Friday, June 30, 2017

newtonian mechanics - Why an accelerometer shows zero force while in free-fall


I'm trying to explain the effect in the title a non-physics person (of chemistry background) and I found it very hard to explain in simple, intuitive ways. I'm not a physicist either, so following the famous quote If you can't explain it simply, you don't understand it well enough, I decided to ask :)



I was playing with a 3-axis MEMS accelerometer on a device and I had developed a PC program to visualise different sensor values in real time. The accelerometer proved very fun to play with - as you are rotating the device around, the gravity-induced force transfers between the three axes, but also responding to motions, etc.


Then I tried throwing the device in the air and noticed that while the device was in flight, all 3 axes showed zero - as if the connection was off and I was reading nothing. In fact, the connection was fine and the values I read were very close, but not exactly zero. Then I thought - "but of course, in free fall you don't feel any acceleration, so the accelerometer shows zero". This is the part I find hard to explain to the other person - why the device has force acting on it (gravity), but the sensor inside "sees" no force applied.



An example of an intuitive type of answer I was trying to construct are round soap bubbles - they become squished while resting on a surface (same as an accelerometer laying on a table shows 1g in the relevant axis), but when allowed to fall freely, they become round, no squishing (0g on the accelerometer). But the soap bubbles experience a ton of air resistance, as well as wobbling, so it isn't a very good example.



I'm trying to get an explanation that doesn't involve reference frames, as we both don't have an intuition about them. Please excuse me if I'm not using the right terms, as I said I'm not well-versed in physics.



Answer



Since you've asked for an explanation without involving reference frames, I'll base my explanation on more familiar concepts of force and acceleration.


A typical accelerometer could be modeled as a mass on a spring. The position of the mass, when the spring is not compressed or stretched, is defined as zero and corresponds to zero reported acceleration.


When the spring is compressed or stretched, the displacement of the mass from its zero position is measured (using one of many available techniques) and reported as a positive or negative acceleration.


So, what is reported is a displacement of the mass, which may or may not correspond to the acceleration of the accelerometer that we would measure externally.


The diagrams below illustrate how it works in x or y directions, where gravity does not play a role.


enter image description here



Whenever the body of the accelerometer accelerates, it pushes or pulls the spring, which, in turn, pushes or pulls the mass. The force required to accelerate the mass, ma, will cause the spring to compress or stretch, and the resulting displacement of the mass will be measured and reported as a positive or a negative acceleration.


For z direction, illustrated on the diagrams below, the dynamics are different.


enter image description here


At rest, the spring is compressed or stretched to counter the weight of the ball.


In a free fall, when all parts of the accelerometer experience the same acceleration, the mass accelerates due the gravity - not due to the push or the pull of the spring. Therefore, the spring does not experience any forces, hence, there is no compression or stretching and no displacement, hence, the reported acceleration is zero.


This description would not be entirely accurate, if the resistance of the air was taken into account. Since such resistance would slightly reduce the acceleration of the falling accelerometer, the spring wold have to compress or stretch a bit to adjust (reduce) the acceleration of the mass accordingly.


newtonian mechanics - Why does this object periodically turn itself?


See below gif image taken from here.



enter image description here


Or see this Youtube video about 30 sec in.




  1. Is this a real effect?




  2. Why does it seem to turn periodically?





  3. Can it be explained by classical mechanics alone?




  4. Is there a simple equation that models this behaviour?





Answer



It's a classical mechanics effect for sure although a really interesting one. Following links on "Dzhanibekov effect" one gets at Marsden and Ratiu's "Introduction to Mechanics and Symmetry" Chapter 15 Section 15.9 "Rigid Body Stability" treating this with use of the Casimir functions.


From remark 1: A rigid body tossed about its middle axis will undergo an interesting half twist when the opposite saddle point is reached.


Here is another and more profound example under weightless conditions.



http://www.youtube.com/watch?v=L2o9eBl_Gzw


This seems to be a home experiment where a guy throws the spinning object upwards.


http://www.youtube.com/watch?v=3VwS5ykAUHI


And this seems to be a computer simulation.


http://www.youtube.com/watch?v=LR5hkgfRPno


There is a related unstable orbit effect which you can try out easily yourself with a tennis racket. A treatment due to Ashbauch Chicone and Cushman is here:


Mark S. Ashbaugh, Carmen C. Chicone and Richard H. Cushman, The Twisting Tennis Racket, Journal of Dynamics and Differential Equations, Volume 3, Number 1, 67-85 (1991). (One time found at http://math.ucalgary.ca/files/publications/cushman/tennis.pdf which is no longer a working link.)


http://www.youtube.com/watch?v=4dqCQqI-Gis


Which groups can be lattice gauge groups?


Let me state first off that in this question I am most interested in lattice gauge theories, and not necessarily with Fermion couplings. But if Fermions and continuum gauge theories can also be addressed in the same breath, I am interested in them too.



I have seen many different gauge groups for both lattice gauge theories and continuum field theories---SU(N), O(N), and Z_N being the most common---but I do not know whether there is a systematic way to generate a gauge theory for an arbitrary group, or even whether it is sensible to talk about gauge theories for arbitrary groups. So my first question is: Can any group be a gauge group for some theory? If not, why? If so, to what extent is the structure of the gauge theory determined by the group?


Related to the last point, is there a "canonical" choice of action (or Boltzmann weight) associated to a given gauge group? For example, in this paper (eqns. 5 and 7) I see that they define a Boltzmann weight for an S_N lattice gauge theory in terms of a sum of characters of the group. Is such a prescription general?



Answer



Yes, any group can be a gauge group. To each oriented edge of the lattice you assign a group element, with the opposite orientations of an edge associated to inverse group elements. The observables are conjugacy classes of products of group elements around a loop of oriented edges. The most basic such observable is the product of group elements around an elementary square (a "plaquette") of the lattice. The action is typically some function of the conjugacy class of the product around a plaquette, summed over all plaquettes in the lattice. The only "canonical" action I know of is the "delta action", where the conjugacy class of the group identity element has one value and all other conjugacy classes have another value (this works because in any group the identity element is in its own conjugacy class).


References:


Phase structure of non-Abelian lattice gauge theories


Monte Carlo study of Abelian lattice gauge theories


Phase structure of lattice gauge theories for non-abelian subgroups of SU(3)


electromagnetism - What's a good reference for the electrodynamics of moving media?


The answer to a previous question suggests that a moving, permanently magnetized material has an effective electric polarization $\vec{v}\times\vec{M}$. This is easy to check in the case of straight-line motion, using a Lorentz boost.


I suspect this formula is still correct for motion that is not in a straight line, but I'm not interested in reinventing the wheel. Does anyone know of a textbook or journal article that derives this $\vec{v}\times\vec{M}$ term? Even better, does anyone know of experimental observation of this effect?



EDIT:
Followup question: What is the electric field generated by a spinning magnet?



Answer



To close the loop, Andrew, the answer to your newest question is:


The best and most famous reference about the electrodynamics of moving bodies is



Einstein, Albert (1905-06-30). "Zur Elektrodynamik bewegter Körper". Annalen der Physik 17: 891–921. See also a digitized version at Wikilivres:Zur Elektrodynamik bewegter Körper.



The English translation, "On the Electrodynamics of Moving Bodies", is here:




http://www.fourmilab.ch/etexts/einstein/specrel/www/



The content of this paper became known as the special theory of relativity. I am just partly joking because for uniformly moving media, the Lorentz boost to the rest frame is still the most natural way to proceed.


optics - Why are most metals gray/silver?


Why do most metals (iron, tin, aluminum, lead, zinc, tungsten, nickel, etc.) appear silver or gray in color? (What atomic characteristics determine the color?)


What makes copper and gold have different colors?




vacuum - How does space affect the human body (no space suit, no space craft)


How does "outer space" affect the human body? Some movies show it as the body exploding, imploding or even freezing solid.


I know space is essentially a vacuum with 0 pressure and the dispersion of energy makes it very cold. So are the predictions above accurate?



Answer



There have actually been cases of (accidental!) exposure to near-vacuum conditions. Real life does not conform to what you see in the movies. (Well, it depends on the movie; Dave Bowman's exposure to vacuum in 2001 was pretty accurate.)


Long-term exposure, of course, is deadly, but you could recover from an exposure of, say, 15-30 seconds. You don't explode, and your blood doesn't immediately boil, because the pressure is held in by your skin.


In one case involving a leaking space suit in a vacuum chamber in 1965:



He remained conscious for about 14 seconds, which is about the time it takes for O2 deprived blood to go from the lungs to the brain. The suit probably did not reach a hard vacuum, and we began repressurizing the chamber within 15 seconds. The subject regained consciousness at around 15,000 feet equivalent altitude. The subject later reported that he could feel and hear the air leaking out, and his last conscious memory was of the water on his tongue beginning to boil




(emphasis added)


UPDATE: Here's a YouTube video regarding the incident. It includes video of the actual event, and the test subject's own description of bubbling saliva.


