}^j} dt \qquad
\mbox{where}\; \dot{x}^i\equiv
\frac{dx^i}{dt}$$
| I think you should take that as the definition of the word "length". I wouldn't try to derive it at all.
It is basically saying that if, for example, you want to know the length of the unit circle in the first quadrant, set
$$x^1 = \cos t$$
$$x^2 = \sin t$$
$$\dot{x}^1 = -\sin t$$
$$\dot{x}^2 = \cos t$$
and do
$$\int_0^{\pi/2} \sqrt{(-\sin t)^2 + (\cos t)^2}dt = \pi/2$$
| {
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Simple Harmonic Motion - What are the units for $\omega_0$? I'm trying to understand the units in:
$$mx''+kx=0$$
And the general solution is $$x(t)=A \cos(\omega_0 t)+B \sin(\omega_0 t).$$
Let $\omega_0 =\sqrt{\frac{k}{m}}$ - the unit for the spring constant $k$ is $kgms^{-2}$
or $Nm^{-1}$, where $m$ is in $kg$, so that the units of $\omega_0$ seem to be "per second" (i.e) $1/s$.
But, later we put $\omega_0$ in to the $cos$ and $sin$ functions which will return dimensionless ratios. So, The constants $A,B$ must be in $m$, since $x$ is in $m$.
What I don't understand is why my book says $\omega_0$ has the unit $rad/s$, I get that the input for cosine is $rad$ or some other angle measure, but where did the radians come from?
My analysis of units only proved $1/s$ as the actual units..!
I have just been informed that radians are dimensionless. So, that answers part of this question, yet I still don't know why we can't say that dimensionless one in degrees or rotations..? How do I know what kind of cosine and sine table to use with this dimensionless number?
| I find it helpful to use rad in units to understand these mechanical properties:
The unit for spring constant k for a torque spring (clockwork spring) is N.m, but call it (N.m)/rad and it is easier to understand: "Torque per angle".
Also N.m is "formally" the same as J, and this can make it hard to distinguish from torque, especially when the work is associated with a rotational movement. It helps then to think of Joule as N.m.rad, meaning "torque times angle turned"
analog to the way work from linear motion is "Force times distance traveled".
I would never call Joule anything else than J in documentation, it just helps to think about it that way.
Just a thought, rad is of course "1" and can be left out at will.
I do not see how leaving out rad would lead to confusion with degrees. In engineering, degrees is used as a unit all the time, but never without the little degree sign °
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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?
| When I was a schoolboy, our teacher of physics asked once one of our brilliant students (a girl), something like: "What is the nature of gravity?". She thought for a moment and answered: "I do not know. And what is it? Our teacher replied: "If I had known..."
As long as gravity is not derived from other nature features, it should be considered as a fundamental law. We may study its properties but we may not ask such questions - just by definition of a fundamental law. It is as an axiom - it is given as such.
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How is thermodynamic entropy defined? What is its relationship to information entropy? I read that thermodynamic entropy is a measure of the number of microenergy states. What is the derivation for $S=k\log N$, where $k$ is Boltzmann constant, $N$ number of microenergy states.
How is the logarithmic measure justified?
Does thermodynamic entropy have anything to do with information entropy (defined by Shannon) used in information theory?
| The reason for the logarithm in the statistical definition of entropy is that it makes the entropy additive and the number of microstates multiplicative. This means that if you have subsystems A & B, total number of microstates is the product of the subsystem’s number of microstates, and the entropy is additive over subsystems due to the properties of the logarithm. A way to see the relation of thermodynamic entropy and statistical entropy is to picture the volume occupied by a system of classical particles as partitioned in cells of very small volume. Then you start counting the number of ways to distribute N particles in the cells. The number of microstates goes like the numbers of cells to the power of the number of particles. The number of cells is the volume divided by the number of partitions. Taking the logarithm of this expression you get number of particles multiplied by the logarithm of the volume over the number of partitions. Multiplying by k you get an expression for the entropy. Then the change of entropy, for a process where volume changes, is obtained by subtracting the final entropy from the initial entropy. The number of partitions is cancelled from the equation by the properties of logarithms. So this is another reason for the logarithm in the expression. What you get finally is the Boltzmann constant multiplied by the number of particles and the logarithm of final volume over initial volume. The way this result is obtained makes clear that the change in entropy is path independent. The same expression for the change in entropy can be obtained, using thermodynamics, for the reversible isothermal expansion of an ideal gas, by dividing the heat absorbed over the temperature. If a process is not isothermal then you can break it in small nearly isothermal infinitesimal paths, then integrate the change in entropy. This is the easiest way to answer your question. Similar arguments can be used for quantum systems but they are more complex.
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What are the properties of two bodies for their collision to be elastic? For example, must the shock wave in each body be of a particular form which influences the shape and material properties of the bodies?
I suspect part of the the answer is that the objects must be spherical, and the round-trip of the shock wave in each body must be the same, but this is just a guess.
| Extended reply to Anonymous Coward:
Perfect, I'll argue that gaining rotation in a collision doesn't make it inelastic. The kinetic energy is still there, it's just in the form of rotational kinetic energy. But yes, that is strictly semantics. You pretty much covered the question.
Generally I would claim that no macro objects can collide fully elastically but elementary particles can. Of course, elementary particles only do so in the absence of an interaction and the "collision" is though a force like Coulomb repulsion. Molecules, atoms, and other things can blur the line by having a meaningful chance of colliding fully elastically as well as a meaningful chance of some energy conversion taking place with kinetic energy. Like nuclear cross sections, it is possible to provide numeric quantification of the probabilities of different interaction types.
For fun consider an equilibrium gas. Collisions should be on average net-zero kinetic energy $\Delta$ but it's possible that collisions frequently (almost always even) convert binding energy or some other internal stored energy into kinetic energy. I imagine this would be the case for gases with a large molecule size, it could even be of relevance to a chemist.
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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.
| Unless the cyclist is leaning forward most of the weight is on the back tire. Therefore it has more contact with the road and when you turn there is more lateral force on the back. You do more braking whith the back so you don’t go head first. And finally you are applying your entire forward force on it so that contributes to more wear.
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how does a helicopter get forward thrust? Just passed a helicopter on my way to work. We have read in some detail how an airplane gets forward thrust and lift by deflecting air. How does a helicopter with horizontal fans achieve that ?
| In general, the forward thruswt is achieved by tilting the entire helicopter forward. This converts some of the lift produced by the main rotor into a forward component of force. The same is true for turns.
Yes, the swashplate may be used to create a thrust imbalance; but this imbalance does not provide any horizontal forces. It merely creates a force imbalance which tips the helo slightly, so that a component of the upward force of the main rotor gets converted into a lateral force.
Take a look at any helo moving with any kind of rapid forward velocity and you will see that the entire craft is tipped noticably forward. Same with sharp turns: the helo tilts noticably in the direction of the turn.
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Batman spotlight in the sky I have noticed that obstructing a spotlight typically results in a blurred shadow unlike the crisp batman symbol in the comics of batman. Is there a way to create a spotlight with a crisp batman symbol?
| You can use a Laser. Just track it whit moving mirrors like the beam of a TV tube. There are computers that do an outline but its totally possible to track horizontal lines that appear and disrepair at the proper moments and the smoke from fire woks can make a perfect screen. O ya and a BIG laser.
