Saturday, August 31, 2019

wordplay - Oh boi! Dis aint Engliz! #1


First some Engliz:


These weird combination of letters are actually words translated in some different languages, and concatenated.


Each line has one word translated in different languages. These nine words form three groups of three words each, with groups separated on the basis of a common relation.



What you have to do:



  • Find the words

  • Find the languages

  • Find the groups


Since this question is the first part of a series of questions I'll be posting, so this one has a couple of easy workarounds.




  1. rukamapaditupalapaainadaimntawvqhia


  2. makrateliguraizeakjiandaokawilitxiab

  3. kaghaldipaperazhipepantawv

  4. mzeeguzkiataiyanglahnub

  5. varskvlaviizarxinghokuhnubqub

  6. dedamitsalurradiqiuhonuantiajteb

  7. pankariarkatzaqianbipenikalamemhluav

  8. kompasiiparrorratzaluopanpapunilibkoobqhia

  9. zghvisitsasoahaikekaihiavtxuv






Knao sum non Engliz:

Da magick wond laees heir



Abrakedavra hocus pocus



Remememememember : Thou shant weight four hintz. Asck four'em.


Eeven partial ansirs ar olso welkome.




The second puzzle in this series can be found here.




Answer



Languages



1. Georgian
2. Basque
3. Chinese
4. Maori/Hawaiian
5. Hmong



Translated words appear in this order of translation.



Words





  1. Map
    - რუკა (ruka)
    - mapa
    - 地图 (ditu)
    - palapaaina (?)
    - daim ntawv qhia







  2. Scissors
    - მაკრატელი (makrateli)
    - guraizeak
    - 剪刀 (jiandao)
    - kawili
    - txiab







  3. Paper
    - ქაღალდი (kaghaldi)
    - papera
    - 纸 (zhi)
    - pepa
    - ntawv







  4. Sun
    - მზე (mze)
    - eguzkia
    - 太阳 (taiyang)
    - la (ra?)
    - hnub







  5. Star
    - ვარსკვლავი (varskvlavi)
    - izar
    - 星 (xing)
    - hoku
    - hnubqub







  6. Earth
    - დედამიწა (dedamitsa)
    - lurra
    - 地球 (diqiu)
    - honua
    - ntiajteb







  7. Pencil
    - ფანქარი (pankari)
    - arkatza
    - 铅笔 (qianbi)
    - penikala
    - memhluav







  8. Compass
    - კომპასი (kompasi)
    - iparrorratza
    - 罗盘 (luopan)
    - papuni (?)
    - libkoobqhia







  9. Sea
    - ზღვის (zghvis)
    - itsasoa
    - 海 (hai)
    - ke kai
    - hiavtxuv (hiav txwv)






Groups



- Scissors, Paper, Pencil are items used for arts/crafts or school supplies
- Map, Compass, Sea are all related to Sea Navigation
- Sun, Star, Earth are all objects in Space



I'm missing the relation, they look like they're all related to




arts and sciences.



solar system - Meteorites from Mars?


So I've heard of meteorites "originating from Mars" (e.g. AH84001), but the phrase confuses me. I'm interested in what this means - have these rocks somehow escaped Mars' gravity and ended up here; or were they part of the material that Mars formed from, but did not end up as part of the red planet? Or another explanation?



Answer



These are chunks of rock that existed as part of the crust of Mars but were ejected into interplanetary space by a very powerful impact and then eventually impacted the Earth. It wasn't until we had sent probes to Mars and began to understand the composition of martian minerals and atmosphere that we started realizing some of the meteorites we had already found were from Mars. The clincher has been the analysis of trapped gasses within meteorites. Nearly 100 Martian meteorites are known to have been found.


Interestingly, most Martian meteorites fall into only 3 mineralogical categories (Shergottites, Nakhlites, and Chassignites, or SNC meteorites) and had been identified as being unusual as meteorites even before they were confirmed to have originated on Mars.



Similarly, there are likely some number of Earth meteorites on Mars.


Friday, August 30, 2019

electric circuits - Voltage drop = more electrons on one side of resistor


I have been asking myself this question for a long time now. Suppose we have two resistors in series connected to a voltage source. Simply put, does the voltage drop on each resistor mean that there are more electrons on one side of the resistor compared to the other side? Because if an analog voltmeter is connected in parallel to that resistor, current will flow through the voltmeter.



