Saturday, November 30, 2019

general relativity - length contraction in a gravitational field




As space-time is distorted in a gravitational field, relativistic effects such as time dilation and length contraction take effect.


Time dilation is explained simply enough: closer to the source of gravity, slower the time passage.


However, space contraction gives no such clear answer. for example, a single thread on the matter contains the following:



  • gravitational field would produce a transverse expansion of distances

  • gravitational fields produce increased distances in the direction perpendicular to the field

  • [gravitational fields] produce decreased distances in the radial direction


So, to pose the question as bluntly and directly as I can:


Ignoring all other effects (relativistic or otherwise), which is shorter: a measuring stick closer to the source of gravity or a farther one?



If the orientation of the stick plays a part, try to answer for both the case where it's pointing towards the source of gravity, as well as perpendicular.


Similarly, if the position of the observer relative to source of gravity plays a part try to take that into account (but it doesn't seem to in the case of time dilation so I don't expect it here either)



Answer



Your question isn't well defined, because there is no way to compare the two measuring sticks.


I can take two clocks, put them at different heights, then bring them back together and I will find that they now show different times due to time dilation. However if I take two rulers, put them different distances from a black hole then bring them back together I will find that they are still the same length. The reasons for this are covered in the answer Ben Crowell linked above.


However there is a sense in which radial distances are stretched not contracted (as rockandAir8747 says in their answer). But to understand what this means you need to understand what we mean by the Schwarzschild radial distance. The Schwarzschild coordinates use a radial coordinate $r$ that we naively interpret as the distance from the centre of the black hole. However we cannot measure the distance to the centre of the black hole because there's an event horizon in the way. So what we do is lay out our measuring tape in a circle centred on the black hole and measure the circumference of this circle. Because we know that in flat space the circumference is $2\pi r$ we can take our measured circumference and divide by $2\pi$ to get the Schwarzschild radial distance $r$.


But this $r$ coordinate is calculated assuming space is flat, so it shouldn't be any surprise that it is different from the distance you'd measure if you let down a measuring tape towards the event horizon. There's a nice image showing this difference in the question Difference between coordinate and proper distance in Schwarzschild geometry:


Radial distance


The $dr$ in the bottom diagram is the distance in Schwarzschild coordinates while the $dr$ in the top diagram is the distance you'd measure with a ruler, and it's obviously bigger than the Schwarzschild $dr$.


I suppose in this sense you could claim that the ruler is length contracted near to the black hole, because it takes more rulers than you expect to cover the distance $dr$.



If you want the gory details, including an equation for how to calculate lengths measured towards the event horizon then see the questions:



The measured distance, s, between the two Schwarzschild radii $r_1$ and $r_2$ is given by:


$$ s = \left[ r\sqrt{1-\frac{r_s}{r}} + \frac{r_s}{2} log \left( 2r \left( \sqrt{1-\frac{r_s}{r}} + 1 \right) - r_s \right) \right]_{r_1}^{r_2} $$


Do you have my spare key?


I have three bicycle locks, ones like this:


Picture taken from https://commons.wikimedia.org/wiki/File:Bike_cord_%26_lock_26d06.jpg


Because they look so much alike I etched a unique number into each lock and the same number into the keys that unlock it. Unfortunately I seem to have lost all my spare keys! (That also have the corresponding numbers etched into them.)


You tell me that you have one of my spare keys, but you don't want to tell me which one. Of course, I don't believe you.


How do you convince me that you have one of my spare keys, without letting me get any information whatsoever about which one you have?


I still have the keys to all the locks and I am willing to cooperate.


Disclaimer: This puzzle was inspired by a stackexchange answer.




https://crypto.stackexchange.com/questions/57674/how-do-i-explain-zero-knowledge-proof-to-your-7-year-old-cousin/57678#57678



Correct solutions to this problem have been given, however I was hoping someone would come up with a solution that



does not involve any props like tall trees or poles.



If someone comes up with a solution like this, which I would find more elegant, in a reasonable time span I will accept their answer, otherwise the green checkmark will go to Keelhaul.


(Picture taken from https://commons.wikimedia.org/wiki/File:Bike_cord_%26_lock_26d06.jpg)



Answer





Lock the three chains together in the form of the Borromean Rings, and hand me the linked set, then walk away so that you can't see which one I unlock. I unlock the one that I have the key for, separating all the rings. Then, leaving them all unlinked, I relock the chain that I unlocked, and return three unlinked chains to you.

It also works in "reverse", where you give me the unlinked chains, and I return them Borromean-linked.



word - The arrow of time flies in reverse



Here is the puzzle:



Arrow of time


(Click on the image to see a larger image.)



Overview:


This is a word graph. The nodes are short words, the edges are long words.


Whenever an edge (i.e., a long word) connects two nodes (i.e., two short words) it means that the long word can be reversed and decomposed into the two short words. Another way to say it: The two short words can be reversed and interleaved in some way to produce the long word.



Objective:



Fill in each of the numbered nodes (1-18) with the appropriate short word. Optionally, label each of the edges with the appropriate long word. (Although it's not much of an option because everything is interdependent, so you will have to figure most of them out anyway.)


The grayed-out nodes and edges are also part of the graph. They are additional connections provided to assist you by giving you more information.



Connecting mechanism:


For illustrations of the connecting mechanism, you can look at the shaft of the arrow which is already complete. Let's look at how BONELESS connects SLOB and SEEN:


Start with the two short words:

SLOB SEEN



Reverse them:

BOLS NEES


Interleave them (you can space the letters out however you wish,
but you must maintain their order):

BOLS
NEES



BO L S
NE E S


BONELESS

Obviously, you can reverse the process to decompose BONELESS into SLOB and SEEN.




Computer usage:


Consider this puzzle to have a partial NO-COMPUTERS tag attached. You may write computer programs or use existing tools for pattern matching or regular expressions in order to find short word and long word candidates. However, you may not write a computer program to solve the entire puzzle brute force.


In your answer, please briefly indicate your methods.



Precaution:


You will be doing a lot of thinking in reverse for this puzzle. I would advise you to mark your current position in space-time so that you can find your way back afterward.








A few notes about @DrXorile 's solution:



Node #3 is EMIT (+NOON = NOONTIME, +NAPS = TIMESPAN, +TRAP = PARTTIME)

Node #10 can be RUT or TANG or even TIDE again

Node #18 really is DECOR




Answer



I fear that your dictionary must be more extensive than mine!


However, here are some answers that seem to work:



1. stop (+deter = protested, + mien = nepotism)
2. noon (+stop = pontoons, + gild = noodling (based on hint from OP. I originally discounted this one because I couldn't find an answer for 3.))

3. emit (I don't know why I didn't get this one; +noon=noontime, +naps=timespan, + trap=parttime (neither timespan nor parttime were in my dictionary)) I then worked backwards a bit from 8 to get to 4
8. stir (+stop=protists, + said=diarists, +forte=retrofits)
7. lace (+stir=recitals, +rots=sectoral) (it could have been "at" or "lace", but lace worked better at 6)
6. sips (+mace=escapism,+lace=specials)
5. deed (+sips=despised,+name=demeaned (tricky one))
4. trap (+lane=parental,+deed=departed)
14. tide (+seen=neediest,+slam=medalist)
9. seats (+edit=stateside,+tide=steadiest)
13. sear (+tide=readiest,+lung=granules,+tapes=separates,+grab=barrages)
12. ruse (+sear=erasures) (I reject tide since it's already at 14)

11. seam (+tide=mediates, +ruse=measures,+sort=maestros)
10. rut (+seats=statures, +seam=matures) (could also have been tang (magnates,stagnates))
15. roam (+tide=mediator, +lain=manorial)
16. need (+roam=demeanor) (there were several possibilities here so I needed to explore a bit to find a reasonable bridge to "taro")
17. deep (+need=deepened, +taro=operated)
18. decor (+tale=relocated, +deep=proceeded, +evince=reconceived, +sure=resourced (my dictionary had neither "reconceived" nor "resourced" so I found these by mixing the two words in all possible ways and scanning the list!))



So still two gaps...



arrow of time updated




Methods: I have a short python program.


The first part does a subtraction of a small word from a big word. It works recursively to produce a list of possible results. e.g. "lelabc" minus "lab" = ["elc","lec"]. "lelabc" minus "lba" = [].


Subtract("lelabc","lab")
Out: ['elc', 'lec']

Then the second part takes an input of a small word. It finds all longer words in my dictionary and does a reverse subtraction. It checks whether those reverse subtractions (if any) are in my dictionary. If they are, add them to a set. Return the set. E.g.:


findAll("lane")
lane iv veinal
lane id denial

lane it entail
lane iv venial
lane reg general
lane sit entails
lane rte eternal
lane ret eternal
lane deb enabled
lane reb enabler
lane trap parental
lane deem enameled

lane muon noumenal
lane tarp prenatal
lane tiro oriental
Out: 'deb','deem','id','it','iv','muon','reb','reg','ret','rte','sit','tarp','tiro','trap'}

findAll("evince")
Out: set()

Finally, I take two small words and take the intersection of the small words in the set. I don't know if this is cheating, but the computer is definitely better at finding intersections of two sets than I am.


I also wrote a short script to mix two words together to scan for ones that might not be in my dictionary!



When is the "minus sign problem" in quantum simulations an obstacle?