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Non-Linear Density Shell Problem I'm trying to understand Newton's Shell Theorem (Third)
http://en.wikipedia.org/wiki/Shell_theorem
However this applies to a sphere of constant density. How is this formulated for sphere of varying density, e.g., a ball of gas bound together by gravity?
Actually, this requires another question: how does the density of ball of gas bound by gravity diminish with radius?
EDIT
as DMCKEE pointed out "You can't answer the problem ignoring thermodynamics", so I've removed that proviso.
| A really simple way of thinking about the shell theorem is that it applies to an infinitesimally thin spherical shell of constant surface density. Then, any spherically symmetric ball of stuff (even if it varies in density with distance from the center) can be built of these shells (with different densities for each shell).
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How does stuff glow in the dark? Many things have glow in the dark properties (glow sticks, paint, toys ..), and I am wondering what is the physics behind them. How do these materials store light energy and emit it later when dark?
What dictates the wavelength(s) of the glow? Is there a limit on how much energy can be stored and recovered in these materials? Do they give out heat as well as light?
| When one excites fluorescence with light (e.g. blue light), the incoming photon pushes an electron into a higher state. Generally one always excites vibrational states (movement of the molecules' atoms) as well. Therefore the emitted photon has lower energy (e.g. green light). The lost energy (internal conversion) heats the solvent or lattice. The shift in frequency (from blue to green) is called Stokes shift.
People plot these energy levels in Jablonski diagrams. I stole this from
http://biophys.med.unideb.hu/old/pharmacy/fluorescence_print.pdf
You can only excite as many electrons as there are in the material. This means that absorbing laser goggle are not safe for high power pulsed laser. The early photons of the pulse will saturate all the electrons in the goggles and the rest of the pulse can advance into the eyes and do damage.
In a glow stick the excitation is done by chemical energy. Hydrogen peroxide splits some organic molecule and one of the produced excited carbon dioxide molecules can transfer its energy to a fluorophore. I once saw an experiment where the chemicals from the glow stick were added to tea and the tea's chlorophyll started glowing in red.
The only reason I can think of right now, why anyone would want to store energy like this, is to be able to access it with the speed of light. In high power pulsed lasers special crystals are pumped into the higher state and the energy can suddenly be converted into photons. That way it is possible to create pulses with Petawatts of power.
| {
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Detection of the Electric Charge of a Black Hole By the "No Hair Theorem", three quantities "define" a black hole; Mass, Angular Momentum, and Charge. The first is easy enough to determine, look at the radius of the event horizon and you can use the Schwarzschild formula to compute the mass. Angular Momentum can be found using the cool little ergosphere Penrose "discovered". However, I don't know how to determine the charge of the black hole.
How can an electromagnetic field escape the event horizon of a Reissner-Nordström black hole? Is there any experiment we could theoretically do to a black hole to determine its charge?
| I will say that this question has a bit of a misconception--you don't need to interact with the horizon or extreme-limit GR effects like the ergosphere in order to measure the parameters of the black hole. The metric tensor is different for black holes with different masses and charges, and have different $\frac{1}{r^{n}}$ falloffs for different values of the parameters. In principle, you can measure $M$, $Q$ and $a$ solely based on observations of the orbital parameters of three different planets in orbit around the black hole.
| {
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How come an anti-reflective coating makes glass *more* transparent? The book I'm reading about optics says that an anti-reflective film applied on glass* makes the glass more transparent, because the air→film and film→glass reflected waves (originated from a paraxial incoming wave) interfere destructively with each other, resulting on virtually no reflected light; therefore the "extra" light that would normally get reflected, gets transmitted instead (to honor the principle of conservation of energy, I suppose?).
However, this answer states that "Superposition is the principle that the amplitudes due to two waves incident on the same point in space at the same time can be naively added together, but the waves do not affect each other."
So, how does this fit into this picture? If the reflected waves actually continue happily travelling back, where does the extra transmitted light come from?
* the film is described as (1) having an intermediate index of refraction between those of air and glass, so that both the air-film and film-glass reflections are "hard", i.e., produce a 180º inversion in the phase of the incoming wave, and (2) having a depth of 1/4 of the wavelength of the wave in the film, so that the film-glass reflection travels half its wavelength back and meets the air-film reflection in the opposite phase, thus cancelling it.
| The thickness of the AR coating is chosen such that the reflections from the two interfaces cancel out (at the wavelength for which the AR coating was designed):
See Anti-reflective coating in Wikipedia.
As endolith points out in the comments, to explain how the transmission is enhanced, you have to draw a few more rays in the diagram. Here's another illustration, from the Wikipedia article for Fabry–Pérot interferometer, which shows a few higher-order reflections:
For the anti-reflective coating, you choose the thickness such that R1 and R2 cancel while T1 and T2 constructively interfere. Note that this is dependent on the wavelength, the angle of incidence, and the index of refraction of whatever is being coated. With other thicknesses, you can make a high-reflectivity coating, or a coating of whatever reflectivity you want.
| {
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Why is it that when driving in a car, and a lightning bolt strikes, my AM radio gets cut off for a while, but FM stays on? I noticed this one day, a lightning/thunder occurred and my Fabulosa Spanish music died for a second. But not FM?
| For a more complete answer, one should also consider the basic difference between the two transmission schemes. AM - i.e Amplitude Modulation - encodes a signal by varying the amplitude of a radio wave of a single frequency. FM - i.e. Frequency Modulation - encodes a signal by varying the frequency itself within a narrow range about a central value. When lightning strikes, the electrical discharge generates a lot of radio noise at a set of frequencies; if one of those matches the AM frequency you are tuned to, the amplitudes add, messing up the whole signal. By contrast, since FM monitors variation in frequency, it is fairly insensitive to noise in any particular frequency in the monitored range. (that insensitivity is actually what motivated the invention of FM in the first place)
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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.
| If you are observing the inside of the sphere, you are absorbing light to make the observation. The light would get dimmer and dimmer very quickly until you could see nothing at all.
| {
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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.
| Because refractive index is a function of wavelenght, every lens experiences chromatic aberration, and so does a human eye.
This could have severe effect on human vision. However, eye has tuned spectral bandwidth - it is well known eye is most sensible to visible light which has wavelenght of 550 nm. Relative luminosity is also a function of wavelenght (in which more than 70% of the luminous energy is confined to a defocus range of less than 0.25 D defocus on either side of focus (if eye is optimally focused at 550 nm)). The exact spectrum to which we are sensitive, depends on light levels. Rods saturate at moderate and high light levels, so in this region spectral sensitivity is governed by cones. At lower light levels cones are no longer as sensitive and rods dominate the response. All this results in different spectral sensitivity for various luminosity and in this way eye is able to filter out some effects of chromatic aberration.
| {
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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?
| Free spin 3/2 fields cause no problems; see Weinberg's QFT book, Volume 1.
The problem with elementary spin 3/2 fields is the difficulty of accounting for the interaction with the electromagnetic field. The Rarita-Schwinger field equations with the standard minimal coupling via the covariant derivative violate causality, as they allow superluminal signalling - already on the single particle level.
Nonrenormalizability is another issue, but could be handled in the sense of effective field theories if the other defect were absent.
