Wednesday 31 December 2014

riddle - Yet another backwards hangman


Insprired by the previous backwards hangman (created originally by Alex, see here), which was solved so quickly, I though I'd make another one. Hopefully slightly harder! Three solutions in a minute suggests more demand is there and more supply needed. I've made the clues a little more cryptic:


Fill in an "x" - Grows good for your car
Fill in a "g" - Your just deserts, and what to do to war!
Fill in a "k" - Emerges only after they're gone
Fill in an "l" - Land of the dragon
Fill in a "r" - Pottery offered for sale
Fill in a "v" - Greetings are forces of nature!

Fill in a "d" - Struggles through
Fill in an "n" - As new, but changes and diminishes.


The answer will be in the form



W_LL



followed by an explanation...



Answer



The question was:




WA_ES
waxes : good for your car to protect the coat; grows like the waxing moon
wages : you earned them by working, and you wage war
wakes : "emerges" like awakens, and "after they're gone" like the wake of a boat. – credit @Charles Koppelman
Wales : has a dragon on its flag
wares : wares are things for sale; pottery = earthenware
waves : a wave of the hand, a wave on the ocean
wades : wading through a bog is quite a struggle
wanes : diminishes like the waning moon; "As new, but changes" points to "wanes" being an anagram of "as new". – credit @Zandar




In quantum mechanics, why do the probabilities of the possible outcomes of a measurement add up to 1?


The question assumes the standard formalism with projector-valued measures rather than POVMs. Suppose a measurement has two possible outcomes, and the corresponding probabilities are greater than 0 and less than 1. Neither outcome is therefore certain. Then why is it certain that either outcome is obtained (as it seems, if the probabilities add up to 1)?


Added after four answers: All the answers provided so far elaborate on the comment by @Vladimir: "It is not a 'quantum mechanical' feature but a consequence of probability definition." @Lubos and @Mark cast the question into a quantum-mechanical form, e.g., why do the absolute squares of the amplitudes associated with the possible outcomes of a measurement add up to 1? They also explain why the sum remains equal to 1. (However, for a decaying particle the probability of finding it decreases, while the probability of finding its decay products increases. So the "conservation of probability" has something to do with the proper conservation laws.) @David makes it clearest why these answers are insufficient.


Keep in mind that no actual measurement is perfect. While theorists may ignore this, experimenters know well enough that in many runs of a given experiment no outcome is obtained. (The efficiency of many real-world detectors is rather low.) This means that in order to make the probabilities add up to 1, one discards (does not consider) all those experiments in which no outcome is obtained.


So let me follow up with another question.





arithmetic - Longest word worth at most a million



An old and popular puzzle, recently revived on Twitter by Alex Bellos, Chris Smith, and others, asks to take the first 26 primes, relate each to letters of the alphabet (A = 2, B = 3, C = 5, ..., Y = 97, Z = 101), and use this code to find the (legitimate) English word that comes closest to a million, where the word's numeric value is given by the product of its letters. CAB is thus valued 5x2x3=30.


With the same code, what is the longest English word valued at most a million?




buoyancy - Volume of melted polar ice caps



If (or rather when) the polar ice caps melt, will they occupy more, the same or less volume?


Somehow all options seem possible to me. I don't assume they are just big ice cubes floating on the poles. So, if a huge mountain of submarine non-floating ice melts, wouldn't it occupy less space? But if the mountain is above the sea level, that's a new mass of water entering the sea.


Obviously, in a real case scenario other factors would be relevant too, like warmer water occupying more space than cold water. But I'm interested in this factor isolated from the rest.




What is more basic thing to explain phenomenons : Energy or Force?



You know many physical happening are explained by considering forces or we can explain these phenomenon using energy (like energy conservation, etc.).


so i was wondering that explaining the physical phenomenon using energy method is more fundamental or explaining through considering forces?


Explaining things differently helps us to view thing differently. So either one of the methods(energy and forces) can gives us more broader vision, i.e. explanation of one phenomenon (through one of the method) can also be applied for the explanation of some other phenomenon(with some minor changes as required) and this is what i mean by fundamental.


thanks.




Does time travel violate conservation of mass/energy?



Imagine I exist at time $t_1$ and my mass is $m$. At time $t_2$ I time travel back to $t_1$. At time $t_1$ there is now a net increase of mass/energy in the universe by $m$.


At time $t_3 = t_2 - x$ where $x < t_2 - t_1$, I travel back to t1 again. The net mass in the universe has now increased by $2 \times m$.


Properly qualified, I can do this an arbitrary $n$ number of times, increasing the mass in the universe by $n \times m$. This extra mass, of course, can be converted to energy for a net increase in energy.


Does this argument show that traveling back in time violates the conservation of mass/energy?



Answer



Conservation of Energy is a consequence of Time-translational Symmetry of the system. If this symmetry is broken, there'd be no Conservation of Energy.


Tuesday 30 December 2014

Is there a thought experiment which brings to light the contradiction between General Relativity and Quantum Mechanics?


I've been told that GR and QM are not compatible, is there an intuitive reason/thought experiment which demonstrates the issue? (Or one of the issues?)



Answer



The simple thought experiments are not related to the naive reason people give for the incompatibility--- people say that they are incompatible because of renormalizability. This is not easy to explain, and it is in fact false, because N=8 supergravity is with some scientific confidence renormalizable. But it's not a good theory.


The reason for the incompatibility is the behavior of black holes--- the fact that they have an entropy which scales as the area, not the volume. In quantum field theory, the fields near the horizon have infinite entropy, as shown by 't Hooft. So you need a different kind of quantum theory, one which is nonlocal enough to allow black holes to have entropy which goes as area.


This is the only incompatibility, since string theory is exactly such a quantum theory, and it is a consistent theory of gravity. So there is no further incompatibility left.


word - Is this selfmade wordplay solvable and strong enough?


Here is a question I have created myself and I want to use it in a self-created game. But I want to check if others can solve it. It is designed to be a little bit challenging.



I am standing around
All night All day
Have multiple guys in me
forever they will stay
no possibility to say
"Please go away"
Some of them sometimes leave for a while

But they are back by no later than closing time
The guys can be used for drawing
Do you know my name?



EDIT:


Maybe the closing time-part is confusing. The closing time for a office is meant.


Edit2 : Maybe this could help: enter image description here



Answer



Based on @RadoslavHristov's answer.


You're a




Pen/Pencil holder



Im standing around All night All day



You are, on your desk for example



Have multiple guys in me



Refers to the pens/pencils




forever they will stay no possibility to say "Please go away"



Obviously, they don't speak



Some of them sometimes leave for a while



They are used to write



But they are back by no later than closing time




We always put them back in their place



The guys can be used for drawing



Obvious when you're a pen!



condensed matter - Reconciling topological insulators and topological order


We make an important distinction between the topological insulators (which are essentially uncorrelated band insulators, "with a twist") and topological order (which covers a variety of exotic properties in certain quantum many-body ground states). The topological insulators are clearly "topological" in the sense of the connectedness of the single particle Hilbert space for one electron; however they are not "robust" in the same way as topologically ordered matter.


My question is this: Topological order is certainly the more general and intriguing situation, but the notion of "topology" seems actually less explicit than in the topological insulators. Is there an easy way to reconcile this?


Perhaps a starting point might be, can we imagine a "topological insulator in Fock space"? Would such a beast have "long range entanglement" and "topological order"?


Edit:


While this has received very nice answers, I should maybe clarify what I'm looking for a bit; I'm aware of the "standard definitions" of (symmetry protected) topological insulators and topological order and why they are very different phenomena.


However, if I'm talking to nonexperts, I can describe topological insulators as, more or less, "Berry phases can give rise to a nontrivial 'band geometry,' and analogous to Gauss-Bonnet there is a nice quantity calculable from this that characterizes instead the 'band topology' and this quantity is also physically measurable" and they seem quite happy with this.


On the other hand, while the connection to something like Gauss-Bonnet might be clear for topological order in "TQFTs" or in the ground state degeneracy, these seem a bit formal. I think my favorite answer is the adiabatic continuity (or lack thereof) that Everett pointed out, but now that I'm thinking about it perhaps what I should have asked for is -- What are the geometric properties of states with topological order from which we could deduce the topological order with some kind of Chern number (but without starting from a Chern-Simons field theory and putting in the right one by hand ;) ). Is there anything like this?



Answer




As you have mentioned, topological insulators (TI) are "topological" because they can not be smoothly connected to trivial band insulators without closing the band gap (and without breaking certain symmetry). Simply generalize this to the many-body case, we may say that the topologically ordered states are called "topological" because they can not be smoothly connected to the trivial product state without closing the many-body gap.


To gain a better understanding, one should realize that "topology" is a complement to "geometry". By geometry, we mean that there is a sense of measurement of the distance and angle etc., and the shape and the size of the object matters. While by topology, we mean that one can continuously deform the object, and the shape or the size does not matter. So the topological properties are those properties that can endure continuous deformation of the state (by continuity we mean without encountering a quantum phase transition). To protect the topological properties against deformation, a gap between the ground states and the excited states is always required. So topological property is only defined for gapped quantum matters (both TI and topological order are within this scope). On the other hand, gapless quantum matters do not have topological properties, and their properties are geometrical.


