Thursday, May 31, 2018

cosmology - Why is the Planck/WMAP estimate of the age of the universe preferred?


A recent physorg article is titled "The measurements of the expansion of the universe don't add up". The article says



The current analysis of the variable brightness of cepheids with space telescopes such as the Hubble, along with other direct observations of objects in our cosmic environment and more distant supernovae, indicate that the H0 value is approximately 73.9 kilometres per second per megaparsec (an astronomical unit equivalent to about 3.26 million light years).



However, measurements based on the early Universe provide an average H0 value of 67.4 km/s/Mpc. These other records, obtained with data from the European Space Agency's Planck Satellite and other instruments, are obtained indirectly on the basis of the success of the standard cosmological model (Lambda-CDM model) ...



I thought I could learn more about these different methods from wikipedia, but the age of the universe article only talks about the Planck data and WMAP, and an age ~ 13.7B years. The wiki talk page does have a section that refers to the Hubble's Law page, which does talk about different methods and history, but also says



More recent measurements from the Planck mission published in 2018 indicate a lower value of 67.66±0.42% although, even more recently, in March 2019, a higher value of 74.03±1.42% has been determined using an improved procedure involving the Hubble Space Telescope.[60] The two measurements disagree at the 4.4σ level, beyond a plausible level of chance.[61] The resolution to this disagreement is an ongoing area of research.[62]



So why is the Planck/WMAP estimate of the age of the universe preferred over similar alternatives, such as estimates based on Hubble observations?




forces - How does the shape (form; not cross-section) of a spring impact performance?


Cylindrical compression springs are everywhere and then some applications choose other forms like rectangular or unique polygonal form. What impact does the form of a compression spring have and how do you calculate an optimal solution?




Cryptic crossword clue: Seblings?


Seblings? (5,8).


This is a cryptic crossword clue. An easy one for me to start. What is the answer to it?


EDIT: Oof! Clearly not the way to post a question. I'm not sure what else I could say, though. It's a cryptic crossword clue, and if you were doing a cryptic crossword, all you would be given would be "Seblings? (5,8)."


For those of you not familiar with this, the clue is the word "seblings", the question mark shows you what kind of clue it is, and the 5 and 8 show that you the answer is two words, the first word five letters long, and the second is eight.


If this isn't enough, just let me know, and I'll try to present it another way.




Answer



I think the intended answer is probably



Mixed blessing



because



well, seblings is blessing, mixed (anagrammed). Though this doesn't fit the standard format for a cryptic clue.



thermodynamics - Why is it so inefficient to generate electricity by absorbing heat?


When I turn on a heater, it's supposed to be roughly 100% efficient. So it converts electricity to heat with great efficiency, but why can't we do the reverse: generate electricity by absorbing heat? I have been searching the internet and from what I have read it seems completely pointless because it is so inefficient, like ridiculously inefficient, as in 10% efficient. So why can't we do the reverse? I get that energy is lost when converting from one form of energy to another but how can we get such great efficiency going from one form but have horrid efficiency going back?


I also read online that one way to cool the earth down could be to radiate the heat off the planet. Anyways, sorry about my mini debate, can anyone answer how we could potentially cool the earth, because to me it would seem funny if we couldn't, and if we could then global warming wouldn't be as bad of a thing as it is now, would it?



Answer



tl;dr- Current technology absorbs temperature gradients, not heat. As temperature gradients become arbitrarily large, their information content nearly approaches the heat's information content, such that the apparent thermal efficiency,$$ {\eta}_{\text{Carnot~efficiency}}~~{\equiv}~~\frac{E_{\text{useful}}}{E_{\text{heat}}}~~{\approx}~~1-\frac{T_{\text{cold}}}{T_{\text{hot}}} \,,$$nearly approaches unity, showing that we can almost absorb heat while a temperature gradient is sufficiently large.




Hypothetical/future technology: Absorbing heat for energy


You could harness heat with near-perfect efficiency! Just requires finding Maxwell's demon. Maxwell's demon can be tough to find, but Laplace's demon could tell ya where it's at.


The fun thing about Maxwell's demon is that it likes to separate stuff out based on its highly precise perception and movement:



            .


So, you basically tell Maxwell's demon to let out high-speed particles when they're at nearly-tangential velocities to power a dynamo. And, bam! Electricity.


One trouble with this scheme is that we don't really know what heat is. I mean, we get the gist that particles are bouncing around and such, but we don't know all of the exact locations and velocities and such for all of the particles. And given that ignorance, we're basically unable to do anything with heat.


Except, of course, when our ignorance isn't complete. At the macroscopic level, we can appreciate stuff like temperature gradients; the larger the temperature gradient, the more information we have about relative motion of the particles at different temperatures.


And we can exploit this information, up to the point at which we've drained it away. For example, we can use heat to boil water, producing steam and thus raising pressure, using that pressure to turn a turbine. As the steam turns the turbine by going from a region of higher pressure to lower pressure, we again lose discriminating information about the system until our ignorance is again complete; but, we get useful energy out of the deal.


Conceptually, it's all about information. Whenever we have information about something, we may be able to turn that information into effect until the point at which we cease having information. Though we might say that we don't necessarily lose all of the information, as the energy that we get out of the deal isn't so much actually "energy" quite so much as it's a system that we have relatively more information about, and thus can exploit more readily.


Maxwell's demon and Laplace's demon are powerful critters because they have tons of information. By always having information, they can always construct systems that they can exploit for the extraction of energy. By contrast, humans tend to be limited in what information we have.


And that's the problem with just arbitrarily absorbing "heat": heat is a vague description about stuff moving around. In fact, even knowing a temperature is fairly useless information by itself; rather, we need temperature gradients, i.e. discriminating information, to knowingly construct a system that behaves how we want it to, e.g. a power generator.


In real life, there's interest in creating molecular machines, like observed in the classical example of ATP synthase, as a future technology. As @J... pointed out, Maxwell's demon in the above is acting as a thermal rectifier which are currently being researched (example).





Current technology: Absorbing temperature gradients, not heat



Why is it so inefficient to generate electricity by absorbing heat?



The above describes a system for generating electricity from heat. However, current technology never does this.


With current technology, we absorb temperature gradients. This may sound pedantic, but the fact that we're absorbing gradients and not heat itself is precisely why we can't get the energy equal to the heat out of the process.


Since we absorb the gradients, the Carnot efficiency tends to increase with the size of the gradient,$$ {\eta}_{\text{Carnot~efficiency}}~~{\approx}~~1-\frac{T_{\text{cold}}}{T_{\text{hot}}}. $$


Conceptually, the reason for this is that, as the temperature gradient$$ {\Delta}T~~{\equiv}~~T_{\text{hot}}-T_{\text{cold}} $$becomes arbitrarily large, the information contained in knowing the temperature gradient approaches the information that Laplace's demon would know, at which point efficiency would approach unity:$$ \lim_{{\Delta}T{\rightarrow}\infty}{\left(1-\frac{T_{\text{cold}}}{T_{\text{cold}}+{\Delta}T}\right)}~~{\rightarrow}~~1, $$i.e. 100% efficiency.


This is, sure, you wouldn't know the exact velocities of all of the particles, but what you don't know is dwarfed by what you do know, i.e. the extreme relative temperature gradient.


Wednesday, May 30, 2018

riddle - The Perpetual Teapot (no.28)


So, as I stated in my first Teapot riddle, I was challenged by a fellow user to create a Teapot riddle, and thus I have created two (numbers 26 and 27). However, to put my traditional stamp on it, I've decided to create one more. Welcome to the Perpetual Teapot!





The Rules



  • I have one word which has several (2 or more) meanings.

  • Each of the meanings is a teapot (first, second ...).

  • You try to figure out the word with my hints.

  • I will post a new hint everyday; with each hint making it easier to solve.




The Puzzle


Since this one will probably expand quite a bit, I'm going to use a different format to keep the post as short and clean as possible.





  • My first teapot is a small stream.

  • My second teapot is the average type of something.

  • My third teapot is a rapid series of notes forming a scale.





Bonus Reputation Opportunity


I will award $50$ reputation if you can explain why I chose the word.





Good luck to you all and have fun!



Answer



Is it a ...



Run



First teapot -



Run is the name of small streams in North America and is also a name of a stream in the Dutch province of Noord-Brabant.




Second teapot -



Going for a run is an average type of exercise to get in shape with. One might say its a pretty common exercise.



Third teapot -



A run in musical terms is a rapid series of ascending or descending musical notes often closely spaced in pitch forming a scale.



Edited to remove initial guess of...




Headwaters



general relativity - How to prove a symmetric tensor is indeed a tensor?