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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?
| Make sure that you have some sort of life skill when you're done that isn't research, too. Teach yourself to program, or work on an experiment where you build things. It will enhance your research, and make you more employable should reasearch not work out.
| {
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Does a slide in a projector act as a diffuser? I was wondering whether or not a slide in a slide projector acts as light diffuser? So when I have a light source that does not have a parallel beam on the slide can I expect that the other side of the slide is illuminated with a Lambertian of similar reflectance?
I am wondering this because I want to know if I have to raytrace the light path of a projector from the slide (assuming Lambertian reflectance of the slide) or from the light source.
| The slide will not act as a diffuser, but there will often be a diffusing element upstream near or adjacent to the slide, so treating it as a diffuse source is probably a good approximation.
| {
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Roughly how many atoms thick is the layer of graphite left by a pencil writing on paper? Actually I can't expand much as the question pretty much explains the query. I would be interested in the method of estimating an answer as well as a potential way to measure it experimentally. Thanks.
P.s. I'm not sure what tags to include for this one - for those that can, feel free to edit them as you see fit.
| We run an experiment on my A Level Physics course to answer this question.
*
*Expose the graphite in the pencil you wish to use at either end. Measure the length of the graphite, its diameter (then calculate its cross-sectional area) and the electrical resistance along its length (either by direct measurement using a multimeter or by passing a current through and using Ohm's law to calculate it)
*Calculate the resistivity of the graphite using resistivity = (Resistance x Cross section area) / length
*Scribble a thick pencil line on a piece of paper. Measure its length, its width and its electrical resistance.
*The cross sectional area of your pencil line will be its width x its thickness. So the resistivity equation can be rewritten as: Resistivity = (Resistance x Width x Thickness) / Length
*Rearrange for Thickness and substitute in your measurements to get a value. Divide this value by the diameter of a carbon atom to get a rough estimate.
| {
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Trying to understand the EPR paradox So I keep reading all these articles on the EPR paradox, and I follow them pretty easily right up until it gets to the most important matter.
Assuming you are trying to measure x and y spin,
Wikipedia and others say that when you measure x-spin on the first particle, it suddenly becomes impossible to measure the y-spin on the other particle.
But no one really goes on to say what this means in a physical sense.
Let's say you have 2 actual detectors. When the first particle hits the x-detector, now x-spin is measured for both particles. When the second particle hits the y-detector, now y-spin is measured for both particles. But all these articles say the second detector is unable to measure y-spin. So what happened? Did the detector just explode or something?!
| As reading the wikipedia article, we can clearly see a way to transfer information faster than the speed of light, therefore we know there is an error in the article.
Let's say entangled particle pair has identical x-spins. We measure y-spin of one particle. The x-spin of the measured particle becomes random, the x-spin of the other particle is not affected.
Bohr debunked Einstein's, Podolsky's and Rosen's claim that, by using entangled particles, it is possible to measure both momentum and position accurately, by saying that measuring position causes the laboratory to have somewhat uncertain momentum, which cause the momentum measurement in the laboratory to be uncertain.
So don't you think Bohr's way to solve the EPR-paradox is the right way to solve the EPR-paradox?
| {
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Physical Interpretation of a Scalar Quantity Related to Currents/Conservation Laws Let $Q_{ab} = (\psi_{;a})(\psi_{;b}) - (1/2)g_{ab}|\nabla \psi|^2$ be the energy-momentum tensor of the wave equation in some space time. I will use semicolons to refer to covariant differentiation and $\partial$s to refer to coordinate differentiation. Let $\pi_{ab}$ be the deformation tensor for some fixed vector field $X$. In deriving "almost conservation laws" one uses the identity.
$(Q_{ab}X^b)^{;a} = (\psi^{;a}_{;a})(X^a\partial_a\psi) + (1/2)Q^{ab}\pi_{ab}$
Does there exist a physical interpretation of the scalar $Q^{ab}\pi_{ab}$ that is not based on the above formula? I am most interested in how one should think about this quantity in relativistic contexts, e.g. black hole geometries.
P.S. Just in case anyone is tempted, I am looking for something more than "$Q^{ab}\pi_{ab}$ vanishes if the flows of $X$ are isometries."
| Here we will not give a physical interpretation per se, but just derive an alternative covariant expression for the sought-for scalar.
There are given a scalar field $\psi$ and a metric
$$ds^2~=~g_{ab} ~dx^a dx^b.$$
Let $\nabla$ be the corresponding Levi-Civita connection. The Lagrangian scalar function is:
$$ \Lambda~:=~ \frac{1}{2}\langle \nabla\psi , \nabla\psi \rangle ~=~ \frac{1}{2}\nabla_a\psi ~ g^{ab}~ \nabla_b\psi ~=~ \frac{1}{2}\partial_a\psi ~ g^{ab}~ \partial_b\psi . $$
The energy-momentum tensor reads:
$$ Q_{ab} ~:=~ \nabla_a \psi ~\nabla_b \psi - g_{ab}~\Lambda. $$
We also have a vector field $X$. Let us make a co-vector (=one form) by lowering the index:
$$ \eta_a = g_{ab}~ X^b. $$
The deformation tensor $\pi$ wrt. the vector field $X$ is defined as the Lie derivative of the metric:
$$\pi_{ab} ~:=~ ({\cal L}^{}_X g)_{ab}~=~ (\nabla_a \eta)_b + (\nabla_b \eta)_a, $$
$$\pi^{ab} ~:=~ g^{ac} ~\pi_{cd}~g^{db}~=~ -({\cal L}^{}_X g)^{ab}~=~ g^{ac}\partial_c X^b +g^{bc}\partial_c X^a - X^c\partial_c g^{ab} . $$
The sought-for scalar is:
$$S~:=~ \frac{1}{2}\pi^{ab} Q_{ab}. $$
Another vector field $Y$ is defined as:
$$ Y^a~:=~g^{ab}~ Q_{bc}~X^c~=~g^{ab}~\nabla_b \psi ~X[\psi] - \Lambda X^a, $$
$$ Y[f]~=~X[\psi]~\langle \nabla\psi , \nabla f \rangle - \Lambda X[f].$$
Then it is a bit tedious but straightforward to reproduce the formula in the question formulation (v3):
$$S~=~\mathrm{div}_g Y - \Box_g \psi ~ X[\psi], $$
and also to derive another covariant expression
$$S ~=~ \langle \nabla\psi , \nabla[ X[\psi]] \rangle - \mathrm{div}_g( \Lambda X).$$
| {
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Books that every layman should read To add to Books that every physicist should read:
A list of popular physics books for people who aren't necessarily interested in technical physics.
(see also Book recommendations)
| *
*'One Two Three ... Infinity' - By George Gamow
*'Mr. Tompkins in paperback' - By George Gamow ( combines Mr. Tompkins in Wonderland with Mr. Tompkins Explores the Atom )
*'Mathematics for the Million: How to Master the Magic of Numbers' - Lancelot Hogben
*'Relativity Simply Explained' - By Martin Gardner
| {
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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]
| It's not that the earth is expanding; instead the primary effect is that the earth's surface is shrinking. The effect occurs because the earth's surface doesn't remain flat. Instead, it gets tilted and folded. Some parts of the crust get subducted; once they disappear, what's left appears smaller.