Another incident:



The experiment of exposing an unpressurized hand to near vacuum for a significant time while the pilot went about his business occurred in real life on Aug. 16, 1960. Joe Kittinger, during his ascent to 102,800 ft (19.5 miles) in an open gondola, lost pressurization of his right hand. He decided to continue the mission, and the hand became painful and useless as you would expect. However, once back to lower altitudes following his record-breaking parachute jump, the hand returned to normal.



If you attempt to hold your breath, you could damage your lungs. If you're exposed to sunlight you could get a nasty sunburn, because the solar UV isn't blocked by the atmosphere (assuming the exposure happens in space near a star). You could probably remain conscious for about 15 seconds, and survive for perhaps a minute or two.


The considerations are about the same in interstellar or interplanetary space, or even in low Earth orbit (or a NASA vacuum chamber). The major difference is the effect of sunlight. As far as temperature is concerned -- well, a vacuum has no temperature. There would be thermal effects as your body cools by radiating heat, but over the short time span that you'd be able to survive, even intergalactic space isn't much different from being in shadow in low Earth orbit.


Reference: http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970603.html


Thursday, June 29, 2017

optics - Frequency Chirp, Instantaneous Frequency and Photons


I understand ideas coming from Optics, such as the concept of frequency chirp and instantaneous frequency, and their use in nonlinear optics. However, I am struggling in giving them an intuitive quantum interpretation in terms of photons, linking the classical description (that is typical, for example, of high intensity ultrashort laser pulses) with a quantum interpretation of the same phenomena.


As far as I know, a laser pulse can be written as: $$ E(t) =A(t) e^{iω_0t}$$ where $A(t) $ is the envelope of the pulse and $ω_0$ the carrier frequency. Fourier transforming this expression, we can obtain the spectrum of the pulse: it will be centered around the frequency $ω_0$ and its shape will be determined by the Fourier transform of the envelope. In a short pulse, therefore, many different frequencies can be present and, in my intuitive mental representation of this phenomenon, the pulse will be composed by many different photons, each one with a frequency belonging to the spectrum, and the number of photons at that frequency will be proportional to the intensity of that spectral component.


The first problems arises when I consider the concepts of frequency chirp and of transform limited pulse. As far as I know, a transform limited pulse is the shortest pulse that can be obtained with a certain spectrum, while in chirped pulses the components at different frequency will have different arrival times, determining a delay between the various components that will make the pulse considered longer than the associated transform limited one. As a consequence of these considerations, I am thinking that in a transform limited pulse the photons at all the different frequencies of the spectrum will be spreaded over the whole pulse (that will be shorter), while if the pulse is chirped we will have regions of the pulse (both in time and in space) in which photons at a certain frequency are more present (are measured with an higher probability?) with respect to photons at a different one, thus determining a longer pulse. Is this representation meaningful and/or correct? Is there any way to obtain an intuitive quantum description that is coherent with the classical one?



The second problem comes from the concept of instantaneous frequency, that is defined as the temporal derivative of the phase of the pulse, thus being intimately related to the fact that we are considering a pulse with a certain envelope in time. If the previous consideration on chirped pulses is correct, I would like to intuitively represent the idea of instantaneous frequency as the frequency associated to the photons that would be more probable to be measured in a given instant of time in the pulse. Photons at a certain frequency, in a chirped pulse, will thus be measured with an higher probability in the portion of the pulse in which their frequency corresponds to the instantaneous frequency. Is this representation correct? What is the physical meaning of the instantaneous frequency? How is the concept of carrier frequency be related to this intuitive quantum representation?


This question arises from the fact that in university courses on nonlinear optics several phenomena, such as sum frequency generation, difference frequency generation, second harmonic generation, four-wave mixing and optical Kerr effect are represented in terms of photons at a given frequency (usually, the carrier frequency of the beam considered; is it because it is the more probable one for that pulse?), while to demonstrate them we usually use quantum arguments involving quantities such as the frequency chirp and the instantaneous frequency.



Answer



A recent post explicitly solves a model of a massless quantum scalar field (a proxy for the quantum EM field) driven by a classical current. The current is an arbitrary function of time and space except that it's limited to a finite time-interval, so the solution includes waveforms with time-varying frequency as a special case. The following conclusions are based on that solution, and some highlights from the math are outlined at the end. The conclusions described here agree with the earlier answers by Andrew Steane and S. McGrew.


I'll use units in which the speed of light and Planck's constant are both equal to $1$.



...in my intuitive mental representation of this phenomenon, the pulse will be composed by many different photons...



Yes. This part is correct, with the caveat that the pulse may be more accurately described as a quantum superposition of different numbers of photons. This comment, like all of the following comments, assumes that the simple model cited above is an adequate model for whatever light-producing devices are assumed in the question.




...the pulse will be composed by many different photons, each one with a frequency belonging to the spectrum...



It is more accurate to say that each individual photon spans all the frequencies belonging to the spectrum. A single photon may be in a quantum superposition of many different wavenumbers $\mathbf{p}$, and hence in a quantum superposition of many different frequencies $\omega=|\mathbf{p}|$. Every photon has the same "profile" in the wavenumber domain (aka momentum domain), and the special superposition of different numbers of these identical photons is what accounts for the essentially-classical character of the pulse. This is quantified below.



...the number of photons at that frequency will be proportional to the intensity of that spectral component.



This can be a correct statement if it is interpreted carefully. The number of photons detected at a given frequency, using a frequency-selective detector, will be proportional to the intensity of that spectral component. Prior to detection, each individual photon spans a broad bandwidth. This is analogous to the statement that, in a double-slit experiment, each individual photon goes through both slits, even though an individual photon will only be detected at one of the slits if detectors are placed within the slits.


Of course, the simple fact that real detectors must be localized in space and must effectively "integrate" over a finite time-interval implies that the detector itself can only distinguish different frequencies down to some finite resolution. This limitation is already present in classical physics, and it is still present in the quantum picture.



...if the pulse is chirped we will have regions of the pulse (both in time and in space) in which photons at a certain frequency are more present (are measured with a higher probability?) with respect to photons at a different one...




As explained above, this statement can be made accurate by modifying it slightly, like this: "if the pulse is chirped we will have regions of the pulse (both in time and in space) in which a frequency-selective photon-counter photons would register more photons at some frequencies than at others."



Is there any way to obtain an intuitive quantum description that is coherent with the classical one?



Yes. This is the subject of the highlights shown below from the exact solution in the previously-cited post.



The second problem comes from the concept of instantaneous frequency...



Since an individual photon does not have an individual frequency, the notion of "instantaneous frequency" is no more (and no less) problematic than it is in the classical picture. An individual photon has a time-dependent profile, much like a classical wave does; and in fact the properties of the classical wave may all be recovered from the properties of a single photon, with the understanding that the essentially-classical wave involves many identical copies of that photon. This is quantified below.




Photons at a certain frequency, in a chirped pulse, will thus be measured with a higher probability in the portion of the pulse in which their frequency corresponds to the instantaneous frequency.



The comments already made above apply again here. The concept of photons at a certain frequency is not really accurate, just like it's not accurate to describe a classical chirped waveform (say) as being at a certain frequency. With several caveats, though, this can be turned into an accurate statement. Some of these caveats are described after the next excerpt...



What is the physical meaning of the instantaneous frequency? How is the concept of carrier frequency be related to this intuitive quantum representation?



After the relationship between the classical and quantum pictures is reviewed below, these questions will be implicitly answered: the concepts of instantaneous frequency and carrier frequency apply just as well to each individual photon as they do to the overall essentially-classical pulse, with the understanding that any application of measurement can only produce one of the possible outcomes defined by that particular measurement. If we apply a measurement that asks what the photon's frequency is, then we'll get some frequency as the answer. That does not imply that the photon was restricted to that frequency prior to the measurement; we know this because of experiments like those described in another post.


Even for a classical wave, the concept of "instantaneous frequency" needs to be carefully defined. For a mathematically idealized definition, we could separate the classical wave function into its positive- and negative-frequency parts (defined using a Hilbert transform), then defining the "instantaneous frequency" to be the time-derivative of the phase of either one of these two complex-valued parts. A more practical definition is based on the fact that any real frequency-selective photon counter is going to have some finite size and will effectively integrate over some finite time-interval. In view of this, we can define "instantaneous frequency" in a more practical way as a king of average frequency in that region of space during that time interval. With this (loose) definition, the connection suggested in the excerpt is at least qualitatively correct: the frequency-selective photon-counter will tend to register more photons with frequencies near the "instantaneous frequency" in whatever spatial region and at whatever time the counter is operating.





To describe the relationship between the classical and quantum pictures in more detail, here are a few highlights from the post cited above (https://physics.stackexchange.com/a/443761).