The topological/geometrical distinction is also reflected from the mathematical tools we used to study the physics. For gapped quantum matters, we use topological tools like homotopy, cohomology, K-theory, category theory etc. For gapless quantum matters, we use geometrical tools like gravity theory (AdS/CFT).


optics - Fresnel distance and Geometrical limit



I read about the geometrical limit of wave theory. The source from where I read had a slightly different explanation to provide than here(The more rigorous answer is too complicated for me to understand). Though I do not understand completely the easier method in the linked question either, I would like to understand what the source form where I read is trying to say:-



An Aperture of size $a$ illuminated by a parallel beam sends diffracted beam (the central maxima) in angular width approximately $\lambda/a$. Travelling a distance $z$, it aquires the width $z\lambda/a$ due to diffraction. The distance at which this width equals the size of the aperture is called the fresnel distance. $z_F=a^2/\lambda$. It is the distance beyond which the divergence of beam of width $a$ becomes significant. At distances smaller than $z_F$, spreading due to diffraction is smaller than the width of the beam, and at distances greater than $z_F$, spreading due to diffraction dominates over that due to ray optics($a$ is the width of the aperture)



I fail to see the meaning behind this line of argument. Is ray optics valid for distances less than the fresnel distance? Is it not that the validity holds when all objects are comfortably larger, and not smaller, than the wavelength of light? How is the "divergence" due to diffraction, and its equivalence to the aperture width, related to the geometric limit of wave theory?



Answer



To understand this explanation, you need to understand Fourier decomposition of the electromagnetic field.


In any homogeneous medium, any electromagnetic field can be thought of as a linear superposition of plane waves, all in different directions. Because they run in different directions, the phase delays they undergo in propagating from, say, your aperture to another, parallel plane are all different. Therefore the wavefront gets "scrambled" owing to these direction-dependent phase delays. This interference between the different plane wave components of the electromagnetic field is what we commonly call "diffraction". I further explain this idea, as well as draw some diagrams in this answer here as well as this one here.


So with this introduction in mind, let's look at your paragraph. For simplicity, assume only one transverse direction and one axial (in the direction of propagation) direction. Let's also assume scalar optics, i.e. that the electromagnetic field is well represented by the behaviour of one of its Cartesian components, so that we can do Fourier optics on scalar field.


So we have a uniformly lit aperture of width $a$. Its transverse profile is therefore the function ${\rm rect}(2 x/a)$ where ${\rm rect}(x) = 1;\,|x| \leq 1$ and ${\rm rect}(x) = 0;\,|x| > 1$. We take a Fourier transform to find the superposition weights of each plane wave component, because each such component has a transverse variation $\exp(i\,k_x\,x)$ where $k_x$ is the Fourier transform variable with units of reciprocal length. The Fresnel distance is, as the paragraph says, simply the axial distance needed for this spread to double the beam width. So it is a rough measure of how quickly the light spreads.



So this is how the "divergence" arises from diffraction, i.e. the interference between an optical field's plane wave components as they propagate. Also $\sin\theta = k_x/k$ where $k = 2\pi/\lambda$ defines the angle that this plane wave component makes with the axial direction. We take the Fourier transform, we find that there is a spread of $k_x$ values such that the plane wave components most skewed to the axial direction make an angle without direction of roughly $\lambda/a$. So, owing to these skewed components, the field's energy spreads out.


The beam width diverges slowly at first and then, after an axial distance of several Fresnel distances, the divergence speeds up so that the propagation becomes well modelled by the cone of rays diverging from the centre of the aperture. Indeed if you plot contours of constant intensity, they are hyperbolas which begin at right angles to the aperture but bend so that their asymptotes are the cone defined by ray theory. The Fresnel distance defines how far the "knee" of the hyperbol is from the aperture.


For your question:



Is it not that the validity holds when all objects are comfortably larger, and not smaller, than the wavelength of light?



This is in general right, but it breaks down near focuses and in situations like this where we are near and aperture and if the aperture is comparable to the light wavelength. In this case you should be able to understand from the Fourier analysis the reciprocal relationship between the aperture width and the angular spread.


What is the definition of a quantum integrable model?


What is the definition of a quantum integrable model?



To be specific: given a quantum Hamiltonian, what makes it integrable?




Does the material of the slits affect the pattern of the electrons in the double-slit experiment?


The question comes from thinking about the way how electrons act on the slits. Do material of the slits affect electrons motion? Assume the radius or length of the slits are set in this case.




Monday 29 December 2014

statistical mechanics - Critical exponents and scaling dimensions from RG theory


In most books (like Cardy's) relations between critical exponents and scaling dimensions are given, for example $$ \alpha = 2-d/y_t, \;\;\nu = 1/y_t, \;\; \beta = \frac{d-y_h}{y_t}$$ and so on. Here $y_t$ and $y_h$ are scaling dimensions of scaling variables $u_t$ and $u_h$ related to (reduced) temperature $t$ and field $h$. This is always discussed in the context of the Ising model. I am confused about what $y_t$ and $y_h$ are in general? In general you have scaling dimensions $y_1, y_2\dots y_n$ for the scaling fields $u_1, u_2,\dots u_n$, where it's not clear what $y_t$ and $y_h$ are. In cases where the RG is 'diagonal', $t$ and $h$ themselves are scaling variables, the problem does not exist, but that is not the general situation.


For example, the RG equation of the XY model in $d=2+\epsilon$ is $$\frac{dT}{dl} = -\epsilon T+ 4\pi^3 y^2,\;\; \frac{dy}{dl} = \left(2-\frac{\pi}T\right)y,$$ where $T$ is the temperature and $y$ is related to vortex fugacity. There is a finite temperature fixed point for $T^\star=\pi/2$. Imagine we want to calculate $\nu$ and $\alpha$ at this non-trivial fixed-point. By linearizing the above at the fixed point, we get some two dimensions $y_1$ and $y_2$ for two scaling variables $u_1$ and $u_2$. These variables are both linear combinations of $T$ and $y$. How can I know which scaling dimension/eigenvalue corresponds to the thermal eigenvalue $y_t$? What are the values of $\alpha$ and $\nu$?



Answer



The OP is right that the couplings $g_1$, $g_2$,... parametrizing the field theory are in general combinations of the scaling field of the RG fixed points $u_1$, $u_2$,...



In the Ising model, the two most relevant scaling field $u_1$ and $u_2$ can be associated with the temperature $t$ and the magnetic field $h$. However the OP is wrong when saying that $u_1$ and $u_2$ are diagonal (meaning that $u_1=t$, $u_2=h$). Indeed, in general, every coupling that respect the Ising symmetry will have a projection onto $u_1$, whereas all couplings that break the symmetry will have a component onto $u_2$. This means for instance that one can in principle drive the transition not by changing $t$ in the bare action, but by changing the interaction $\lambda$. In that case, the correlation length will diverge as $|\lambda-\lambda_c|^{-1/y_t}$. However, this is usually not what happens for microscopic models, and is thus not discussed in most books.


The confusion arises because one usually work in perturbation theory, where very few coupling constants are kept, implying a projection of the whole flow onto a very small subspace of the true space of all coupling constants.


Concerning the flow equation of the XY model close to two dimensions, one should notice that here we only have one relevant field (which one naturally associates with the temperature, as it is the experimentally tunable parameter), corresponding to $u_1$ as it preserves the XY symmetry $y_t$ is thus the value of the positive eigenvalue associated with deviations to the fixed point (one then compute $\nu$ and $alpha$ from the equations given by the OP). There are no symmetry breaking field here, so one does not see the effect of $u_2$, and its eigenvalue $y_h$. Instead, the other direction is an irrelevant one (also associated with a XY symmetric field), the eigenvalue only contributing to correction to scaling.


event horizon - What is exactly the density of a black hole and how can it be calculated?



How do scientists calculate that density? What data do they have to calculate that?



Answer



Black holes are really hard to get a density. Basically, they are so dense that there is no known mechanism for providing sufficient outward force to counterbalance the inward pull of gravity, so they will collapse into an infinitesimally small size. Of course, that doesn't seem likely, it seems likely there is something that will keep the volume from being 0, but it is extremely dense.


An alternative method of measuring the volume of a black hole is to take the radius beyond which light can't escape, also commonly known as the Event horizon. Wikipedia has a great article on potential black hole sizes and masses, using the event horizon. Here's a few example values:


Stellar black hole: mass = 2$\times$10$^{31}$ kg, volume = 3.4$\times$10$^{12}$ m$^3$. The density would then be mass/volume, or 6$\times$10$^{18}$ kg/m$^3$.


Galactic sized: Mass is 2$\times$10$^{39}$ kg, volume= 10$^{37}$ m$^3$, density= 200 kg/m$^3$.


It seems that the larger they are, the less dense they would be, but only if you consider the event horizon as the limit. Of course, we don't know what is beyond an event horizon, so...


cipher - The Twenty Doors (ROOM 4)


This is part of The Twenty Doors series.
The previous one is The Twenty Doors! (ROOM 3)
The next one is The Twenty Doors! (ROOM 5)




You now enter room 4. This one has a load of A's and B's carved into the wall, along with a message. AAAAA ABAAABAAAB ABBABABBAAAABAA AAAAAABBAAAAABB AAAAB ABAAABAAAB BAABABABAAAABAAABBAABAABABABBA AABABABBABBAABBBAAAA ABAAAABBAA BAABAAABBBAABAA ABABBAAAAAABAAAABBAA AAABABAAAABABBAABBBABAABAABBABAABBABAAAAAAAAAABABB I may not be distinct but I've always loved bacon.



On the floor, as usual, there is a bit of paper.



Bmmp rum gq rfc uyw rm qydcrw!



But this time, there are three doors (not counting the one you just came through). You turn over the paper, and there is something written on the back



This time, press the button on your chosen door to open it, but if you choose the wrong one, prepare to DIE!!!



Hmm... This one involves two cryptograms. Damn. Anyway, you might as well try.


Which door should you go through?



The next door will be added when this door is solved!



Answer



I just joined StackExchange. This is my shot at the answer:


The door you have to pick is



Door two.