Our professor defined a rank $(k,l)$ tensor as something that transforms like a tensor as follows:


$$T^{\mu_1' \mu_2'...\mu_k'}{}_{\nu_1'\nu_2'...\nu_l'} ~=~ \Lambda^{\mu_1'}{}_{\mu_1}...\Lambda^{\mu_k'}{}_{\mu_k}~\Lambda^{\nu_1}{}_{\nu_1'}...\Lambda^{\nu_l}{}_{\nu_l'}~T^{\mu_1\mu_2...\mu_k}{}_{\nu_1\nu_2...\nu_l}$$


Where $\Lambda$ are the Lorentz transformation matrices (translations, rotations, or boosts). I'm not sure if this is only for SR or if also for GR since we've only been talking about SR thus far, though GR is something we'll be covering soon.


He wrote on the whiteboard: if $S_{\mu\nu\rho}=S_{\nu\mu\rho}$ then $S$ is symmetric in $\mu$ and $\nu$.


But let's just talk about a rank (2,0) symmetric contravariant tensor for a second, denoted $S^{\mu\nu}$ and equals $S^{\nu\mu}$. How would we prove this is a tensor? Our book uses $R$ in place of $\Lambda$ in their formulations above, where $R$ might just be rotations. I'm sure General Tensors would have any jacobian and inverse jacobians are matrices rather than just the Lorentz transformations. This is question in Prof. Zee's, "Einstein's Gravity in a Nutshell", Chapter I.4 Exercises 2.


Also if you want to give a student like myself who is new to tensor some advice on learning tensors, and some tensor properties, and how to work with them, be my guest :)


Also, are all transformations homogeneous linear transformations? - These can be read about at: http://www.math.ucla.edu/~baker/149.1.02w/handouts/e_htls.pdf


http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/rank-two-tensors/ This lecture gives a nice matrix form of what a symmetric (2,0) tensor looks like. I think this may help, thinking of these tensors as matrices themselves, visually. Basically they're symmetric matrices of the form $A^T=A$. Also we can think of $\Lambda$ if it's a rotation matrix having the property $\Lambda^{-1}=\Lambda$.


Does this picture below do anything for the problem? enter image description here





thermodynamics - Do gravitational waves have entropy?


We know, according the current understanding of black holes and General Relativity, as well as quantum fields in General Relativity, that black holes have an entropy proportional to the area of the event horizon. No problem with that. My question is simpler, but something I am not clear on. Does the gravitational radiation, eg that produced by the binary black holes that merged that carried away about 3 solar masses of energy, also carried away entropy? (There is no problem with the black hole entropies, the final black hole still had more than the initial ones added up)


I am not referring to Hawking radiation nor to any entanglement issue. This is more basic. The question is what statistical or microscopic property of the gravitational wave can be described as representing that entropy? The different ways it could have been produced? Some statistics on the (linearized theory) gravitons? Or does the question or concept of entropy of a gravitational wave not make sense? I've read up what I could find, no clear answer.



Answer



As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical gravitational wave defined by a full solution of the Einstein Field equations has an entropy of zero just as a full solution of Maxwell's equations does; if you know at the outset that the wave has come from a lone black hole whose state is known, then measurement of the amplitude, polarization and arrival time alone will fully define the wave (the six independent, modulo gauge, components of the metric tensor at your position).


But from these perfectly defined states, gravitational wave and light wave systems can take on "imprints" from their interactions with the World around them in many ways, so that any set of macroscopic measurements of a wave leaves much about the wave's state that is unknown:





  1. As in Lawrence Crowell's Answer, the source could have an unknown configuration. There may be a complicated system of gravitating black holes generating the waves, so our ignorance of this configuration means that we cannot infer the full state of the wave from macroscopic measurements. There could even be some advanced society of creatures modulating the waves for communication purposes; the message that they encode has Shannon entropy that helps compose the wave's total entropy;




  2. Waves scatter from objects; unless the scattering is very simple, the scattering will lead to changes in the full wave state that cannot be gleaned from macroscopic measurements alone. The Optical Grasp of light scattered from rough surfaces increases as properties of the surfaces become encoded into the light's full state which are inaccessible to a macroscopic observer. In theory, gravitational waves are perfectly analogous: their grasp will be increased by interactions with complicated matter systems;




  3. Gravitational waves, like light, can in theory thermalize, so that gravitational black body radiation is in theory possible. One could imagine gravitational waves bouncing back and forth and interacting with vast regions of space filled by black holes and hot gas.





However, I suspect in practice the entropy of gravitational waves will be much lower than that of light. The interaction between gravitational waves and matter is vastly weaker than that between light and matter, simply by dent of (1) the weakness of the gravitational force's action on matter in comparison with that of the electromagnetic force and (2) the fact that gravitational wave sources are quadrupolar and higher order unlike light sources which can be dipolar. Therefore, the thermalization and increase of grasp theorized above are probably just that: theoretical possibilities that seldom if ever arise in our Universe, at least over timescales of the order of the Universe's present age.


electromagnetism - formula for transparency of very thin film of metal


Is there formula that gives transparency of very thin film of given metal (tens of nanometers) to the visible light/light of given wavelength ? Which properties of metals are needed for the formula ?


I need to know which thickness of aluminium has 40% transparency to visible light. Thanks.




mathematics - 21sums - more constraints


My initial question had lots of answers, as we can practically choose the middle square to be anything we want, and work outwards to the exterior edges, with the corners providing the flexibility needed.


So, as a bonus, I have added some more constraints. Same rules as before, with the outside sums only counting 2 sets of minuses.





8 7 3
--- --- --- ---
5 13 14 11 5
--- --- --- ---
7 11 10 13 7
--- --- --- ---
8 11 9 14 8
--- --- --- ---

5 6 9

Can you now solve the grid?



Answer



I think I have the answer




8 7 3
3 5 2 1
5 13 14 11 5

2 3 4 4
7 11 10 13 7
5 1 2 3
8 11 9 14 8
3 2 4 5
5 6 9

Reasoning



The 8 at the top left hand corner of the grid must be made by the sum of 3 and 5 or 4 and 4. If 5 is on the left, this forces the digit below to be 0 (which is not allowed). If 4 is on the left it forces the digit two below to be 6 (also not allowed). Hence, it must be 3 on the left and 5 on the right. From there, we can easily follow the grid around the edge and then deduce the inside part from remainders.




fluid dynamics - Why pouring milk from a height makes a chain shape?



I have noticed a chain-like shape when milk is poured from a height into a cup.


The chain pattern repeats itself after some distance till it reaches the milk in the cup.


Any reason behind this? enter image description here




Tuesday, May 29, 2018

optics - How do Optically Active Compounds Rotate Plane Polarized Light?


I am not sure if this is more of a Chemistry or a Physics question, but in my Organic Chem class we discussed that chiral molecules will rotate plane polarized light. However, my professor did not discuss the mechanism at all. She also said that the only way to determine the angle of rotation was experimentally. I am really curious to know how and why this process occurs and if there is any way to calculate or even estimate the angle of rotation based on the structure of the compound.




electrostatics - Is it correct to say that electro-static potential of a charge is the energy of a motionless charge?


Is it correct to say that electro-static potential of a charge is the energy of a motionless charge?


I ask this to better understand this (great) answer;



Answer




Is it correct to say that electro-static potential of a charge is the energy of a motionless charge?




If you are speaking about electrostatic potential energy,
Than a motionless charge, as well as a charge in motion, can have an electrostatic potential energy. Just like a motionless mass $m$ a height $h$ above and close enough to the surface of the earth that the acceleration due to gravity is constant, can have a gravitational potential energy of $mgh$.


If you are speaking about electrostatic potential,
Than that would be the electrostatic potential energy per unit charge, or in the case of gravitational potential energy, the gravitational potential energy per unit mass.


That said,
The reason why your linked session says "practically we can't measure the (electrostatic potential) energy of a charge" is that they are talking about an absolute value for this energy. In general there is no absolute potential energy of a charge unless an absolute value of potential energy (or potential) is arbitrarily assigned a value of zero to some position.


Consider the gravitational potential energy analogy. If a motionless mass $m$ is a height $h$ above the floor in a room, it is said to have gravitational potential energy of $mgh$ with respect to the floor of the room. But if the floor of the room is also a height $h$ above the ground outside the building containing the room, the mass has a gravitational potential energy with respect to the ground outside of $2mgh$. If there is a surface of table in the room a height $1/2$ h from the floor, the mass has a gravitational potential energy of $(1/2)(mgh)$ with respect to the surface of the table and the floor of the room.


I think you get the idea.


Hope this helps.