Answer





Simply put, does the voltage drop on each resistor mean that there are more electrons on one side of the resistor compared to the other side?



Yes, the charge density at one end of the resistor must differ from the other if there is a current through.


Consider, for simplicity, a resistive element of length $L$, area $A$ and resistivity $\rho_r$.


Joined to each end are conductors with area $A$ and resistivity $\rho_c$.


Assume a constant current density of magnitude $J$ along the length of the conductors and resistive element.


The electric field, in the direction of $J$, within the conductors and resistive elements is given by $E_c = J \rho_c $ and $E_r = J \rho_r $ respectively.


Then, at the boundary between the conductor and resistive element, the electric field abruptly changes in value by


$$\Delta E = \pm J\left(\rho_r - \rho_c \right) \approx \pm J\rho_r\;,\quad \rho_c \ll \rho_r$$



This implies a charge density at each end of the resistive element.


See, for example:


enter image description here


For another approach, consider that the slope of the electric potential changes abruptly at the interface.


For a steady current and assuming essentially ideal conductors, the electric potential along the conductors is constant but begins changing with distance inside the resistive element (it must since there is potential difference between the ends of the resistor).


Again, this implies an abrupt change in the electric field at the boundary which requires a charge density at the boundary.


grid deduction - Double feature: Braaains


This puzzle is part 1 of the Double feature series. The series will continue in "Double feature: Russian desman".





enter image description here


Rules of Akari (also known as Light Up)1



  • Insert light bulbs into any number of empty cells.

  • Black cells are walls and cannot contain light bulbs.

  • Numbers in black cells indicate how many light bulbs are directly adjacent to that cell (vertically and horizontally).


  • A light bulb illuminates its own cell as well as every cell visible from it in all four directions (up/down/right/left), continuing until a wall comes in the way.

  • Every white square must be illuminated by at least one light bulb.

  • No light bulb may be illuminated by another light bulb.


Across
2. Otherwise heartless woman of Beethoven's dedication (4)
5. Major city identified in articles about John Fogerty's band (5)
7. A major criminal organization? (5)
8. Martial artist striking left and right to draw blood (3)
9. Rating below "unlimited" (3)

11. Chapeau smuggled in chateau (3)
13. 0.0000001% of an iPod (4)
15. Uranium extracted from unopened, rusty pen (3)
16. Primary meal ingredients: mostly soy soup from Japan (4)
17. "Whatever!" – Warhol after losing final bid (3)
19. Disagreements follow key goods being held by third parties (7)


Down
1. Jimmy and Jack – sort of same (5)
3. Some fail at choosing door fastener (5)
4. Force out during surefire victory (5)

5. Enthusiastic interstellar hitchhiker not hurt, amazingly (6)
6. Sri Lanka once got rid of mathematical constant – Number Six, maybe (5)
10. Zealous leader leaving lethargic place (3)
12. In the morning, a space station is out of order (5)
14. One's first impression is what can be heard in 15 across (4)
15. Central Missouri is left bitter (4)
18. Almost double what can be heard in 5 across (3)


1 Paraphrased from the original rules on Nikoli.



Solve both puzzles to answer the question: What are zombies?




Answer



What are zombies?



SCARIEST MONSTERS



Solutions to the cryptic:




Across
2. Otherwise heartless woman of Beethoven's dedication (4) ELSE = für EL(i)SE

5. Major city identified in articles about John Fogerty's band (5) ACCRA = A + CCR + A
7. A major criminal organization? (5) TRIAD (dd)
8. Martial artist striking left and right to draw blood (3) (Bruce) LEE = (b)LEE(d)
9. Rating below "unlimited" (3) ELO = (b)ELO(w)
11. Chapeau smuggled in chateau (3) c_HAT_eau
13. 0.0000001% of an iPod (4) NANO (dd)
15. Uranium extracted from unopened, rusty pen (3) STY = (r)(u)STY
16. Primary meal ingredients: mostly soy soup from Japan (4) MISO = M_ I_ + SO(y)
17. "Whatever!" - Warhol after losing final bid (3) ANY = AN(d)Y
19. Disagreements follow key goods being held by third parties (7) ESCROWS = ESC + ROWS