To decrypt the cryptogram on the wall



Use the Baconian cipher with I=J and U=V to decrypt it. That gives you the plain text: 'A IS ONE AND X IS TWENTY FOUR IN THE MAIN CRYPTOGRAM'.




To decrypt the one on the paper



It's encrypted with the affine substitution cipher. Decoding it gives you the plain text: 'Door two is the way to safety!'



On to Room 5 (I hope)!


electrostatics - What will happen if a conductor is introduced instead of a dielectric medium in parallel plate capacitor?


If a conductor like copper is placed between two plates of a parallel plate conductor,neither touching any of them, what will happen to the capacitance of the capacitor?



Answer



Suppose you have a parallel plate capacitor:


Parallel plate capacitor


So we have a capacitance:


$$ C = \frac{\varepsilon_0 A}{d} $$


and a charge $Q$ on the plates given by:


$$ Q = CV $$



Suppose we now insert a sheet of copper in between the plates as you describe:


Capacitors


The electrons in the copper plate are free to move, so they flow towards the positive plate and you end up with two capacitors with an increased capacitance:


$$ C' = \frac{\varepsilon_0 A}{d'} $$


The combined capacitance is obtained by using the equation for two capacitors in series:


$$ \frac{1}{C_{tot}} = \frac{1}{C_1} +\frac{1}{C_2} $$


So in this case the new capacitance is:


$$ C_{tot} = \frac{\varepsilon_0 A}{2d'} $$


And since the copper sheet has a thickness greater than zero $2d' \lt d$ and therefore when you increase the copper sheet the capacitance increases.


Sunday 28 December 2014

mathematics - Display a number using a scientific calculator with most keys are stuck


Your have a scientific calculator such that most of the keys are unable to be pressed. The only keys that work are those for the functions $$ x^2 \;\; \sqrt{x} \;\; x!\;\; \exp\;\; \ln\;\; \sin\;\; \cos\;\; \tan\;\; \cot\;\; \sec\;\; \csc\;\; \arcsin\;\; \arccos\;\; \arctan $$


When a key is pressed it is applied to whatever appears on the screen and the screen changes to the result. The calculator currently displays at $0$. How can we turn the display to read $2015$?


My try was that press $\cos$, we will get 1. Then press $\arctan$, we will get $\pi/4$. Press $\sin$, we will get $\sqrt{2}/2$. Press $x^2$, we will get $1/2$. Press $\arcsin$ then $\csc$, we will get 2. We can keep using the $x^2,x!$ and $\exp$ to get a larger number, but I don't know it is possible to get $2015$? I know $2015 = 5 * 13 * 31$. Can someone help me?




homework and exercises - Beat frequency for 3 waves



Consider 3 waves of frequency 101, 103, 106 hz, and of same intensity. What should be the beat frequencies?



Now I can calc it for 2 waves, and i know how to write the combined equation of the two. But the addition of a third causes lots of problems...the equation is getting cumbersome. Can you please help me here? I dug around online, and some responses were neglecting the third wave as it was too close to one of the waves. If I want to avoid that, is there any other solution?




classical mechanics - Why do rockets have multiple stages?


What is the advantage for rockets to have multiple stages?


Wouldn't a single stage with the same amount of fuel weigh less?


Note I would like a quantitative answer, if possible :-)


alt text



Answer




As Omega Centauri wrote, it is mainly about removing unused tank mass; for numbers, see Wikipedia article about Ciołkowski's equation, especially the example there.


special relativity - What is the meaning of a (1/2, 1/2) representation?


A spin-1 representation is equivalently a (1/2, 1/2) representation of the Lorentz group.


Does this mean we are summing together two irreducible representations labelled by the 'quantum number' a 1/2 each; or are they also another irreducible representation which is labelled by the quantum numbers (1/2, 1/2)?


I'm guessing the latter, as otherwise we're getting a boson (spin-1) by putting two fermions (spin-1/2) together, which doesn't on the face of it seem right; and also particles are irreducible representations.



In which case what are the 1/2's in (1/2, 1/2) mean?



Answer



The labelling of finite-dimensional irreducible representations of (the universal cover of the connected component of) the Lorentz group $\mathrm{SO}(1,3)^\uparrow $ by two half-integers $(s_1,s_2)$ arises as follows:


The complexified Lorentz algebra $\mathfrak{so}(1,3)_\mathbb{C}$ has $\mathfrak{su}(2)\times\mathfrak{su}(2)$ as a compact real form, and the finite-dimensional unitary representation theory of these algebras is equivalent, and moreover equivalent to the complex representation theory of $\mathfrak{so}(1,3)$. The actual rotation algebra $\mathfrak{su}(2)\subset\mathfrak{so}(1,3)$ of physical rotations embeds diagonally into the $\mathfrak{su}(2)\times\mathfrak{su}(2)$ algebra. Qmechanic discusses a lot of the mathematics around the relation of $\mathrm{SO}(1,3)^\uparrow$ and $\mathrm{SU}(2)\times\mathrm{SU}(2)$ e.g. in this answer.


The finite-dimensional irreducible representations of $\mathfrak{su}(2)$ are well-known from quantum mechanics to be the spin representations labelled by half-integers $s$, so representing two copies of this algebra gives representations labelled by $(s_1,s_2)$. Since the true rotation algebra embeds diagonally, one says that $s_1+s_2$ is the 3-dimensional spin one associates to such a representation. However, one must note that if one looks at the $(1/2,1/2)$ representation as a representation of $\mathrm{SO}(3)$, it is not irreducible - it decomposes as the $1\oplus 0$ representation in terms of spins, or the $\mathbf{3}\oplus\mathbf{1}$ representation in terms of dimensions.


Also, since you said "particles are irreducible representations", note that these representations are finite-dimensional representations on the target space of the classical fields, not unitary representations on the Hilbert space of quantum states. There are no faithful finite-dimensional non-trivial unitary representations of the Lorentz group since the Lorentz group is not compact (i.e. the rapidity can take on all values). The classification of the unitary representations has nothing to do with the finite-dimensional $(s_1,s_2)$ representations and is called Wigner's classification.


experimental physics - What exactly do we see on the famous neutrino image of the sun?


An answer to the question If we could build a neutrino telescope, what would we see? contains a link to a neutrino image of the sun by the Super-Kamiokande neutrino detector.


neutrino image of the sun


There it says that the image actually covers a large part of the sky of about 90x90 degrees. As the diameter of the sun from earth is around one half of a degree, it must be that many of the neutrinos didn't come straight at us. This seems surprising (to me), as neutrinos should hardly interact with the atmosphere. Maybe the central few pixels of the image are extremely much brighter than the others, but this image doesn't show the difference between those and the surrounding pixels? Or is something else going on?



Answer



The detector that took that image--Super Kamiokande (super-K for short)--is a water Cerenkov device. It detects neutrinos by imaging the Cerenkov cone produced by the reaction products of the neutrinos. Mostly elastic scattering off of electrons: $$ \nu + e \to \nu + e \,,$$ but also quasi-elastic reactions like $$ \nu + n \to l + p \,,$$ where the neutron comes from the oxygen and $l$ means a charged lepton corresponding to the flavor of the neutrino (for energy reasons always an electron from solar neutrinos, but they also get muons from atmospheric and accelerator neutrinos---Super-K is the far detector for T2K).



Then you reconstruct the direction in which the lepton was moving (which is correlated with but not identical to the direction the neutrino was going). This indirect pointing method accounts for the very poor angular resolution of the image.


Saturday 27 December 2014

Can you tell me what is the name of this scientific instrument?


Hello i need help to identify this old instrument.

I only have this photo.
Can you help me?
enter image description here



Answer



I think it is a thermometer which works as follows:


There are two large bimetallic spirals, one in the front and one in the back. The two different metals are clearly visible in the spirals. Both have brass at the inside and steel at the outside, so they will curl less as temperature raises because the brass expands more than the steel.


When temperature increases, the open-end of the front spiral pushes to the right and the open end of the back spiral goes to the left.


The open-end of the back spiral holds the pivot of the lever which goes up to the dial mechanism. The bottom tip of the level leans against the adjustment-screw at the open-end of the front spiral.


When temperature increases, the pivot point goes slightly to the left and the bottom tip is pushed to the right. The long end of the lever is 9 or 10 times as long as the bottom end.


The top of the lever, which is behind the dial, goes about 20 times as much to the left as the back-spiral moves to the left while the front spiral tip moves to the right.



The part which is behind the dial is probably this:


A wire is fixed at the top of the vertical level at the left. We can see the free end of that wire behind the left of the dial. The wire is pulled to the right over a small wheel behind the dial and continues down to the small weight visible slightly to the right.


When temperature increases, the wheel behind the dial turns counter-clockwise. I assume there is some more gearwork behind the dial to amplify the rotation and to reverse the direction to have a clockwise turn of the hand when temperature increases.


There is something in the picture which remains unexplained: There seems to hang a loose wire on the glass on the left side. Because it goes all the way left, I assume it is on the outside and therefore irrelevant.


What does $frac{1}{sqrt{1 - frac{v^2}{c^2}}}$ mean with respect to special relativity?


What does the following mean with respect to special relativity?


$$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$



Answer



Colin McFaul's answer is accurate. If you're wondering why it's in that weird form, it has to do with a fundamental postulate of special relativity.