The "minus sign problem" in quantum simulation refers to the fact that the probability amplitudes are not positive-definite, and it is my understanding that this leads to numerical instability when for example summing over paths, histories or configurations since large amplitudes of one phase can completely negate other large contributions but with the reverse phase. Without controlling this somehow, the sampling of alternative configurations has to be extremely dense, at least for situations where interference is expected to be important.


Assuming I haven't misunderstood it so far, my main question is when this is a show-stopper and when it can be ignored, and what the best workarounds are.


As an example, let's say that I want to simulate Compton scattering or something similar, numerically, to second order. I could evaluate the Feynman diagrams numerically to a certain resolution and sum them. I assume this won't work well. In Lattice QCD, complete field configurations are randomly generated and the action calculated and amplitude summed I guess (I have only superficial knowledge of Lattice QCD unfortunately).




Friday, November 29, 2019

grid deduction - I'm Bad at Naming Things, so Just Try to Solve this Patterned Masyu Puzzle!


$\hskip1in$ enter image description here




The original Masyu rules apply.



  1. Make a single loop with lines passing through the centers of cells, horizontally or vertically. The loop never crosses itself, branches off, or goes through the same cell twice.

  2. Lines must pass through all cells with black and white circles.

  3. Lines passing through white circles must pass straight through its cell, and make a right-angled turn in at least one of the cells next to the white circle.

  4. Lines passing through black circles must make a right-angled turn in its cell, then it must go straight through the next cell (till the middle of the second cell) on both sides.





Thanks to @Deusovi for playtesting this puzzle! :D




Answer



Haven't done these before, but I think I have a solution:



enter image description here



wick theorem - Klein factors and Conformal Field Theory


Consider the mode expansion of a (chiral) scalar field confined to a disc with circumference L: $$ \phi(x) = \phi_{0} + p_{\phi} \frac{2\pi}{L} x + \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} e^{-(k_{n}a)/2} \left(e^{-ik_{n}x} \ b_{n}^{\dagger}+e^{ik_{n}x} \ b_{n}\right) $$ with $k_{n}=\frac{2\pi n}{L}$ , $\phi_{0}$ some "zero-mode", $p_{\phi}$ some "conjugate momentum" and $a$ some short-distance cut-off. The operators fulfill the following bosonic commutation relations $$ \left[b_{n}^{\dagger} , b_{n'}\right]=\delta_{n,n'} \quad\text{and}\quad \left[\phi_{0},p_{\phi}\right]=i $$


(Fermionic) Vertex operators are defined by $$ V_{\alpha}(x)=:e^{i\alpha\phi(x)}: $$ with $: \ ... \ :$ denoting normal ordering. Inserting the mode expansion of $\phi(x)$ into the definition of the vertex operator yields to lowest order in $\frac{a}{L}$:


$$ :e^{i\alpha\phi(x)}: = \left(\frac{L}{2\pi a}\right)^{\Delta(\alpha)} e^{i\alpha\phi(x)} $$ with the "scaling dimension" $\Delta(\alpha)=\frac{\alpha^{2}}{2}$. The pre factor on right side in front of the exponential is sometimes called "Klein factor".


Now here are my questions (They may really be "Newbie"-CFT-questions;) ) :





  1. Since the right hand side is only an approximation of of $:e^{i\alpha\phi(x)}:$ to lowest order I am wondering whether the left hand-side reproduces the correct (say) fermonic commutators in all cases and whether hand side only partially reproduces the correct fermonic commutators?




  2. If the right-handside only indeed only partially reproduces the correct commutation relations how can we say that a certain product of fermionic operators (say a product of 3 fermionic operators) indeed obeys the correct sermonic commutators when written in the "bosonized language"?




  3. What is the importance of the higher-order terms in $\frac{a}{L}$ in the "expansion" of the vertex operator?





  4. Is all this a more general construction in CFT?




I am looking forward to your responses!




thermodynamics - Why does the Boltzmann factor $e^{-E/kT}$ seem to imply that lower energies are more likely?


I'm looking for an intuitive understanding of the factor $$e^{-E/kT}$$ so often discussed. If we interpret this as a kind of probability distribution of phase space, so that $$\rho(E) = \frac{e^{-E/kT}}{\int_{0}^{\infty}e^{-E'/kT}dE'}$$ then what, precisely, does this probability correspond to? Is its physical significance obvious? Specifically, why is it largest for energies close to zero?


Edit: I'd like to know how it is possible that for increasing $T$ the lower energy states grow more populated. The naive thing to think is "If I stick my hand in an oven and turn up the temperature, I'm assuredly more likely to be burned." This raises the question, exactly what does it mean for the lower energy states to grow more populated relative to the higher energy ones?




Answer



The Maxwell-Boltzmann distributes $N$ particles in energy levels $E_i$ such that the entropy is maximized for a fixed total energy $E=\sum E_i N_i$.


The probability that a particle is in the energy level $E_i$ is proportional to the number of particles in the energy level $E_i$ in this particular arrangement of particles in which entropy is maximized (the Maxwell-Boltzmann distribution), which is $N_i$. It so happens that when we distribute the particles such that entropy is maximized, more particles populate the lower energy levels.


I don't follow the above argument @Ben Crowell - we want to show that more particles are distributed in the lowest energy level. In the above, we write $\sum E_i = N_0 E_0 + E_R$ conclude that probability is maximized if the energies of the particles in the lowest energy state, $N_0 E_0$, is minimized, which occurs for $N_0 = 0$ - the opposite of what was desired.


I'm not sure how to intuitively explain the solution to distributing the particles such that entropy is maximized. If we agree that by distributing the particles as evenly as possible in the energy levels, we will maximize the entropy, we can try:


Suppose we require $\sum N_i E_i \equiv \hat{E}$ and that we have plenty of particles $N$, and that $E_i$ increases with $i$:



  1. Starting from $E_0$, put one particle in each energy level whilst $\sum E_i < \hat{E}$.

  2. We need to distribute the remaining particles. Starting from $E_0$, again put one particle in each energy level whilst $\sum N_i E_i < \hat{E}$. The lowest energy level, $E_0$, now has 2 particles

  3. Repeat 2., until all particles are distributed.



We can see that the lowest energy level will be most populated, and that $N_i$, and hence the probability that a particle is in state $E_i$, decreases with $i$. This algorithm won't exactly reproduce the Maxwell-Boltzmann distribution of particles in the energy levels, but it might help with an intuitive feel of why the lower energy levels are more probable.


logical deduction - How do you solve a word puzzle when you have no idea where to start?


When you're looking at a logical word puzzle, what sorts of clues should you try to look for to solve them? There have been quite a few puzzles posted to this site so far that I spent a few minutes pondering, and I just had no idea where to start. Once I saw the answers they usually made sense, but I don't seem to have a good gameplan for trying to figure it out on my own (which is the fun part!) I'd like to get better at solving puzzles like this, and I imagine there are techniques I'm unaware of to finding the important information and where to begin (because part of the trick to puzzles like these is that they give you extra information you don't need to throw you off, right?)


So here's a good example of one such puzzle. (Source; solution also displays at link).



You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.


The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.


You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.


You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.


What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?




I thought about it for a little while, and I just had no idea where to start. All I could think is that, since you have 24 hours and it takes less than 24 hours to die of the poisoning, each of the 1000 people should test one bottle of wine. But that's clearly not the best answer, because that's obvious; they're asking what's the smallest number of testers. I reread the problem several times trying to pick out the piece of information that should be my starting point, but I was unsuccessful.


Once I read the solution it made sense, but it would never have occurred to me. How do you figure out where to start solving a puzzle like this when you're completely flummoxed?



Answer



As I've said in many other questions, puzzles like the one you posted are what I call information puzzles – that is, puzzles where you perform some trial to get as much information as possible about a certain unknown fact.


In this case, you can notice that each slave you use can either die or not die, which is 2 possible outcomes. If you then use $n$ slaves, then there are a total of $2^n$ possible outcomes, and the smallest $n$ for which $2^n \ge 1000$ is $10$. If you think about it in this fashion from the beginning, you will probably realize that each prisoner can sample from more than one bottle in the process, because that's the only way this method can really work.


There is a similar problem in which you have two days to figure out which bottle out of 27 is poisoned, which only requires three slaves instead of the five it would require if you only had one day. Why is this?



Because each slave now has three possible states - not dead, dead in one day, dead in two days. So $3^3 = 27$, as required. But even knowing that a prisoner can sample from more than one bottle, it's still somewhat difficult to devise an algorithm for which bottles each slave should drink each day.




P.S. I don't personally like the formulation of "anywhere between 10 to 20 hours, and you don't know how long" to imply that you can't use a timing attack on the test. I prefer instead that "anyone who drinks the poison dies at the stroke of midnight, regardless of when they drank it".




As for the more general problem of how to solve word problems (which is what your question is about), there is only really a general set of principles you can follow.


First, identify the problem. This puzzle was an information puzzle. Generally, information puzzles will ask you to "find out which [something] [has some property]" – in this case, figuring out which bottle of wine was poisoned. There might be some puzzles that are about direct calculation or traversing a graph or solving an equation.


Then, identify common steps to solve problems like these. In the case of information puzzles, you'll want to identify how many possible results you want, and how many possible outcomes the trial will give you.