Down
1. Jimmy and Jack - sort of same (5) JAMES = J + SAME*
3. Some fail at choosing door fastener (5) LATCH = fai_L AT CH_oosing
4. Force out during surefire victory (5) EVICT = surefir_E VICT_ory
5. Enthusiastic interstellar hitchhiker not hurt, amazingly (6) ARDENT = AR(thur*) DENT
6. Sri Lanka once got rid of mathem. const. - Number Six, maybe (5) CYLON = CE(y)LON
10. Zealous leader leaving lethargic place (3) LAY = LA(z)Y
12. In the morning, a space station is out of order (5) AMISS = AM + ISS
14. One's first impression is what can be heard in 15 across (4) OINK! = O(ne) + INK

15. Central Missouri is left bitter (4) SOUR = (m)(is)SOUR(i)
18. Almost double what can be heard in 5 across (3) TWI = TWI(n)?

Filled-in grids:



Shine a light on the crypticism



enigmatic puzzle - (5 of 11: Slitherlink) What is Pyramid Cult's Favorite Person?


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: #4 Hitori | Next page: #6 Yajilin




enter image description here




Rules:



  1. Connect adjacent dots with 3-directions (parallel with one of the board sides) lines to make a single loop.

  2. The numbers indicate how many lines surround it, while empty cells may be surrounded by any number of lines.

  3. The loop never crosses itself and never branches off.





Special thanks to chaotic_iak for testing this puzzle series!




Answer



Completed slitherlink



enter image description here



Their favourite person



The letters on the outside of the loop spell FRIEND.



Slightly convoluted reasoning




First of all, any pairs of 0 and 2 can be filled since the 2 only has two sides available. Also, the 1/1 in the bottom right corner cannot reach the corner without using two sides of the 1, so fill in the only possibility left. Also, all pairs of 2 and 2 have to have the line between them filled (otherwise completing the 2s would make a closed loop). Fill these in first.

enter image description here

Now, let's look at the top corner. The right-hand side of the loop can not touch the 0, so only one way to go. After that, the left-hand side only has one option to avoid touching the already-completed 1 below it or creating a loop with the right-hand side. After that, the right-hand side needs to continue straight in order to avoid making a loop.

enter image description here

On the left side of the top corner, the line coming from the 0/2 has to connect to the upper loop's left-hand side in order to avoid filling a second side of the 1. Also, we can continue the left-hand side of that same line downward because the only other option would hit a 0.

enter image description here

In the bottom-left corner now, the left-hand side of the line has to continue up. Also, the right-hand side only has one option to avoid hitting the 0.

enter image description here

In the bottom-right corner, both sides need to continue along the border to avoid filling more sides of the 1. Also, whenever we approach the corner of a 2 like we do here, we can fill in the opposite side. (If that side was not filled, we would be branching the loop into two directions at that point.)

enter image description here

Neither of the lines marked red can be filled, because either of them would make it impossible to complete the two 2s.

enter image description here

That leaves only one way to continue the line from the bottom-left corner.

enter image description here

The 1 near the bottom-left corner only has one option to be filled without touching the filled line at the bottom. From there, the line only has one option to avoid the 0 and the already-filled 2 next to it.

enter image description here

The pair of 2s near the bottom-left corner can be filled now, and this leaves only one way to complete the 2 above and to the right of it.

enter image description here

Near the top corner, the line passes through one corner of the 1 below it. Since we can not draw a third line into that same corner, we have to fill in the opposite side of the 1. This leaves only one way to fill in the 2 below.

enter image description here

The 1 to the left of the previous ones only has one option available. After that, the line can only continue in one direction, and that is to join with the line below it.

enter image description here

The left-hand line of the top corner can only continue in one direction without making a 3-way intersection. From here we see that the 2 below has only one option available, so the lines must join here.

enter image description here

There is no way to reach the circled point without either hitting the line above or filling two sides of the 1 below.

enter image description here

Since we can't get to this point, we only have one option to fill the 1. This line has only one option in both directions, so fill those in as well. Now we've solved the entire bottom-left corner.

enter image description here

Let's look at the bottom-right corner. The 2 can only be completed in one way. That leaves only one way to continue the upper of the two line ends. And after filling that in, the bottom-most line now only has one option.