In 2D, say you're given the endpoint of a line that you know starts at the origin, and you want to find its length. According to the diagram below, if you're given a vector in $xy$ coordinates, $(a, b)$ (represented by the red line), you can look at it in an arbitrary coordinate system by rotating your head, but physically, the length of the arrow must remain unchanged. If the coordinates in this other coordinate system are $(c,d)$, we can phrase this with the pythagorean theorem: $a^2+b^2=c^2+d^2=\mbox{distance}$



If we have a special coordinate system drawn, $x'y'$, then $d=0$ and we just have $a^2+b^2=c^2$. In this new frame, its length is just the position on one axis.


diagram of euclidean coordinate system


In relativity, we have one axis $ct$ (where $c$ is the speed of light), and another axis $x$. $ct$ and $x$ are both measured in meters. Instead of a Euclidean distance relation, the factor that remains unchanged from "rotations" is $(ct)^2-x^2$. This is the crux of relativity. It means you can "rotate" space into time and vice versa.


In one frame, the particle may have a constant velocity, so $x=v t$ and the law is conservation of $(ct)^2-(v t)^2=(c^2-v^2)t^2$. If we choose a primed reference frame so that the speed is zero, then we must have $(c^2-v^2)t^2=(ct')^2$. With some algebra we find that this equation implies that $$t=t' \frac{1}{\sqrt{1-\left( \frac{v}{c}\right)^2 }}$$


This is written as $t=\gamma t'$. It all stems from the invariance of $(ct)^2-x^2$. You can do analogous things when talking about $x$ and $x'$.


electricity - Question regarding Drift velocity in general?


The derivation of drift velocity in case of electrons is equivalent to the case of an charged ionic gas and therefore all the arguments also apply there. Now for an ideal "ionic" gas which interacts with each others only as perfect spheres would (for arguments sake), when we apply an external electric field, they would just show drift, that means that they "favour" one direction more than the others where drift velocity is given by $v_d=(eE/m)\tau$ where $\tau$ is the average relaxation time. Hence the net velocities of the electrons wont vectorially add to zero, and moreover the speed as given by kinetic theory of matter would not depend only on the temperature but also depend on electric field. Does this not contradict two fundamental postulates of kinetic theory?


Moreover in case of electrons in a circuit with many series resistors, the relaxation time would be different. The resistivity as defined by $E=\rho J$ and $\rho=m/(ne^2\tau)$ where j=current density, n=number of electrons per unit volume and e=electronic charge, should be different for all the series resistors. But then will not the drift velocity be different for all the resistors if the top equation is correct since the resistivities might defer, causing the relaxation time to differ. Doesn't this contradict the postulation of a steady current which requires constant drift velocity. Moreover suppose we use the same material of different cross section, since there is no area dependence in the drift velocity, it shouldn't change but then again we don't get a steady current as J remains a constant and I is simply J times A.



Answer




Alternate method to derive drift velocity:


Consider a field $\vec{E}$ inside the conductor. Using equations of motion we can say that for every charge inside the conductor, $$\vec{v_1}=\vec{u_1}+\frac{\vec{E}e}{m}t_1$$ $$\vec{v_2}=\vec{u_2}+\frac{\vec{E}e}{m}t_2$$ $$.$$ $$.$$ $$.$$ $$\vec{v_n}=\vec{u_n}+\frac{\vec{E}e}{m}t_n$$
where $t_1,t_2,...t_n$ are the times until each of them collide with another particle respectively.


Summing them and dividing by the number of charge particles($N$) we get, $$\sum_{i=1}^n\frac{\vec{v_i}}{N}=\sum_{i=1}^n\frac{\vec{u_i}}{N}+\frac{\vec{E}e}{m}\sum_{i=1}^n \frac {t_i}{N}$$


We can say that $$\sum_{i=1}^n\vec{u_i}=0$$ as the charges initially are in perfect random motion. Thus substituting $\sum_{i=1}^n\frac{\vec{v_i}}{N}$ as $\vec{v_d}$, the drift velocity and $\sum_{i=1}^n \frac {t_i}{N}$ as $\tau$, the relaxation time for each particle we get $$\vec{v_d} = \frac{\vec{E}e}{m}\tau$$


strategy - Solving Rubik's cube with one algorithm



I was wondering if it's possible to solve a Rubik's cube just iterating a single algorithm. It doesn't matter how complex the algorithm is (as long as it is less than 1 million moves), nor the time required to solve the cube; I just want to know if there exists a "periodic" sequence of moves that allows to turn any valid scrambled configuration to the solved configuration. Once you start executing the algorithm, you can't stop it (unless you finish, of course), rotate the cube and start it on another face.


If the question is still unclear, let's make an example. You have to find a sequence like this one, to be repeated an unlimited number of times, that lets you solve any configuration:


Example: U D R' U' F L' U F' R' U' L D' R  to be repeated 12412183213 times!

Note: The above sequence is just an example, 99,999% it's not working!


I just ask you to prove whether such sequence exists or not. If it exists, I would appreciate you to post it, unless it consists of more than 1000 moves.




Answer



Does an algorithm exist?



Yes. Consider every valid state of the Rubik's cube. It can be brought to the solved state in 20 moves or less. For each state, apply the sequence of moves followed by its inverse. This giant algorithm is guaranteed to solve any cube.



Now, does a reasonable-length algorithm exist?



No. I will show that any such algorithm should have length at least 34326986725785600.



Proof




An algorithm of length L, even if applied an infinite number of times, will only go though 1260L states at most. Applying the algorithm once takes it to at most L states. The important thing is that every element in the Rubik's cube group has order 1260 or less. As a result, even if you apply the algorithm any number of times, you will not be able to reach more than 1260L states.


Now, there are 43252003274489856000 valid states of the Rubik's cube. This means that any algorithm which passes through all these states should have length at least 43252003274489856000/1260=34326986725785600



quantum field theory - $(-1)^{2j}$ in the Spin-statistics theorem - Weinberg/Novozhilov/etc



I am attempting to understand Weinberg's formulation of the spin-statistics theorem as presented in his book "The quantum theory of fields: foundations" pages 233-238. I have at my disposal all three of his Phys rev papers on "Feynman rules for any spin I-III," as well as Novoshilov's book on particle physics (1975, relevant pages 60-77 chapter 4), Streater and Wightman's "PCT, Spin and Statistics, and All That" (1989), Duck and Sudarshan's "Pauli and the spin-statistics theorem" (1998), and Pauli's 1940 paper "The connection between spin and statistics" (Phys rev 58, 716 1940).


Suffice to say either my interpretation of these references, or my understanding is getting stuck. My main problem is with the introduction of the $(-1)^{2j}$ term in the expression for the (anti)commutator relationship between fields: $$\left[ \psi_{ab}(x),\tilde{\psi}^\dagger_{\tilde{a}\tilde{b}}(y) \right]_{\mp}=\left[\kappa\tilde{\kappa}^*\mp (-)^{2A+2\tilde{B}}\lambda\tilde{\lambda}^*\right]P_{ab,\tilde{a}\tilde{b}}(-i\nabla)\Delta_+(\textbf{x}-\textbf{y},0) +\left[\kappa\tilde{\kappa}^*\pm (-)^{2A+2\tilde{B}}\lambda\tilde{\lambda}^*\right]Q_{ab,\tilde{a}\tilde{b}}(-i\nabla)\delta^3(\textbf{x}-\textbf{y})\tag{5.7.19* in Weinberg QtOF:I} $$


Or from Novoshilov page 77:


$$ \left[ \psi_{\sigma}(x),\psi^+_{\sigma'}(y) \right]_{\mp}= \frac{1}{(2\pi)^3}\int{\frac{d^3p}{2p_0}D^J_{\sigma\sigma'}}\left(\frac{p}{m}\right)\{e^{ip(x-y)}\pm\eta\eta^*e^{-ip(x-y}\} $$


$$ =\frac{1}{(2\pi)^3}D^J_{\sigma\sigma'}\left(\frac{-i\partial}{m}\right)\int\frac{d^3p}{2p_0}\{e^{ip(x-y)}\pm(-1)^{2j}\eta\eta^*e^{-ip(x-y}\} $$


In this latter case, the explanation for the appearance of $(-1)^{2j}$ is given as "where we have used $m\alpha\alpha^+=p$ and $D^J\left(-1\right) = (-1)^{2j}$."


In Weinberg's case, the form of the fields $\psi_{\sigma}(x)$ requires that $\left[ \psi_{\sigma}(x),\psi^+_{\sigma'}(y) \right]_{\mp}$ include terms where the coefficient functions are multiplied by their complex conjugates (as $u_{ab}(\textbf{p},\sigma)\tilde{u}^*_{\tilde{a}\tilde{b}}(\textbf{p},\sigma)$ below):


Ie: if $$\psi_l(x)^+ = \sum_{\sigma, n}(2\pi)^{-3/2}\int{d^3p*u_l(\textbf{p},\sigma,n)*e^{ip*x}*a(\textbf{p},\sigma,n)}$$ $$ \psi_l(x)^- = \sum_{\sigma, n}(2\pi)^{-3/2}\int{d^3p*v_l(\textbf{p},\sigma,n)*e^{-ip*x}*a^\dagger(\textbf{p},\sigma,n)} $$ $$ u_{ab}(\textbf{p},\sigma)=\frac{1}{\sqrt{2p^0}}\sum_{a',b'}\left(e^{-\hat{\textbf{p}}*\textbf{J}^{(A)}\theta}\right)_{aa'}\left(e^{\hat{\textbf{p}}*\textbf{J}^{(B)}\theta}\right)_{bb'}\times C_{AB}(j\sigma;a'b')\tag{5.7.14} $$ and $$ v_{ab}(\textbf{p},\sigma)=(-1)^{j+\sigma} u_{ab}(\textbf{p},-\sigma)\tag{5.7.15} $$