The exact solution in the cited post shows that after the current has turned off, the resulting quantum state of the field is the coherent state \begin{align} |\psi\rangle &\propto \sum_{n\geq 0} \frac{\big(A^\dagger \big)^n}{n!}\,|T\rangle \\ &= |T\rangle + A^\dagger|T\rangle + \frac{1}{2!}\big(A^\dagger \big)^2|T\rangle + \frac{1}{3!}\big(A^\dagger \big)^3|T\rangle +\cdots \tag{1} \end{align} where $|T\rangle$ represents the vacuum state at times $t>T$ after the current $J$ is turned off. Each application of the operator \begin{equation} A^\dagger \equiv \int\frac{d^3p}{(2\pi)^3}\ a_J(\mathbf{p}) a^\dagger(\mathbf{p}) \tag{2} \end{equation} to the vacuum state creates a single photon whose "profile" is described by the complex-valued function $a_J(\mathbf{p})$, because each application of the operator $a^\dagger(\mathbf{p})$ to the vacuum state creates a single photon with wavenumber (or momentum) $\mathbf{p}$. So, for example, the state $A^\dagger|T\rangle$ has a single photon with profile $a_J(\mathbf{p})$, and the state $$ \big(A^\dagger \big)^n|T\rangle \tag{3} $$ has $n$ identical photons with this same profile. The coherent state (1) is a quantum superposition of different numbers of photons, all with this same profile. The creation operator $a^\dagger(\mathbf{p})$ is the adjoint of the annihilation operator $a(\mathbf{p})$. Each application of the annihilation operator removes one photon (with the indicated $\mathbf{p}$) from the state, and if no such photon is present, the result is zero. (Not zero photons, but just plain zero, which means that this term no longer contributes to the overall quantum superposition at all.)


To relate the photon picture to the classical-wave picture, we can use the field-amplitude observable $\phi(t,\mathbf{x})$, whose relationship to the creation/annihilation operators is shown in the other post. The expecation value of this observable in the state (1) is \begin{equation} \langle \psi|\phi(t,\mathbf{x})|\psi\rangle=\phi_J(t,\mathbf{x}), \tag{4} \end{equation} where the real-valued function $\phi_J(t,\mathbf{x})$ is related to $a_J(\mathbf{p})$ by \begin{equation} \phi_J(t,\mathbf{x}) =\int\frac{d^3p}{(2\pi)^3}\ e^{i\mathbf{p}\cdot \mathbf{x}}\, \frac{e^{-i\omega t}a_J(\mathbf{p}) +e^{i\omega t}a_J^\dagger(-\mathbf{p})}{ \sqrt{2\omega}} \tag{5} \end{equation} with $\omega\equiv |\mathbf{p}|$. As explained in the other post, if the magnitude of $a_J$ is large enough, then the state (1) describes an effectively-classical wave with time-varying amplitude given by the function $\phi_J(t,\mathbf{x})$.


Equation (5) gives an explicit relationship between the effectively-classical wave $\phi_J(t,\mathbf{x})$ and the single-photon profile $a_J(\mathbf{p})$ in equation (2). To relate this to the behavior of a localized, frequency-selective photon-counting device, we need to construct an observable corresponding to such a device. One way to do this is to let $\phi^+(t,\mathbf{x})$ denote the part of the field-amplitude observable $\phi(t,\mathbf{x})$ that involves only the creation operators $a^\dagger(\mathbf{p})$, and to let $\phi^-(t,\mathbf{x})$ denote the part of the field-amplitude observable $\phi(t,\mathbf{x})$ that involves only the annihilation operators $a(\mathbf{p})$. Then we can use an operator of the form \begin{equation} D(t) = \int d^3x\,d^3y\ f(\mathbf{x},\mathbf{y}) \phi^+(t,\mathbf{x})\phi^-(t,\mathbf{y}) \tag{6} \end{equation} as the observable corresponding to a quasi-localized photon counter. It depends on time $t$ because this formulation uses the Heisenberg picture, where all time-dependence is carried by the observables rather than by the state-vector. The function $f$ may be adjusted to choose where the counter is situated and what range of wavenumbers it is sensitive to.


I said "quasi-localized" because although the field-amplitude observables $\phi(t,\mathbf{x})$ are localized by definition, the operators $\phi^\pm(t,\mathbf{x})$ are not strictly localized: they do not commute with the field-amplitude operators at spacelike separation. As a result, the operator (6) represents a quasi-localized photon counter. An operator that is strictly localized within any finite region of space cannot annihilate the vacuum state (the Reeh-Schlieder theorem), so a strictly localized photon-counter cannot be noise-free. The photon-counter represented by (6) is noise-free but not strictly localized.


The conclusions expressed verbally in the main part of the answer are all based on this mathematical formulation.


covariance - What is a covariant derivative in gauge theory?


I've been studying electroweak theory and you need to keep the Lagrangian covariant by introducing covariant derivatives. What is a covariant derivative? And what does it mean to keep the Lagrangian covariant?


Also, in electroweak symmetry breaking, the gauge bosons attain their masses via the action of a 'covariant derivative' on the Higgs field. What does this mean in physical terms?




riddle - What am I? I am here


I am here,
yet most people don't care,



I kill people,
if you are not careful.


I can be created,
but it's rather useless.


If you can see me,
It's because of Him.


What am I?


Hint:



Him does not refer to a scientist or God




Hint 2:



The first 3 clues directly describe the answer. The last clue describes the answer, but not in the same way as the previous clues.



Hint 3:



"Him" is a Mexican.




Answer




I think It could be



Gravity



I am here, yet most people don't care,



Gravity is everywhere, but most people don't actually think about it



I kill people, if you are not careful.




you can fall off a cliff or something can fall onto you



I can be created, but it's rather useless.



Gravity can be created (http://en.wikipedia.org/wiki/Artificial_gravity)



If you can see me, It's because of Him.



The movie "Gravity" was written by Alfonso Cuarón, a Mexican.




newtonian mechanics - How multiple objects in contact are resolved in an inelastic collision, when edge normals don't "line up"


In a case I understand, let's say I have an object A moving at velocity V toward 3 objects in contact B, C, and D:


enter image description here


The momentum of A is the mass of A times its velocity. To figure out how the inelastic collision ends up when A hits B, I sum the masses of A, B, C, and D, and I divide the old momentum of A by the sum of those masses. That is the velocity all 4 objects end up with. Easy and extendable!


But when objects are in contact along edges with normals not parallel to the original momentum:


enter image description here


Such as here, with C mucking up the nice neat solution I had for the system above. I know that when objects collide, they are pushed away along the normal of the edge of contact. So D is going to move up and right at an angle, and presumably C and E need to get pushed downward to conserve momentum. Further, I still have to deal with my momentum from A pushing things to the right. I'm pretty sure D will be moving to the right slower than A, B, and C. And E won't be moving to the right at all, assuming there's no friction. Right?



What is the summed mass to use in the equation I used to solve my first problem? Can I simplify this into equations dealing with the "branching off" of the "main path" (D branching off, and C + E branching off?)? I need to understand this as a general case, not simply the solution to this one problem... Any possible set of convex polygons in contact being hit. How does this work?


I'm also wondering how to deal with this with partially elastic collisions... Seems like it'd get pretty crazy, especially with systems more complicated than my 2nd example.



Answer



I am going to ignore rotations in order to simplify the problem for your understanding. You have to enforce a series of inelastic relationship of the form


$$\vec{n}_{k}^\top (\vec{v}_i^+-\vec{v}_j^+) = 0 $$


where $\vec{n}_k$ is the normal direction of the $k$-th contact, and $i$, $j$ are the bodies this contact affects. The superscript $\phantom{c}^+$ denotes condition after the impact. You enforce this relationship with a series of $k$ impulses $J_k$ such that


$$ \vec{v}_i^+ = \vec{v}_i + \frac{\vec{n} J_k}{m_i} $$ $$ \vec{v}_j^+ = \vec{v}_j - \frac{\vec{n} J_k}{m_j} $$


Since it all has to happen at the same time it is best to form the problem with matrices.


Consider a Contact matrix $A$ where each column $k$ has +1 in the $i$-th row and -1 in the $j$-th row. For example $A = \begin{pmatrix}0&-1 \\ -1 & 0 \\ 0 & 0 \\ 1 & 1 \end{pmatrix}$ means there are two contacts, one between body 2 and body 4 and another between body 1 and 4. (actually each 0 and 1 are 2×2 for 2D or 3×3 for 3D zero and identity matrices).