Over any appreciable distance, rock has good compressive strength but negligible tensile strength. Consequently, when rock is pushed together it becomes thicker and taller (thereby decreasing the surface area; see the demonstration here: http://en.wikipedia.org/wiki/Mountain_building ). But when rock is pulled apart it instead tends to form cracks that are obviously new crust (and so are not counted when trying to determine if the earth's surface is expanding).
The overall effect is that the size of the older portion of the earth's surface is smaller than the actual current surface of the earth but this is only an effect of standard geology.
| {
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Which Friedmann equation is redundant? For flat FLRW cosmology, we can write down two Friedman equations and one matter equation. Namely,
\begin{align}
H^2 & =\frac{8 \pi G}{3} \rho, \tag{1} \\
\frac{\ddot{a}}{a} &= -\frac{4 \pi G}{3} (\rho +3p), \tag{2} \\
\dot{\rho} &= -3 H (\rho + p). \tag{3}
\end{align}
It is well known that, given an equation of state, $(1)+(3)$ implies $(2)$ [also, $(1)+(2)$ implies $(3)$].
My Question:
Why are $(2)$ and $(3)$ (plus equation of state) not a complete set of equations?
I'm sure it is probably completely obvious to some of you, but not to me.
| I suppose a lot of what constitutes the "Friedmann Equation(s)" is just up to definitions. However, with the 3 equations you listed, there will be redundancy.
Here is the normal derivation I usually see (feel free to skip the first two paragraphs if you aren't familiar with tensors/GR):
Given einstien's equation in the form $R_{\mu\nu}=-8 \pi G S_{\mu \nu}$, (where $S_{\mu \nu}$ is related to the stress energy tensor by $S_{\mu \nu}=T_{\mu \nu} - \frac{1}{2}g_{\mu \nu}T^\lambda_\lambda$) and with some playing around with the spatial portion of the Robertson Walker Metric (see Weinburg Cosmology), we can get a Riemann Tensor $R_{ij}=\tilde{R}_{ij}-2\dot{a}^2 \tilde{g}_{ij}-a\ddot{a}\tilde{g}_{ij}$ (tilde means the spacial metric and it's curvature tensor) to $R_{ij}=-[2K+2\dot{a}^2+a\ddot{a}]\tilde{g}_{ij}$ (where $K$ is the curvature constant (-1,0,+1)).
We then decide on a stress energy tensor, using the principles of homogeneity and isotropy (we don't want there to be some strange asymmetry), to get one of the form $T_{00}=\rho$, $T_{i0}=0$, and $T_{ij}=a^2p\tilde{g}_{ij}$. Which gives us $S_{ij}=\frac{1}{2}(\rho-p)a^2\tilde{g}_{ij}$ and $S_{00}=\frac{1}{2}(\rho+3p)$.
Using all this, plugging back into the EFEs, we get two equations (one for i=j=0, and one for the rest):
(1) $-\frac{2K}{a^2}-2\frac{2\dot{a}^2}{a^2}-\frac{\ddot{a}}{a}=-4\pi G (\rho - p)$
(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$
Then, cause the first equation is sorta unwieldy, we can add three times the first equation to the second to get the nicer (and more familiar):
(3) $\dot{a}^2+K=\frac{8}{3}\pi G a^2$
We can also get the following equation from (1) and (2):
(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$
Which is really no surprise, as this conservation law is found in any solution to the EFEs.
So really what equations you decide to call the "Friedman Equations" is up to whatever your doing. (1) and (2) are the direct consequences of derivation, which you can then use to derive (3) and (4). It just turns out for most cosmology calculations, we don't want to use (1), and the last 3 ((2)(3)(4)) are more useful. However there will be redundancy within these 3 equations (after all, they came from just two equations)!
EDIT: Just to adress the OPs question:
Lets take (2) and (4):
(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$
(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$
Lets multiply (2) by $\frac{2a\dot{a}}{3}$.
$\frac{2a\dot{a}}{3}[\frac{3\ddot{a}}{a}]=\frac{2a\dot{a}}{3}[-4\pi G(3p+\rho)] => 2\dot{a}\ddot{a}=-\frac{8}{3}\pi G(3p a\dot{a} + \rho a\dot{a})$
Some tricky algebra here:
$=>2\dot{a}\ddot{a} = \frac{8}{3}\pi G(-3 a\dot{a}(\rho+p) + 2\rho a\dot{a})$
We now substitute in the conservation of energy equation.
(5) $2\dot{a}\ddot{a} = -\frac{8}{3}\pi G(\dot{\rho}a^2+2\rho a \dot{a})$
But hey! This looks sorta like what you would get if you differentiated the Friedmann Equation (3), note that $K$ is not time dependent (if it were, that would be crazy!). Since your integrating instead of differentiating you will have constants of integration, but you can just tie that into the $K$ term (which is a sorta interesting way to view what $K$ means).
EDIT: Doing a similar process will show you a way to derive (4) to begin with.
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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?
| Well, of course you have to pick the quantities in your dimensional analysis right.
Example: Use dimensional analysis to estimate the potential energy of a star, hold together only by gravitation.
Solution: Newtons gravitational constant $G$ better show up somewhere. This requires us to include something with units $kg^2 / m$. We can get this by inserting $M^2$, with $M$ the mass of the star, and by inserting $1/R$, with $R$ the radius of the star. Thus, the potential energy is estimated to be
$$E_\text{pot} \approx -G \frac{M^2}{R}$$
which is off by a factor $3/5$.
I could, of course, have inserted the mass of a hydrogen atom and then everything would be off by many orders of magnitude...
There is no general guarantee that the constant is of order one. But:
Why it often comes out as "order of 1" is, I believe, the following:
In many cases, the true solution involves an integral over a variable $x$ where the function to be integrated is of the form $f(x) \sim x^n$. In physics, $n$ is small in most cases, so the integration gives a factor $\frac{1}{n}$. Other corrective factors that get ignored in dimensional analysis are $\pi$, $2\pi$ or $4\pi$, i.e., some small integer multiples of $\pi$.
A counterexample... the fine-structure constant $\alpha = \frac{1}{137}$, maybe? But this constant itself can be obtained through dimensional analysis, so I am not sure.
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Photons in expanding space: how is energy conserved? If a photon (wave package) redshifts (stretches) travelling in our expanding universe, is its energy reduced?
If so, where does that energy go?
| Since you say you're talking about what happens locally (in a small volume), I'll answer from that point of view. The usual formulation of energy conservation in such a volume is that energy is conserved in an inertial reference frame. In general relativity, there are no truly inertial frames, but in a sufficiently small volume, there are reference frames that are approximately inertial to any desired level of precision. If you restrict your attention to such a frame, there is no cosmological redshift. The photon's energy when it enters one side of the frame is the same as the energy when it exits the other side. So there's no problem with energy conservation.
The (apparent) failure of energy conservation arises only when you consider volumes that are too large to be encompassed by a single inertial reference frame.
To be slightly more precise, in some small volume $V=L^3$ of a generic expanding Universe, imagine constructing the best possible approximation to an inertial reference frame. In that frame, observers near one edge will be moving with respect to observers near the other edge, at a speed given by Hubble's Law (to leading order in $L$). That is, in such a frame, the observed redshift is an ordinary Doppler shift, which causes no problems with energy conservation.