enter image description here

Since the circled point cannot be revisited, there is only one way to complete both of the 1s touching it.

enter image description here

We need to continue upwards on the right-hand border because the 1 below is already complete. From there, the line can't go to the centre of the three 0s because it has no legal way to continue from there. So we need to continue up the right-hand border.

enter image description here

The upper part of the small line below the top corner can only go right. After that, we can't go right because we'd leave the line above uncompletable. We also can't go down and to the right because there's no way to go after that (except again by leaving the line above hanging). So we need to go up and to the right to connect with the line above. This leaves only one way to continue the line just below on the right-hand border, and that connects with the other line near the middle.

enter image description here

There is only one way to complete the 2 near the bottom-right corner, and only one way to proceed from there. This completes both of the remaining 1s as well.

enter image description here

The line cannot go down and to the left, because we'd make two separate loops. So we go to down and to the right, leaving exactly one way to connect the lines at the bottom. Then just connect the remaining ends in the only way possible, and we get the final solution.

enter image description here



newtonian mechanics - Bernoulli's principle on a curve ball


I've seen a few excellent answers here on the Magnus force, which explains why balls with a spin will curve. However, my intuition is still telling me that the Bernoulli's principle would push it the opposite way and I need help understanding why my reasoning is flawed.


Imagine that you kick a soccer ball on the left side so that it's spinning clockwise (viewing the ball from above) and the ball will curve to the right. Since the left side of the ball is spinning against the air, wouldn't this mean faster relative speeds and thus a lower pressure than the right side? And wouldn't this lower pressure on the left side cause it to curve left instead of right?



Answer



What produces lift is circulation, which causes the airflow to be deflected in one direction, causing an equal reaction in the other direction.


If you want to think in terms of Bernoulli on your soccer ball, the air on the left side is being slowed, while that on the right side is being accelerated by the spin of the ball.



Ion acoustic wave regime


I am fairly new to plasma physics and am making a project on the Ion Acoustic Wave. We have discussed Landau-damping in a plasma where we took the electrons to have a Maxwellian distribution and considered ions as immobile. The idea of the project is to generalise this to the case where we also take ions into account.


In my attempt to do so, I have used Python to solve the dispersion relation numerically. I find that for a temperature ratio $T_{el}/T_{ion}= 10^4$ and mass ratio $m_{el}/m_{ion}= 1/2$ the damping effect is weaker for the IAW than when the ions were considered immobile. How could I understand this result more intuitively? I was expecting the wave to damp faster as there are now two species absorbing energy from the wave. I also found that for heavier ions, the IAW gives almost identical results as the initial plasma model.



enter image description here




quantum mechanics - Rigorous mathematical definition of vector operator?



In standard quantum mechanics textbooks, the concept of operators is often introduced as linear maps that map a Hilbert space $H$ onto itself: $$ \hat{O}: H \rightarrow H \, . $$


However, directly after, we use the position operator $\hat{\vec{x}}$, which isn't of the said shape, but instead is like a triple of operators, e.g. $\hat{\vec{x}} = (\hat{x}, \hat{y}, \hat{z})$. What I thought now is that maybe, one could treat the position operator as a linear map $$ \hat{\vec{x}}:H \rightarrow H \otimes R^3 \, . $$


Is it wrong to say the that the position operator is that kind of map? Does it make sense?


Edit: I know that you can treat the position operator as three independent operators $\hat{x}$, $\hat{y}$, $\hat{z}$. I just want to know if my way of treating the position operator is an equivalent way to see things, or if it is wrong to treat it like that. If it is wrong, why is it wrong?



Answer



So there's a natural isomorphism


$$ \varphi: H^{\oplus 3} \to H \otimes \mathbb R^3\\ (a, b, c) \mapsto a \otimes e_0 + b \otimes e_1 + c \otimes e_2 $$


where $e_i$ is a basis for $\mathbb R^3$. The definition you're objecting to is


$$ \hat{\vec{x}}: H \to H^{\oplus 3}\\ v \mapsto (\hat x(v), ~ \hat y(v) , ~\hat z(v)) $$


You can convert it to your version by composing with the above isomorphism.



$$ \tilde x \equiv \varphi \circ \hat{\vec{x}}:~~ H \to H \otimes \mathbb R^3\\ v \mapsto \hat x(v) \otimes e_0 + ~ \hat y(v) \otimes e_1 + ~\hat z(v)\otimes e_2 $$


So your definition makes complete sense.