The inelastic relationships are



$$N^\top A^\top v^+ =0$$ with the contact normal block diagonal matrix $$ N = \begin{pmatrix} \vec{n}_1 & 0 & \cdots & 0 \\ 0 & \vec{n}_2 & & 0 \\ \vdots & & & \vdots \\ 0 & 0 & \cdots & \vec{n}_K \end{pmatrix} $$ and $$ v = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{pmatrix} $$


The momentum exchange is described by the relationship


$$ M v^+ = M v - A N J $$ where $M$ is the block diagonal mass matrix $M=\begin{pmatrix}m_1& & & \\ &m_2 & & \\& & \ddots & \\ & & & m_N\end{pmatrix}$ and $J$ the vector of impulses $J^\top=(J_1\,J_2\,\cdots J_K)$


To solve the problem we combine the momentum with the inelastic collisions to get


$$ v^+ = v - M^{-1} A N J $$ $$ N^\top A^\top \left(v - M^{-1} A N J \right) = 0$$ $$ \left(N^\top A^\top M^{-1} A N\right) J = N^\top A v $$


$$ \boxed{ J = \left(N^\top A^\top M^{-1} A N\right)^{-1} N^\top A^\top v }$$


Example


With $A$ as above (4 2D bodies, 2 contacts) and $\vec{v}_i = (\dot{x}_i,\dot{y}_i)^\top$, $\vec{n}_1=(1,0)^\top$, $\vec{n}_2 = (0,1)^\top$ then


$$ A = \left(\begin{array}{cc|cc} 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \hline -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ \hline 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \hline 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1 \end{array}\right) $$


$$ N = \left(\begin{array}{c|c} 1 & 0\\ 0 & 0\\ \hline 0 & 0\\ 0 & 1 \end{array}\right) $$



$$ M = \left(\begin{array}{cc|cc|cc|cc} m_{1} & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & m_{1} & 0 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 0 & m_{2} & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & m_{2} & 0 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & 0 & m_{3} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & m_{3} & 0 & 0\\ \hline 0 & 0 & 0 & 0 & 0 & 0 & m_{4} & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & m_{4} \end{array}\right) $$


$$ v = \begin{pmatrix} \dot{x}_1 \\ \dot{y}_1 \\ \hline \dot{x}_2 \\ \dot{y}_2 \\ \hline \dot{x}_3 \\ \dot{y}_3 \\ \hline \dot{x}_4 \\ \dot{y}_4 \end{pmatrix} $$


$$ N^\top A^\top M^{-1} A N = \left(\begin{array}{cc} \frac{1}{m_{1}} + \frac{1}{m_{4}} & 0\\ 0 & \frac{1}{m_{2}} + \frac{1}{m_{4}} \end{array}\right) $$ $$ N^\top A^\top v = \begin{pmatrix} \dot{x}_4-\dot{x}_2 \\ \dot{y}_4 - \dot{y}_2 \end{pmatrix} $$


$$ J = \begin{pmatrix} \frac{\dot{x}_4-\dot{x}_2}{\frac{1}{m_2}+\frac{1}{m_4}} \\ \frac{\dot{y}_4-\dot{y}_1}{\frac{1}{m_1}+\frac{1}{m_4}} \end{pmatrix} $$


Then


$$v^+ = v - M^{-1} A N J = \begin{pmatrix} \dot{x}_1 \\ \frac{m_1 \dot{y}_1 + m_4 \dot{y}_4}{m_1+m_4} \\ \frac{m_2 \dot{x}_2 + m_4 \dot{x}_4}{m_2+m_4} \\ \dot{y}_2 \\ \dot{x}_3 \\ \dot{y}_3 \\ \frac{m_2 \dot{x}_2 + m_4 \dot{x}_4}{m_2+m_4} \\ \frac{m_1 \dot{y}_1 + m_4 \dot{y}_4}{m_1+m_4} \end{pmatrix} $$


Appendix


To include rotations follow the guidelines here:



Wednesday, June 28, 2017

quantum mechanics - Partition function of a hydrogen gas


Hi I have a doubt (I'm not very expert in statistical mechanics, so sorry for this question). We consider a gas of hydrogen atoms with no interactions between them. The partition function is: $$ Z=\frac{Z_s^N}{N!} $$ where $Z_s$ is the partition function of one atom. So we write $$ Z_s =Tr\{e^{\beta \hat H}\} $$ and we must consider sum to discrete and continous spectrum. So I'd write for the contribute of the discrete spectrum: $$ Z_{s_{disc}}=\sum_{n=1}^{+\infty}n^2 e^{\beta \frac{E_0}{n^2}} $$ but this serie doesn't converge. For continuum spectrum, I'm not be able to write the contribute to the sum, because I have infinite degeneration, so where's my mistake?


I have thought that the spectrum tha I considered for energy values is for free atom in the space, and partition function maybe is defined for systems with finite volume.


So, this doubt however would me ask a problem. Can whe prove that operator $e^{-\beta \hat H}$ always has got finite trace for halll phsical systems?




Answer



This is quite a subtle problem:



  • Notice that you have the same problem if you have only a single hydrogen atom.

  • You've ignored all the unbound states of the hydrogen atom, which make thing even worse (they are higher in energy, and have even larger degeneracies).

  • Divergent partition function is actually not really a problem --- all physical properties will depend on derivatives of the logarithm of it.

  • However, you will find that at any finite temperature things like entropy (which are physical) does indeed diverge.

  • This is again not so bad if you realise that you've effectively got an infinite system --- the entropy per volume should still be finite.


Concretely, you should regularise things by considering a finite box of volume V, with N particles. This will modify the relevant states (and their energies) in correct (but incredibly hard to compute) ways to render all physical properties finite.



Intuitively the effect is due to the n'th bound state having a radius $O(n^2)$. Thus you can approximate the effects of a finite boundary by simply removing all states above a certain size --- the divergences are so slow that in practice physical quantities will only depend very weakly on the cut-off.


References:



  1. http://scienceblogs.com/builtonfacts/2011/01/the_hydrogen_partition_functio.php

  2. http://pubs.acs.org/doi/abs/10.1021/ed043p364


acoustics - Modelling noise with distance



I was wondering about the relation between noise with distance, assuming a point source, using sound as the method for communication and air as the medium of communication. Obviously as the distance from the sound source increases, noise should increase- but what is the nature of this relationship? Is it linear or non-linear?


Any ideas? Is there any other way noise can be modeled (in the context of communication between individuals)? Thanks!




Does greater number of lines of force around the magnet imply greater magnetic field strength around it?



enter image description here


If we sprinkle iron fillings on a sheet of glass placed over a short bar magnet. The arrangement of iron fillings will be similar to the one shown above.



Why do some of the iron fillings arrange in a particular pattern when sprinkled around a magnet, instead of getting attached to the magnet?
All the filings are resting on a surface where there is friction. If the force due to the magnet does not exceed this friction, the filings won't accelerate to the magnet (only the closest ones experiencing the largest forces do). Iron is also ferromagnetic, which means it can concentrate magnetic fields. So filings that are too far to be affected by the magnet itself can be attracted to adjacent filings which are concentrating the magnetic field better. This is why most of the filings seem to be more attracted to each other than the magnet itself. But essentially all the magnetic field is caused by the magnet.



After this good answer by gregsan to the above question, I got a doubt here. Considering gregsan reason to be true, for gaps between the fillings around the magnet. I thought, if the reason for gaps in between the fillings is due greater field strength at those gaps (caused by displacement of fillings towards the magnet), and if the reason for those lines in between the gaps is due to lesser magnetic field at those locations (causing the fillings to remain at those same position). I thought iron fillings lines around the magnet exist if magnet can't exert more force to accelerate them against the frictional force, on the other hand in the gaps, fillings are more accelerated towards the magnet, so they get attached to the magnet causing gaps. Now, I assumed the lines around the magnet to be the locations where force due to magnet is less. So, if more is the lines around the magnet, greater will be locations where force due to magnet is less. Considering this, I thought, greater number of lines of force indicate less intensity of magnetic field around the magnet, but I have been taught in my school that, field strength to be more if greater is the lines of force around the magnet. Is it that magnetic field strength will be concentrated more at a shorter distance, if lines of force is more? I don't know whether I have misunderstood the answer of gregsan or is there any reason which would account for greater number of lines around the powerful magnet? If any is the case, please explain.




classical mechanics - How long would it take for an upright rigid body to fall to the ground?


Let's suppose there is a straight rigid bar with height $h$ and center of mass at the middle of height $h/2$. Now if the bar is vertically upright from ground, how long will it take to fall on the ground and what is the equation of motion of the center of mass (Lagrangian)?



Answer



A falling tree is basically an inverted pendulum.


The period of a pendulum of length $h$ for small oscillations is $2\pi \sqrt{h/g}$, with $g$ the acceleration due to gravity, about $10\ m/s$. For an inverted pendulum near the top of its arc, there is no period, but the quantity $\sqrt{h/g}$ does represent a characteristic time scale for this system. The tree will take a few of these characteristic times to fall. $h$ for a tree is an "effective height", and depends on the mass distribution of the tree. If all the mass is at the top, $h$ is the height of the tree. If the tree is uniform, $h$ is $2/3$ the true height.



For a tree with $h = 40\ m$, the characteristic time is $2\ s$. For small angles, the angle the tree makes with the vertical will be multiplied by $e$ in this time. Let's start the tree at $1^{\circ}$ so that it needs to multiply its angle by $90$ to fall. $\ln(90) = 4.5$ so the tree takes about $9\ s$ to fall.


This is mathematically an underestimate because the characteristic time increases slightly as the tree falls, but not too much. Give it a nice round $10\ s$ and you get something that matches the first YouTube video I found.


quantum mechanics - how molecules radiate heat as electromagnetic wave?



an object of higher temperature radiate infrared rays as a way to decrease the temperature. how a molecule produce a electromagnetic wave? in atoms electromagnetic radiation is caused by electrons. what is responsible in molecules?