If you want more detail, David Hogg and I wrote about this at considerable (perhaps even excessive!) length in an AJP paper.
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Does the wavelength always decrease in a medium? I was studying a GRE Physics Test problem where optical light with a wavelength of 500 nm travels through a gas with refractive index $n$.
If we look at the equations for wave motion and index of refraction
$$c=\lambda_0\nu\quad\text{(in vacuum)}$$
$$v = \lambda\nu\quad\text{(in medium)}$$
$$n = c/v$$
we see that, if the frequency is constant, the wavelength decreases in the medium compared to vacuum. Is this a consistent property at all frequencies and for all mediums with refractive index real and greater than 1?
Are there dielectrics which change the frequency (still for n > 1), and is there an example of that?
| As for your second question, dielectrics which change the frequency, any medium which changes the frequency must be a nonlinear medium. There are many of them. They do not change the frequency continuously, but rather they generate higher order harmonics, an integral number times the original frequency.
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What are the details around the origin of the string theory? It is well-known even among the lay public (thanks to popular books) that string theory first arose in the field of strong interactions where certain scattering amplitudes had properties that could be explained by assuming there were strings lurking around. Unfortunatelly, that's about as far as my knowledge reaches.
Can you explain in detail what kind of objects that show those peculiar stringy properties are and what precisely those properties are?
How did the old "string theory" explain those properties.
Are those explanations still relevant or have they been superseded by modern perspective gained from QCD (which hadn't been yet around at the Veneziano's time)?
| Well I believe the original clue was Regge trajectories. It was observed that if you plotted mass squared vs. angular momentum for strongly interacting resonances, they tended to follow straight lines. This could be explained as the spectrum of rotating strings connecting massless particles.
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With potential $V(x)= ax^6$ the quantized energy level $E$ depends on which power of $n$? A particle in one dimension moves under the influence of a potential $V(x)= ax^6$, where $a$ is a real constant. For large $n$, what is the form of the dependence of the energy $E$ on $n$?
| For large $n$, the semiclassical approximation is valid and for bound states we may use the Bohr-Sommerfeld quantization condition:
$n = \frac{1}{h}\oint p dq$, where $n$ is the principal quantum number, $h$, the Planck constant,
$q$, the position and $p$, the momentum on the classical trajectory.
In our case due to the conservation of energy:
$\frac{p^2}{2m}+ax^6 = E \rightarrow p = \sqrt{2m(E-ax^6)}$,
where $m$ is the mass and $E$, the total energy.
By substitution, we obtain:
$n = \frac{1}{h}\int_{-(E/a)^{1/6} }^{(E/a)^{1/6} }\sqrt{2m(E-ax^6)} dx$,
where the integration is between the two turning points. By scaling the integration variable:
$x =( \frac{E}{a})^{1/6}y$
We obtain:
$n = \frac{\sqrt{2m}}{h}{(\frac{E}{a})}^{2/3}\int_{-1 }^{1} \sqrt{(1-y^6)} dy$.
Thus:
$E \propto n^{3/2}$
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Positive Mass Theorem and Geodesic Deviation This is a thought I had a while ago, and I was wondering if it was satisfactory as a physicist's proof of the positive mass theorem.
The positive mass theorem was proven by Schoen and Yau using complicated methods that don't work in 8 dimensions or more, and by Witten using other complicated methods that don't work for non-spin manifolds. Recently Choquet-Bruhat published a proof for all dimensions, which I did not read in detail.
To see that you can't get zero mass or negative mass, view the space-time in the ADM rest frame, and consider viewing the spacetime from a slowly accelerated frame going to the right. This introduces a Rindler horizon somewhere far to the left. As you continue accelerating, the whole thing falls into your horizon. If you like, you can imagine that the horizon is an enormous black hole far, far away from everything else.
The horizon starts out flat and far away before the thing falls in, and ends up flat and far away after. If the total mass is negative, it is easy to see that the total geodesic flow on the outer boundary brings area in, meaning that the horizon scrunched up a little bit. This is even easier to see if you have a black hole far away, it just gets smaller because it absorbed the negative mass. But this contradicts the area theorem.
There is an argument for the positive mass theorem in a recent paper by Penrose which is similar.
Questions:
*
*Does this argument prove positive mass?
*Does this mean that the positive mass theorem holds assuming only the weak energy condition?
| Given that your argument requires moving in an accelerating frame and considering its Rindler horizon, I wonder if what you stated is more similar to statements about asymptotically hyperbolic hypersurfaces in a space-time. In which case, that the analogue of the positive mass theorem can be derived using only the weak energy condition has been shown before (for example, see this paper of Anderson, Cai, and Galloway).
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Solid objects inside the event horizon - can they remain "solid"? So, once something is inside a black hole's event horizon, it can only move towards the center. This is fine for a point-object. But 3D solid objects rely on molecular forces to stay in one piece. These forces act in all directions inside the solid. But there could not be any interaction between atoms at different distances from the singularity, right?
So, what happens to a solid (let's assume a crystalline lattice, for simplicity) hypothetically placed inside the event horizon? Does it get cleaved in ultra-thin concentric layers, centered on the singularity?
| No solid can be placed inside an event horizon because any solid matter under the horizon means information about it is lost.
Also a solid object inside the BH would allow to transfer information from inside, say, by changing its gravitational field by rotating it.
The both things are impossible thus no body containing information can be inside the BH.
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How can it be that the beginning universe had a high temperature and a low entropy at the same time? The Big Bang theory assumes that our universe started from a very/infinitely dense and extremely/infinitely hot state. But on the other side, it is often claimed that our universe must have been started in a state with very low or even zero entropy.
Now the third law of thermodynamic states that if the entropy of a system approaches a minimum, it's temperature approaches absolut zero.
So how can it be that the beginning universe had a high temperature and a low entropy at the same time? Wouldn't such a state be in contradiction to the third law of thermodynamics?
| I had puzzled about that as well. But while the temperature is high the mass/energy density is extremely uniform (as ilustrated by the uniformity of the cosmic microwave background radiation 380,000 years later in the evolution of the universe). And gravity changes everything. Uniform density is very low for systems dominated by gravity (in the relative sense that clumpy distributions have higher gravitational entropy). So I think this is one way entropy is relatively low.
Another reason may be that the phase space of micro states grows as the universe expands.
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Plotting a wave function that represents a particle The problem is this:
A particle is represented by the wave function $\psi = e^{-(x-x_{0})^2/2\alpha}\sin kx$. Plot the wave function $\psi$ and the probability distribution $|\psi(x)|^2$.
This the problem 2.1 in the book Fundamental University Physics Volume III by Marcelo Alonso and Edward Finn. The thing is I don't know what values $k, \alpha$ and $x_{0}$ should have. Probably I don't know what $\psi$ really represents in this case.
| Then plot first with $x_0=0$, $\alpha=1$, and $k=1$.
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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?
|
I get the physical significance of vector addition & subtraction. But I don't understand what do dot & cross products mean?
Perhaps you would find the geometric interpretations of the dot and cross products more intuitive:
The dot product of A and B is the length of the projection of A onto B multiplied by the length of B (or the other way around--it's commutative).