Finally, try and solve the problem using those methods. In the case of information puzzles, this involves trying to map each desired result to an outcome of the trial. In the case of the prisoners having two days to sample wine bottles, you need some way of configuring the bottles on the first day, so that on the second day, there are still enough prisoners alive that you can carry on with the bottles you still don't know aren't poisoned.


If these don't work, generally the most you can do is keep thinking about it until you do find some way to solve it. Sometimes, if you go looking for a solution, you won't even understand it or how it solves the problem.


As an exercise, try to devise the schedule that three slaves would follow for sampling wine if they had two days to test 27 bottles.


classical mechanics - Understanding Poisson brackets



In quantum mechanics, when two observables commute, it implies that the two can be measured simultaneously without perturbing each other's measurement results. Or in other words, the uncertainty in their measurements are not coupled.


But in classical mechanics, in an analogous way we have the Poisson brackets instead, where when two functions/variables of the systems have a vanishing Poisson bracket with each other, we say they are in "involution".


I understand that when a variable $j$ has a vanishing Poisson bracket with the Hamiltonian of the system, means that $j$ is a constant of motion, conserved in time. But for the general case:



If the Poisson bracket of $f$ and $g$ vanishes ($\{f,g\} = 0$), then $f$ and $g$ are said to be in involution.





  • What does the above mean physically? Is it in some sense related to measurement results as it is with commutators in QM, meaning two variables in involution are completely independent of each other?





  • Finally, is there a simple classical example, where one can see 2 observables are in involution, but 2 other (same system) may not be (i.e. counter-example for involution)? I think it would help better understand what in "involution" means.





Answer



The classical poisson bracket with the generator of any transformation gives the infinitesimal evolution with respect to that transformation. The familiar


$$ \partial_t f = \{H,f\}$$


means nothing else than the time evolution of any observable is given by its Poisson bracket with the Hamiltonian, which is the generator of time translation. More generally, given any element $g$ of the Lie algebra of observables on the phase space, the infinitesimal evolution under the transformation that it generates (parametrized by an abstract "angle" $\phi$) is given by


$$ \partial_\phi f = \{g,f\}$$


What is meant by this is that every observable $f$ gives rise (by Lie integration) to a symplectomorphism $\mathrm{exp}(\phi f)$ on the phase space (since the true Lie integration of observables projects down surjectively onto the symplectomorphisms). More precisely, the statement above therefore reads



$$ \partial_\phi( f \circ \mathrm{exp}(\phi g))\rvert_{\phi = 0} = \{g,f\}$$


so that vanishing Poisson bracket implies $f \circ \mathrm{exp}(\phi g) = f$, i.e. invariance of the observable under the induced symplectomorphism.


Thus, if the Poisson bracket of $f$ and $g$ vanishes, that means that they describe an infinitesimal transformation that is a symmetry for the other.


For example, an observable is invariant under rotation if its Poisson bracket with all $L^i = \epsilon^{ijk}x^jp^k$ (components of $\vec L = \vec x \times \vec p$) vanishes.


For $x$ and $p$, the non-vanishing $\{x,p\}$ means the rather trivial insight that $x$ is not invariant under the translations generated by the momentum.


Fun fact: Wondering what the actual Lie integration of these infinitesimal symmetries might be leads directly to the quantomorphism group, and is a natural starting point for geometric quantization, as discussed in this answer.




Responding to the tangential question in a comment:



Considering the invariant under generation idea, so when trying to apply Noether's theorem, one only has to apply the Poisson bracket of the Lagrangian (e.g.) with $x$, and if it vanishes momentum is conserved for that system?




No, it means that the "Lagrangian" (you cannot really take the Lagrangian, because it is not a function on the phase space) is invariant under varying the momentum of the system (since $x$ generates "translations" in momentum). The Hamiltonian equivalent of Noether's theorem is simply that $f$ is conserved if and only if


$$ \{H,f\} = 0$$


since conservation means invariance under time evolution.


Thursday, November 28, 2019

cosmology - Is Dark Energy A Constant?



My understanding of dark energy is that it's like hot air bouncing around in a balloon, except the air is tiny subatomic particles that can't be seen or even detected in any way except for their apparent effect on the universe's expansion, and the balloon is gravity.


I question whether they can really tell if the universe is expanding faster now than it was a billion years ago, there are so many variables and margins of error, but if it is, wouldn't it slow down later?


The propulsion effect of dark energy would taper off like a rocket running out of steam as it cools down and gets sucked into black holes, and then no matter how far or fast the universe had expanded, over an infinite amount of time, gravity would be all that was left, wouldn't it?


I don't believe in black hole radiation, I don't think anything could ever escape the huge gravity of a massive black hole as everything has gravity, because everything distorts space by it's existence, but even if you believe that the universe ended with nothing but photons flying around, then wouldn't their gravity be the same anyway, and enough to cause the big crunch, the only logical cause for the big bang?




quantum mechanics - $SO(3)$, orbital angular momentum, vector product


I have a big confusion with group theory terminology. I know that orbital angular momentum (OAM) is $\mathrm{SO}(3)$-symmetric in 3D-space. Let's define QM orbital angular momentum (OAM) conventionally:


$$\pmb{L} = -i \pmb{r} \times \pmb{\nabla}$$


This definition can also be written using a set of $\mathrm{SO}(3)$ generators:


$$L^{\mu} = -i r_i \; S_{ij}^{\mu} \; \nabla_j$$


where $\mu = \{x,y,z\}$ for 3D space, and $S_{ij}^{\mu}$.


So... generators stand for the definition of a vector product in given space, essentially, definition of orthogonality? Or this is only in this case, I suppose, in which case why such a coincidence?




If I proceed with this:



$$\pmb{r} e^{-iS^{\mu} \phi} \pmb{p}= \pmb{r} \cdot \pmb{p} - i \delta \phi \; \pmb{r} S^{\mu} \pmb{p} + \cdots = \mathrm{const} \; e^{- i \pmb{r} \cdot \pmb{p}} + \delta \phi L^{\mu}$$


Matter wave in zeroth order and OAM in first? Does it have any interpretation?




Wednesday, November 27, 2019

strategy - Clear board in Othello


Is there a valid set of moves in Othello that allows a win score of 64-0 for either player?


And is there record of any such game in the past?



Answer



Yes, I found this video on YouTube that has this "perfect game". Terrible music by the way


https://www.youtube.com/watch?v=prWG1OFgVqg



quantum mechanics - Help Simplifying a Commutator Equation



For the SHO, our teacher told us to scale $$p\rightarrow \sqrt{m\omega\hbar} ~p$$ $$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$ And then define the following $$K_1=\frac 14 (p^2-q^2)$$ $$K_2=\frac 14 (pq+qp)$$ $$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$ The first part is to show that $$Q \equiv -K_1^2-K_2^2+J_3^2$$ IS a number. My approach: $$16Q=J_3^2-K_1^2-K_2^2=(p^2+q^2)^2-(p^2-q^2)^2-(pq+qp)^2$$ $$=p^4+q^4+p^2q^2+q^2p^2-(p^4+q^4-p^2q^2-q^2p^2)-((pq)^2+(qp)^2+pqqp+qpqp)$$ $$=2p^2q^2+2q^2p^2-pqpq-qpqp-pqqp-qppq$$ At least point, I am unsure of how to simplify any further. A lot of these look like the form of anticommutators, which does not seem to provide any useful information in turning Q into a number. Any help would be appreciated!


EDIT::


This is how far I have gotten.
enter image description here




quantum mechanics - What is the name of the principle saying it is meaningless to talk/ask questions that can not be measured/tested?


Watching quantum mechanics lectures and it was mentioned that it is pointless/meaningless to try to talk/question things that can not be tested/measured.


Is this a principle? And if so what is it's name?


Also does this apply to questions other than Quantum mechanics? E.g. does it make sense to ask if earth was the only object in universe if gravity would still exist? Although it seems intuitive to answer yes, yet it is something that can never be tested/verified. It seems almost as meaningless as to ask if the universe was only made of a unicorn would it have gravity?



As an analogy consider glottogony, at some point it was banned as it seemed to be an unanswerable problem. http://en.wikipedia.org/wiki/Origin_of_language



Answer



It's not really a single principle - it's a philosophy and in the context of philosophical discussions about science, it is usually known as positivism.



http://en.wikipedia.org/wiki/Positivism



As any philosophy, it cripples the penetrating power of science if it is extended too far - and every philosophy ultimately fails. The thought experiment about the Earth in the Universe is just one among millions of examples. Positivism could have a problem with the whole concept of thought experiments.


While it was very useful and important for the development of quantum mechanics to realize that science doesn't have to talk about things that can't be measured, i.e. that theories that deny the existence of things that can't be observed are just fine, it still remains true that science also can talk about concepts that can't be observed, such as quarks.


It's up to the scientific method to decide whether an auxiliary concept or a theory that isn't accessible to observations has an explanatory power that justifies its validity - and the answer may be different in each individual situation.


experimental physics - What does the LHC do afterwards?


Astronomical telescopes are now mega projects and cost $1Bn and although they are pitched to solve the current interest of the day they are general purpose machines and with upgrades and new instruments have a life of perhaps 50years.


It seems that large accelerator projects are built to answer one question, to find one particle. But since the design must be based around the particle having a particular energy and the cost and timescale being so large - you have to be pretty damn sure that you expect the particle to exist and at the predicted energy. It almost seems that if you had a good enough estimate to build the accelerator then you don't really need to!


Is something like the LHC a one trick pony? You turn it on and confirm the Higgs or if not - build a bigger one?