Tuesday, June 27, 2017

general relativity - Spacetime around a Black Hole


If we consider the sun, then space-time is curve around it. My question is that what is the kind of curvature of space and time around the black hole. Is that space and time more curved around the black hole. Can anyone explain me about that?



Answer



Space-time, as you said curves around a body in space, and the greater the mass of the object the greater the curvature, this, in the simplest way can be described using light cones and Schwarzschild. Black-holes create many odd situations in space in which odd events take place. Black holes, although often though of as places in themselves, they have their own geography. There are two checkpoints to look for when one falls in a black hole these are called, the Horizon and the Singularity. The Singularity is the easiest to explain, in terms of description.


Black Hole


As you can see the singularity is the center point of the black hole, it also shows us the relative place of the Horizon, our second checkpoint. The Horizon is defined by the size of the black hole. The calculation in how to work out the Schwarzschild radius, (where to place the Horizon) is stated below:



Equation


The important thing about the Horizon is that NOTHING SPECIAL happens if one were to pass it, on e would feel no special sensation and may not even know they had passed the horizon unless they had a way of measuring it. The only interesting thing that happens at the Horizon is that when someone else watch one they would see an odd situation, but that would be going on to a completely different discussion point. The Singularity is the only place in which one would feel anything, and that thing would a infinite tidal forces, ripping one apart because gravity would be infinite. This is where we come back to your original question, gravity is NOT a force, it is only talked of as a force so that it is simple to imagine, but a curvature of space, and therefore curvature would be infinite. This is only at the singularity though, at the horizon the curvature would not be quite as intense. If you want to know the exact equation then look up Ricci/Riemann Curvature equations as I have chosen to describe this is in a non-mathematical sense. Below I have left a diagram of a black hole to help you visualize it.


Black hole...


(original source)


I hope I helped and did not ramble too much. If you found my explanation helpful and not too boring then look up more and even try listening to Leonard Susskind, or Brian Greene, both really interesting and have videos on YouTube. It is really interesting so dive as deep as you can, then go even deeper!


pattern - What is a Spaced Out Word™?


This is my first try at a puzzle, so hopefully it at least lasts a few hours before someone figures it out :)


This is in the spirit of the What is a Word/Phrase™ series started by JLee with a special brand of Phrase™ and Word™ puzzles.


If a word conforms to a special rule, I call it a Spaced Out Word™.



Use the following examples below to find the rule.


enter image description here


And, if you want to analyze, here is a CSV version:



Spaced Out Words™,Not Spaced Out Words™
HUMILIATE,DEGRADE
HEXAGONALLY,OCTAGONALLY
DUNGEON,CATACOMB
NEFARIOUSLY,HEINOUSLY
NONSPHERICAL,NONCUBIC

WOOZINESS,DIZZINESS
PRECOGNITION,CLAIRVOYANCE
EXTENSION,ANNEX

UPDATE:


It occurred to me that, even if a puzzler were to reverse-engineer my criteria from the words above, they may not be able to fully determine the rules. To help, here are some additional words that are Not Spaced Out Words™ to narrow down the criteria:



 CELEBRATION
WEATHERMEN
MANIPULATING

ADORINGLY

It's less of a hint and more of a clarification, but I put them in spoiler text for those who may want to solve the puzzle as it was originally written (and maybe refer back to these after, for verification).


ADDITIONAL WORDS


Below are a few more Spaced Out Words™ and Not Spaced Out Words™:



 Spaced Out Words™,Not Spaced Out Words™
AROUSING,EXCITING
ADJOURN,RECESS

And a few more if you need more. I think this just about exhausts the list...




 Spaced Out Words™,Not Spaced Out Words™
POVERTYSTRICKEN,IMPOVERISHED
UNCOPYRIGHTED,PUBLICDOMAIN

(yes, that last Not Spaced Out Word™ is technically two words)


Answer



The solution is that Spaced Out Words™ contain the letters for



Homonuclear gases at room temperature.

Word™ Element
----------------------------
HUMILIATE HELIUM

HEXAGONALLY OXYGEN
DUNGEON NEON
NEFARIOUSLY FLUORINE
NONSPHERICAL CHLORINE
WOOZINESS OZONE.
PRECOGNITION NITROGEN
EXTENSION XENON
AROUSING ARGON
ADJOURN RADON
POVERTYSTRICKEN KRYPTON

UNCOPYRIGHTED HYDROGEN



newtonian gravity - Why centrifugal force (not centripetal force) is considered while deriving the effect of rotation of Earth on "g"?


As I was trying to work out the expression for the apparent "g"-in case of the rotation of Earth-I was considering the centripetal force (m$w^2$r) and the weight (mg) all the time, thinking these were the forces that were acting on the body.



enter image description here


But later I found out that it should have had used centrifugal force instead of centripetal force. But why? We should be considering the forces acting on the body right? And it's the centripetal force that is acting on the body (with direction towards the center of the small circle).


enter image description here



Answer



You're right, but you've forgotten one thing: the reference frame. When you are in an inertial frame of reference the only real force in a circular movement is the centripetal force $\vec F_c = - \frac{mv^2}{r} \vec û_r$


But, ¿what happens when you are in a non-inertial reference frame ? Remember that Newton's second law works perfectly on inertial reference frames, but it doesn't work on non inertial reference frame. So, if you want the law to remain valid you need to add some fictitious or inertial forces. That is, any observer located in a non-inertial reference system will need fictitious forces to explain correctly the movement (is the "trick" to making the second law works) In a circular movement (Earth) you are not in a inertial reference frame = Earth is a non-inertial reference frame, for this reason, for you (who are in this reference frame) doesn't exist any centripetal force, you just feel a force (that doesn't exist really) pushing you out . If you are inside of the Earth this is your non inertial reference frame, then, you need to consider inertial/fictitious forces, like centrifugal force and of course Coriolis force. Now get out of the Earth and you won't have to include these forces, naturally.




Final comment, as you can see in the second image you've put, the centrifugal force causes a little change in the weight direction $mg$ an it does not point exactly to the center of the Earth


Hope that helps


J.



mathematics - Black and white queens on an 8x8 chessboard


What is the largest number of queens that can be placed on a regular $8\times8$ chessboard, if the following rules are met:





  1. A queen can be either black or white, and there can be unequal numbers of each type.

  2. A queen must not be threatened by other queens of the same color.

  3. Queens threaten all squares in the same row, column, or diagonal (as in chess). Also, threats are blocked by other queens.



Bonus:



Would this number change if rule 1 was changed to enforce equal numbers of black and white queens?





Answer



I'll guess



32 queens.

Every other row is filled with alternating queens. Starting queens on each filled row alternate.

W B W B W B W B
- - - - - - - -
B W B W B W B W
- - - - - - - -
W B W B W B W B
- - - - - - - -

B W B W B W B W



I think this is ideal, since for a single queen, cutting off all lines of sight would require:



B - B - B
- - - - -
- B W B -
- - - - -
B - B - B

Filling white queens into the spaces and repeating the pattern gives us the 32 solution.




mathematics - Does this strategy work?


I'm thinking about the following strategy for Fastest way to collect an arbitrary army:




  1. When a soldier decides to go to some house he "reserves" it.

  2. Once a soldier is free (has delivered the news to another house) he takes the closest house from all unreserved and goes there. If there are several closest houses he chooses at random.


My question is "Is there a house placement example for which this strategy gives a time bigger than $2+\sqrt{2}$?".



Answer



No I don't think that strategy works.


enter image description here


If we consider a clusters of very close houses such that every house will be reserved before they are reached, it is easy in imagine arbitrarily long paths like the one in the top left of my picture. The arrangement in the diamond would fail yours but pass mine. Each number represents a very dense arrangments of that number of houses. At least 1 house will not be reached before your due date much less get back to the castle.


Monday, June 26, 2017

nuclear physics - $α$-decay and electron


I'm a Web developer and I'm working on a physic interactive Web page to represent how atoms interact. I'm reading information about α-decay and here is what I understand.


Let's say we have a isotope 222Rn that will change, after more than 3 days, into a 218Po because it will release a "small 4He nuclei" at 5 % the speed of light.. I've got 2 questions.



First, what is going on in those 3 days, what is happening. Does the nuclei is "unstable", vibrating ?


Second, where does the electron go?




statistical mechanics - Exorcism of Maxwell's Demon


I am possessed! Yes, with the thinking that if there is actually a Maxwell's Demon, then it would open the negligible weighted door which would ultimately make the second law invalid. But really can second law be invalid? This is not my question. It is a universal law. So, what should be the logic that the Demon would fail?? Please don't say that work must be done to open the door. The door is so light that negligible work is done.



Answer



The resolution to Maxwell's demon paradox is mostly understood to be through Landauer's principle, and it is one of the most compelling applications of information science to physics. Landauer's principle asserts that erasing information from a physical system will always require performing work, and particularly will require at least $$k_B T \ln(2)$$


of energy to be spent and eventually released as heat. The concept of 'erasing information' is relatively tricky, but there are some pretty solid foundations to think that this principle is right.