Specific electron energy gap values $E_{i+1}-E_i$ vs. photons with arbitrary energy $hbar omega$


The energy levels of electrons in an atom are quantized $E_i$. A photon of a specific momentum $\vec p$ and energy $$\omega=(E_{i+1}-E_i)/\hbar$$ hits an atom and gets absorbed. Okay now say the energy of the photon is $$\omega'=(E_{i+1}-E_i)/\hbar+\Delta E/\hbar,$$ where the value $\Delta E$ is some energy, which is not even near the value of any electron state gap energy. hence the quantity $\Delta E$ cannot be absorbed by means of excitation of the other particle. What happens in this case? Does this energy-wise non-fitting photon just go through the material? Does the photon "spit" in a way that the part $E_{i+1}-E_i$ of the photon energy gets absorbed and a new photon with energy $\Delta E$ will move away? Microscopically, has this anything to do with soft radiation?



Okay well, if we model the atom as only being able to suck up energy by changing it's energy levels, this would violate momentum conservation. I guess the atom must start to move too, or if we view it as part of a bigger solid state system, the photon will make the thing wobble a little. From that perspective, we'd have to consider the kinetic energy distribution of the particles composing the material and this would be able to suck up any real energy or momentum value.


If we leave the microscopic description (in terms of which I'd like to see an answer to the first question - after all, I'm speaking of the behaviour of a single photon here) and go on and view this as a "gray body" problem, and in case the answer really is that after the photoelectric effect the rest energy gets termalized, then I'd like to know the relation (for a class of materials) between the absorption per frequency and material characterizing parameters like (electron/atom density? emissivity? ...).




mathematics - Robbers - The ultimate compass challenge



Please read the Cops post for all details.



Points are scored on the basis of how long a challenge remains unsolved before you solve. You may solve any number of challenges. You cannot solve your own challenges. Points scored by robbers and by cops are independent, and you can participate in both posts separately.




Monday, May 28, 2018

enigmatic puzzle - (7 of 11: Fillomino) What is Pyramid Cult's Favorite Shape?


Dear PSE users and moderators,
I’m new here in PSE, but I really need your help. There was this person who gave me a black envelope consisting 10+1 pages of puzzles, and also a scribble saying: “Find our favorites and you will be accepted to join our ‘pyramid cult’. Feel free to ask for help from your beloved friends on PSE. They will surely guide you into all the truth.” I’m also a newbie on grid puzzles, so, could you please give me any hint to solve these? It’s getting harder and harder later on..

- athin


Jump to the first page: #1 Numberlink | Previous page: #6 Yajilin | Next page: #8 Ripple Effect




enter image description here



Rules:



  1. Fill in all empty cells with numbers under the following rules.

  2. Divide all of the board into blocks. Fill each block with the same number in 3-directions (parallel with one of the board sides).

  3. Each block contains as many cells as the number in the block.


  4. Same sized blocks cannot touch each other, in 3-directions.





Special thanks to chaotic_iak for testing this puzzle series!



Answer



The answer is



SPHERE, obtained by taking the letters in the regions labelled 1




I was doing a step by step solution, then had to go, then there was Sconibulus' answer when I came back - anyway, I'll drop it there.



Easy first steps enter image description here
Then the first tricky part: if you assume that the two 8s on the right are not connected, you have an impossible situation in the top left (red question mark). enter image description here
Therefore: enter image description here Now the 7 cannot block the 6 underneath, which implies enter image description here The 6s on the left cannot be connected: enter image description here Again the 6s on the left cannot be connected, and neither can the 3s nearby: enter image description here The 6s must be separated by a 1, which clears the 6 on the left. The 4s cannot be connected, and therefore the 5s must be connected. enter image description here The central left 8 can't be connected with the topmost 8, which yields two more 1s and the rest is clear. enter image description here



No Cheating - Cipher challenge


NO CHEATING


Today’s cipher is a famous cryptographic method known for its economy, ingenuity, and difficulty in cracking. Its inventor created all sorts of cool things, including musical instruments, a timepiece especially useful at the North Pole, and the earliest ancestor of the Oculus Rift. Aside from this cipher, which does not bear his name, he is most famous for co-inventing a means of communicating at long distances.


The key to unlocking the mystery is his name. Good luck!


The ciphertext is:


WXYCNW LAHWNZ WONSNI LEXAPE OSMWOW RCSRQC AZ


This is a cross-post from my puzzle blog. (Link in my profile.)


Hint:




No J




Answer



The message is:



A murder has just been committed at Salt Hill.



Which is...




The first line of an early telegraph that helped catch a murderer. It was the first arrest ever made using a telegraph. The inventor of this cipher method is also the co-inventor of the telegraph.



The method of encryption is ...



... the Playfair cipher. The hint says "No J", which hints at an alphabet of 25 letters, where the regular English 26-letter alphabet has one letter omitted. Usually J is represented by I. These letters are arranged in a 5×5 grid. There are several ciphers that use such a grid, and the Playfair cipher is one of them.



The key is ...



... Charles Wheatstone. The cipher is named after Lord Playfair, who promoted its use, but it was invented by Wheatstone, a well-known scientist and inventor.

The key is used to transpose the alphabet: The square is filled with the unique letters of the key in order of appearance, followed by the unused letters. This gives the following square:




C H A R L
E S W T O
N B D F G
I K M P Q
U V X Y Z

The cipher ...



... encodes pairs of letters. If the message has an odd number of letters, the message is padded with an X. The cipher cannot encode pairs of the same letters, so these second letter is usually replaced with an X.

The decoded message reads:


AM UR DE RH AS GU ST BE EN CO MX IT TE DA TS AL TH IL LX



It turns out that the letter G is used as replacement for J.

The title is a clue because...



"No Cheating" is another way of saying "Play Fair."



This inventor created all sorts of cool things, including...


...musical instruments



The English Concertina and many others.




...a timepiece especially useful at the North Pole



The Polar Clock



...and the earliest ancestor of the Oculus Rift.



The Stereoscope



He is most famous for co-inventing a means of communicating at long distances.




Telegraphy



symmetry - Noether's current expression in Peskin and Schroeder


In the second chapter of Peskin and Schroeder, An Introduction to Quantum Field Theory, it is said that the action is invariant if the Lagrangian density changes by a four-divergence. But if we calculate any change in Lagrangian density we observe that under the conditions of equation of motion being satisfied, it only changes by a four-divergence term.


If ${\cal L}(x) $ changes to $ {\cal L}(x) + \alpha \partial_\mu J^{\mu} (x) $ then action is invariant. But isn't this only in the case of extremization of action to obtain Euler-Lagrange equations.


Comparing this to $ \delta {\cal L}$


$$ \alpha \delta {\cal L} = \frac{\partial {\cal L}}{\partial \phi} (\alpha \delta \phi) + \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \partial_{\mu}(\alpha \delta \phi) $$



$$= \alpha \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi \right) + \alpha \left[ \frac{\partial {\cal L}}{\partial \phi} - \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \right) \right] \delta \phi. $$


Getting the second term to zero assuming application of equations of motion. Doesn't this imply that the noether's current itself is zero, rather than its derivative? That is:


$$J^{\mu} (x) = \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi .$$


I add that my doubt is why changing ${\cal L}$ by a four divergence term lead to invariance of action globally when that idea itself was derived while extremizing the action which I assume is a local extremization and not a global one.




Sunday, May 27, 2018

quantum mechanics - Schrodinger equation from Klein-Gordon?


One can view QM as a 1+0 dimensional QFT, fields are only depending on time and so are only called operators, and I know a way to derive Schrodinger's equation from Klein-Gordon's one.


Assuming a field $\Phi$ with a low energy $ E \approx m $ with $m$ the mass of the particle, by defining $\phi$ such as $\Phi(x,t) = e^{-imt}\phi(x,t)$ and developing the equation



$$(\partial^2 + m^2)\Phi~=~0$$


neglecting the $\partial_t^2 \phi$ then one finds the familiar Schodinger equation:


$$i\partial_t\phi~=~-\frac{\Delta}{2m}\phi.$$


Still, I am not fully satisfied about the transition field $\rightarrow$ wave function, even if we suppose that the number of particle is fixed, and the field know acts on a finite dimensional Hilbert Space (a subpart of the complete first Fock Space for a specific number of particles). Does someone has a reference to another proposition/argument for this derivation?


Edit: for reference, the previous calculations are taken from Zee's book, QFT in a Nutshell, first page in Chapter III.5.



Answer



I think you are mixing up two different things:




  1. First, you can see QM as $0+1$ (one temporal dimension) QFT, in which the position operators (and their conjugate momenta) in the Heisenberg picture play the role of the fields (and their conjugate momenta) in QFT. You can check, for instance, that spatial rotational symmetry in the quantum mechanical theory is translated into an internal symmetry in QFT.