Comments


So this answers the question as stated. Judging by your comments on other answers I think you're looking for a nicer way to formalize the annoying statement 'blah operator transforms under blah group like a vector'. Whether this definition yields what you want should be the subject of another question.


Namely I think you're hoping somehow conjugation by a representation of the group will factor through your tensor product and have the group action on $\mathbb R^3$ just act on the second factor. Spent a couple of minutes trying to naively get this to work without much luck, will update if there's progress, or at least give a better explanation of why not.


Edit


I'm going to describe how this construct makes the "covariance under conjugation by some representation of some action on $\mathbb R^3$" condition more explicit, as OP had hoped.


For clarity let's start by giving a name to the operation of conjugating this "vector operator".


Let $G$ be some group that admits an action on $\mathbb R^3$. Let $U: G \to H$ be a unitary representation of $G$ on $H$. Define, for any $R \in G$


\begin{align} C_R: \mathrm{Mor}(H, H^{\oplus 3}) &\to \mathrm{Mor}(H, H^{\oplus 3})\\ f(\cdot) &\mapsto (U(R) f(\cdot)^i U(R)^\dagger)_{i \in (0, 1, 2)} \end{align}



where $\mathrm{Mor}(V, W)$ is collection of linear maps $V \to W$. This is just a name for component-wise conjugation of this tuple of operators.


Now the main claim


Proposition The statement


$$ C_R(\hat{\vec{x}}) = \left(\sum_{j}R_{ij}\hat{\vec{x}}(v)^j\right)_{i \in (0, 1, 2)} $$


is equivalent to


$$ \varphi \circ C_R(\hat{\vec{x}}) = (1 \otimes R)(\varphi \circ \hat{\vec{x}}) $$


Proof For brevity we give the $\implies$ direction, the other direction should be obvious from the same computation


\begin{align} \varphi\left(\left(\sum_{j}R_{ij}\hat{\vec{x}}(v)^j\right)_{i \in (0, 1, 2)}\right) &= \sum_i\left(\sum_{j} R_{ij} \hat{\vec{x}}(v)^j\right) \otimes e_i \\ &= \sum_{ij} \hat{\vec{x}}(v)^j \otimes R_{ij}e_i\\ &=\sum_{j} \left(\hat{\vec{x}}(v)^j \otimes \sum_i R_{ij}e_i\right)\\ &= (1 \otimes R)\left(\sum_{j} \hat{\vec{x}}^j(v) \otimes e_j \right)\\ &=(1 \otimes R)\left(\varphi(\hat{\vec{x}(v)}) \right) \end{align}


quantum field theory - Compact QED and Non-compact QED - Polyakov textbook


This question is related with Polyakov, "Gauge Fields and Strings" section 4.3


Firstly, Polyakov define a QED on a lattice





  1. Compact QED \begin{align} S = \frac{1}{2} \sum_{x, \alpha, \beta} (1-\cos(F_{x,\alpha\beta}) ) \end{align} where $F_{x,\alpha\beta} = A_{x,\alpha} + A_{x+\alpha,\beta} - A_{x+\beta, \alpha} - A_{x,\beta}$ with $- \pi \leq A_{x,\alpha} \leq \pi$




  2. Non compact QED \begin{align} S = \frac{1}{4 e_0^2} \sum_{x,\alpha\beta} F_{x,\alpha\beta}^2 \end{align} here $- \infty \leq A_{x,\alpha} \leq \infty$




In the textbook, Compact version (periodic) is Natural version of QED which is related with charge quantization.


Here i have a few basic question.


1 : Why periodicity gives compactness?


2 : What is the physical difference or usefulness of compact or Non-compact QED? (Is this compact concept is only related with the lattice theory?)



3 : Is periodicity of $A_{x,\alpha}$ really related with charge conservation?


My guess from equation 4.32 in textbook, which is \begin{align} q_0 = \frac{1}{2\pi} \oint_{L} A_{x, \delta} \end{align} The periodicity of $A_{x,\alpha}$ gives some number after loop integral. $i.e$, (ration for certain loop)



Answer



1 : When $A$ is restricted to $[-\pi,\pi)$, the group of gauge transformations is the compact $\mathrm{U}(1)$ group. (Roughly speaking, to be compact, a group needs to be bounded.) In the nonperiodic theory, the gauge group is the group of real numbers under addition, which is noncompact.