To apply it to the demon, you should realize that the demon consists of (at least) two parts: a sensor to detect when particles are coming, and an actuator to actually move the door. For the demon to work correctly, the actuator must act on the current instruction from the sensor, instead of the previous one, so it must forget instructions as soon as a new one comes in. This takes some work: there is some physical system encoding a bit and it will take some energy cost to flip it.


Now, there are some criticisms of Landauer's principle, and it is not completely clear whether it is dependent on the Second Law of Thermodynamics or if it can be proved independently; for an example see this paper (doi). Nevertheless, even if it is a restatement of the Second Law, it carries considerable explanatory power, in that it clarifies how the Second Law forbids the demon from operating.


Quantum entanglement and 'perfect' synchronized measurements



Given two entangled particles A and B barely separated by a very tiny small distance such as the communication delay between both observers Alice and Bob willing to measure the state of A and B respectively is almost null, assuming the classical communication channel between Alice and Bob is reliable and the synchronization between Alice and Bob is perfect (using an atomic clock for instance), that is if Alice measures particle A, Bob will do the same for B INSTANTLY, I really wonder what would be the outcome of this experience ?




Sunday, June 25, 2017

quantum mechanics - Why is it OK to keep the quadratic term in the small $hbar$ approximation?


I am following this[link broken] set of notes:



Riccardo Rattazzi, The Path Integral approach to Quantum Mechanics, Lecture Notes for Quantum Mechanics IV, 2009, page 21.



I am having some issues to understand the small $\hbar$ expansion.


Consider the path integral in quantum mechanics giving the amplitude for a spinless particle to go from point $x_i$ to point $x_f$ in the time interval $T$ $$ \int D[x]e^{i\frac{S[x]}{\hbar}}=\ldots $$ where $$ S[x]=\int_{0}^{T}dt\,\mathcal{L} $$ let's assume now that the action has one stationary point $x_0$. Let's change the variable of integration in the path integral from $x$ to fluctuations around the stationary point $$ x=x_0+y $$ $$ \ldots=\int D[y]e^{i\frac{S[x_0+y]}{\hbar}}=\ldots $$ Let's Taylor expand the action around $x_0$ $$ S[x_0+y]=S[x_0]+\frac{1}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}y(t_1)y(t_2)+\ldots $$ which leaves us with $$ \ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2\hbar}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}y(t_1)y(t_2)+\ldots}=\ldots \tag{1.65} $$ this is where the author considers the rescaling $$ y=\sqrt{\hbar}\tilde{y} $$ which leaves us with $$ \ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}\tilde{y}(t_1)\tilde{y}(t_2)+\mathcal{O}(\hbar^{1/2})} \tag{1.66} $$ and we "obviously" have an expansion in $\hbar$, so when $\hbar$ is small we may keep the first term $$ e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}\tilde{y}(t_1)\tilde{y}(t_2)} $$ I do not like this rationale at all. It's all based on the rescaling of $y$ we have introduced, but had we done $$ y=\frac{1}{\hbar^{500}}\tilde{y} $$ we wouldn't have obtained an expansion on powers of $\hbar$ on the exponent. What is the proper justification for keeping the quadratic term?




tensor calculus - Symmetry in terms of matrices


When we encounter a problem in physics which can be expressed in terms of matrices or tensors, why do we decompose the tensor in terms of its symmetric and antisymmetric or trace components? What is the physics motivation behind doing so?




Answer



OP is basically asking:



Why do we decompose (reducible) group representations in irreducible group representations?



Partial answer:




  1. To classify the (reducible) representation.





  2. Irreducible representations can not be further truncated without destroying the group symmetry.




  3. Because certain irreducible sub-representations of the given (reducible) representation may be forbidden by e.g. selection rules, other physical principles, etc, and this is always useful information.




planets - Does a star need to be inside a galaxy?


Must a star belong to a galaxy, or could it be completely isolated?


In case it can be isolated (not belong to a galaxy), could it have a planet orbiting around it?



Answer



No, stars do not need to be inside a galaxy. It is estimated that about 10% of stars do not belong to a galaxy [1]. While most intergalactic stars formed inside a galaxy and were ejected by gravitational interactions, stars can form outside of galaxies as well [2].


We assume that such stars could have planets, just like stars in a galaxy, although no specific examples have been detected yet.



[1] "Detection of intergalactic red-giant-branch stars in the Virgo cluster", Ferguson et al. Nature 391.6666 (1998): 461.


[2] "Polychromatic view of intergalactic star formation in NGC 5291", M. Boquien et al. A&A, 467 1 (2007) 93-106.


thermodynamics - Is there a mechanism for time symmetry breaking?


Excluding Thermodynamic's arrow of time, all mathematical descriptions of time are symmetric. We know the arrow of time is real and we know the equations describing physics are real so is there any insights into what breaks the symmetry between real time and the mathematical description?




geometry - Can you solve those laser puzzles? (puzzles created by the community)



So, I was bored and I made a small game for this site! In this game the user who solves the previous puzzle gets to create the next one and they will always be more complex. Website link and string down below!


How it works:



  • You paste in the current puzzle string in the input box and press reset just to be sure. This will setup the game and place "invisible" mirrors on the board of which there are 8 types.

  • You shoot lasers with a click on one of the small boxes on the side of the board the green color tells you where the laser shot came from and blue tells you where it exited. If there is only a blue box and no green box then it means that the laser entered and exited in the same box.


  • You place the mirrors with first clicking on one and then clicking on a square on the board, to delete a mirror click on it again or overwrite it with another


  • The 8 mirrors are:



    • Vertical and horizontal reflection mirrors (worth 1 point). As a example, let's take the horizontal mirror. If the laser comes from above or below it gets reflected back in the opposite direction. If it comes from left or right it passes through the mirror.

    • 45° degree mirrors (worth 3 points). I think this is pretty self explanatory as to what happens.

    • Rotating vertical and horizontal mirrors (worth 5 points). These behave the same way as their standard counterparts with the added ability that when they get hit they turn into their opposite. So if the laser comes from above at a horizontal rotating mirror it gets bounced back up and the mirror now becomes a vertical one, which then again turns into a horizontal one if it gets hit again etc etc. The rotation occurs also if the laser passes through.

    • Rotating 45° degree mirrors (worth 7 points). These are the same as the standard 45° mirrors with the rotation ability





  • Your goal in this game is to figure out at which position what mirror is under. You got some things to help you with this:



    • The first is the point counter which shows what the total sum of points of all mirrors on the board, so if there is only 8 avaliable points and you are sure that there is one rotating 45° mirror somewhere, then that only leaves 1 point left for one basic mirror.

    • The second is the number of trys, this is a counter which tells you how many times you send a laser through from any position since innitializing the board

    • And the last thing is a counter which tells you how many squares the laser has traveled in the current shot.



  • The reset button resets the board to its initial state, this doesnt delete the mirrors you placed on the board but just puts the "invisible" mirrors back in their starting position



Some additional things to consider:



  • The outer layer of the board can't have mirrors in them

  • You need to find out the initial state of the board so if there is only one rotating mirror on the board and you hit it once with the laser the correct solution will still be to place the initial position of the rotating mirror on the board.


When you get the correct solution to the puzzle a message will inform you of your victory and then you will be able to create the next puzzle where you will be able to place a total points of mirror worth $n+2$, where $n$ is the previous puzzle's maximum points. The first puzzle posted by me will have 3 maximum points, so the next will have 5, then 7 etc (Scoring is now different in the new version). To create a new puzzle you simply place mirrors on the board. When you place exactly $n+2$ points worth of mirrors you get the next puzzle string. When you are creating a new puzzle you will also be able to see the actual laser traveling, and if the laser travels over a box multiple times the color will get darker.


You should also probably sort this thread after date to see the latest puzzle. The next valid puzzle in the chain will be earliest one with the correct solution to the previous one. The last thing I want to impose is the correct format of a valid answer. Edit: check out the current answers!