The magnitude of the cross product is the area of the parallelogram with two sides A and B. The orientation of the cross product is orthogonal to the plane containing this parallelogram.
Why can't vectors be divided?
How would you define the inverse of a vector such that $\mathbf{v} \times \mathbf{v}^{-1} = \mathbf{1}$? What would be the "identity vector" $\mathbf{1}$?
In fact, the answer is sometimes you can. In particular, in two dimensions, you can make a correspondence between vectors and complex numbers, where the real and imaginary parts of the complex number give the (x,y) coordinates of the vector. Division is well-defined for the complex numbers.
The cross-product only exists in 3D.
Division is defined in some higher-dimensional spaces too (such as the quaternions), but only if you give up commutativity and/or associativity.
Here's an illustration of the geometric meanings of dot and cross product, from the wikipedia article for dot product and wikipedia article for cross product:
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How can area be a vector? My professor told me recently that Area is a vector. A Google search gave me the following definition for a vector:
Noun: A quantity having direction as well as magnitude, esp. as
determining the position of one point in space relative to another.
My question is - what is the direction of area? I can relate to the fact that velocity is a vector. The velocity of a moving motorbike for example, has a definite direction as well as a definite magnitude assuming that the bike is moving in a straight line & not accelerating.
My friend gave me this explanation for the direction of Area vector. Consider a rectangular plane in space. He argued that the orientation of the plane in space can only be described by considering area as a vector & not a scalar.
I still wasn't convinced. Suppose the plane was placed such that its faces were perpendicular to the directions, North & South for example. Now the orientation of the plane is the same irrespective whether the so called vector points to north or to the south. Further what is the direction of a sphere's area?
Does considering area as a vector have any real significance? Please explain.
Thanks in advance.
| There is an especially picturesque example of the Law of Pythagors in three dimensions applied to the areas of a simplex. (Where by "simplex" I believe I mean a section of space bounded by three orthogonal planes and one arbitrary plane.) The sum of the squares (of the areas) of the three small faces are equal to the square of the area of the oblique face. It is easily explained by the pressure/flow type arguments put forward in the other answers posted here, plus the obvious physical condition that an undisturbed fluid is in equilibrium with itself.
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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?
| First, you state a few things that aren't quite right in your question. While the view that's generally talked about is that Phobos and Deimos are likely captured asteroids, dynamically it's a pretty difficult problem (you generally need a third (in this case fourth?) body to take away the extra energy, and it's hard to get a circular orbit around the equator). See for a bit more on that.
In terms of Phobos' demise, there are two things that make this problem very difficult to estimate. First, Phobos' orbit evolves as it orbits around Mars, so you can't just take a linear approach and say, "It's moving towards Mars at 18.3 cm/year so it's going to hit in about 50 million years." It's more complicated and non-linear.
But besides that, there's the Roche Limit to consider, whereby the moon will break up due to tidal forces before it would actually hit. The problem there is that Phobos is already within the Roche Limit, meaning that it's only being held together now by the physical strength of the rock it's made of. And since we don't really know what it's made of inside (though we can make educated guesses and I'm sure there are models out there for its strength), these unknowns make it somewhat difficult to estimate.
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Rainbow around Sun From the perspective of a person, a rainbow is formed when the Sun is behind the person, and there is a critical angle made by the rainbow.
However, on several occasions, usually at noon when the Sun is higher, I saw a ring around the Sun made of the colors of the rainbow. Is that a rainbow? Is within the definition of a rainbow? And how is it possible?
| The other answers describe a rainbow-like phenomenon involving ice crystals, which may very well be what you saw. However, there is another possibility.
A normal rainbow occurs when light enters a spherical drop, refracts at the curved surface (dispersing the colors), reflects off the back of the drop, and then leaves the drop, refracting again. The angle between incident and outward-going light is about $42^\circ$, and so you see rainbows $42^\circ$ from the point directly opposite the Sun.
No one said, though, that the light had to undergo exactly one total internal reflection before leaving the drop. It can reflect multiple times, coming out at different angles. A second-order rainbow is at a slightly different angle from the normal first-order one. More interestingly, third- and fourth-order rainbows can be found circling the Sun (not circling the point opposite the Sun), simply due to the geometry. Wikipedia has some information, though unfortunately I cannot find any good diagrams for this effect online. Third-order rainbows are very hard to see, but they have been documented and are in a sense more "true" to the definition than phenomena involving ice.
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Does the proton have an equatorial bulge around its spin axis? And if so, can we observe a difference in the electron scattering cross section with transversely polarized VS longitudinally polarized protons?
P.S. Let me make my question more precise. Consider the charge shape of the proton. In the rest frame of a proton with spin in +z direction, what's the spatial dependence of the expectation of the electric charge operator $j^0(x)$, at some particular renormalization scale? Can this question be sufficiently answered by the currently available data from polarized proton scattering experiments?
| At the level of individual Proton interactions I reckon it is far more common to use the parton model, ie it is a collection of three quarks, $uud$, and associated gluons.
Also the 'spin' axis is a quantum physical term and is best not converted to classical terms.
nb. This answer was given to a previous, simpler vesion of the question.
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Surface tension of solutions and mixtures The inspiration for this question is over on cooking.stackexchange, asking more about actual measurements for commonly consumed liquids, but I'm interested more generally as well.
What determines the behavior of surface tension for solutions and mixtures with respect to concentration? Of course, I expect that the answer depends on the liquid, since different liquids have different causes of cohesive forces. I would be interested both in quantitative answers (very approximate/general, probably) and qualitative ones.
Additionally, what is the dependence of surface tension on temperature? Does that have any interaction with dependence on solution/mixture concentration?
(I know this question borders on chemistry, but there's no chemistry stackexchange, and besides, I'm sure at least some surface tension properties admit explanations from statistical mechanics!)
| The wikipedia section on the influence of temperature and solutes on surface tension is actually pretty decent with some good reference to basic text on this topic.
That said, there is no straightforward answer to your question because it all depends a lot on the liquid(s) that you are looking at. If for example you have a solution of water and surfactant molecules, you will decrease surface tension, but only to the limit of the critical micelle concentration. Moreover, the surfactant (as the word says) will collect at the surface of the liquid and therefore have a greater effect then expected based on the bulk concentration.
About the temperature effect. For most liquids the answer of @Herve that the surface tension decreases with increases temperature is true, but there are some exceptions. Mainly metallic substances can also show an increase in surface tension with temperature.. This is in particular important in the study of welding, where the surface tension gradients due to temperature have a large effect on the quality of a weld. In this case indeed the interaction between temperature and solutes will determine the surface tension (see the article for one of the possible reasons)
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How do you prove $S=-\sum p\ln p$? How does one prove the formula for entropy $S=-\sum p\ln p$?
Obviously systems on the microscopic level are fully determined by the microscopic equations of motion. So if you want to introduce a law on top of that, you have to prove consistency, i.e. entropy cannot be a postulate.
I can imagine that it is derived from probability theory for general system. Do you know such a line?
Once you have such a reasoning, what are the assumptions to it?
Can these assumptions be invalid for special systems? Would these system not obey thermodynamics, statistical mechanics and not have any sort of temperature no matter how general?