Then, we can write


$$ (2p^0)^{-1}\pi_{ab,\tilde{a}\tilde{b}}(\textbf{p}) \equiv \sum_{\sigma}u_{ab}(\textbf{p},\sigma)\tilde{u}^*_{\tilde{a}\tilde{b}}(\textbf{p},\sigma) = \sum_{\sigma}v_{ab}(\textbf{p},\sigma)\tilde{v}^*_{\tilde{a}\tilde{b}}(\textbf{p},\sigma)\tag{5.7.20} $$ as $$ \pi_{ab,\tilde{a}\tilde{b}}(\textbf{p})=P_{ab,\tilde{a}\tilde{b}}(\textbf{p},\sqrt{\textbf{p}^2+m^2})\tag{5.7.22} $$



and regroup terms to turn this into a function of $\textbf{p}$ only:


$$ \pi_{ab,\tilde{a}\tilde{b}}(\textbf{p})=P_{ab,\tilde{a}\tilde{b}}(\textbf{p})+2\sqrt{\textbf{p}^2+m^2}Q_{ab,\tilde{a}\tilde{b}}(\textbf{p}) $$


Where


$$ P(-\textbf{p})=(-)^{2A+2\tilde{B}}P(\textbf{p}) $$ $$ Q(-\textbf{p})=-(-)^{2A+2\tilde{B}}Q(\textbf{p})\tag{5.7.26} $$


But in all these cases I do not see how we can preferentially multiply $(-1)^{2j}$ to the $e^{-ip(x-y)}$ term alone. In the case of Novoshilov, because


$$ \hat{p}\equiv-i\partial $$


His "page 77" simply reads to me as:


$$ \left[ \psi_{\sigma}(x),\psi^+_{\sigma'}(y) \right]_{\mp}= \frac{1}{(2\pi)^3}\int{\frac{d^3p}{2p_0}D^J_{\sigma\sigma'}}\left(\frac{p}{m}\right)\{e^{ip(x-y)}\pm\eta\eta^*e^{-ip(x-y)}\} $$


$$ =\frac{1}{(2\pi)^3}D^J_{\sigma\sigma'}\left(\frac{p}{m}\right)\int\frac{d^3p}{2p_0}\{e^{ip(x-y)}\pm\eta\eta^*e^{-ip(x-y)}\} $$


$$ =\frac{1}{(2\pi)^3}D^J_{\sigma\sigma'}\left(\frac{-i\partial}{m}\right)\int\frac{d^3p}{2p_0}\{e^{ip(x-y)}\pm(-1)^{2j}\eta\eta^*e^{-ip(x-y)}\} $$



Wherein this $(-1)^{2j}$ term simply appears on the inverse exponential as if by magic. So too, does Weinberg's proof run into difficulty. The statement $(5.7.19*)$ only makes sense if the form of the integral in the (anti)commutator returns $Q(-\textbf{p})$ and $P(-\textbf{p})$ for the $Q(p)*e^{-ip(x-y)}$ and $P(p)*e^{-ip(x-y)}$ terms only. But I do not see how this occurs. Why would not both $e^{\pm ip(x-y)}$ terms simply act as $F(p)e^{\pm ip(x-y)}$ and not one preferentially as $F(-p)$?


In other words, why does the $(-1)^{2j}$ term only survive on one component of the commutator or anticommutator?


In Streater and Wightman's treatment, where as best I can tell the issue comes down to the number of dotted and undotted indices in the irreducible-Lorentz-representation spinors, this same sort of "preferential" action is expressed in $(4-51)$, where the authors write that "[...this result] is a consequence of the transformation law of [the holomorphic function] $\hat{W}$ under the group $SL(2,C)\otimes SL(2,C)$..." And this is borderline unintelligible to me.


Does anyone know why what seems to be a violation of the associative property is allowed here? I am likely missing something specific, and I would be greatful for any and all help towards the right direction.




What is the need of complex functions in wave analysis?



It is commonly known that waves can be express in terms of sine or cosine function.
But when I study further, I seen that for analyising the waves, it is common to use complex functions in the form
$$y=y_{_0}e^{i(kx-\omega{t})}$$
where $y_{_0}$ is the amplitude, $k$ is the wave number, $\omega$ is the angular velocity and $x$ & $t$ are position an time respectively. Ofcourse, I know that the function $e^{ix}$ can be written in the form $\cos{x}+i\sin{x}$ and so it is a periodic function with period $2\pi$ but my question is for what purpose we define it in terms of complex numbers? It seem to be more convenient to use real functions for real variables such as amplitude, electric and magnetic field of an electro magnetic wave, and also in quantum mechanics. What actually this interpretation means or what is the advantage of such functions?




electric circuits - Is Ohm's law obeyed in power transmission?


We learnt in high school that according to Ohm's law $V/I=R$. We also learnt that during power transmission in an electric line $P=VI$ and that in order to minimize loss voltage is raised. As a consequence current is reduced. This contradicts with the Ohm's law which states that $V$ is proportional to $I$. What is the difference?



Answer



The usable voltage out of a battery or a generator is $V = E - i R_{wire} $ where $E$ is the rated voltage (being 12V or 768kV), $i$ is the current and $R_{wire}$ is the resistance of the wire. The current is usually found by the load on the circuit as $i=\frac{V}{R_{load}}$.


So alltogether with power $P=i V$ we have


$$ V = \frac{E R_{load}}{R_{load}+R_{wire}} \\ P = \frac{E^2 R_{load}}{(R_{load}+R_{wire})^2} $$


Only when the wire resitance is negligible you get $$V=E \\ P=\frac{V^2}{R_{load}} $$


Friday 26 December 2014

quantum field theory - Are Goldstone bosons necessarily spin-0 particles?


EDIT: Bosonic fields with spin $s>0$ transform non-trivially under Lorentz transformation. Hence, if any of them acquires a VEV, that would violate Lorentz invariance as I learnt from the posts 1 , 2 , and 3. Does it mean that Goldstone bosons, obtained after spontaneous symmetry breaking, are necessarily spin-0 particles in a theory which respects Lorentz invariance? If yes, does it mean that one can have $s>0$ bosonic Goldstone particles in non-relativistic field theories such as condensed matter systems?



Answer



No. Magnons are spin-1 Goldstone bosons.


quantum mechanics - Is double-slit experiment dependent on rate at which electrons are fired at slit?


I am a mathematician and I am studying string theory. For this purpose I studied quantum theory. After reading Feynman's book in which he described the double-slit experiment (Young's experiment) I was wondering if I send one electron per day or per month (even more), could I see the interference pattern?



Answer




Yes, the interference pattern will occur, although you'll have to wait a while to be able to see it. As long as the average arrival time between photons is markedly greater than the travel time from slit to detector, the actual rates don't matter - each photon interacts with the slits by itself.


This URL shows such an experiment, in which a laser beam was so attenuated that the separation between photons was in the kilometer range, while the target-detector distance was in the meter range, and an image intensifier was used to detect photon positions. After about 500,000 photons had been detected, the result was


enter image description here


definition - What is the degree of freedom?


In here, https://en.wikipedia.org/wiki/Degrees_of_freedom_(mechanics),
the degree of freedom is defined as "the number of independent parameters that define its configuration." So, if $N$ particles are in the system, the degree of freedom is $3N$.


But here, https://en.wikipedia.org/wiki/Degrees_of_freedom_(physics_and_chemistry),
defined as "The set of all dimensions of a system is known as a phase space, and degrees of freedom are sometimes referred to as its dimensions." In this sense, the degree of freedom is $6N$.


What is the definition of the degree of freedom?




Answer



A degree of freedom is basically a system variable that's unbound (free).


We say "degrees of freedom" rather than just "variables" to clarify that we're referring to that freeness of the system rather than a specific count of variables.


For example, consider a 2-D grid with a particle at $\left(x,y\right)$. We can also refer to that particle's location in terms of polar coordinates, $\left(r,\theta\right)$. So that's 4 variables: $\left\{x,y,r,\theta\right\}$; however, at most we can only fill in 2 of them. This is what we mean by the system having "2 degrees of freedom": sure there're more than 2 variables, but only 2 of them are free.


Example: $3n$ vs. $6n$ from the question


If you have a system of $n$ particles, then their positions have $3n$ degrees-of-freedom:




  • 1 for each $x$ coordinate;





  • 1 for each $y$ coordinate; and




  • 1 for each $z$ coordinate.




But what if you want to include their velocities? Then you need $3n$ more for the components of velocity: $v_x$, $v_y$, and $v_z$. That brings it to $6n$.


However, neither $3n$ nor $6n$ is particularly fundamental or worth memorizing. You'll generally want to think out the number of degrees of freedom every time you consider a physical situation.


angular momentum - Force responsible for increasing a spiraling object's tangential velocity



Suppose we have a small mass attached to a string that has been fed through a hole in the friction-less table on which the mass is rotating. Pulling the string downwards thus decreases the radius of the mass's circular motion.


In this scenario, several things seem to be true:



  1. The system's angular momentum is conserved.

  2. The kinetic energy of the mass increases as the radius of its circular motion decreases.

  3. Pulling on the string does work, which is presumably translated into the kinetic energy gained by the mass.



It would appear that the object is in circular motion at any given instant, which implies that the tension force provided by the string is always perpendicular to the velocity of the mass. But what force or force component is then responsible for increasing the tangential velocity of the mass as the radius decreases?



Answer




It would appear that the object is in circular motion at any given instant, which implies that the tension force provided by the string is always perpendicular to the velocity of the mass.



Mostly true, but not quite. If the string is held steady, then the object is in circular motion. But by pulling the string harder than that, it moves off of circular by a tiny amount.