  2. Secondly, you can take the "non-relativistic limit" (by the way, ugly name because Galilean relativity is as relativistic as Special relativity) of Klein-Gordon or Dirac theory to get "non-relativistic" Schrödinger QFT, where $\phi$ (in your notation) is a quantum field instead of a wave function. There is a chapter in Srednicki's book where this issue is raised in a simple and nice way. There, you can also read about spin-statistic theorem and the wave function of multi-particle states. Let me add some equations that hopefully clarify that (I'm using your notation and of course there may be wrong factors, units, etc.):




The quantum field is: $$\phi \sim \int d^3p \, a_p e^{-i(p^2/(2m) \cdot t - p \cdot x)}$$


The Hamiltonian is:


$$H \sim i\int d^3x \left( \phi^{\dagger}\partial_t \phi - \frac{1}{2m}\partial _i \phi ^{\dagger} \partial ^i \phi \right) \, \sim \int d^3p \, \frac{p^2}{2m} \,a^{\dagger}_p a_p$$


The evolution of the quantum field is given by:


$$i\partial _t \phi \sim [\phi, H] \sim -\frac{\nabla ^2 \phi}{2m}$$



1-particle states are given by:


$$|1p\rangle \sim \int d^3p \, \tilde f(t,p) \, a^{\dagger}_p \, |0\rangle $$


(one can analogously define multi-particle states)


This state verifies the Schrödinger equation:


$$H \, |1p\rangle=i\partial _t \, |1p\rangle$$ iff


$$i\partial _t \, f(t,x) \sim -\frac{\nabla ^2 f(t,x)}{2m}$$


where $f(t,x)$ is the spatial Fourier transformed of $\tilde f (t,p)$.


$f(t,x)$ is a wave function, while $\phi (t, x)$ is a quantum field.


This is the free theory, one can add interaction in a similar way.


quantum field theory - What is Euler Density?


Can someone please explain to me what Euler Density is? I have encountered it in Weyl anomaly related issues in various articles. Most of them assumes that its familiar, but I couldn't find any accessible paper or a book discussing that. So, it would be nice if I can understand what it is physically and mathematically and also find a reference where I can look it up.


Also related to that it would be nice to find a reference where people have derived $\langle T_i^i\rangle$ in curved background which involves Euler Density, $W^i$ etc.



Answer



Euler density is simply the integrand in $2n$ dimensions of the integral that is equal to the Euler characteristic. The Euler characteristic may be written as the integral of the following Euler density in $2n$ dimensions: $$E_{2n} = \frac{1}{2^n} R_{i_1 j_1 k_1 l_1} \dots R_{i_n j_n k_n l_n} \epsilon^{i_1 j_1 \dots i_n j_n} \epsilon^{k_1 l_1 \dots k_n l_n} $$ Note that for $n=1$ i.e. in two dimensions, it is linear in the Riemann tensor - and therefore also in the Ricci scalar (because the Riemann tensor is fully determined by the Ricci scalar in 2D). In four dimensions, the Euler density is quadratic in the Riemann tensor, and so on.


The Euler character - a "regularized number of points in a manifold" - may also be calculated in many other ways, e.g. for polytopes by adding the number of faces, subtracting edges, adding vertices, etc. For nice manifolds, it's only nonzero for even-dimensional manifolds. For closed orientable two-dimensional Riemann surfaces, it is given by $2-2h$ where $h$ is the number of handles (the genus also known as $g$). One may construct a general open/closed orientable/unorientable two-dimensional manifold by adding $b$ (circular) boundaries i.e. holes and $c$ crosscaps (holes with identified antipodal points, creating an unorientable manifold) and the total Euler characteristic is then $$ \chi = 2-2g - b - c.$$ You should imagine that if there is a function $L(\sigma^i)$ depending on the manifold's coordinates sigma such that $L$ has units of $U$, the path integral $\int DL(\sigma^i)$ has units of $U^\chi$ where $\chi$ is the Euler characteristic: that's what I meant by saying that $\chi$ is the regularized number of points.


So the Euler characteristic (or character) is arguably the most important and most elementary topological invariant of a manifold. The fact that the integral of $E_{2n}$ is a topological invariant may be seen by calculating its variation which vanishes (for any variation of the metric) - one reduces the derivative to some of the standard identities for the Riemann tensor, especially the two Bianchi identities involving antisymmetrization (and, in one case, one derivative).


The derivation of the trace of the stress-energy tensor is done for $d=2$ in Polchinski's "String Theory", Volume I. Equation (3.4.31) says $$T^a_a (\sigma) = -\frac{C}{12} R(\sigma)$$ where $R$ is the Ricci scalar, also interpretable as a multiple of the Euler density. The $C$ ends up being a definition of the central charge. I don't know the general form of a similar equation in $d$ dimensions but its exact form - at least the parameters - do depend on the theory. I guess that in general, the trace is equal to some linear combination of the Euler density and perhaps some other generators besides the stress-energy tensor.


Is there an equation for the residual strong nuclear force?



First of all, note the qualifier "residual". The present question is not the same as that asked, and answered, in the StackExchange question "Is there an equation for the strong nuclear force?" which was really addressed as though it asked "Is there an equation for the strong interaction between quarks?"


If there is no such equation, is there a numeric simulation program?




pressure - Water solidifies near the bottom of a hypothetically deep oceanic trench


How deep (kilometers) would an oceanic trench on planet Earth have to be in order for the water to become a solid because of the pressure from the water column? For simplicity, maybe assume chemically pure water, and a constant temperature of 300 kelvin of the entire body of water. An answer with realistic amounts of salt and with temperatures at each depth corresponding to the ambient lithospheric temperature would be welcome too, of course.


It can be assumed that the trench is so narrow that the change in global sea level is negligible.



I considered asking the same question for an air-filled well, in contact with the atmosphere, but I realized that the critical temperature for air is only 133 kelvin, so there would be no depth at which a phase transition would occur.




Saturday, May 26, 2018

general relativity - What is the evidence for Inflation of the early universe?


The theory of Inflation explains the apparent consistency of the universe by proposing that the early universe grew exponentially for a 1E-36 seconds. Isn't a simpler explanation that the universe is just older and so the homogeneousness comes from a slower more steady growth? Is there any evidence that rules out a slow growing universe and supports Inflation theory?



Answer



I can not go into much detail here but let me say that exponential growth brings many things that we see around us right now: absence of magnetic monopoles, a homogeneous universe in which no section is a "preferred" section i.e. has more matter density, and many more observable quantities.


Actually after a brief search I found a wiki article stating most of the things i said above and much, much more: http://en.wikipedia.org/wiki/Cosmic_inflation#Observational_status


cipher - Oh boi! Dis waas never Engliz! #2


This puzzle is second in the series, the first one being this. But these are related in no way.


Below are some weird combinations of letters, which are actually words translated in some different languages, and concatenated.






  • Each line has one word translated in different languages.




  • These 22 words lead to new groups of words. For example if two of the lines translate to Mount and Everest, the new group will be "Mount Everest".




There can also be words like "eel", "on" , and "must", which on arranging, sound or look like a group, which is Elon Musk in this case.


There will be 10 groups of two words, and two groups as individual words, to have a total of 12 groups (Let this set of groups be called Class A)





  • The groups from class A are related in a way, and can further be grouped, making a total of 4 B-classes, with each having 3 members from class A.




  • There is also a superclass C, which contains everything in class A and B.





What you have to do:





  1. Find the languages

  2. Find the 12 members of class A

  3. Find the 4 members of class B

  4. Find the only member of superclass C





Words below:





  1. tartiblashukuhlungavargakirimasortuvannya

  2. sirrahasyasekretimfihlo

  3. kusaginnayiochlikindlalaholodni

  4. qishebusikasitakalayazyma

  5. honcharumbumbikumbalkarukulol

  6. vihiluvakijokerhazillashadiganzhartivnyk

  7. zalizotemirinsimbiyakada

  8. haritayashilrangdaoluhlazazelenyy

  9. lyudynakishiminisaumuntu

  10. dveridoraeshikumnyango


  11. likhtarilangachiroqvesakkuduva

  12. pryntsisikhulushahzodakumaraya

  13. dyvnogalatiokungajwayelekileamutu

  14. kridaihryimidlalooyinlar

  15. simyettihataisikhombisa

  16. yonibturadiukukhanyasaeneliyaspalakh

  17. phuthumashoshilingikmankarannapospishayte

  18. kapelyukhisikhwamashapkatoppiya

  19. shifokorlikaryaudokotelavaidyavarayek

  20. khubekadampledumbletushibqolish


  21. askarsoldatisoshasoldaduvek

  22. manobirozkancanekuda



Da magick wond laees heir



Abrakedavra hocus pocus





Even partial answers are also welcome!