Version 1 (use this version for level 1 to level 4): https://preview.c9.io/vajura/blackbox/ver1/index.html


Version 2 (use this for level 5 and on, in this version mirrors cost more and the difficulty only increments for 1 point as a bonus the puzzle strings are 100 times smaller) https://preview.c9.io/vajura/blackbox/ver2/index.html


Level 5 is not yet updated so if you want to solve it use Version 1, if you solve Level 5 in the old version before Josh update his puzzle to the new version you are free to create the Level 6 puzzle from this link(just start placing mirrors the points will update and when you placed a total worth of 13 the new puzzle string will be shown) https://preview.c9.io/vajura/blackbox/ver2.1/index.html



Level 1, made by OP



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

Level 2, made by BmyGuest



C44IiCsMjgi606BiGkKKXd7O0ZFVNjAAamkCekOlNdQxtgzT1hS7HN0lby8Or+ZB9otEplgYGmENnUfwwVuAqUTD/VnygVesX7K9zGDFZphqwHw8Ji/r1KuKYLeIrxDHO54aOG880XX9LO4FSAPXxCC/azcx8zSfpg+ei5b2N4TIhLzerp8eaPgGwi0n0zQiSjm/m6ou2d5g+E4+1Jtanaz4/2p3h9eSp7bNcKg2aH0sPRxqOeQFzHqSmgQH5nzj8UFkGghqAAAeNbn2buJQDaWh4yyVVfks66wTEOD4D2xQnVikuGgFvwJ6Icgo2TEGuHq6LU3+67cFFuOcBLioQupZj9pHI8toYbD9/vdzTODEoihC9Yb0DGO94CSht2Gj18Z0v25jsmHBZ5bus999/Haz/RyyrRPK+dwUcadnZUeCVzVdZ3CLbGWmO7N2aQydTwBPlPYtXc3HrqXbp8Z0wbzHplcR6Z8e+U7Cb0qH7O++v5JMQOq8UNcqn+JUXRL3bMwPVC1V1HWh7xkgTvoOTgabTiCcBJNXJ3584SqK2NhYPlG+S5XT3MwJ1y1W52P5qSLkEpztjH1nO9XKp7P3RA3I9gIOkw6i7G3gHxtwbQsncmBFzmVswVU++wa55Ve3X0dfD2kjMUCvw22iifmuf7STzwkLIq1li0dpJo11jeEetz24QuCY5GA1FA0VqIbxfWynX060vsyIYVB0TGyd1TNwuPWzTmjbi/7F0yaY8CSG9fGBzVSz+jV4PktafHCiB4Ske/EtttzLPIL3F+2/O9nN7M+nVZa3kMvEGii8JqxxkXpGEfIEShUxzMnWCnU7CxnQgp0W3ReffQy3wAgJh7MJ+j5b5PjvAczcEJuxBRaGtrHYXzzzhAE/oL9CyOVvONyrApqYbnPl2m5QwZ8dE+HQjQVeMaDuzsTbU9QACu4Ard9ytfULZRjWWRKXgSbuR54xKhHd1t4la6b5xXoYFaaHmtUM+ZaY6OqrbMQp44EjFYTaMFZRV6+sAYx6Wmo4mfLPJquBn0OdURzxJTKb7cMvy7Q2GbxigjZmpowZ2H8rH2SfaMBTtb+osERkOqbIaKb17tlt9wA2TcI2MWLuYe8JhBx6265P6hGgNAzR2pFzYJTuo+wbKAJwYLhkp19rCaoMwy/YtitfaYuBavf+jCZ5DeJ+wfpgDGANBMVWWiobxYdlqH0z+7YWMTC5S9PzEqnQ1IqcjU5k+yZTAMNH0JYWZ8A9u20+gtm3G6ye6m/2DezwafNp4pgYT1nd1v/u6gjBfOZvk4KjTU2JppJQfGwVfRn5KXOYZb7KbR5kKtm9sg2TaTYKwJiuOyqXP9SaYh57mCk+5ALJKMKaII1m2sfWZFiSrs/7q3TTWdKoZu8PFSbN3z07Mw3DhFdtg6eM/eeyUoFj0LoL/X4/5E25ry4rSE3YoT6V+HPFM2lEXv3/G+/Q/17yYs1DZbGg/w5AyMI8xPg9EpWLJgocyRvjWN69ydnGnkVsnr3HLPuSsJ1Noeoa/Pj9xDmhURHHH0h7xY5M5z7H55B8QTZ1ZHh/dDkyvh/erz8wQ8kgsyZux5M04NtcmpFOdPXYQ+Eix2SO

Level 3, made by Crispy



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

Level 4, made by tttppp




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

Level 4.5, puzzle made by COTO (but originally numbered "level 4" by accident)



u0S6f3vkLrci2hA9fY4sPEZkMfdZsq7/cjNoy0wL3S7utpfX8rsX6B6/GYBxMtn6GuSBXXQIUCsnOrYze7KGtxgS0ywnQlptbvqfryNGYDep0LFk/Ij/iRTocU3GwEME82Z0RA7cC5KH1d/wNM7jMWPC7wqjNk44u+oziAra+FMIDkFwKjtVNIFBnEKACHB0z4gKzcgSmtFV6ziXjzk51Ukvfjc4g1bFapNIWvos6q1ARkkw44nNuxXUnmus/QB8jhN32SiuyhQkzYBiNO02TBTS+B43M7zJHL95iFgTuF326VLkA79QMwKbpfEONv/xFeL3+Ck3jDvc2YjtknOv7aDGc2kgcsQkmiXEOBcVzkQjQDCKEa4p+z+ZHhHJy83uOEu8Iwnp7+h427pvXGZn9UYZv2gl0V0MxYTwx6OLv3rZf/3Whdrv29uGtjOy01z537QuRk0XXytECq2zOfmO6JD7NaJZTM2VL9qshizSmQ/b//yWugJl2Ojadsxc43BvNaSAi6gyIwJudYP8FyaZru2Lst9tvGEyf7jyN/UkMpOmqYfU4fsXMCiE9w1WHULjnpx5vgRhx4Kdf+3Zm0k6smoqY/l4uxGf3rhHddWjTuy+UX+42gsEyc/ynQ/STfVXfeQfVAFcwexWZmDy/dhP42rjSBEYZymbAKkEnOXlkuXQvSn5N0kb51c4kzUVFA0+tLsSqh+j7K/zYkUuul+L/K8FT/f6CvQA2yK8Ah1a5bTRNUYgwBe2Slz6tPMpASeXaXsyq42PVH3XpVReVsqvNgk6HpJTCDUmzl5P/qMicEOcj2yyTsM9Mgm1xTGr8bTpW+I7+63jADlT9gyzNZ2/Lq+oCS06h05DhG7llTo9pkjj+4ovhMjigY/Z8vDKWLlHnY1uUYK2DmdgLHz05DykEdEtXfgG0BN16VEHxbBf4eXcAzRQ1KzfsmR9Tbb8JguelrtSht+lODLuuwhMimzsJ2dRTOfhX3gwNhMZ+JR4c3eXoLpy7QBfHoCkGTl2cVBgZNjB0fZPmJp4mkuuZnUCcz+0d9WRZNTRSfnPQSpr2m5j5v//3IXDi4AwHox5PSFUaTsLBUw/+p+rC7lDlxpPycWz6kzgub5nhqKISUCHckvZGo6aSMmwS9VgDMMLh7GhSjDnY+T1Bb8izrij6dJvnbcye4OcfcptTyVPiE3j0Pr3hwhV8YHBIqQgtOfc53ciyh1OS3QfPehr5BxVSyuViXjfYBad3eTAqiHd7FaLnCCNUjsTj3ZZHhxK1okR3uheFVfyg6EHi33ifRNRngg41E0YzC8gk7WywuBkNaebbtiqNfF85RK6hL8vQli39RItm/vniU6wJTzdlxO9uPcvyQGsPlBLqG8iBAVc9bd521uYnIP2QdaPe57MX9lJJwYn9A0GHZF2o84eeAnfHRx7YYfk8PtFRka+pF+32dAI89UERIvrR4KKHcq7HaV4IG7pqHghi056glHT3uZU52Fqoq2tos1iKe3V3kDaNZk0Q+fz+Ib99zlp07+0opn5oFJkvoPVtonZVViK+q62SSkj/cx6DDaH293GXEmidAu5kBv6LNYjEXTqXsci8I0vwXap

Level 5, puzzle made by Josh



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

As I know many people here are smart and I have not really spent any time on making this secure. I don't doubt that someone will decode the puzzle strings or crack the whole thing if they want too.



Also sorry for any bugs!


BONUS: Crack the code used! Achieved



Answer



For the people who want to crack it by using code (and to get my bonus points):



The OP used Tea crypt with the password "vaj123":



so



s2 = Tea.decrypt(inputS, "vaj123"); decrypts the string (in inputS) into a map of values which it saves in s2. The first value has to be discarded, the rest of the values are used to make up a 20 by 20 grid. -1 is nothing placed, and 0-7 are the diffirent mirrors used.

You can execute the above command in the browser's console on the Black Box page.



And if you want the reverse-engineerd stuff:



Full code of the generation of the puzzle can be found here: http://jsbin.com/civuvetezi/1/watch?js,output



Yes I know, I am a party pooper.


Good luck!


thermodynamics - How many degrees of freedom does a spring have?


I'm currently learning about thermodynamics and heat capacities. We were told that the theoretical molar heat capacities of all solids should be $3R$. I was told this is because there are 6 different vibrational degrees of freedom for each atom in a solid. Since theses atoms are stuck in place, they can't rotate and translate so I tried to figure out exactly what these 6 degrees of freedom look like. Here are my thoughts:



Firstly, I like to think of degrees of freedom as places where energy can go. For example, in a gas, energy can be transferred into a particle as kinetic energy. This kinetic energy can make it translate on each axis and rotate on each axis (assuming it's nonlinear polyatomic). Of course, each particle also has a potential energy associated with it (another place energy can go), but we can ignore it because the average potential energy of the system won't change (for fixed volume).