If thermodynamics/stat.mech. are completely general, how would you apply them the system where one point particle orbits another?
| The functional form of the entropy $S = - \sum p \ln p$ can be understood if one requires that entropy is extensive, and depends on the microscopic state probabilities $p$.
Consider a system $S_{AB}$ composed of two independent subsystems A and B. Then $S_{AB} = S_A +S_B$ and $p_{AB} = p_A p_B$ since A and B are decoupled.
$$
S_{AB} = - \sum p_{AB} \ln p_{AB} = -\sum p_{A} \sum p_B \ln p_A -\sum p_{A} \sum p_B \ln p_B
$$
$$
= -\sum p_{A} \ln p_A - \sum p_B \ln p_B = S_A + S_B
$$
This argument is valid up to a factor, which turns out to be the Boltzmann constant $k_B$ in statistical mechanics: $S = - k_B \sum p \ln p$ which is due to Gibbs, long before Shannon.
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Why is the string theory graviton spin-2? In string theory, the first excited level of the bosonic string can be decomposed into irreducible representations of the transverse rotation group, $SO(D-2)$. We then claim that the symmetric traceless part (i.e. the 35 rep) is the spin-2 graviton - but isn't the label "spin-2" intrinsically 3+1 dimensional? I.e. it labels the representation under the little group $SU(2)$?
| It is traditional to label massless (and some massive) states in higher dimension by their 3-d "spin" counterparts, even thought he label is completely inaccurate, as you say. All antisymmetric forms are "spin-1", the symmetric two-index object is "spin-2", a fundamental spinor is "spin 1/2" and a vector of spinors is "spin 3/2". These labels refer to the maximum helicity of the associated massless particle, although the number of components is completely different than in 4d. For learning the higher dimensional rotation group, there is an article by Scherk from the 1970s.
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Reference area of a parachute I am trying to do an aerodynamic drag equation on a descending parachute (the round variety) and have no idea what the reference area on one would be. I know for a sphere, you can use radius*radius*PI to get the reference area. Is that the same for a parachute?
| Poking around on Google with various search terms that included "parachute shape" i came upon "The Parachute Manual" by Dan Poynter. Table 8.1.7 from that book catalogs empirical data on a host of parachute shapes. Assuming that the parachute is round, as you say, and does not have any holes (apparently, many designs purposefully include gaps to improve stability), the first section of that table (on page 457) is the one to look at. To interpret the terms of the table, I tried reading the glossary. It gives the following three definitions:
*
*nominal diameter: the diameter of a circle made with the same amount of cloth as the parachute.
*projected diameter: the diameter of the real parachute when it is inflated.
*constructed diameter: the diameter of the real parachute when it is not inflated.
From this it is clear that the table says "nominal diameter" but means "constructed diameter." The table is in fact using a nominal diameter of 1 for all the parachutes it lists.
Using this correction, the table reveals that the ratio between projected diameter and nominal diameter for round parachutes varies between 0.6 and 0.7 This means that given the projected radius, the area is, roughly speaking, somewhere between $2\pi r^2$ and $3\pi r^2$.
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Is visible light and radio waves made of the same thing? I understand that there is such a thing as the electro magnetic spectrum, and that light and RF are both on it, so dose that mean that they are made of the same thing? Just at different frequencies.
| you see, the point here is waves are just oscillating energy vectors.
Do you consider a male voice and a female voice as same? Then yes, visible light and RF waves are same!
Its only a matter of how much energy each of them is carrying.
While, RF is having lower photon energy than visible light attributed because of the frequencies.
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Why are the even and odd Regge trajectories degenerate? This is an old classic which I don't think ever got a clear answer. The Gribov-Froissart projection that gives the relativistic version of Regge trajectories treats even angular momentum differently from odd angular momentum. The trajectory functions in general are separate for even and odd angular momentum.
But in QCD, the odd trajectories interpolate the even trajectories in every case--- the two trajectories are degenerate. Is there a way of understanding why the even-odd trajectories are degenerate? Is is a symmetry argument? Can you find a natural system where they are not degenerate?
A quick google search revealed this reference, which doesn't answer the question, but gives an experimental signature for the degeneracy: http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-0576.pdf , There is probably excellent data by now on this.
| Once upon a time, I asked an experienced phenomenologist who worked on particle physics in the 60s why even and odd signatured trajectories lie on top of each other.
He said the phenomenon was called 'exchange degeneracy' and that so far no one has an explanation.
I'm looking back at my notes on Dual Resonance Models, and it looks like by introducing isospin a la Chan and Paton, the isospin multiplicity appears to be correlated with signature, and exhibit exchange degeneracy.
Sorry about replying to a 1-year old topic.
Edit
Earlier this summer while visiting the Caltech campus, I had the wonderful opportunity to speak with S. Frautschi (an early pioneer of the application of Regge theory to particle physics). At some point, I asked him about this phenomenon. He kindly reminded me that the origin of signature is due to exchange forces in relativistic scattering that generate the left-hand cut in the complex $s$-plane, absent in potential scattering. He told me that his interpretation of the phenomenon of 'exchange degeneracy' is that the exchange forces are 'especially weak'. For example, in $\pi^+ \pi^-$ elastic scattering, the exchange channel is exotic, and hence subdominant. To date, this is the best answer I have heard as it provides a dynamical bases for the phenomenon.
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Nonextensive statistical mechanics I know that the Tsallis($S_q$) entropy is called nonextensive information measure in the sense that if $P$ and $Q$ are two probability distributions then $S_q(P\times Q)=S_q(P)+S_q(Q)+(1-q)S_q(P)S_q(Q)$. My question is what is meant by nonextensive statistical physics? What is its connection with Tsallis entropy maximisation?
| Normally entropy is seen as an extensive thermodynamic coordinate, i.e. proportional to the mass of particles: Volume increases by considering a larger amount of gas in the "same state", while pressure will stay the same. So for simple systems entropy of a system that is a combination of 2 systems will be sum of individual entropies. Here it is less. This means that not the whole direct product of states of A and states of B is accessible to the combined system.
Consider a state space of 3x3 pixels for a particle A alone: 9 states. consider B to be an identical system: also 9 states. Now assume you can put A and B in the same 3x3 pixel space and allow them to also sit in the same pixel: 9x9=81 states, however suppose that they interact such that they can not sit in the same pixel:9x9 - 9=72 states so in this second example entropy is not additive for combination of systems. while it is in the first...
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Would a generator in vacuum/space provide electricity endlessly? At it's simplest, electricity generation is achieved by induced voltage due to a changing magnetic field. In a vacuum in the absence of friction, would the initial spin imparted to the rotor of a generator ever come to a halt?
i.e. Would a traditional generator in space generate electricity perpetually (notwithstanding component failure etc)?
| No, because the process of transferring the voltage into a useable form or device will reduce it. In other words, under perfectly idealized conditions (which are impossible), yes it might spin forever, but as soon as you try to use your generator to power a device, you'll slow it down.
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How would you store heat? Um .. naive question perhaps but if somebody wanted to store heat, how would they go about it? Can heat be stored?
I'm told that decomposing kitchen waste in a closed vessel results in a rise in temperature on the body of the vessel. I'm just wondering whether it could be stored for later use.
| Building off of the comments to the question.