Is the LHC really a more general purpose experiment but the Higgs gets the press attention or is it just that the nature of discover in HEP is different and you need to build a single one shot experiment?



Answer



The LHC was envisioned as a "discovery" machine, a multipurpose one. The Higgs gets the press but the expectations is that new physics will become accessible with the higher energy available for center of mass collisions.


The Z was discovered in the SPS the proton antiproton previous generation collider. The previous machine in the same tunnel as the LHC, LEP, an electron positron collider was needed to establish with great accuracy the parameters of Z itself and the standard model.


In general leptonic collisions probe elementary interactions with many less assumptions than proton proton or proton antiproton machines. This is because one is throwing balls of three quarks with their gluons at each other and measures the debris, in order to study the interactions. New physics, because of the high energy, will appear, but will be in a complexity unprecedented up to now. Hopefully the next generation will be a lepton machine that will allow to establish the appropriate models unequivocally.



Now on the question of one off detectors: CERN is practically using all the accelerators built up to now as increasing energy stages to feed the end machine, the LHC. Nothing is wasted. In addition a lot of experiments are approved and running in beam lines that are not in the mainstream but may prove valuable or have unexpected theoretical repercussions.


Thus one expects that the LHC will open the window to the new physics that is tantalizing us,strings and unification of all forces at the moment, and the next generation machines will be leptonic ones to allow accurate measurements of parameters and decide between models.


quantum mechanics - Can closed loops evade the spin-statistic theorem in 3 dimensions?


The famous spin-statistics result asserts that there are only bosons and fermions, and that they have integer and integer-and-a-half spin respectively. In two-dimensional condensed matter systems, anyons are also possible. They avoid this result due to a topological obstruction.


Suppose that "loops" were fundamental, rather than point-particles. In addition to normal relative motion, it would be possible for one loop to go through the hole of another, and a path of this sort would not be continuous with one that doesn't. Would this provide enough of a topological obstruction to be able to avoid the spin-statistics theorem?



Answer



The answer to your question is yes, your intuition is 100% correct. It all boils down to the topology of the configuration space $\mathcal C$, mainly the first homotopy group $\pi_1(\mathcal C)$ (which is non-zero in your example). See problem set 1, problem 3 from this course at Oxford. This exercise is precisely about loops in 3+1D! One has to argue that for any type of point-particle statistics in 2+1D (representations of the Braid group) there exist a corresponding loop statistics in 3+1D.


Note however that these loops will only lead to non-trivial statistics in 3+1D, in higher dimensions there will not be any topological obstruction. This is related to the fact that in higher dimensions, you can always untie knots.


More generally you can think about many different ways of getting non-trivial statistics. You can give your object more complicated internal structure than just point-particles (loops are just one example) or you can put your objects on topologically non-trivial manifolds. See for example this paper about so-called "projective ribbon permutation statistics", which is a way of having non-trivial statistics in higher dimensions but with "defect" that have some internal structure.


EDIT: This is an answer to the question asked by Prathyush in the comments.



Well, yes and no. If you are interested in more general statistics for point-particles, you have to go to 2+1 dimensions where you have anyons. Under an exchange of two (abelian) anyons, the wave function changes by a phase $e^{i\pi\alpha}$. Here $\alpha = 1$ correspond to fermions, $\alpha=0$ correspond to bosons while for any phase $\alpha\in[0,1]$ you have any-ons. So in the sense of exchange statistics, anyons interpolate between fermions and bosons.


There is however another approach to take. In a famous paper, Haldane suggests the so-called exclusion statistics, which defines particle statistics in terms of a generalized Pauli exclusion principle (as you suggest). A natural question is then, does the interpolation of anyons between fermions and bosons lead to a interpolation of exclusion statistics? Murthy and Shankar seem to have tried to answer this question, and they find the corresponding exclusion parameter for the $\alpha$ anyon (equation (16)). However, I don't know enough about exclusion statistics and the state of the field to give many details. But you can learn a lot from reading some of the papers which cite Haldanes paper.


statistics - How do I calculate the experimental uncertainty in a function of two measured quantities


I am performing an experiment where I'm measuring two variables, say $x$ and $y$, but I'm actually interested in a third variable which I calculate from those two, $$z=f(x,y).$$ In my experiment, of course, both $x$ and $y$ have experimental uncertainties, which are given by the resolution of my measurement apparatus among other considerations. I am also considering doing multiple runs of measurement to obtain good statistics on my measurement of $x$ and $y$, and therefore on $z$. I don't really know how the statistical spread will compare to my calculated (resolution-induced) uncertainty, though.


I would like to know what the final uncertainty for $z$ should be, and I am not very familiar with the error propagation procedures for this.



  • What are the usual ways to combine the experimental uncertainties in measured quantities?

  • When should I use the different approaches?

  • How do I include statistical uncertainties when they are present?

  • What happens if the statistical spread of a variable is comparable to the instrument's resolution, so that I can't neglect either contribution?


  • What are good references where I can read further about this type of problem?


I would also appreciate answers to cite their sources - and particularly to use 'official' ones - where possible.




special relativity - What speeds are "fast" enough for one to need the relativistic velocity addition formula?


In this question the accepted answer says:



For objects moving at low speeds, your intuition is correct: say the bus move at speed $v$ relative to earth, and you run at speed $u$ on the bus, then the combined speed is simply $u+v$.


But, when objects start to move fast, this is not quite the way things work. The reason is that time measurements start depending on the observer as well, so the way you measure time is just a bit different from the way it is measured on the bus, or on earth. Taking this into account, your speed compared to the earth will be $\frac{u+v}{1+ uv/c^2}$. where $c$ is the speed of light. This formula is derived from special relativity.




What is "fast" in this answer? Is there a certain cutoff for when it stops being $u+v$ and becomes $\frac{u+v}{1+ uv/c^2}$?



Answer



For simplicity, consider the case $u=v$. The "slow" formula is then $2u$ and the "fast" formula is $\frac{2u}{1+(u/c)^2}$. In the plot you can see these results in units of $c$. The "slow" formula (red/dashed) is always wrong for $u\ne0$, but it is good enough [close enough to the "fast" formula (blue/solid)] for small $u/c$. The cutoff you choose depends on the accuracy required. When $u


A series expansion about $u=v=0$ shows the "slow" formula as the first term and that the corrections are small for $uv \ll c^2$:


$$ \frac{u + v}{1+uv/c^2} = (u + v)\left[1-\frac{uv}{c^2} + \left(\frac{uv}{c^2}\right)^2 + O\left(\frac{uv}{c^2}\right)^3\right] $$


quantum field theory - Arrow and flow of charge in fermion propagator



The momentum-space fermion propagator in the free Dirac theory is given by





The arrow on the fermion propagator is said to represent the flow of charge.


How can we derive this statement quantitatively from the Dirac Lagrangian?


What is the quantitative form of the charge being referred to here?



Answer



The arrow is not really about charge. Neutrinos are neutral, and you need arrows in their propagators anyway. The more correct statement is that arrows represent the flow of fermion number.


Moreover, the arrows are not directly related to the Dirac Lagrangian, because you also need arrows for charged bosons. In fact, you need arrows whenever you have a particle with non-trivial quantum numbers.


In my opinion, the most intuitive way to understand the arrows is to go back to the basics: Wick contractions. When you want to evaluate the $n$-point function $$ \langle\phi_1\phi_2\phi_3\cdots\phi_n\rangle $$ where $\phi\in\{\psi,\psi^\dagger\}$, Wick's theorem tells you to calculate all possible contractions: $$ \langle\phi_1\phi_2\phi_3\cdots\phi_n\rangle=\sum_\mathrm{contractions}\pm\langle\phi_i\phi_j\rangle\cdots\langle\phi_k\phi_\ell\rangle $$



In principle, there are four types of contractions to consider: $$ \langle\psi\psi\rangle,\qquad\langle\psi\psi^\dagger\rangle,\qquad\langle\psi^\dagger\psi\rangle,\qquad\langle\psi^\dagger\psi^\dagger\rangle $$ but a simple calculation shows that $$ \langle\psi\psi\rangle=\langle\psi^\dagger\psi^\dagger\rangle=0 $$


Indeed, if $\psi$ has any non-trivial quantum number, then so does $\psi|0\rangle$; and therefore $\langle0|\psi$ will have the opposite quantum number. This implies that these states are orthogonal and $\langle 0|\psi\psi|0\rangle=0$. For the same reason, $\langle 0|\psi^\dagger\psi^\dagger|0\rangle$ vanishes as well.


Therefore, we introduce the following diagrammatic rule: we always draw an arrow coming out of the $\psi$ fields, and one entering the $\psi^\dagger$ fields. Any diagram where the directions of the arrows is not conserved includes a factor of $\langle\psi\psi\rangle$ or $\langle\psi^\dagger\psi^\dagger\rangle$, and therefore it vanishes. This means that one need only consider diagrams where the direction of the arrows is conserved.


Tuesday, November 26, 2019

kinematics - What is the best path for a given initial and final state?


I am trying to calculate an efficient acceleration curve given starting and final positions and velocities. I'm assuming no friction, and that the acceleration can be applied in any direction at any time.


Given:



  • $p_0$ = starting position

  • $v_0$ = starting velocity

  • $p_f$ = final position

  • $v_f$ = final velocity


  • $T$= total time


I want to find a nice $a(t)$ function that will produce final conditions.