Do not trust Google translate if it translates the whole combination to a word. It might be misleading. Confirm it by translating the words separately. For example, it translates qishebusikasitakalayazyma to alchemist, but it isn't.


Some hints:



#20 - Your eyes are belong to me. It do not need a translation.


#3 - Brother of #10 is always against me.


#7 - I've felt fear twice while you were reading this. Well I feel it was thrice


#9 - What quite a lot of superhero's names have


#22 + #12 - C'est une traduction Française



Edit: Assume you have 3 'The' s available.




Answer



The languages are



Uzbek, Sinhala, Ukrainian, and Zulu



Here are the words I have so far:



1. Sorting (tartiblash / ukuhlunga / varga kirima/ sortuvannya)
2. Secret (sir / rahasya / secret / imfihlo)
3. Hunger (kusaginnayi / ochlik / indlala / holodni)

4. Winter (qish / ebusika / sita kalaya /zyma)
5. Potter (honchar / umbumbi / kumbalkaru / kulol)
6. Joker (vihiluvak / ijoker / hazillash / adiganzhartivnyk)
7. Iron (zalizo / temir / insimbi / yakada)
8. Green (harita / yashil rangda / oluhlaza / zelenyy)
9. Man (lyudyna / kishi / minisa / umuntu)
10. Door (dveri / dora / eshik / umnyango)
11. Lantern (likhtar / ilanga / chiroq / vesak kuduva)
12. Prince (prynts / isikhulu / shahzoda / kumaraya)
13. Strange (dyvno / galati / okungajwayelekile / amutu)

14. Game (krida / ihryi / midlalo / oyinlar)
15. Seven (sim / yetti / hata / isikhombisa)
16. Flash (yonib turadi / ukukhanya / saeneliya / spalakh)
17. Hurry (phuthuma / shoshiling / ikman karanna / pospishayte)
18. Hat (kapelyukh / isikhwama / shapka / toppiya)
19. Doctor (shifokor / likarya / udokotela / vaidyavarayek)
20. Dumble (khubeka / dample / dumble / tushib qolish)
21. Soldier (askar / soldat / isosha / soldaduvek)
22. Little (mano / biroz / kancane / kuda)




Class A:



We know 20 of these are pairs and two remain alone, for 10 + 2 = 12 members of Class A.
Class A members are: Sorting + Hat, Winter + Soldier, Hurry (Harry) + Potter, Iron + Man, Green + Lantern, Dumble + Door, Doctor + Strange, Flash, Joker, Hunger + Games, Little + Prince, Secret + Seven.



Class B:



Winter Soldier, Iron Man, and Doctor Strange are all Marvel characters.
Green Lantern, Joker, and Flash are all DC characters.
Sorting Hat, Harry Potter, and Dumbledore are all Harry Potter characters.

Little Prince, Secret Seven, Hunger Games are all Children/Young Adult books.



Superclass:



These are all fiction.



chess - Can you stop the rambling rook?


This is a chess puzzle someone showed me a long time ago. If you already know this puzzle please give the others a chance to solve it first. This puzzle was originally created by Otto Gallischek and it was published in Weser-Kurier on 1960-02-25. The problem can be seen here on the website Yet Another Chess Problem Database. It is also featured in this Chess Base article.


There is one small change though-this puzzle has a White queen instead of a rook on h3, as the original problem has.





Above you see a position where white is obviously on the winning side. The queen is one step from giving a checkmate. And the heroic h pawn made his way from h2 to c7 and is waiting for a promotion. However, black saw a last chance and moved Rf3+ in the last turn. The rook can't be taken, because black couldn't move anymore which is a stalemate (= draw). If the king moves away the rook will continue to give checks.



Can you stop the rambling rook and find a way for white to win the game anyway?


PS: Computers are not allowed, and no cheating.



Answer



If white can ever stay out of check for one turn, then it can promote its pawn and put black in checkmate. So in order to play perfectly, black must attempt to put white in check with every move. In turn, white should make sure that black has only one option for a check in the next move, or else the rook will "break free" and have much more influence over white's moves.


White's overall game plan is to sacrifice pawns and maneuver the king until black falls into a simultaneous capture+checkmate.


enter image description here



1. e3 Pawn moves forward to block rook.
1. ... Rxe3+ Rook captures pawn and places king in check.
2. c3 Pawn moves forward to block rook.

2. ... Rxc3+ Rook captures pawn and places king in check.
3. Ka2 King begins a "serpentine" maneuver, which will continue until move 9.
3. ... Ra3+ Rook takes only available move to put king in check, continuing until move 9.
4. Kb1 Ra1+
5. Kc2 Rc1+
6. Kd3 Rc3+
7. Ke2 Re3+
8. Kf1 Re1+
9. Kg2 Rg1+
10. Kf3 Rxg3+ Rook captures pawn in row 3 and places king in check.

11. Ke2 King begins serpentining back to column A.
11. ... Re3+ Rook continues to pursue.
12. Kd1 Re1+
13. Kc2 Rc1+
14. Kb3 Rc3+
15. Ka2 Ra3+
16. Qxa3# Queen takes rook and places black king in checkmate.

The theory of strings stretching between intersecting D-branes


I am trying to understand various aspects of intersecting D-branes in terms of the gauge theories on the worldvolume of the D-branes. One thing I'd like to understand is the worldvolume action for strings stretching between the D-branes. One thing I have considered is $M+N$ D-branes initially coincident but then an angle $\theta$ developing between $M$ and $N$ of them. The gauge symmetry is broken from $U(M\times N)$ to $U(M) \times U(N)$ with the off-block-diagonal terms of the gauge field becoming massive. I anticipate that $\theta$ is the vev of some Higgs field that mediates this transition.


What is the action of this Higgs field and where in the string spectrum does it come from?



Answer




First of all, if a stack of $M$ branes is rotated relatively to a previously coincident stack of $N$ branes, it's clear that the degrees of freedom that encode the relative angle $\theta$ are nothing else than the transverse scalars determining the position/orientation of these two stacks. Any D-brane or any stack of D-branes may be rotated in any way and the quanta of the scalar fields that remember the positions are just open string modes attached to these D-branes with both endpoints. If you study the location/orientation of a D-brane or a stack of D-brane, it's its own degree of freedom that has nothing to do with the behavior of other D-branes.


So the Higgs fields arise from the normal scalars determining the transverse positions of these stacks of D-branes and the action for these D-branes is still the same D-brane Dirac-Born-Infeld action.


Now, you apparently want to see how this degree of freedom that you call $\theta$ – it's just an awkwardly chosen "degree of freedom" that can't be invariantly separated from other degrees of freedom determining the shape of the D-branes – break the $U(M+N)$ symmetry down to $U(M)\times U(N)$ and your wording makes it sound like you believe it is just ordinary Higgs mechanism in the whole space.


However, it's important to realize that this breaking of the gauge symmetry doesn't occur uniformly in the whole space. In fact, near the intersection of the stacks, the gauge symmetry is approximately enhanced to the original $U(M+N)$. What's important is that the off-block-diagonal blocks transforming as $(M,N)$ under $U(M)\times U(N)$ arise from open strings whose one end point sits at one stack and the other end point sits at the other stack. The distances between the stacks go like $\theta \cdot D$ where $D$ is the distance between the intersection. So the open string modes get an extra mass $\theta D T$ where $T$ is the string tension.


To summarize, the symmetry breaking is always described by the ordinary (non-Abelian) Dirac-Born-Infeld action. Your "rotation of stacks relatively to each other" only differs from the ordinary "separation of parallel stacks in the transverse dimension" by the fact that the distance/separation between the stacks depends on the location along the branes. It is meaningless to ask for any new actions because the (non-Abelian) Dirac-Born-Infeld action always describes all the low-energy dynamics of similar systems. The stringy/D-brane dynamics is always governed by the same laws and one should only learn it once.


Let me mention that whenever the distance between the stacks exceeds the string length, all the off-diagonal open string modes are string-scale heavy and it is inconsistent to keep them unless you also keep the excited string harmonics in the spectrum. So a derivation of "effective field theory" that would neglect the stringy tower but that would just freely describe the higgsed large string-scale masses of the off-block-diagonal modes would be inconsistent.


logical deduction - A "Letter" Slope puzzle?


Your brother has been forcing you to some read some of his frankly ridiculous poetry, so, knowing that he was never a numbers man, you decide to vex him with a Number Slope puzzle. In fact, you are so confident that he won't be able to solve it that you bet all your PSE rep on it.


The next day, as you work through one of your countless puzzle books, you find a sheet of paper inside. It's not another poem this time, thank goodness, but it is slightly concerning...