Since it starts at one distance and reaches a closer point, it cannot be moving in circles but in a spiral. So the path has a (small) radial component. That means the radial string and the not-exactly-tangential velocity vector are no longer exactly perpendicular.


electromagnetic radiation - x-ray interaction with atmosphere


Why x-ray are stopped by atmosphere while they are more energetic than UV or IR?



They certainly interact with atmosphere but I can't understand which phenomenon stop them.




Thursday 25 December 2014

cipher - The shifted library


I'm keeping an alphabetically sorted (somehow) collection of books in my library. Unfortunately somebody shifted things around a bit and then deleted some stuff in all of the resulting pairs. Only the 26th pair is a correct match, I believe, although that's been scrambled even more and has become completely unrecognizable now. Can you figure what it used to be? The numbers might help you.


Here are the 25 shifted pairs:



{3,8} $\implies$ CHL DC - FT
{6,1} $\implies$ MLYB - GAXPCAS
{9,9} $\implies$ JAMJOYC - WUTHRINGHIGHT
{4,3} $\implies$ FODOR DOTOVK - UL
{9,2} $\implies$ JLG - CMPUHMT
{9,5} $\implies$ DLDFO - THCTCHRTHRY

{1,9} $\implies$ ????? - BINSNCUS
{1,11} $\implies$ MAMLVLL - TGTY-FOU
{9,11} $\implies$ GERGE ELT - MBY-DCK
{1,6} $\implies$ JONTN SWFT - MDDLEMRC
{4,1} $\implies$ MKWN - GULLVE'SVELS
{1,4} $\implies$ W - THDVNTUFHUKBYFNN
{4,5} $\implies$ BMSTK - LICINWNDLND
{14,17} $\implies$ MIGE E EVNTES -
{5,8} $\implies$ JRRLK - DQUX
{8,2} $\implies$ AYPW - RDFRIGS

{17,1} $\implies$ GBRLGRRQZ - DNTOTHSOFT
{3,7} $\implies$ P - ONUNDDYSOFSOITUD
{7,2} $\implies$ JEUSE - OKILLMOCKIGBIRD
{4,7} $\implies$ LOTOLSTOY - PRIDANDPRJUDIC
{5,3} $\implies$ THUCODOYL - KI
{1,9} $\implies$ M - THEHUFTHESEES
{13,14} $\implies$ ERNES HEMNGWY - LOL
{4,12} $\implies$ XRU - THOTH
{4,7} $\implies$ AYLLY - TTUKT




And the 26th is:
{10,N/A} $\implies$ TTKJTKZU - DFJ



Answer



Continuing from @Roger's answer, which shows how to get meaning out of the groups of letters,



the numbers refer to letter positions in the author names.

The first numbers end up spelling the alphabet (3rd in Charles Dickens is A, 6th in Emily Bronte is B, ...)

The second numbers, still taken from the author, spell out
DECODEWITHMISSINGLASTNAME

The "missing last name" refers to the 7th entry, which is George Orwell; it was missing (or just question marks) in the original list.
Using a Vigenere cipher with ORWELL as the key, we can decode the 26th entry as:
FCOFIZLD - HBY
which can be filled in to give:

F SCOTT FITZGERALD - THE GREAT GATSBY



What is meant by quantum coherence in the context of a two-level atomic system?



What is meant by quantum coherence in the context of a two-level atomic system?



Answer



When you prepare a pure quantum state of a two-level system, $| \psi \rangle = a |0 \rangle + b |1 \rangle$, the associated density matrix will be $$\rho = \begin{pmatrix} |a|^2 & a b^* \\ a^* b & |b|^2 \end{pmatrix}.$$ The non-diagonal terms are usually called the 'coherent' terms, which come from having a pure state instead of a statistical distribution of $|0\rangle$ with probability $|a|^2$ and $|1\rangle$ with probability $|b|^2$.


Interactions with the environment wash away these terms and make you lose coherence (they make the non-diagonal terms $\to 0$). The time scale associated is what is called the coherence time.


Personally, I think a better characterization of this idea is purity, which is given by $\text{Tr}(\rho^2)$ and is basis indepdendent.


general relativity - What is the relationship between the formal definition of a tensor and the frequently discussed notion of a "higher order matrix"?


I've been doing some self study on the principles of tensors & manifolds in preparation for a first course in general relativity. I tend to learn better when presented with the full mathematical formalism of a topic, and to that end I've been pursuing more mathematically rigorous texts (Geometrical Methods of Mathematical Physics by Schutz, Tensor Analysis on Manifolds by Bishop, and the first few chapters of Wald's General Relativity).


All of these books define a tensor as follows (with some slight variation):



Definition 1 Let $V$ be a vector space over the field $\mathbb{F},$ and let $V^*$ be its dual space. A tensor on $V$ is a multilinear map $T : V^* \times V^* \times \dots \times V^* \times V \times V \times ... \times V \rightarrow \mathbb{F}$.



This makes perfect sense, and the following math is understandable using this definition.



The question is this: I frequently see people explain tensors as "like higher order vectors." I have seen more than once the following claim:



"Definition" 2 A scalar is a 0th-order tensor, a vector is a 1st-order tensor, a matrix is a 2nd-order tensor, and you can keep going from there, thinking of tensors as an extension of the concept of a matrix.



So the question is how are these two definitions related?


Using the first definition, a 1st order tensor would be, presumably, a linear map $T : V \rightarrow \mathbb{F}$. And I suppose such a linear map $T : V \rightarrow \mathbb{F}$ is a vector in the dual space of $V$, so that does match up with the second definition's claim that a first order tensor is a vector. But it doesn't sit quite right with what the second definition seems to be implying.


This second definition raises all sorts of other questions: certain sets of matrices can be made into vector spaces. So what distinguishes a matrix and a vector in the wording of the second definition?


Is this second definition perhaps a special case, where vectors and matrices of real numbers correspond to tensors over specific vector fields? Or is it something else entirely?




acceleration - How does dark energy allow the universes expansion to accelerate?


I was wondering how dark energy is effecting the rate at which the universes expansion occurs.



Answer




How does dark energy allow the universes expansion to accelerate?




I hope that it is clear to the questioner and the readers that the horse pulling dark energy is the experimental observation that the expansion of the universe is accelerating. Dark energy is proposed as the reason why the expansion is accelerating. It is called "dark" because it is not interacting with normal matter, but only with the space time structure.


Take a three dimensional explosion in space. The fragments will fly off and steadily expand from each other. If one observed that their expansion was accelerating, it would mean that extra energy was appearing in the system ( explosives in the fragments? ). The same reasoning applies. The geodesics in space were supposed to follow the Big Bang (after the inflationary period) expanding at a steady rate imposed by the initial impetus, but in an accurate model decelerating slowly because of the weak effect of gravity which is attractive. The observation that the expansion is accelerating introduced the simple concept of extra energy entering the local four dimensional space , and called "dark energy". It is still a matter for research.


classical mechanics - When/why does the principle of least action plus boundary conditions not uniquely specify a path?



A few months ago I was telling high school students about Fermat's principle.


You can use it to show that light reflects off a surface at equal angles. To set it up, you put in boundary conditions, like "the light starts at A and ends at B". But these conditions by themselves are insufficient to determine what the path is, because there's an extra irrelevant stationary time path, which is the light going directly from A to B without ever bouncing off the surface. We get rid of this by adding in another boundary condition, i.e. that we only care about paths that actually do bounce. Then the solution is unique.


Of course the second I finished saying this one of the students asked "what if you're inside an elliptical mirror, and A and B are the two foci?" In this case, you can impose the condition "we only care about paths that hit the mirror", but this doesn't nail down the path at all because any path that consists of a straight line from A to the mirror, followed by a straight line to B, will take equal time! So in this case the principle tells us nothing at all.


The fact that we can get no information whatsoever from an action principle feels disturbing. I thought the standard model was based on one of those!


My questions are



  • Is this anything more than a mathematical curiosity? Does this come up as a problem/obstacle in higher physics?

  • Is there a nicer, mathematically natural way to state the "only count bouncing paths" condition? Also, is there a "nice" condition that specifies a path in the ellipse case?

  • What should I have told that student?




Answer



Multiple classical solutions to Euler-Lagrange equations with pertinent/well-posed boundary conditions (such solutions are sometimes called instantons) are a common phenomenon in physics, cf. e.g. this related Phys.SE post and links therein.


In optics, it is well-known that already e.g. two mirrors can create multiple classical paths.


cosmology - Can space expand with unlimited speed?


According to this article on the European Space Agency web site just after the Big Bang and before inflation the currently observable universe was the size of a coin. One millionth of a second later the universe was the size of the Solar System, which is an expansion much much faster than speed of light. Can space expand with unlimited speed?



Answer



Yes, the expansion of space itself is allowed to exceed the speed-of-light limit because the speed-of-light limit only applies to regions where special relativity – a description of the spacetime as a flat geometry – applies. In the context of cosmology, especially a very fast expansion, special relativity doesn't apply because the curvature of the spacetime is large and essential.


The expansion of space makes the relative speed between two places/galaxies scale like $v=Hd$ where $H$ is the Hubble constant and $d$ is the distance. When this $v$ exceeds $c$, it means that the two places/galaxies are "behind the horizons of one another" so they can't observer each other anytime soon. But they're still allowed to exist.


In quantum gravity i.e. string theory, there may exist limits on the acceleration of the expansion but the relevant maximum acceleration is extreme – Planckian – and doesn't invalidate any process we know, not even those in cosmic inflation.


Understanding scattering cross section


I think I might have a serious misunderstanding of some concepts to do with scattering cross sections and would really appreciate any help.