So far I have the following solution that works, but produces an incredibly inefficient $a(t)$ curve:


First I calculate the constant acceleration required to get from $v_{0}$ to $v_{f}$ :


$$ a_v = \cfrac{v_f - v_{0}}{T} $$


Then I calculate the change in position this acceleration will create over $T$ :


$$ p_v = \cfrac{1}{2} a_v T^2 $$


Next I calculate the average velocity required to get from $p_{0}$ to $p_f$ and to counteract $p_v$ :


$$ v_p = \cfrac{p_f - (v_{0} + p_f ) }{ T } $$



Next I calculate the acceleration needed to produce this average velocity over the total time:


$$ a_p= \cfrac{2 v_p}{ T} $$


Finally, I add twice that acceleration to the first half of my acceleration function, and subtract twice that acceleration from the second half. This produces the net position change that I want, but has a net $0$ velocity/acceleration change so $v_f$ stays correct:


$$ a(t) = \begin{cases} a_v + 2 a_p & t \leq \frac T 2 \\ a_v - 2 a_p & t > \frac T 2 \end{cases} $$


While this solution provides a result, it can cause the simulated objects I'm working with to move backward before moving toward their final goal, along with other weird behavior. I think the ideal solution would minimize total acceleration applied over time (and thus force, since the mass of the object will stay constant over this time).


I know that the constraints on this problem are that the integral of $a(t)$ must equal $v_f - v_{0} $, and that the integral of that integral must equal $p_f - p_0$ . I just don't know how to setup the problem to solve for those constraints. I don't even really know what I should Google for to try and solve this problem. Any help would be greatly appreciated.



Answer



Here's my messy approach. It's not ideal, but it may work reasonably well:


Approximate the position function $x(t)$ as a fourth degree polynomial:


$$x(t) = at^4 + bt^3 + ct^2 + dt + e$$



This way, you have one extra degree of freedom to manipulate, which you can use to minimize unnatural motion. Let's assume the motion starts at $t = 0$ and ends at $t = T$.


Then $x(0) = p_0 = e$ and $x'(0) = v_0 = d$, so we can write:


$$x(t) = at^4 + bt^3 + ct^2 + v_0 t + p_0$$


For the remaining three variables, we write a matrix system. Here's the augmented matrix:


$$\begin{bmatrix} T^4 & T^3 & T^2 & p_f - p_0 - v_0 T \\ 4 T^3 & 3 T^2 & 2 T & v_f - v_0 \end{bmatrix}$$


With Mathematica, I row-reduced the above to


$$\begin{bmatrix} 1 & 0 & -1/T^2 & \frac{3p_0 - 3p_f + 2 v_0 T + v_f T}{T^4} \\ 0 & 1 & 2/T & \frac{-4p_0 + 4p_f - 3 v_0 T - v_f T}{T^3} \end{bmatrix}$$


For simplicity, I'll define $\alpha = \frac{3p_0 - 3p_f + 2 v_0 T + v_f T}{T^4}$ and $\beta = \frac{-4p_0 + 4p_f - 3 v_0 T - v_f T}{T^3}$.


This system tells us that as long as we ensure $a = \alpha + c/T^2$ and $b = \beta - 2c/T$, any value of $c$ will result in a polynomial $x(t)$ that matches your conditions.


You can then choose $c$ based on any number of criteria. For example, minimizing acceleration yields $c = -\alpha T^2$ and Floris' result, although this may sometimes result in retrograde motion.



enigmatic puzzle - Oh no! The mods are gone! (and Rand al'Thor, too!)



Dear Puzzling Users,


Oh dear, where start? I guessed it started with my feelings that the mods are way too overbearing. And then I realized, "Hey, I don't have to put up with this sh*. I can just get rid of them all!" So that's what I did: I kidnapped all your moderators plus Rand al'Thor just for kicks. Now, I get that I have to leave "clues" or whatever bull crud I have to put up with, so FINE! HERE ARE YOUR RUDDY CLUES Y'ALL! BUT I'LL SOON TAKE OVER ONCE I CAN GET THE MODS TO HAND OVER THEIR POWER HAHAHAHAHAHAHAHA Have fun!!!!!!!!


Deusovi:

So the other day, my sis and I were playing poker. She's a notorious cheater though, so she decided to cheat by throwing some parsley at me. Yeah, I know. She's kinda dumb though. Anyhow, it landed on my eye, and in return, I landed a nice slap on her back. Didn't go well with my mom, though, when my sister started to wail, crying, "Whale!" Kinda random. Then my mom started going berserk saying, "Show me the card again! Show me the card again!" Then she said in a weird Southern accent that my sister was the winner! What the heck?


Oh you want Deusovi's location? Here: "Read and puncture the beginning, heart of a German man named Ing" (3, 7)


Rubio:
enter image description here enter image description here enter image description here


GPR:
Ontop of a very boring rainbow,
Lays Gentle Purple Rain.
He was like the 16 French kings:
Dead.
Oh the misery of his passing,

So clueless was he to wander so close to Rubio
And yet so far away, Ontop a very boring rainbow
Ontop of an entranced doorway,
On a very rainy day,
Laid Gentle Purple Rain,
Shivering in the rain.


Gareth:
enter image description hereenter image description here



^See edit note on the bottom^




Rand'AlThor:
You know me quite well, I hope,
After all, I warn you of the hags
By singing you a song.
And if the sky was clearer,
You could see the whole city from up here.
What am I?


Bet you could never figure out where they are. Ha! Rescuing them - that's funny. I'll see you later when I'm dictator of Puzzling SE


With Love,

North



Hint 1:



What's that hidden behind's Gareth's picture? What are hags? 16 French kings?



Hint 2:



Deusovi's is a giant wordplay and play-on-words. Also, every single mod besides Deusovi (because he's special) is in, on, below, around, at etc. a landmark. Regarding Rubio's — three of them are wordplay. If you can figure it out, the rest should become easier to deduce.




Hint 3:



In GPR's puzzles, there are only a few important key words to lead you into the answer.In Rand's riddle, everything is of pertinence.



Hint 4:



These are massive hints I am about to reveal:
Rubio's puzzle are puns on states or monuments/places found in those states.
Gareth's puzzle is anoter wordplay. Read the lines carefully, and almost attempt to take everything literally. Consider what the Reverse-Flash says about himself, and it should give you the location.
Rand is somewhere in Britain.

Parsley sounds a lot like this one place found in a bay somewhere mentioned in that ridiculous story. Just remove the l in parsley. The italisized give you indicators of homophones as discovered by one of the commentors.



Hint five: Several edits. A photo has been added to the Rubio's puzzle.



Deusovi is in Wales.
Reverse-Flash. Reverse flash. Reverse flash flash's name in reverse?
Hags ---> witches. Britain. Song. Tall building







Note: Don't worry about the CC. It was just an attempt at a practical joke — it was supposed to be answered as RED HERRING. Of course, the CC was not a very good one, and is not necessary to solve this puzzle by any means.


Edit: I'm really, really, sorry, fellow Puzzling SE. I for some reason, the image will not decode itself when imbeded into the question. There is some kind of mishap that happened, I am not quite sure how this happened. Please forgive me. In the meantime, the image that shoud've popped up when decoded should've been the Reverse-Flash picture.




Answer



Deusovi is



In Cardigan, Wales. I just swooped in and stole this from the hints and all of Strawman's work. "Card Again" -> Cardigan, "Whale" "wail" -> Wales. and the hint saying it's in Wales and not a monument.



Rand is probably




On Big Ben. Since it is in Britain, is tall sings a song "by chiming". The reference to witches is probably a warning about the witching hour. It is also one of the most well known monuments in Britain.



Gareth is



At Eastern Wall in Jerusalem Wally West in reverse is East(ern) Wall[y]



For Rubio is:



At the USS Utah memorial at Pearl Harbor.




Because:



1000 Canons = Grand Canyon Arizona
No place like home = Kansas
Penny Wood = Lincoln tree in Sequoia National park (Nevada) penny + sylvan = Pennsylvania
Frying pan = Rose Hill, North Carolina, home of the worlds largest frying pan? Alaska Oklahoma which is shaped like a big frying pan (thanks @thecoder16)
Should have left a virgin = West Virginia
Red is the new gold => El Dorado = city of gold => city of color = Colorado
Golden gate bridge is Red -> California (thanks @ffao)
Definitely a pyramid => Del operator + aware = Delaware. Very Mountain = Vermont Luxor Pyramid in Las Vegas, Nevada
Elephant Socks => Mass wearing Chausettes (french for socks) > Massachusetts.




OK NV KS
AZ WV PA
?? CA MA

As the discussion in the comments have seen, these States seem to be about US battleships. USS Okalahoma and USS Arizona were completely destroyed, Nevada, West Virginia, and California were damaged, Kansas and MA were not present, and PA was also not involved. The remaining destroyed battle ship is the USS Utah, so Rubio is at the USS Utah Memorial at Pearl Harbor.

quantum field theory - Distance and time measurement in the famous Superluminal Neutrinos Experiment


I tried to understand the technical aspects of the OPERA/CERN experiment, but apparently it takes some professional experience. Therefore I would like to ask someone better acquainted with such experiments to give some details concerning the setup in plain English.