Thought you could defeat me with your pitiful numbers? Pah! You'll be soiling yourself!


grids



What on earth is that supposed to mean? Why didn't he just send the completed grid? Or is he teasing you, hiding the solution in a puzzle of his own? Is all your precious, hard-earned rep now gone? What is your brother really saying?





Note:



The message, as well as the rule which provides the enumeration of the letters SOIL, have been correctly found. However, there is one more unidentified, intentional, confirming detail in the puzzle related to the additional letter. I will leave this question open until that aspect of it is found.

Edit: I have edited the image to provide a hint to the aforementioned detail.




Answer



To bare the mechanics of the self-inflicted insult suffered by our brother, as uncovered in TheGreatEscaper’s solution:





Row 1.   Taking the Number Slope™ puzzle hint assumes that S, O, I and L are equivalent to 1, 2, 3 and 4, in a consistent order observed within each set of linear squares within a tetromino.   The circled cells determine that O is 1 or 4 while S and L are in the middle. Thus I is also 1 or 4.



Row 2.   This is the direct result of that partial ordering. The circled cell group determines the order as O S L I or its reverse.


Row 3.   The 4×4 grids proceed to specify all but two inconsequentially ambiguous areas. Letters at dotted corners can now be copied to the 5×5 grid.


Row 3 -T.   Giant upside-down T  tetromino sighting — as the dotted corners are brought together.


Row 4.   A new mystery letter ▯ presumably is equivalent to 5 in the order.   The circled subsequence L S ▯ establishes the complete ordering as...



                    ... I L S O ▯.



The 5×5’s bottom row’s direction is unknown initially but S is in the middle either way.   This forces L into the center cell and the rest follows readily enough.   At the top of the 5×5 grid we at last see our brother’s embarrassing confession.



                    “ I   L O S T





Subtle reinforcing hints.


Why is the mystery letter T when E could just as well spell an equivalent message?   Continuing on TheGreatEscaper’s reasoning, there are 5 tetromino shapes known to science, T being the one not in play here, while polyominologists simply have not yet discovered an E-shaped tetromino or even pentomino.   As subliminal reinforcement, a gigantic upside-down T  tetromino is formed within the 8×8 combined grid of row 3 -T above.


How were numbers and letters matched up?   The 4 tetrominoes in play enumerate the 4 known grid letters, appropriately enough, by counting slopes.





This order is also found, in reverse, among the circled letters in row 3 above. But the puzzle’s poser has mentioned that yet one more reinforcing detail hides in plain sight among the grids:  




The letters in the dotted corners of each 4×4 grid are I, L, S, and O in that order. (This means that the grids are ordered by the number of slopes of their corresponding letters.) But the 5×5 grid seems to be "paired" with the second grid: the space between them is shortened!

Just as the 5×5 grid is an "exception", the letter T is an exception to the rule about number of slopes: it has two slopes, like the L it is paired with, but it is the fifth letter in the sequence. Also, it must be moved to the end of the solving order, just as its letter must be moved to the end of the sequence order.



mathematics - Probability of Seeing a Car in 10 Minutes & 30 Minutes



On a deserted road, the probability of observing a car during a thirty-minute period is 95%.
What is the chance of observing a car in a ten-minute period?



Hint: To clarify the question we are saying the probability of seeing any other cars in 30 minutes is 95% or more clearly, and more usefully, the probability of not seeing any other cars is 5%.



Answer



The answer is:



100% - 5%^(1/3) (cube root of 5%), which is about 63%



Why?



Because the probability of not seeing a car in thirty minutes is equal to the probability of not seeing a car for ten minutes to the third power. That is, not seeing a car for ten minutes three times in a row is like not seeing a car for thirty minutes




Or, with a formula:



If $P_{not30}$ is the probability of not seeing a car for 30 minutes and $P_{not10}$ is the probability of not seeing a car for ten minutes, $P_{not30}$= $P_{not10}^ 3 \Rightarrow$ $P_{not10} = \sqrt[3]{P_{not30}}$



Friday, May 25, 2018

electromagnetism - Electron Electric Field Mass?


I am confused of whether or not the expected electromagnetic field generated by the point-like electric charge of the electron distributed smoothly across space as a probability distribution creates the presence of an effective field mass.


I know that the probability current of the electron is $\langle \gamma^{0k} \rangle$, which is conserved. Multiplying the probability current of the electron by its total charge $q$ gives the charge current density $J^k = q\langle \gamma^{0k} \rangle$. From the current I can calculate the expected four-potential field as \begin{align} A^k(r,t) = \int \dfrac{J^k(r',t-c/r')}{|r-r'|} d^3 r' . \end{align}



From $A^k$ I can calculate the expected electromagnetic fields as $F^{jk} = \partial ^j A^k - \partial^k A^j$. Finally I can calculate the expected electromagnetic energy as \begin{align} U_{\text{eff}} = \dfrac{1}{8\pi} \int F^{jk}F^{jk} d^3 r . \end{align} Applying Einstein's relation energy is proportional to mass, I get the following effect field mass of the electron as \begin{align} M_{\text{eff}} = \dfrac{U_{\text{eff}}}{c^2} . \end{align}


Is $M_{\text{eff}}$ a real observable? I was unable to find an analytic solution for $M_{\text{eff}}$, however I did compute $M_{\text{eff}}$ for Gaussian distributed probability functions for the electron with varying standard deviation (spacial localization), and I noticed that for standard deviations of the order of $10^{-10}$ meters (lattice size), $M_{\text{eff}}$ was very small in comparison to the electron's rest mass. When the standard deviation was $10^{-15}$ meters (nucleus size), the $M_{\text{eff}}$ was comparable in size to the rest mass of the electron.




general relativity - curvature tensor component capable of doing work on $T_{mu nu}$


I'm wondering what part of the curvature tensor is able to do work (and hence transfer energy) in matter. I'm wondering if this tensor: http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor satisfies that property


I want to understand the generic assertion that GR doesn't conserve energy, and which scenarios do conserve it



Answer



Here is a FAQ entry I wrote for physicsforums.


How does conservation of energy work in general relativity, and how does this apply to cosmology? What is the total mass-energy of the universe?



Conservation of energy doesn't apply to cosmology. General relativity doesn't have a conserved scalar mass-energy that can be defined in all spacetimes.[MTW] There is no standard way to define the total energy of the universe (regardless of whether the universe is spatially finite or infinite). There is not even any standard way to define the total mass-energy of the observable universe. There is no standard way to say whether or not mass-energy is conserved during cosmological expansion.


Note the repeated use of the word "standard" above. To amplify further on this point, there is a variety of possible ways to define mass-energy in general relativity. Some of these (Komar mass, ADM mass [Wald, p. 293], Bondi mass [Wald, p. 291]) are valid tensors, while others are things known as "pseudo-tensors" [Berman 1981]. Pseudo-tensors have various undesirable properties, such as coordinate-dependence.[Weiss] The tensorial definitions only apply to spacetimes that have certain special properties, such as asymptotic flatness or stationarity, and cosmological spacetimes don't have those properties. For certain pseudo-tensor definitions of mass-energy, the total energy of a closed universe can be calculated, and is zero.[Berman 2009] This does not mean that "the" energy of the universe is zero, especially since our universe is not closed.


One can also estimate certain quantities such as the sum of the rest masses of all the hydrogen atoms in the observable universe, which is something like 10^54 kg. Such an estimate is not the same thing as the total mass-energy of the observable universe (which can't even be defined). It is not the mass-energy measured by any observer in any particular state of motion, and it is not conserved.


MTW: Misner, Thorne, and Wheeler, Gravitation, 1973. See p. 457.


Berman 1981: M. Berman, unpublished M.Sc. thesis, 1981.


Berman 2009: M. Berman, Int J Theor Phys, http://www.springerlink.com/content/357757q4g88144p0/


Weiss and Baez, "Is Energy Conserved in General Relativity?," http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html


Wald, General Relativity, 1984


mathematics - A game with 52 cards


Alice and Bob play the following game with two (identical) standard decks of $52$ cards.



  • First Alice secretly arranges one deck of $52$ cards in a long row on the table. All cards are face-down, and Bob has no knowledge of the positions of any of the cards.

  • Then the game goes through several rounds. In every round, Bob first pays $1$ Euro to Alice. Then Bob arranges the second deck of cards in a long row on the table, parallel to Alice's row and with all the cards face-up. Then Alice tells Bob all the face-up cards in his row that agree with the corresponding face-down cards in her row. Then the next round starts.

  • The game ends, as soon as Bob knows the positions of all the face-down cards in Alice's row. Bob then receives $x$ Euros from Alice.




Question: What is the smallest value of $x$, for which Bob can still avoid (with absolute certainty) to lose any money?





situation - Bank Teller and his mother



I found this on 9GAG, albeit without an answer. I am really curious on what it is. It's copied word for word from 9GAG, hence the terrible english.