Now let's take a hypothetical solid that is only 1 atom thick. Its lattice will look like this:enter image description here


My teacher would say this has 4 degrees of freedom because each atom in the solid can vibrate up/down or left/right. But using my definition of degrees of freedom, I would say there should only be 2 degrees of freedom. The potential energy associated with the x direction and the potential energy associated with the y direction. The reason I'm discounting kinetic energy is because when $E_{kinetic}=0$: $$E_{mechanical} = E_{potential}$$


Therefore, we can define the amount of energy added simply by what it does to the maximum potential energy. This is why I think a normal spring only has one degree of freedom (its potential energy). So why does the system depicted above have 4 DOF's?




english - What is the smallest set of letters that can spell any integer?



Today my 9-year-old nephew told me that he can spell any integer in English using only 9 letters. This is how he's doing it:


    ...
-3: MINUS ONE MINUS ONE MINUS ONE
-2: MINUS ONE MINUS ONE
-1: MINUS ONE
0: ONE MINUS ONE

1: ONE
2: ONE PLUS ONE
3: ONE PLUS ONE PLUS ONE
...

The letters he's using are E, I, L, M, N, O, P, S and U.


Can we do better and spell any integer with less than nine letters? I think so!


Just please be sure to briefly explain any maths you are using in your method, so that my young nephew (and his uncle) can understand.




Saturday, June 24, 2017

combinatorics - 15 Distinct Weights' Sorting


There are $15$ balls. Each of them has a different weight. You want to sort them according to their weights.
You have a friend who will help you with his scale to do this. At each weighing process, you will give $3$ balls to him. Then he will specify the order of weights of these $3$ balls. But, he will not say the weights of the balls. (for example: For three balls, he will say only order: $a \gt b \gt c$ )


At least how many weighing are needed to ensure that all the balls are correctly ordered by their weights?




dimensional analysis - Functional derivative and units


The both sides of below equation don't give the same units, e.g. $$ \frac{\delta}{\delta \phi (\tau)}\int_a^b \phi (\tau') d\tau'=1\;. $$ where $a<\tau



wordplay - Words that are anagrams of themselves


Can you think of a 6+ letter word that is an anagram of itself? The rule is that the letters of the word must be able to be rearranged to spell the original word, but none of the letters are in their original position.


4 letter words like dodo mama and papa are examples, but how many can you think of with 6 or more letters - obviously the number of letters will always have to be even. What's the longest self-anagamatic word?




fluid dynamics - Can a very thin sheet of any material float inside a liquid?


When a body is immersed in a liquid,buoyancy is the net force of all the forces acting on it. Now the forces are equivalent to those which will act on the same volume of liquid.Rightly so,but considering the chaotic motion of the molecules of the liquid,for every force acting on that particular volume of liquid,there will be another force in the opposite direction but of different magnitude.The magnitude differs because of the difference in height. Now let's consider a very very thin sheet immersed in the liquid.The opposite forces now will be of the same magnitude as difference in height is negligible and hence will cancel each other!so the only force acting will be its weight. So the thin sheet will never be balanced!So does that imply that no thin sheet will ever float inside a liquid??



Answer



You seem to be considering an (idealised) infinitely thin sheet. You're right that such an object would have no volume and thus displace no fluid --- but it would also have no mass, since its mass is equal to its density times its volume, which is zero. I guess you could conclude that it should have neutral buoyancy, because both its weight and the buoyancy force are zero.


However, one should be careful about this conclusion, because, any real sheet of material will be made of atoms. It will have a finite mass and a finite volume, and its buoyancy will depend on the ratio of the two, as it does for any material.


Friday, June 23, 2017

word problem - The telltale code


The other night, while writing this Java class that could be serialized into a socket, I fell asleep and had a weird dream.


In my dream, when I ran the program, the socket and the class object suddenly came alive. I tried to walk away from them but they followed behind.


As I desperately tried to change the program, I found the following strange piece of code in it, having no apparent rhyme or reason:


charge();
queue(factorial((26 * 275903).toAscii()));


h = m_socket[0].object;
assert( h.locale == "en_GB" );

if (true)
{
b = h.extractInternalStructure();
g = hash(b);
r = b.read(g);
// m_socket[1].send(r);
}


Which story had I been dreaming?



Answer



I don't have time to flesh out every detail right now, but the story is clearly



Jack and the Beanstalk.



Some of the clues that give it away:



A charge is a fee, queue is FIFO, and the ascii gives fum!, i.e. "Fee fi fo fum!".

The variable h represents Jack, an Englishman (locale = en_GB).
Be he alive or be he dead (always true), I'll grind his bones (extract internal structure and hash it) to make my bread (b.read(g)).



Some more:



"m_" is a coding convention to denote a "member". m_socket[0] denotes the nose and the "h" is the "object" of its sense. m_socket[1] denotes the mouth, and the comment line implies that the result would be eaten.



In the back-story:



A "jack" is a type of socket and a Java "bean" is a special type of Java class/object that can be serialized (among other properties). In the story, the two follow behind, i.e. "stalk" the author. Thus: Jack and the bean stalk!




word - Grandma's Simple Puzzle Box


One day, after a long morning walk, my Grandma came home with a box in her hand and told us the story of how she got it.


She said that she met six women in the park who were sitting together in a line. Five of them wore shirts of two different colors, but the last one wore only a single color. The group greeted her and placed the box in her hand, along with a piece of paper. They said that they had been trying to open the box for the last few days but hadn't managed it yet.


When Grandma showed me the box, I noticed some buttons to enter a word. The note she'd been given had this sequence of numbers written on it:


1   12   6   4   10   36   14   48   9   20   55

This must be a clue to identify the word. I'm sure it's a meaningful English word. Can someone help me open the box?


Hint 1



You need to play with transformed digits and find the word hidden in this puzzle. Each number will transform individually and the nature of the numbers will help you in that.





Answer



Few easy steps to solve it:



Original sequence:
1 12 6 4 10 36 14 48 9 20 55
Which i will divide with the simplest sequence:
1 2 3 4 5 6 7 8 9 10 11
=
1 6 2 1 2 6 2 6 1 2 5

And use hint: "Five of them wears the two colors T-shirts but the last one wears the single color".
16 21 26 26 12 5
And that makes word:
p u z z l e



general relativity - How is the classical twin paradox resolved?


I read a lot about the classical twin paradox recently. What confuses me is that some authors claim that it can be resolved within SRT, others say that you need GRT. Now, what is true (and why)?



Answer



To understand this paradox it's best to forget about everything you know (even from SR) because all of that just causes confusion and start with just a few simple concepts.


First of them is that the space-time carries a metric that tells you how to measure distance and time. In the case of SR this metric is extremely simple and it's possible to introduce simple $x$, $t$ coordinates (I'll work in 1+1 and $c = 1$) in which space-time interval looks like this


$$ ds^2 = -dt^2 + dx^2$$


Let's see how this works on this simple doodle I put together



alt text


The vertical direction is time-like and the horizontal is space-like. E.g. the blue line has "length" $ds_1^2 = -20^2 = -400$ in the square units of the picture (note the minus sign that corresponds to time-like direction) and each of the red lines has length zero (they represent the trajectories of light).


The length of the green line is $ds_2^2 = -20^2 + 10^2 = -300$. To compute proper times along those trajectories you can use $d\tau^2 = -ds^2$. We can see that the trip will take the green twin shorter proper time than the blue twin. In other words, green twin will be younger.


More generally, any kind of curved path you might imagine between top and bottom will take shorter time than the blue path. This is because time-like geodesics (which are just upward pointing straight lines in Minkowski space) between two points maximize the proper time. Essentially this can be seen to arise because any deviation from the straight line will induce unnecessary space-like contributions to the space-time interval.


You can see that there was no paradox because we treated the problem as what is really was: computation of proper-time of the general trajectories. Note that this is the only way to approach this kind of problems in GR. In SR that are other approaches because of its homogeneity and flatness and if done carefully, lead to the same results. It's just that people often aren't careful enough and that is what leads to paradoxes. So in my opinion, it's useful to take the lesson from GR here and forget about all those ad-hoc SR calculations.


Just to give you a taste what a SR calculation might look like: because of globally nice coordinates, people are tempted to describe also distant phenomena (which doesn't really make sense, physics is always only local). So the blue twin might decide to compute the age of the green twin. This will work nicely because it is in the inertial frame of reference, so it'll arrive at the same result we did.


But the green twin will come to strange conclusions. Both straight lines of its trajectory will work just fine and if it weren't for the turn, the blue twin would need to be younger from the green twin's viewpoint too. So the green twin has to conclude that the fact that blue twin was in a strong gravitational field (which is equivalent to the acceleration that makes green twin turn) makes it older. This gives a mathematically correct result (if computed carefully), but of course, physically it's a complete nonsense. You just can't expect that your local acceleration has any effect on a distant observer. The point that has to be taken here (and that GR makes clear only too well) is that you should never try to talk about distant objects.


classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...