The paper says they conducted a "high-accuracy geodesy campaign that allowed measuring the 730 km CNGS baseline with a precision of 20 cm". They also describe the clock setting and synchronization, but I'm not sure I understand (there is something about "master clock"). So:




  1. I figure that for some reason it wasn't possible - in order to obtain better accuracy than 20 cm - to run photons first along the track, and calculate the distance based on the velocity $c$ and time measured (or simply compare the times for photons and neutrinos), right? Was it because such "contraptions" do not allow for measuring photons or for some other reason? EDIT: I do realize now it's (practically) impossible.





  2. How was the time measured? Were there two clocks at the beginning and at the end of the track, or was there just one clock hooked up (with the wretched fiber-optics cable) to both ends? If there were two clocks, were they synchronized through a GPS satellite or some other external device, or was there some sort of direct synchronization procedure? (What was it?)




To make sure: I looked at the paper before asking, but I am not really sure what they mean. I am not a professional particle physicist, and I have no knowledge whatsoever concerning the devices used, as well as procedures, technical jargon, etc. That's why I asked for help, and will be much obliged for information given in "plain English".


EDIT: In his answer dmckee said that: "Both the timing and the distance were established with the help of GPS." Does it mean that the margin of error is the same for both measurements - time and distance? If the the distance (of 730 km) was measured with 20 cm accuracy, would there be a margin of error for time (clock synchronization) of the same proportion? After all, time is distance and distance is time.




quantum field theory - EM wave function & photon wavefunction


According to this review



Photon wave function. Iwo Bialynicki-Birula. Progress in Optics 36 V (1996), pp. 245-294. arXiv:quant-ph/0508202,



a classical EM plane wavefunction is a wavefunction (in Hilbert space) of a single photon with definite momentum (c.f section 1.4), although a naive probabilistic interpretation is not applicable. However, what I've learned in some other sources (e.g. Sakurai's Advanced QM, chap. 2) is that, the classical EM field is obtained by taking the expectation value of the field operator. Then according to Sakurai, the classical $E$ or $B$ field of a single photon state with definite momentum p is given by $\langle p|\hat{E}(or \hat{B})|p\rangle$, which is $0$ in the whole space. This seems to contradict the first view, but both views make equally good sense to me by their own reasonings, so how do I reconcile them?



Answer




As explained by Iwo Bialynicki-Birula in the paper quoted, the Maxwell equations are relativistic equations for a single photon, fully analogous to the Dirac equations for a single electron. By restricting to the positive energy solutions, one gets in both cases an irreducible unitary representation of the full Poincare group, and hence the space of modes of a photon or electron in quantum electrodynamics.


Classical fields are expectation values of quantum fields; but the classically relevant states are the coherent states. Indeed, for a photon, one can associate to each mode a coherent state, and in this state, the expectation value of the e/m field results in the value of the field given by the mode.


For more details, see my lectures
http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf
http://www.mat.univie.ac.at/~neum/ms/optslides.pdf
and Chapter B2: Photons and Electrons of my theoretical physics FAQ.


cosmology - Why isn't dark matter just ordinary matter?


There's more gravitational force in our galaxy (and others) than can be explained by counting stars made of ordinary matter. So why not lots of dark planetary systems (i.e., without stars) made of ordinary matter? Why must we assume some undiscovered and unexplained form of matter?



Answer




There is a very precise reason why dark planets made of 'ordinary matter' (baryons - particles made up of 3 quarks) cannot be the dark matter. It turns out that the amount of baryons can be measured in two different ways in cosmology:



  • By measuring present-day abundances of some light elements (esp deuterium) which are very sensitive to the baryon amount.

  • By measuring the distribution of the hot and cold spots in the Cosmic Microwave background (CMB), radiation left over from the early universe that we observe today.


These two methods agree spectacularly, and both indicate that baryons are 5% of the total stuff (energy/matter) in the universe. Meanwhile, various measures of gravitational clustering (gravitational lensing, rotation of stars around galaxies, etc etc) all indicate that total matter comprises 25% of the total. (The remaining 75% is in the infamous dark energy which is irrelevant for this particular question).


Since 5% is much less than 25%, and since the errors on both of these measurements are rather small, we infer that most of the matter, about 4/5 ths (that is, 20% out of 25%) is 'dark' and NOT made up of baryons.


newtonian mechanics - Joule heating for a fluid


Say you have a conductive liquid with a changing magnetic field going right through it, causing an electric current. How exactly does the electric current travel and how could you calculate the effect of Joule heating on the liquid?




Monday, November 25, 2019

quantum mechanics - Is there only radial motion in the Hydrogen ground state?


The ground state of the Hydrogen atom is spherically symmetric. In other words, the wave function Psi depends only on the distance r of the electron from the nucleus.


As a consequence all derivatives of Psi with respect to angles theta and phi yield zero.


Does this imply that the average kinetic energy in the ground state [which can be calculated without difficulty from the wave function] is determined exclusively by the radial motion of the electron?



If so, that would be a rather odd result. Let us say the electron is at position (x, 0, 0). Then the kinetic energy would be the result of motion either away from the nucleus (direction +x) or towards the nucleus (-x), but not from motion perpendicular to the x-axis. So in essence the motion of the electron would be 1-dimensional, like a pendulum.




Global Chern-Simons forms and topological gauge theories


I am reading the classic Dijkgraaf and Witten paper on topological gauge theories and something struck me that I didn't understand. For a trivial bundle $E$ on smooth 3-manifold $M$ with compact gauge group $G$ the Chern-Simons form is of course $$S(A)=\frac{k}{8\pi^2}\int Tr (A\wedge dA + \frac{2}{3}A\wedge A\wedge A).$$


The confusing sentence is the following: "If the bundle $E$ is not trivial, the formula for the action does not make sense, since a connection on a non-trivial bundle cannot be represented by a Lie algebra valued 1-form as in that formula." My concern is that this is wrong - of course we can represent a connection in a non-trivial bundle with a Lie-algebra valued 1-form! In fact, some texts define the connection in exactly that way (my trusty Choquet-Bruhat and DeWitt-Morette has that as one of their three equivalent definitions for a connection)- locally, a connection is represented as a linear map $\omega_p:T_p(E)\to \mathfrak{g}$.


So are we supposed to read the "as in that formula" to mean "globally"? Since of course, if the bundle is not trivial there isn't a global Lie-algebra valued 1-form that we can use for the Chern-Simons action. So that seems like a fine argument, but in QFT we are always writing down actions with sections of bundles (fields $\phi(x)$, say) which are not trivial - we just make sure that the action is invariant under the gauge transformations.


So, what is going on with that statement? Is it really saying we cannot represent a connection in such a way, or is it just stating the action is now local, and the rest of the paper is going to go on and tell me how to fix this with an element of $H^3(BG,\mathbb{R})$, etc.?



Answer



As you say yourself, indeed every connection on a bundle is locally given by a Lie algebra valued 1-form and in general only locally.


Let's say this more in detail: for $X$ any manifold, a $G$-principal connection on it is (in "Cech data"):





  1. a choice of good open cover $\{U_i \to X\}$;




  2. on each patch a 1-form $A_i \in \Omega^1(U_i)\otimes \mathfrak{g}$;




  3. on each double intersection of patches a gauge transformation function $g_{i j} \in C^\infty(U_i \cap U_j, G)$





such that




  1. on each double intersectin $U_i \cap U_j$ we have the equation $A_j = g_{i j}^{-1} A g_{i j} + g_{i j}^{-1} \mathbf{d} g_{i j}$




  2. on each triple intersection $U_i \cap U_j \cap U_k$ we have the equation $g_{i j} g_{j k} = g_{i k}$.





Okay, now you would like to form a Chern-Simons 3-form... something out of this. What you immediately get from the above data is a bunch of local differential 3-forms, one on each patch: $CS(A_i) \in \Omega^3(U_i)$.


To make these 3-forms globally glue together to what is called a 3-form connection we need the evident data of higher gauge transformation




  1. on each patch we have the local 3-form $CS(A_i)$;




  2. on each double intersection there should be a 2-form $B_{i j} \in \Omega^2(U_i \cap U_j)$ which gauge transforms the respective CS-3-forms into each other, by $CS(A_j) = CS(A_i) + \mathbf{d} B_{i j}$;





  3. on each triple intersection there should be a 1-form $\alpha_{i j k} \in \Omega^1(U_i \cap U_j \cap U_k)$ which exhibits a second-order gauge transformation ("ghosts of ghosts"!) between the first order gauge trasformations, in that $B_{i j} + B_{j k} = B_{i k} + \mathbf{d} \alpha_{ i j k}$




  4. finally on each quadruple intersection there should be a smooth function $h_{i j k l} \in C^\infty(U_i \cap U_j \cap U_k \cap U_l, U(1))$ which gauge-of-gauge-of-gauge-transforms the gauge-of-gauge-transforms into each other, in that $\alpha_{i j k} + \alpha_{i k l} = \alpha_{j k l} + \alpha_{i j l} + h_{i j k l}^{-1}\mathbf{d}h_{i j k l}$.




That's the data that makes the local Chern-Simons 3-form into a globally well-defined 3-form field. (For instance the supergravity C-field is of this form, with some further twists and bells and whistles added, as we have discussed here).


In mathematical language one says that this kind of local gauge-of-gauge-of-gauge gluing data for global definition of higher form fields is a "degree-4 cocycle in Cech-Deligne cohomology". This is precisely the right data needed to have a well-defined 3-dimensional higher holonomy as is needed here for the definition, because the Chern-Simons action functional is nothing but the 3-dimensional higher holonomy of this 3-form connection.