You work at a bank and suddenly a gang of robbers shows up. One of them comes at you and tells you to give him all the money or he will shoot you. You tell the robber you don't have access to the deposit.


Suddenly the phone rings. The robbber tells you to answer the phone and don't say a word about this. You pick up the phone and it happens to be your mother.



After you talked to her, 15 minutes passed and the police arrived along with your mother and arrest the robbers.


What did you tell your mother at phone such that she understood you need help and the robbers couldn't notice?


Hints:



  • The phone you used was a ring phone (not cell phone);

  • You used a "little trick" when talking to phone with your mother;

  • You didn't instruct your mother for this emergency, so she didn't know about a secret code for help.



Answer



The trick is:




You ask your mother questions like "Is this an emergency? I'm busy at work" and say "I can come round later and help." but you use the mute button so all your mum hears is "emergency... work... help" etc. The actual words don't matter as long as you can filter in the message.



enigmatic puzzle - Who Stole My Friends' Car?? Help!


Help! Last night my friends' new car was stolen!


I'm pretty sure it was one of the gangs that hang around here, and they've been a bit cocky. Where the car was parked, they spray-painted the following on the ground.


I'm sometimes the last gear, and sometimes the first


But I can't make head or tale of it, and nor can the police. Can you help?




So you found out who stole it, how about trying to find them?




Answer



The gang who stole your friends' car are called



THE BAD LADZ



Thanks to @n_palum for helping in chat




If you



Type it into imgur reversed (The name of the image if you click edit is 'I'm sometimes the last gear, sometimes the first' and the last gear is the gear to REVERSE [Thanks @Techidiot]) you get the following: https://i.stack.imgur.com/3Z0LA.png






enter image description here



Text:



YnJvd3NlcnMud3RmL3B1enpsZS5odG1s



Using Base64 decoder (thanks @n_palum) you get the following website:




http://browsers.wtf/puzzle.html



(The console output is 32543 [Thanks @KritixiLithos] which is the OP's user ID)


Which has



the word 'Of'



The tab name is




Bottom



Giving



Bottom Of 32543



At the bottom of the OP's profile (at least at the time, may be removed in future) is a barcode



barcode




which decodes to



ZALD ABD ETH



which is an anagram of



THE BAD LADZ



Thursday, May 24, 2018

kinematics - How can an object move from point A to point B?


If I was at point $A$ and I wanted to walk directly to point $B$, I would have to walk half way to point $B$, but before that I would have to walk half way to halfway to halfway to point $B$ and half of that again and so on and so fourth. if I halved this distance an infinite amount of times then there would be an infinite amount of actions I would need to perform in order to cross from $A$ to $B$. Therefore, if each action took any quantity of time at all, then it would take me an infinite amount of time to cross from $A$ to $B$, even if $A$ and $B$ were only a few centimetres apart!


Because I am not infinitely old and I can move, there must be a flaw in this logic. Where is it?



Answer



You have stumbled onto one of Zeno's many paradoxes - the so-called Dichotomy paradox.


The resolution lies in the fact that sum of terms in an infinite series do not necessarily add up to produce an infinity, so the basic premise is flawed. In particular, the series


$$S = \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} \ldots$$



is a convergent infinite series. The sum is


$$ S = 1 \, .$$


Since Zeno belonged to the BC era, this paradox stood as a paradox for many years. In fact, Zeno was satisfied with the reasoning than it had an infinite number of steps, but didn't find the sum. However, things changed with the development of Analysis and Calculus in the 19th century, and with these convergence arguments, it is settled now.


statistical mechanics - Canonical averages in a Fermi gas aka generalized Fermi-Dirac distribution



I am in the process of applying Beenakker's tunneling master equation theory of quantum dots (with some generalizations) to some problems of non-adiabatic charge pumping. As a part of this work I encounter thermal averages of single-particle quantities with a fixed total number of electrons. They are pretty straightforward to derive, as recently discussed on Physics.SE, see Combinatorial sum in a problem with a Fermi gas.


I have trouble finding references to prior work for this basic quantum statistics problem. Can you suggest some relevant references in this context?


Progress report:


Writing this stuff up, I realized that I'm basically asking for finite electron number and finite level spacing generalization of the Fermi function:


$\langle \nu_k \rangle = Z_n^{-1} \sum_{n=0}^{n-1} (-1)^{n-m} Z_m e^{-\beta \epsilon_k( n-m)}$ where $Z_n$ is the $n$-electron canonical partition function.


For either $n \gg 1$ or and $\epsilon_{k+1}-\epsilon_k \ll \beta^{-1}$ this reduces to the standard Fermi-Dirac distribution:


$\langle \nu_k \rangle= \frac{1}{1+z_0 e^{-\beta \epsilon_k}}$



The advantage of the new formula is that it only depends on $k$ via $e^{-\beta \epsilon_k}$ with all the combinatorics hidden in $Z_n$. This is unlike the textbook expression summarized in Wikipedia, or Beenakker's implicit function $F(E_p |n)$ (which is $\langle \nu_p \rangle$ in notation).


The question still stands: I don't believe the first formula above is new, what is the right reference?



Answer



A particle-number projection method which is based on Fourier transform of the grand-canonical partition function has been developed by Ormand et al. Phys. Rev. C 49, 1422 (1994) in the context of nuclear quantum Monte-Carlo simulations. This gives a closed form formula which scales quadratically in the number of levels. It has been used by Alhassid et al. Phys. Rev. B 58, R7524 (1998) to describe (numerically) deviations from Fermi-Dirac distribution, see Eq. (140) in Alahssid's review paper.


Since the approach described in the question appears to be new, I've posted it to the arXiv. For equidistant (1d harmonic oscillator) spectrum, generalization of Fermi-Dirac distribution leads to an intriguing series of polynomials and Rogers-Ramanujan partial theta function, see math.SE discussion.


special relativity - Is it possible for information to be transmitted faster than light by using a rigid pole?


Is it possible for information (like 1 and 0s) to be transmitted faster than light?


For instance, take a rigid pole of several AU in length. Now say you have a person on each end, and one of them starts pulling and pushing on his/her end.


The person on the opposite end should receive the pushes and pulls instantaneously as no particle is making the full journey.



Would this actually work?



Answer



The answer is no. The pole would bend/wobble and the effect at the other end would still be delayed.


The reason is that the force which binds the atoms of the pole together - the Electro-Magnetic force - needs to be transmitted from one end of the pole to the other. The transmitter of the EM-force is light, and thus the signal cannot travel faster than the speed of light; instead the pole will bend, because the close end will have moved, and the far end will not yet have received intelligence of the move.


EDIT: A simpler reason.
In order to move the whole pole, you need to move every atom of the pole.
You might like to think of atoms as next door neighbours If one of them decides to move, he sends out a messenger to all his closest neighbours telling them he is moving. Then they all decide to move as well, so they each send out messengers to to their closest neighbours to let them know they are moving; and so it continues, until the message to move has travelled all the way to the end. No atom will move until he has received the message to do so, and the message won't travel any faster than all the messengers can run; and the messengers can't run faster than the speed of light.


/B2S


Wednesday, May 23, 2018

visual - Emoji Story - bad jokes edition


For devices that cannot display all the emoji, Android screenshots are included with each.





  1. ✊✊🚪😕❓👻👻❓🚫😢






joke 1





  1. ❓☠️💲🌽🦌👂






joke 2





  1. 🚶‍♂️🚶‍♂️🍻🚶🦆





joke 3






  1. ❓6️⃣😱7️⃣7️⃣🍴9️⃣





joke 4






  1. ❓👃🚫📏👞





joke 5





  1. ❓❌🐈🐦🥕






joke 6



Answer





  1. ✊✊🚪😕❓👻👻❓🚫😢



    Knock knock. Who's there? Boo. Boo who? You don't have to cry about it.






  2. ❓☠️💲🌽🦌👂



    How much does a pirate pay for corn? A buccaneer.





  3. 🚶‍♂️🚶‍♂️🍻🚶🦆




    Two men walk into a bar. The 3rd man ducks.





  4. ❓6️⃣😱7️⃣7️⃣🍴9️⃣



    Why was 6 afraid of 7? Because 7 ate 9.






  5. ❓👃🚫📏👞



    Why can't your nose be 12 inches? Because then it'd be a foot. (Thanks to [@AustinWeaver][https://puzzling.stackexchange.com/users/41564/austin-weaver])





  6. ❓❌🐈🐦🥕



    What do you get when you cross a cat and a parrot? A carrot.






solid state physics - What happens if a metal wire that was one-atom in thickness was pulled across a finger?