It might be instructive to think carefully about what is being stored when you store "electricity" in a capacitor or battery.
Note that it is not electrons as I can charge a capacitor either positive or negative relative to a floating ground. The "what" in that case is energy in the form of electric fields (capacitor) or chemical potential (battery).
Heat is also energy, in the form of excitation of microscopic degrees of freedom, and the way you store it is by exciting the microscopic degree of freedom in some material and then don't allow it to transmit that energy to other forms---which you do by insulating the hot stuff. Or you can convert the heat to some more tractable form (as in Dan's answer or Georg's comment) and store that.
| {
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"timestamp": "2023-03-29T00:00:00",
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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?
| Most proposed designs of the space elevator are such that the whole structure is under tensile stress from the ground anchor point. In these designs, there are stress limits that constraint the material properties of the ribbon. The calculations (based on geosynchronous height of earth) point to that 130 GPa figure.
There is potentially another design approach in which there is no stress limit required in any point in the structure. In this case, the tensile stress is entirely from the geo synchronous orbit holding up the structure against its weight (rather than the earth holding it up against centrifugal force). You only need to make sure the whole structure is at equilibrium, so the center of mass stay roughly at GEO. So, you start at GEO, and start each level one at a time. after finishing each level, you adjust your center of mass to stay at equilibrium. Then you proceed to build the next level below the previous one, until you reach ground.
In order to the upper levels to be able to hold the weight of the lower ones, the structure will follow a exponential pattern of joints. If whole elevator structure will have $N$ levels, the ground level (Level 0) will have one link. the next level (Level 1) will have $k$ links, which all sustain the weight from level 0 link. Level 2 links will have $k^2$ links, which sustain each of the $k$ links of the level 1 links. The last level will have $k^N$ links.
So at GEO, the stress is the whole weight of the structure by the cross section area of all links. the area grows as $k^N$ while the weight of the whole structure grows as $\frac{1-k^{N+1}}{1-k}$. So asymptotically the stress as GEO stays under parameter control.
The benefit of this approach is that you even can make the whole structure with normal materials (no stress limit required). Of course the tensile strength of the material chosen still affect the number of links and levels required to make the structure sustain its own weight.
| {
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Indirect band gap semiconductor for LEDs? Can someone please explain why Indirect band gap semiconductor can not be used for LED creation. Can you also please give me some reference link for details.
| Indirect bandgap semiconductors do not emit as well as do not absorb light for photon energies close to band gap due to reasons described above. Namely, the main reason is the momentum conservation law. However, it is possible to make photodetectors (absorbers) on such semiconductors utilizing interband electron transitions with energies much higher than the bandgap. In this case, such transitions occur at the center of the Brillouin zone with zero or small changes of the momentum.
| {
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Why is it important that Hamilton's equations have the four symplectic properties and what do they mean? The symplectic properties are:
*
*time invariance
*conservation of energy
*the element of phase space volume is invariant to coordinate transformations
*the volume the phase space element is invariant with respect to time
I'm most inerested in what 3 and 4 mean and why they are important.
| Coordinate invariance guarantees that the phase space $M$ can be endowed with a symplectic 2-form $\omega$ which locally is given by $\omega = dq^i \wedge dp_i$. This form is closed ($d\omega = 0$) and nondegenerate, i.e. $\omega^n$ is a volume form, where $2n$ is the dimension of $M$. The other conditions say that for any Hamiltonian function $H$, the induced flow on $M$ preserves both $H$ and $\omega$. It turns out that proving this is totally trivial using the language of symplectic geometry.
So why bother? By reformulating mechanics in terms of symplectic manifolds, we now have all the modern tools of differential geometry and topology at our disposal.
Example. Using Morse theory, for any (sufficiently nice) Hamiltonian, we can obtain bounds on the number of equilibrium solutions in terms of the Betti numbers of $M$.
Example. Gromov nonsqueezing. I won't state it, but it is a classical analogue of the uncertainty principle. In particular, it shows that certain reasonable phenomena are impossible for Hamiltonian systems.
Example. Reduced phase spaces (symplectic reduction). This is very clean from the geometric point of view, but very messy in local coordinates.
Example. Integrable systems, spectral curves, etc. Again, hard to imagine formulating a lot of this without geometry at our disposal.
Example. Deformation quantization, geometric quantization (and others). This gives some insight into what quantization actually is, or at least should be. Dirac-style canonical quantization might work for lots of simple systems, but there are subtleties involved and for more complicated systems it is not at all obvious what the right quantization is. The geometric point of view also relates quantization to representation theory (see e.g. the work of Kostant).
| {
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"timestamp": "2023-03-29T00:00:00",
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Is an electron/proton gun possible? In the 1944 SF story “Off the Beam” by George O. Smith, an electron gun is constructed along the length of a spaceship. In order to avoid being constrained by a net charge imbalance, it is built to also fire the same number of protons in the other direction, dissipating the mass of the “cathode”.
With current knowledge, is this plausible? That is,
*
*Can a practical (i.e., not built with unobtanium insulators) electrostatic device like an electron gun separate and accelerate electrons and protons in this manner?
*Can it actually disassemble solid matter? If so, how does the composition of the cathode affect the difficulty?
| Both electron gun and proton can be made but they don't work the way you want. Electrons or cathode rays flow from high potential region to a low potential region. To accelerate electrons you need high voltage. I agree with dmckee regarding protons. Protons are made by ionizing the Hydrogen gas because that's the easy way. When we use an electron gun in Electrodynamics applications, generally we have a vacuum tube and a very high potential difference and also we have an anode. Due to this the things work better. If there is no vacuum tube the electrons go to the shortest path to ground around it and in the way ionizes air. Practically if you have made an electron gun using thermionic emission or any other method, the electrons hit the ground immediately.
Its just like thunders. With huge potential difference and massive charge, they hit the highest point on earth. That's a deadly weapon.
| {
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"timestamp": "2023-03-29T00:00:00",
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Apple falls for which of these 2 reasons? Needles to say I am a visitor here. I do not belong to the science world;)
But I have read both of these things before:
*
*Apple falls to the ground because curved spacetime pushes it there (same force as keeps moon in orbit)
*Apple 'falls' to ground because the ground is rushing up to meet the apple (which is actually suspended in space) because of Earth's acceleration through space.
I don't think these can both be true. I'd appreciate any clarification - thank you.
| Number 1 is correct. Number 2 is incorrect. The apple has the same velocity as the Earth since it grew on earth. Also, if the earth were moving into the apple you wouldn't see the apple accelerating unless the Earth is constantly linearly accelerating at 9.8m/s^2, which is impossible.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does a change in temperature affect relative humidity Assume that the air pressure and the amount of water in the air stay constant. How can I figure out how much a change in temperature affects the relative humidity?
| Relative humidity is just the percentage of what the air at a given temperature can hold. This is given by the Clausius–Clapeyron equation, which rises roughly exponentially with temperature doubling approx every 10degrees C. So if your relative humidity is X, and the saturation vapor pressure at the new temperature is Y times the value at the old temperature, your new (constant volume) humidity is X/Y. You wanted constant pressure, so your absolute humidity is changed by the change in volume, i.e. your humidity also scales inversely with volume, although this second effect is much smaller than the first.
| {
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