If you can build it, that is. From the above it is not entirely obvious how to build the 3-form cocycle data $\{CS(A_i), B_{i j}, \alpha_{i j k}, h_{i j k}\}$ from the given gauge field data $\{A_i, g_{i j}\}$.


But this can be done. This is what Cheeger-Simons differential characters were discovered for. An explicit construction that is very natural for the application to Chern-Simons theory we have given in




  • Fiorenza, Schreiber, Stasheff, Cech cocycles for differential characteristic classes, Advances in Theoretical and Mathematical Phyiscs, Volume 16 Issue 1 (2012) (arXiv:1011.4735, web)


Based on this we give a detailed introduction to and discussion of Chern-Simons action functionals for globally non-trivial situations like above in



  • Fiorenza, Sati, Schreiber, A higher stacky perspective on Chern-Simons theory (arXiv:1301.2580, web)


That article gives the local formulas that apply generally, discusses the simplifications that occur when the 3-manifold can be assumed to be bounding, discusses what happens if not, and then explores various other properties of globally defined Chern-Simons theory, such as how to couple Wilson lines to the above story. If you just look at the first part, I think you should find what you need.




edit: In the comments below came up the question why a similar discussion is not also needed when writing down the Yang-Mills action functional, whose Lagrangian is the 4-form $\langle F_A \wedge \star F_A \rangle$ (where $\star$ is the Hodge star of a given metric (gravity) and $\langle -,-\rangle$ is an invariant polynomial, the "Killing form" or trace), or similarly the topological Yang-Mills action functional, whose Lagrangian is the 4-form $\langle F_A \wedge F_A \rangle$.



The reason is that these Lagrangians are built from curvatures evaluated in an invariant polynomial. The very invariance of these invariant polynomials under the adjoint action of the gauge group on its arguments ensures that if $\{U_i \to X\}$ is a good open cover of 4-dimensional space(-time) and if the gauge field is given by the Cech-cocycle data $\{A_i, g_{i j}\}$ with respect to these local patches, that then on double overlaps the two (topological or not) Yang-Mills Lagrangians coming from two patches are already equal


$$ \langle F_{A_i} \wedge F_{A_i}\rangle = \langle F_{A_j}\wedge F_{A_j}\rangle \,. $$


Hence if we write $\nabla = \{A_i, g_{i j}\}$ for the gauge field connection abstractly and denote the (topological) Yang-Mills Lagrangian globally by $\langle F_\nabla \wedge F_\nabla\rangle$, then this is already a globally defined 4-form. Mathematically, this statement is what is at the core of Chern-Weil theory.


Notice that there is nevertheless an intricate relation to the story of the Chern-Simons functional. Namely the local Chern-Simons form $CS(A_i)$ has the special property (essentially by definition) that its differential is the topological Yang-Mills Lagrangian:


$$ \mathbf{d}CS(A_i) = \langle F_{A_i} \wedge F_{A_i}\rangle \,. $$


This means that with the Chern-Simons Lagrangian regarded as a 3-form connection then the topological Yang-Mills Lagrangian is its curvature 4-form. Therefore the relation between the topological Yang-Mills Lagrangian 4-form and the Chern-Simons 3-form is precisely an analogue in higher gauge theory of the familiar relation two degrees down of how the electromagnetic potential 1-form -- which is not globally defined in general -- has a curvature 2-form that is globally well defined.


Mathematically this is why Chern-Simons functionals are called "secondary invariants"


Indeed, this is a bit more than just an analogy: the Chern-Simons 3-form is precisely a doubly higher analog of the electromagnetic field as we pass from the point, via the string, to the membrane.


I have some lecture notes with more along these lines at nLab:twisted smooth cohomology in string theory.


strategy - The lion and the zebras


The lion plays a deadly game against a group of 100 zebras that takes place in the steppe (= an infinite plane). The lion starts in the origin with coordinates $(0,0)$, while the 100 zebras may arbitrarily pick their 100 starting positions. The the lion and the group of zebras move alternately:



  • In a lion move, the lion moves from its current position to a position at most 100 meters away.


  • In a zebra move, one of the 100 zebras moves from its current position to a position at most 100 meters away.

  • The lion wins the game as soon as he manages to catch one of the zebras.


Will the lion always win the game after a finite number of moves? Or is there a strategy for the zebras that helps them to survive forever?



Answer




Zebras win.



Here is the strategy:




The zebras choose 100 vertical strips on the steppe, each of width 300m. The strips do not intersect. Each zebra promises to stay horizontally centered on its own individual strip.





If the lion enters a strip, then the zebra in that strip flees vertically away from the lion until the lion leaves the strip.



It works because:  



The lion cannot kill a zebra on the turn it moves into a strip, because the strips are too wide. And if it is already inside a strip, then the zebra in that strip has just moved 100m away from it, so it cannot catch the zebra.




Sunday, November 24, 2019

quantum field theory - Why are topological properties described by surface terms?


An example are the anomalies in abelian and non-abelian gauge quantum field theories.


For example, the abelian anomaly is $\tilde {F}_{\mu\nu}F^{\mu\nu}$ and the integral over this quantity is a topological invariant which measures a topological characteristic of the gauge field $A_\mu$.


All such quantities can be rewritten as total derivatives and then, using Gauss' law transformed into a surface integral.


What is the intuitive reason that quantities which describe topological properties can always be written as surface integrals?




Best EM/Photon rocket using avalable tech?


Given 10-1000 Watts of electrical power (and no other consumables), what is the current best way to turn it into thrust? Just running it through a heater on an insulating pad would result in an IR thruster, but has bad focus. A laser has good focus but only for a small percentage of the energy.




general relativity - Is local Lorentz + diffeomorphism invariance equivalent to full local Poincaré invariance?


Consider classical General Relativity without the torsion field (the affine connection is already assumed to be symmetric from the start). It is well known that this theory is independent of the coordinates used. The physics stay the same when we perform a passive coordinates change. This is known as general covariance, and any theory could be formulated in this way. It is not a fundamental characteristic of GR.


It is also known (but there's lot of confusion out there about this) that GR is also invariant under active coordinates transformations (also known as diffeomorphisms), which could be interpreted as a kind of gauge transformations, and not just as simple changes of local coordinates. This property describes physics: an active local coordinates transformation "pushes" the spacetime point to another place of the same manifold. Invariance of the theory under such gauge transformations is telling that the physics is the same at any place on the manifold. Any spacetime point is equivalent, to describe the physics, and the theory is background independent. I believe that this is a subtle formulation of translation invariance, in GR.


Please, don't confuse the general property above with some symmetry of the metric (i.e. isometry). I'm considering general spacetimes, not particular solutions with some special symmetry (isotropy, homogeneity, etc).



GR is also incorporating local Lorentz invariance (at any spacetime point), which says that the physics is independent of the local frame used by the observer (accelerating, rotating axes, free fall, ...).


Usually, the full Poincaré group (Lorentz + translations) isn't made local in standard GR: only the homogeneous Lorentz part is made local. I'm wondering if the "missing" translation part is actually subtly included through the diffeomorphism invariance.


So the question is this: Is the local Lorentz invariance + diffeomorphism invariance (active coordinates changes) of classical GR equivalent to the full local Poincaré invariance?


I'm expecting that some will say "NO" to the question above, because full Local Poincaré invariance is supposed to bring torsion into GR (I never saw any convincing proof of this). Torsion (i.e. the antisymmetric part of the affine connection) is usually assumed to vanish trivially from the start in GR, but there is no contradiction in letting it enter the classical GR formulation. I don't see why we need to explicitly "add" the local translation invariance to get torsion. It can already be there in classical GR, and I suspect this is because of the diffeomorphism invariance -- interpreted as a formulation of the local translation invariance.


I never saw that interpretation before, so I need opinions on it. Maybe I'm getting it all wrong!


If the answer is really a big "NO", then how do we describe local translation invariance to explicitly bring torsion into the theory? As far as I know, torsion could be added directly to classical GR, without the need to talk explicitly about local translation invariance.




Does the contact area affect friction forces?


I recall studying a law of friction some years ago, in engineering school. All I remember is that when first approximation was taken, the popular $f=k \times N$ was derived.


What could that be? Or if this does not exist, what is the proof that contact area does not affect friction force?


EDIT:
So, the book is about designing a friction clutch. This device consists of a number z of disc pair, one connected to the input shaft and the other disc from the pair connected to the output shaft.


If I understand the formula and meanings of the variables correctly, the maximum torque, that the device cn transmit is proportional to $F \times z$. F is the force, pushing the disc package together.


Does this change anything?




quantum mechanics - Why particle number operator $hat{N}$ is $hat{a}^daggerhat{a}$ rather than $hat{a}hat{a}^dagger$?


Both $\hat{a}^\dagger\hat{a}$ and $\hat{a}\hat{a}^\dagger$ are Hermitian, how do we know which one represents the particle number?



Answer




We require that the number operator have the following property:


$$\hat n |0\rangle = 0.$$


We know that


$$\hat a |0\rangle = 0$$


and we know that


$$\hat a |1\rangle = |0\rangle$$


and we know that


$$\hat a^{\dagger} |0\rangle = |1\rangle. $$


Thus, it follows that


$$\hat a \hat a^{\dagger} \ne \hat n$$



since


$$\hat a \hat a^{\dagger} |0\rangle = \hat a|1\rangle \ne 0.$$


Now, it remains to be shown that


$$\hat a^{\dagger}\hat a = \hat n. $$


Can you take it from here?