As far as I can tell the differential cross section is basically a measure of what fraction of particles get scattered into some solid angle. To get the total cross section all we need to do is then integrate over all solid angles. However I can't determine why this wouldn't just be equal to 1 i.e. that we are guaranteed to find the scattered particle at some solid angle. An attempt in my mind to fix this would be to not include the infinitesimal on axis solid angle. However, unless the differential cross section was infinite their, it wouldn't make any difference.


Could anyone tell me what's gone wrong with my reasoning? Much appreciated!



Answer



Scattering cross sections can have other parameters besides angle. For example you commonly have cross sections vs. angle and final energy, $d\sigma/d\Omega\, dE_\text{final}$. This might reflect the fact that in a generic elastic scattering process forward-scattered particles tend to retain most of the beam energy, while backwards-scattered particles must deposit a lot of energy and momentum in the target.


If you're doing purely elastic scattering from a crystal, for example Bragg scattering, the cross section $d\sigma/d\Omega$ will essentially vanish except for some very particular angles. It's the values of those magic angles that are useful, and the relative strengths of scattering at one angle versus another. The fact that the entire incident beam has to go somewhere can sometimes be a useful check, but more often the unscattered / forward-scattered beam gets sent to some beam dump and ignored.


enigmatic puzzle - Consecutive Vowels


Part of the Fortnightly Topic Challenge #35: Restricted Title 1




Potentially NSFW: Consecutive Vowels




  • Being a native speaker (6)





  • _____ goin' nowhere! (3 4)




  • Democratic Debbie (7)




  • Major airplane manufacturer (6)





  • Like the people in the linked comic (assuming bald means male and long hair means female) (8)




  • Like Hal or Cortana (2)




  • Go extinct (3 3)





  • A sign of economic prospering (3 6)




  • Victory (7)




  • He ran with Palin (6)




  • Comedy series by Szekely (5)





  • Closest Hawaiian island to Hawaii (4)




  • ______ in flames (2 2)




  • Not weaknesses (9)





  • He wrote rhymes just like these, he wrote rhymes with much ease! (2 5)




  • Possibly a kid's first existential question? (2 1 3)




  • A promise of return (3)





  • Boyfriend (4)




  • In ____ (instead of) (4 2)




  • Cute, like old people (6)




Your answer is an alternate punchline -- or maybe hovertext? -- for the linked comic.




Answer



Each of the answers



has a string of consecutive vowels somewhere in it, and only that one string of consecutive vowels. (There may be just one vowel!) In addition, those strings are always made up of different vowels.



Taking these



substrings, we can make a 5-bit binary number for each word. Take AEIOU, and write a 1 if the vowel is in that string and 0 if it is not. For instance, the first answer, FLUENT, has an E and a U, so the corresponding number is 01001.
enter image description here

Read off the binary numbers using A=1, B=2, ... Z=26 to get the phrase I WANT TO BE TOUCHING YOU - which describes both the vowels and the characters!




geometry - Join all circles together only with 6 lines


In the below image, can you draw 6 straight lines that pass all the circles?
As soon as you start drawing lines you can't take your pen up until you draw all six lines.
hint: you don't have to keep the polyline inside the square.


Edit: Best answer is the one in which straight line passes center of the circles.


enter image description here



Answer




I think this is one possible solution:



solution



gravity - What is the cause of gravitional force of attraction?


Ok,this is a silly question. But for quite long I have been thinking about this. What is gravitation all about?? My book directly writes since force of gravitation is directly proportional to the masses and inversly proportional to the square of the distance ie. $$F = G\frac{M_1 . M_2}{r^2}$$ . So is this gravitation all about? What is the cause of it? The book then says



We are not in the scope in classical approach so as to discuss the cause of this. However,as a matter of fact that if there are more than one mass,then there is gravitation as it is their fundamental property.( Cause of gravitation can be attributed to exchange of non-classical particles between the masses) .



Then it ended the talkings. What are those non-classical particles? Why and how are they exchanged? So far I have studied physics , I never found such weird phenomenon like this. I want to know what is the cause of gravitation and what is the cause of exchange of those so-called non-classical particles? And how do they do it?




Wednesday 24 December 2014

logical deduction - Not enough pylons!


Bob is a temp working with a construction company in repairing roads and barriers. One day Bob got called up to repair a barrier along a strip of highway. Bob will need 100 "Company X" pylons to divert traffic away from his site while he repairs. Being the best employee around, Bob goes into work the day before to round up the pylons he needs. After looking everywhere in the warehouse, Bob is only able to find 50 "company X" pylons! He then realized John must have taken the rest and hidden them from Bob. Bob also found a note that says his supervisor will be driving by each site tomorrow for inspection! If his supervisor sees that Bob is not following protocol of using proper pylons, he will surely be fired!


The next day after Bob was clever and did some tricks he was able to finish the repairs without getting fired.


How did Bob save his job?


Notes:




  • Bob has the rest of the day to prepare, and has the supplies in the warehouse to his disposal (But there are no other "Company X" pylons).





  • If Bob tried to use hand craft fake pylons (or non "Company X" pylons), his supervisor would surely notice and Bob would not be following company rules!




  • Bob has already called John asking for pylons but John said he buried them and encrypted the location. (John won't share the encryption, he's said too much already)




  • There is shipment being made to the warehouse in 2 days with more pylons, but this will be too late for Bob.






Answer



First of all, break the rear-view mirrors of the supervisor's car (admit it, you've always wanted to do that!).
Then, split each pylon into two identical pieces, and place them along the road. Without a rear-view mirror, the supervisor won't notice that the pylons were divided!


enter image description here


Strategy for speedsolving 2048


Once one has mastered winning 2048 in general, it's often an interesting exercise to try to do it as fast as possible.


What strategies are there for speeding up 2048-solving?
For example, how can you pass the boring beginning game quickly (ex. mashing up and left),

and how do you know when to stop so you don't mess up your middlegame?




geometry - Neighboring circles


If we join two circles on a plane, each will have exactly one neighbor.


Neighboring Circles - 1


Given three or more circles, we can build a chain where each circle has exactly two neighbors.


Neighboring Circles - 2


There are also arrangements where each circle has exactly three neighbors, like the one shown here.


Neighboring Circles - 3


Is it possible to arrange a finite number of equally sized, non-overlapping circles on a plane such that each circle has exactly four neighbors?


What if the circles can have different sizes?




Answer



It is not possible when the radii are equal. Consider the topmost circle, C. In order to pack four neighbors around C, none of which are above C, you have to use the below arrangement (where C is the gray circle):


enter image description here


This means there is a circle to the immediate right and left of C. Repeating the same argument over and over shows that there is an infinite line of circles at the same level as C, contradicting the finiteness requirement.


When you allow differing radii, the below arrangement works:


enter image description here


cipher - A not very easy puzzle with an easy answer. Something that a grim puzzler might like


1f411



pirate gibberish


00000010


1f607


.


1f441


1f44f


1f411


.


Cigarette ___


lion's-bane@



ワs


22 2


65 61 73 79


160 141 162 164


"2" & right(getMethod("mixintowater"), 5)


9 (4*2) 2 8 _ (4*3) (7*4) _ 6 (9*3) _ (6*2) 2 6 (3*2)?


Hint:



Unicode, guess, binary, Unicode - Unicode, Unicode, Unicode - guess, guess+symbols,Japanese,chemistry, ASCII, ASCII, code - ...





Answer



The last line is a telephone cipher.


A regular telephone keypad looks like this:


+-----+-----+-----+
| | ABC | DEF |
| | | |
| 1 | 2 | 3 |
+-----+-----+-----+
| GHI | JKL | MNO |
| | | |

| 4 | 5 | 6 |
+-----+-----+-----+
| PQRS| TUV | WXYZ|
| | | |
| 7 | 8 | 9 |
+-----+-----+-----+
| | | |
| * | 0 | # |
| | | |
+-----+-----+-----+


Given a character in the code, if there is just a single number (e.g. 2), you press that key once, to give the first letter on that number (i.e. A). If the character is a number "times" some other n (e.g. "7*4"), you press the key n times to get the nth letter for that number (i.e. S).


Doing so, you get the message:



WHAT IS MY NAME?



Which is, of course, Brent Hackers.


quantum mechanics - Probability conservation in WKB tunneling


Suppose we have quantum mechanical plane waves of energy $E$ incident upon a one-dimensional potential barrier $V(x)$ with sloping sides.


enter image description here


One can compare the WKB solutions in the three relevant regions to asymptotics of the Airy function solutions to the linearized Schrödinger equation near the turning points to obtain matching formulae, relating $A, B$ (the coefficients of the right- and left-propagating waves, respectively, left of the barrier) to $F, G$ (the coefficients of the right- and left-propagating waves, respectively, right of the barrier). One can then construct the relevant scattering matrix $S$, defined by:



$$\left[\begin{array}{c} B\\ F \end{array}\right]=S\left[\begin{array}{c} A\\ G \end{array}\right] \, .$$


I have carried out the above computation and found that


$$S=\left[\begin{array}{cc} i\left(\dfrac{1-4e^{2\gamma}}{1+4e^{2\gamma}}\right) & \dfrac{4e^{\gamma}}{1+4e^{2\gamma}}\\ \dfrac{4e^{\gamma}}{1+4e^{2\gamma}} & i\left(\dfrac{1-4e^{2\gamma}}{1+4e^{2\gamma}}\right) \end{array}\right]$$


where


$$\gamma=\frac{1}{\hbar}\int_{x_{1}}^{x_{2}}\sqrt{2m\left[V\left(x\right)-E\right]}dx$$


measures the "strength" of the reflecting, classically-forbidden region.


This scattering matrix $S$ is in fact unitary, and hence this scattering process conserves the probability current across the barrier. But why should this be true?