What would happen if a metal wire that was one-atom was pulled across your finger? Would it cut off your finger, or would it pass through your finger without harming you? What if the metal one atom thick was "unbreakable"?





optics - What makes a material a good lens for CO2 lasers?



I read on Wikipedia that zinc selenide and germanium make good lenses for $\mathrm{CO}_2$ lasers. My question is, Why?




electromagnetism - Which solution to the electromagnetic wave equation is the most accurate model of monochromatic light?


When a photon is modeled as a monochromatic electromagnetic wave its electric and magnetic components are usually taken to be sine waves (for example here http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html). I believe the practical reason for this is that any solution of the electromagnetic wave equation can be expressed as a sum of sine waves. But physically, when a photon can be interpreted as a wave, how is it best modeled? Do we have empirical evidence to think that it is best modeled by a single sine wave for the E and B fields, or if not which solution of the electromagnetic wave equation would best model it? Are there experiments that could show that light waves resemble more say square waves than sine waves?


Update


To rephrase the core of my question more properly: If a photon / electromagnetic wave travels through a point in space, in vacuum, and we measure the electric and magnetic fields at this point with a very high temporal resolution, would we measure the electric and magnetic fields to fluctuate exactly as sine waves, or as something else? Has such an experiment ever been made?




Tuesday, May 22, 2018

particle physics - Why do neutrons repel each other?


I can understand why 2 protons will repel each other, because they're both positive. But there isn't a neutral charge is there? So why do neutrons repel? (Do they, or have I been misinformed?)


The reason why I'm asking this is because, I've just been learning about neutron stars and how the neutrons are forced (as in, they repel) together according to my teacher (he's a great teacher btw, though what I just said doesn't make it seem so).


So I wondered, why they have to be forced by gravity and not just pushed?



Answer



Neutrons have spin 1/2 and therefore obey the pauli exclusion principle, meaning two neutrons cannot occupy the same space at the same time. When two neutrons' wavefunctions overlap, they feel a strong repulsive force. See http://en.wikipedia.org/wiki/Exchange_interaction .



calculation puzzle - Stranded Nomad Riddle


Background Story


A nomad is injured in the desert. And he can't move, but fortunately an oasis is nearby, and he can stay there for enough time until rescue arrives. But the water is a magical water, that it can't be stored. So it should be drunk immediately or it'll disappear. In other words, one cannot bring the water outside the oasis.


The other nomad friends, know the exact position of the injured nomad, which is $n$ days away from the base. The nomads are going to send delegations to save him, but the problem is that each nomad can only bring food supply for one nomad for $3$ days. The base has arbitrarily large number of nomads inside, so they decided to send a group of nomads to rescue the injured nomad, that is, to go to the place of injured nomad, and then go back to the base. All nomads must survive. But they have weird cultural rules, which is each time a nomad leaves the base, they have to kill a sheep as redemption.


Note that the injured nomad also need to eat while going back to the base.



Question


What is the number of sheep need to be slaughtered in order to save the injured nomad in case n=2? n=3? arbitrary n?


Furthermore, what is the minimum number of days required to save the nomad?


Example


For example, if n=1, the nomad group can just send one nomad. It'll consume 1 day of supply, then arrives at the injured nomad place with 2 days of supply left, then each of them consume 1 day of supply while going back to the base, leaving no supply left when they arrived at the base. Since only one nomad leaves the base, only one sheep is killed. So 1 sheep will be slaughtered, and the rescue time is 2 days.


Notes




Answer



Using Joe Z.'s method for $n=2$, I can think of a strategy for $n=3$ in $6$ days killing $13$ sheep. Pretty sure this is the minimum amount of days because it's $2n$. Not sure if it is the minimum amount of sheep.


I won't do a day by day guide like Joe but my explanation hopefully is clear.



First start off with $2$ nomad's. After $1$ day one nomad gives a day supply to the other and heads back. So, the one returning has just enough to get back and the other has a full three-day supply.


From his point of view it now is only an $n=2$ problem because he is now at distance $1$ with full supply.


So use the $n=2$ strategy here. You can do this by sending an extra nomad for each nomad that needs to be at $1$ day travel with full supplies.


The $n=2$ strategy eventually also returns all these nomad's at this point with no food left so for each of them they need an extra nomad to supply them.


So to sum it up: start sending $2$ nomads every day for $4$ days. half of them return. the other half does the $n=2$ strategy. One day before they are finished send $5$ nomads away so they meet up with the other five when they are finished with their $n=2$ strategy. The now $10$ nomads have exactly enough to return. counting the $4$ that returned in the beginning and not counting the stranded nomad, that's $13$ sheep that were killed.


ADDITION AND CONCLUSION:


you can generalize strategy this by doing same thing for other $n$. You will have then: sheep killed for $n$ is three times the [number of sheep killed for strategy $n-1$] plus one. And days to take is always $2n$.


Given this, this strategy is equally good as Joe's with going to $n=1$ one to $n=2$. because $3*1+1 = 4$ sheep.


And $n=2$ worked out to be:




D0: 2 nomads leave (N1 and N2)


D1: N1 gives 1 supply to N2 and heads back


D2: N2 arrives at N0 and gives him 1 supply and start heading back. N3 and N4 leave.


D3 N3 and N4 meat up with N0 and N1 and they give them each a supply.


D4: they all arrive back with their supply used up.



So wrapping it up. If $F(n)$ is the amount of sheep necessary or a nomad stranded at $n$ days distance.


$$F(1) = 1$$


$$F(n) = 3*F(n-1) + 1$$


And days it takes is always $2n$.



I'm not that good in math but I know there is a simple way to convert this recursive formula to a direct one.


homework and exercises - Identity of Operator Product Expansion (OPE)


I have one more s****d question in Polchinski's string theory book, Eqs. (2.3.14a)


$$ j^{\mu}(z) :e^{ik \cdot X(0,0)}:~ \sim~ \frac{k^{\mu}}{2 z} :e^{ik \cdot X(0,0)}:,$$


where $j^{\mu}_a =\frac{i}{\alpha'} \partial_a X^{\mu}$, $::$ is normal ordered, defined as $$:X^{\mu}(z,\bar{z}): = X^{\mu} (z,\bar{z})$$ $$:X^{\mu}(z_1,\bar{z}_1) X^{\nu}(z_2,\bar{z}_2): = X^{\mu}(z_1,\bar{z}_1) X^{\nu}(z_2,\bar{z}_2) + \frac{\alpha'}{2} \eta^{\mu \nu} \ln |z_{12}|^2 $$. $\sim$ means equal up to nonsingular terms.


I thought I have derived it in analogy of integration by part, but it turns out that not as I thought. Actually how to derive Eq. (2.3.14a)? Eq.(2.3.14b) will be expected in analogous...



Answer



We can use \begin{equation} \begin{split} : F : : G: = \exp \left( - \frac{\alpha'}{2} \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\frac{\delta }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta }{\delta X_{G\mu}(z_2, {\bar z}_2)} \right) :F G: \end{split} \end{equation} This gives \begin{equation} \begin{split} : \frac{i}{\alpha'} \partial X^\mu(z) : : e^{i k \cdot X(w,{\bar w})}: &= \exp \left( - \frac{\alpha'}{2} \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\frac{\delta }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta }{\delta X_{G\mu}(z_2, {\bar z}_2)} \right) \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~: \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &~~~~~~~~~~~~~~~~ - \frac{i}{2} : \int d^2 z_1 d^2 z_2\log|z_{12}|^2 \frac{\delta( \partial X^\mu(z) ) }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta ( e^{i k \cdot X(w,{\bar w})}) }{\delta X_{G\mu}(z_2, {\bar z}_2)}: \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &~~~~~ - \frac{i}{2} : \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\partial \left( \delta^\mu{}_\alpha \delta^2(z_1, z) \right) i k^\alpha \delta^2(z_2, w) e^{i k \cdot X(w,{\bar w})} \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &~~~~~ +\frac{k^\mu}{2} : \partial \left( \int d^2 z_1 d^2 z_2 \log|z_{12}|^2 \delta^2(z_1, z) \delta^2(z_2, w) e^{i k \cdot X(w,{\bar w})}\right) : \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: +\frac{k^\mu}{2} : \partial \left( \log|z-w|^2 e^{i k \cdot X(w,{\bar w})}\right) : \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: + \frac{k^\mu}{2(z-w)} : e^{i k \cdot X(w,{\bar w})} : \\ \end{split} \end{equation} Therefore \begin{equation} \begin{split} : j^\mu(z) : : e^{i k \cdot X(w,{\bar w})}: \sim \frac{k^\mu}{2(z-w)} : e^{i k \cdot X(w,{\bar w})} : \end{split} \end{equation}



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 \...