Saturday, November 23, 2019

special relativity - Rindler motion constant acceleration proper time


I have problem to calculate the proper time for Rindler coordinates: the coordinates in Minkowski space with constant acceleration are given by:$$ \begin{alignat}{7} t' &~=~ \frac{c}{g} \, && \sinh{\left(\frac{g}{c\tau}\right)} \\ x' &~=~ \frac{c^2}{g} \, && \cosh{\left(\frac{g}{c\tau}\right)} \end{alignat} $$


The proper time is $$ \tau_{\text{p}}~=~\int{\sqrt{1 - \frac{v^2 \left(t' \right)}{c^2}}}\, \mathrm{d}t' \,,$$with$$ \begin{alignat}{7} t' & ~=~ \frac{c}{g} \, && \sinh{\left(\frac{g}{c\tau}\right)} \\ \mathrm{d}t' & ~=~ && \cosh{\left(\frac{g}{c\tau}\right)} \,. \end{alignat} $$


How can I calculate $v \left(t' \right)$?





quantum field theory - What is precisely the energy scale of a process?


Coupling constants run with the energy scale $\mu$. But what is exactly this energy scale. My question is, if I have a physical process, how do I compute $\mu$?



Answer




There are at least two answers possible to give, but both, in the end, amount to the same thing: There is no "right" way to fix the energy scale of a process, but that doesn't matter, except that your perturbation theory will probably break if you choose the scale badly.


The old answer: The renormalization scale is arbitrarily defined to fix some parameters of the theory to measured values and get rid of infinities, e.g. a $\phi^4$ scalar coupling $\lambda$ is sometimes fixed by looking at the $\phi^4$ interaction at the channel $s^2 = 4\mu^2, t^2 = 0, u^2 = 0$, but you could as well look at the channel at $s^2=t^2=u^2=\mu^2$. We need to just take any kind of condition to fix our counterterms and get finite answers for our amplitudes.


There is, generically, no real meaning to the renormalization scale, but if you define it at diagram channels like $s^2=t^2=u^2=\mu^2$, then it will reflect the "typical" energy scale of the process. By doing renormalized perturbation theory, the counterterms in the Lagrangian will be fixed by the renormalization conditions, and yield finite answers for amplitudes regardless of how the scale is chosen. However, if you are at processes which have vastly different channels from the scale, the perturbative corrections start to grow large incredibly fast, so that you gain nothing by having amplitudes that are "theoretically" independent of the scale, and this motivates seeking running couplings - by being able to vary the renormalization scale without doing order-by-order perturbation explicitly on the diagram, we save a lot of effort.


The modern, Wilsonian answer: A quantum field theory is viewed intrinsically as an effective theory possessing an inherent momentum cutoff $\Lambda_0$. This isn't a "scale" anymore at which we do some shenanigans because we want to get rid of infinities, it is a property of the theory, telling you how high the excited momenta go. The Fourier modes of the quantum fields are literally cut-off above this scale, and we view the path integral measure (which is "defined" by a limit process on a lattice) as only containing the measures for the momenta below $\Lambda_0$.


"Renormalizing" now means that we can integrate out even more Fourier modes, going to a new cut-off $\Lambda<\Lambda_0$, making them couplings in the Wilsonian effective action rather than dynamical objects of the theory. Then $\Lambda$ is clearly the scale above which we do not expect many modes of the fields to be excited, it is indeed the order of the momenta occuring in the process.


Perhaps a bit disappointing to your question, this way of looking at the cut-off is inherently approximative, and there is no exact way to get the "energy scale of a process" - but again, it doesn't matter, since all the renormalization does is shift the way our perturbation theory works by running the couplings, and we are again in principle free to choose any kind of scale in the process to be our renormalisation scale - but if you take one that deviates much from the intuitive meaning of "order of momenta in the process" too much, you will again have no fun trying to get the counterterms perturbatively.


particle physics - How do we know what the flavour of the neutrino from a beta decay is?


I have read that because of the conservation of the leptonic number, a neutron should decay into $p + e^- + \overline{\nu}_e$. I don't understand this argument because I have also learnt that the leptonic number may not be conserved.


Since the flavour eigenstates of a neutrino are not the same as the propagating eigenstates, it seems to me that we should also consider muonic or tauic neutrinos for this decay (or rather the mass eigenstates), because when one computes the cross section, one assumes that the neutrino is observed a long time after the interaction had taken place, so that flavour oscillating neutrino may occur.


What do you think ?



Answer



You do understand that short-range neutrino measurements show the expected flavor mix, right?


That is if we set up a detector a few meters from a nuclear core we observe neutrino interactions involving electrons. If we set one up just downstream of a neutrino beam-line we see mostly interactions involving muons (with the expected admixture of those involving electrons and anti-muons).


These fact are re-tested sporadically in each new generation of machines because precision neutrino work requires a near detector.


Consequently the flavor states are defined as the states that participates in weak interactions with a flavored lepton. The neutrino-state in a weak interaction involving an electron is what we call "an electron (anti-)neutrino".



So the short answer is, that no, there is no chance that the thing we define to be an electron neutrino is the (different) thing that we define to be a moun neutrino.


Friday, November 22, 2019

quantum mechanics - Domain of symmetric momentum operator vs self-adjoint momentum operator


Is there an example of a function that is not in the domain of the 'naive' symmetric (but not self-adjoint) momentum operator $p:=-i\frac{d}{dx}$ but is in the 'true' self-adjoint momentum operator $p:= \left(-i\frac{d}{dx}\right)^\dagger$.


I am trying to understand the mathematical differences b/w symmetric and self-adjointness and thought that this would be an enlightening example. Could you also show why this example in the domain of the self-adjoint momentum operator using the definition of the adjoint domain: $$D(A^\dagger) := \{ \phi \in {\cal H}\:|\: \exists \phi_1 \in {\cal H} \mbox{ with} \: \langle \phi_1 |\psi \rangle = \langle \phi | A \psi\rangle \:\: \forall \psi \in D(A)\}$$


This question was motivated by reading this fantastic answer by Valter Moretti.



Answer



Define $$ P \equiv -i\frac{d}{dx} \tag{1} $$ and $$ \psi(x) = |x|\,\exp(-x^2). \tag{2} $$ Let $D(P)$ denote the domain of $P$. Clearly, $\psi$ is not in $D(P)$, because it is not differentiable at $x=0$. However, $\psi$ is in the domain of $P^\dagger$, because $$ \langle \psi|P\phi\rangle \equiv -i\int^\infty_{-\infty}dx\ \psi^*(x)\frac{d}{dx}\phi(x) \tag{3} $$ is well-defined for all $\phi\in D(P)$, and $P^\dagger$ is defined by the condition $$ \langle P^\dagger\psi|\phi\rangle = \langle \psi|P\phi\rangle \tag{4} $$ for all $\phi\in D(P)$. The domain $D(P)$ is dense, so $\psi$ can be arbitrarily well-approximated by a function in $D(P)$, just by smoothing out the "kink" in an arbitrarily small neighborhood of $x=0$, but $\psi$ itself is not in $D(P)$, not even after accounting for the fact that vectors in this Hilbert space are represented by functions modulo zero-norm functions. We can't smooth out the "kink" at $x=0$ in $\psi$ by adding any zero-norm function.




Edit:


The the value of $P^\dagger$ acting on the example (2) is $$ P^\dagger\psi = -i\big(s(x)-2x|x|\big)\exp(-x^2), $$ where $s(x)=\pm 1$ is the sign of $x$. This can be checked by checking that it satisfies (4) for arbitrary differentiable functions $\phi$, which is possible because the point $x=0$ can be omitted from the integrand without changing the value of the integral.


classical mechanics - On the Stability of Circular Orbits


Bertrand's Theorem characterizes the force laws that govern stable circular orbits. It states that the only force laws permissible are the Hooke's Potential and Inverse Square Law. The proof of the theorem involves some perturbation techniques and series expansion.


The most natural things that comes to my mind when thinking about such a problem is that the effective force should be a restoring force for circular orbits to be stable.


$f_{\mathrm{eff}}(r) = \dfrac{l^2}{\mu r^3}-f(r) = 0$, for orbit to be circular.


$f'_{\mathrm{eff}}(r)<0$, for orbit to be stable. Assuming a power law, $f=Kr^n$, for the central force, solving it gives me the solution $n>-3$.


This is very weak compared to the statement of Bertrand's Theorem. Could someone explain to me the rationale behind perturbation technique used, and what is missing from my interpretation of 'stable' in my derivation?




Answer



What you just did was to find a condition for attractive power-law forces to have stable orbits where stable means they remain bounded when perturbed around the circular orbit. You got the correct result.


The Bertrand's Theorem though says something different: the only forces whose bounded orbits imply closed orbits are the Hooke's law and the attractive inverse square force. A closed orbit is one which the particle repeats their momentum and position after some finite time - it closes in the phase space.


The idea behind the proof of Bertrand's Theorem is to consider a perturbed orbit and then calculate the periods of the angular revolution and the radial oscillations. If these periods are commensurable then the orbit is closed. You can find the proof here.


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 \...