I know that a solution of the Schrödinger equation will necessarily conserve probability current, but here we have both (1) assumed approximate WKB-type solutions and (2) carried out a complicated asymptotic matching procedure. It is not clear to me why, after this approximation and matching procedure, the resulting solution should still conserve probability current.


Said another way: is there an a priori reason one should expect the WKB-matched solutions to conserve probability current (in spite the fact that they are inexact), or is this simply a happy coincidence that falls out of the computation?




cipher - Decode the message enciphered in these symbols: ↗↑↙↓↘←↑↙↘↘↗↓↗←↙↓↖


↗↑↙↓↘←↑↙↘↘↗↓↗←↙↓↖


The first character of the plaintext is 'P'. The final character of the plaintext is 'E'.


Hint 1;




The second word of the plaintext is DISPUTE.



Hint 2:



NW = 'end of message'. It's an end-of-file marker.





Details (not part of the puzzle):





  • "↗↑↙↓↘←↑↙↘↘↗↓↗←↙↓↖" is the cyphertext. What is the plaintext?




  • The answer is a clue to The Security to the Party [12] (now with party soundtrack!)




  • This type of puzzle is called a Cryptogram. The Google search "How to solve a cryptogram" leads to all kinds of useful resources, hints and tips.





  • The answer is a pair of dictionary words in the English language. The puzzle is not case-sensitive. The plaintext does not contain any slang or acronyms, it does not contain any proper nouns (names), it does not contain roman numerals, it does not contain punctuation or spaces.




  • For anyone who's having trouble with fonts, and for accessibility: the cyphertext is a series of Unicode arrows pointing in different directions. The sequence is: NE N SW S SE W N SW SE SE NE S NE W SW S NW




  • Chat about this puzzle is here: http://chat.stackexchange.com/rooms/18676/pick-up-sticks I'll be popping in and out. Feel free to ask any questions there or in comments.




  • On some screens it looks like some of the arrows are in bold and some aren't. That is not the intention, it's just the way the fonts look. If you use the list of compass points above then you are at no disadvantage.







Update: Argh! I was so busy doing the cipher that I didn't notice the spelling mistake in the plaintext. Well done to nexolute for solving anyway! Future readers beware that the first word of the puzzle is (unintentionally) not a dictionary word, because it has a spelling mistake in it. Are my cheeks red!



Answer



Going in clockwise, ignoring NW, we get the following...


N  : AHOV
NE : BIPW
E : CJQX
SE : DKRY

S : ELSZ
SW : FMT
W : GNU

And let NW be the EOF character. Now this is where I'm bad with, how can we construct the final solution from these?


Using regular expression dictionary, I didn't manage to find anything, so I reduce to search with less characters and get use a little guess and check to get PATERNATY DISPUTE (should it be paternity dispute?). Not so sure if this is right?


particle physics - To what extent are lepton and quark generations tied in the Standard Model?


The Standard Model of particle physics splits both the leptons and the quarks into three generations, with mass and instability going up from the first to the third generation. These are normally displayed together, on the same rows or columns of the table of fundamental particles:



This makes some sense: each charged lepton is tied to its neutrino in most Feynman vertices it appears in, and the quarks are linked to each other by their charges, if nothing else. However, I can't think of any way in which the SM formally links, say, muons and strange quarks. It's there some explicit link with generation-specific interactions? Or is it just coincidence that there's three rungs in both ladders with increasing mass on both?




visualization - Difficulty in visualizing more than three spatial dimension



Is there is any other way than abstract mathematics to visualize higher dimensions. Physicists working in high energy physics live in higher dimensions (pun intended), with their sophisticated mathematical tools they easily work there way out in higher dimensions. Take for a simple example of a $n$-sphere $S^N$ http://en.wikipedia.org/wiki/N-sphere.



  • 0-sphere $S^0$ is a pair of points ${c − r, c + r}$, and is the boundary of a line segment (the 1-ball $B^1$).


  • 1-sphere $S^1$ is a circle of radius $r$ centered at $c$, and is the boundary of a disk (the 2-ball $B^2$).

  • 2-sphere $S^2$ is an ordinary 2-dimensional sphere in 3-dimensional Euclidean space, and is the boundary of an ordinary ball (the 3-ball $B^3$).

  • 3-sphere $S^3$ is a sphere in 4-dimensional Euclidean space.


The problem is how hard I think it is impossible for me to visualize a 3-sphere.




cosmology - The range of light


It occurs to me that the empirical evidence shows that there is a point out in space where light stops coming from.


Putting aside the expansion of the universe for a second, and focusing strictly on the evidence:




  • what would the universe look like if light had a finite range?





  • isn't that what the universe looks like?




Recall Hubble's Law,$ v = H_0D$


The range of light is $H_0 D = c$


Also recall that Edwin Hubble stressed the point that $v$ is apparent recessional velocity. It is the apparent recessional velocity, not the actual recessional velocity. He proposed that rather than Doppler shifts, these redshifts are a "new principle of nature".


Ignoring the theories and strictly examining the empirical evidence, does it make sense that light could have a finite range?




Tuesday 23 December 2014

particle physics - Naive question on quantum mechanics and uncertainty principle


This is a follow up on this question, the answer of which points towards Quantum Mechanics.
As stated I am not a phycisist so please forgive my ignorance.

I will try to understand the issue by going in small steps (questions).
Wiki says:



In quantum physics, a quantum vacuum fluctuation (or quantum fluctuation or vacuum fluctuation) is the temporary change in the amount of energy in a point in space,1 arising from Werner Heisenberg's uncertainty principle



My question here is the following (actually has 2 parts):
1) The statement the temporary change in the amount of energy is what is meant when I read (and in the answer in my previous post) that things pop out of "nothing"?
2) Reviewing the uncertainty principle in wiki the concept (as I am capable of understaning it) is that we can not know both the position and the direction of a particle/object at the same time.How can from this conclude that things can come up out of "nothing ? It is not clear to me.
If we can not know where a particle currently is since we are looking into another property of it i.e. momentum then if we start looking into its position then does that mean that it appears out of "nothing"?
If someone could help me understand this in lamens terms it would be much appreciated.




Answer



Taking the first part of your question first: how do we get the energy-time uncertainty from the position-momentum uncertainty? This turns out to be surprisingly difficult to do rigorously. Even Heisenberg was only able to give an approximate derivation of it (based on a property called Compton wavelength). I found some of rigorous derivations here, here and here, but these are utterly impenetrable for the beginner. Even the Compton wavelength argument is a bit involved, so what I'm going to give is a justification based on dimensional analysis. This doesn't prove the energy-time uncertainty relation, but it shows it is plausible.


The Heisenberg uncertainty principle relates position and momentum. Position has units of distance, e.g. metres, and from basic mechanics ditsance is velocity times time:


$$x = vt$$


Momentum has units of mass times velocity:


$$p = mv$$


So if you multiply together position and momentum (as the Heisenberg UP does) you get:


$$x \times p = vt \times mv = t \times mv^2$$


I've rearranged the right hand side slightly because kinetic energy is $1/2mv^2$, so the right hand side looks like time times energy i.e.


$$x \times p = t \times E$$



NB this doesn't prove that $\Delta x \Delta p = \Delta t \Delta E$ but it shows that it's plausible.


Now onto the second part of your question (assuming I've convinced you that the energy-time UP follows from the position-momentum UP).


First let's ask is the energy-time UP real. Yes it is, and we can observe it fairly easily. You've probably heard that if you excite an atom it will emit light as it returns to it's ground state, and the frequency of the light emitted depends on the energy difference between the excited and ground states. This creates the atomic spectrum, which is routinely used for identifying atoms. Helium was first identified in the atmosphere of the Sun using this technique. Anyhow, the lines in the atomic spectrum don't have a precise frequency. If you measure them carefully you'll find they span a range of frequencies. Part of the broadening is from mundane sources like the doppler shift, but part arises from the E-t uncertainty principle.


So the E-t uncertainty principle is real, and it means we can't be certain about the energy of an atom unless we watch it for an infinite time. But exactly the same argument means that if we take some patch of vacumm we can't be certain about it's energy unless we watch it for an infinite time. That means the energy of the vacuum must fluctuate i.e. energy must spring into existance from nothing.


You may still be a bit unconvinced, but we can actually measure this spontaneous creation in the vacuum using the Casimir effect, so we know it really happens.


Hopefully by now you're convinced about (temporary) creation from nothing, and I guess your next question is precisely what happens when a virtual particle is created. Sadly I can't give you an answer for this. We have mathematical models for the process, like Quantum Field Theory, but whether this is what eally happens, or even if "what really happens" is a meaningful question, I don't know.


Response to comment: this links up with my answer to your other question, What is meant by "Nothing" in Physics/Quantum Physics?, so I thought I'd expand this answer rather trying to put everything in comments.


Anyhow, you ask a fair question. I've taken the position that the vacuum is effectively nothing plus the vacuum fluctuations, and you're asking me how I know it's not something plus the vacuum fluctuations. Actually this is sort of where we came in with your first question in the series.


My answer is that we can do experiments on the vacuum to see what's there. For example we can measure the vacuum fluctuations using the Casimir effect, and we get the answer our theory predicts. We can shine light through the vacuum to see if there's anything there, and we can weigh the vacuum (i.e. see if the vacuum has any gravitational attraction). In all cases we get the results our experiment predicts, and that's why I say the vacuum is effectively nothing plus the vacuum fluctuations.


You could argue that there is something present that we haven't worked out how to detect yet, but without any theoretical backing for this it's like saying there are fairies at the end of the garden that we haven't worked out how to detect yet!



Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...