2 : The most important distinction between the compact and noncompact theories is that the compact one allows ("Polyakov") monopoles, which are topological configurations of $A$. (For the same reason, the XY model has topological defects, called vortices, while a real scalar field does not.) Polyakov monopoles are very similar to Dirac monopoles, but Dirac's argument was that the Dirac string is unobservable because charges are quantized, while Polyakov's is that the string has no energy cost, because $F_{\alpha\beta} = 2\pi$ along its length.


You could define a compact $\mathrm{U}(1)$ gauge theory in the continuum, but monopoles are no longer possible. (Again, the same is true of vortices in the XY model; they have a core at which $\theta$ is undefined.) There could still be a distinction for the model on a space with nontrivial topology.


3 : How is compactness related to discreteness of charge?


Suppose we have a matter field $\psi$ with charge $q$ defined on the sites of the lattice. If you apply a gauge transformation $\psi \rightarrow \psi \mathrm{e}^{\mathrm{i} q \theta}$ on one site, then $A$ changes by $\pm\theta$ on all links connected to this site. If we choose $\theta = 2\pi$, this transformation of $A$ is redundant in the compact theory. Because $A$ is effectively unchanged, in order for the gauge transformation to be a symmetry, $\psi$ must also be unchanged, so we require $q$ to be an integer.


Another way to see this is to use the Dirac argument that the existence of magnetic monopoles requires discreteness of electric charges.





If you have access to it, you might want to look at J. B. Kogut, An introduction to lattice gauge theory and spin systems, Rev. Mod. Phys. 51, 659 (1979) for a detailed review of these ideas.


optics - Is it possible to blur an image in such way that a person with sight problems could see it sharp?


If someone has short or long sight, is it possible to tune image on a computer monitor in such way, that a person could see it sharp as if they were wearing glasses? If not, will 3d monitor make it possible?



Answer




Let's take a simple original picture to look at - just two nearby dots on a white background. If you have bad vision, the dots look blurred.


The way good vision works is to ensure that all the light hitting any particular small area of your retina comes from the same direction in front of you. Conversely, all the light coming from one direction hits one specific spot on your retina.


When you have bad vision, the light from a locus of nearby directions all hits on the same part of your retina, and the light from a particular direction is smeared out over an area on your retina. Hence, blurred vision is an averaging effect. When you look at the dots, you'll see them smear out into each other.


You might try to compensate for this by making a "counter-blurred" image where the source dots are smaller, but if the original dots are close enough that light from the center of one dot is spilling over to overlap light from the center of the second dot, making the dots smaller won't fix that problem. Hence, the dots will always appear blurred. You can't create the impression that the original has for someone with good vision.


A photograph is really just a bunch of nearby dots, and so the same problem applies.


I don't know about the 3D monitor, though. I suppose if it can control the direction of light coming off it, it could be modified to focus the light some and create a sharp image for someone with blurred vision.


Thursday, August 29, 2019

gravitational waves - Is the sensitivity of the LIGO sensitive enough to measure the "expansion of the universe"? What specifically is the numerical ratio of effects?


It would seem that LIGO measures wibbles in the metric (not manifold) of spacetime:


How is it that distortions in space can be measured as distances?


It would seem that the expansion of the universe is an expansion of the metric of spacetime:


If space is "expanding" in itself - why then is there redshift?


Imagine for a moment that LIGO is not an interferometer. (So, it just plain times speed changes, rather than using phase shift of orthogonal directions.)



If the ends of one of the arms was indeed receding away from each other, at a speed consistent with the expansion of the universe, is the sensitivity of real-world LIGO of the needed sensitivity of a machine which could measure that shift?


On other words: ideally I'd like to know these two meters:


(A) The left arm of LIGO is about 4km. It was stretched/shrunk (a few times) for roughly .01 seconds by the gravitational wave. How many meters was it stretched/shrunk in .01 seconds?


(B) Assuming the same abstract LIGO arm was affectable and affected by the expansion of the Universe. How many meters is it stretch/shrunk every .01 seconds?




Note - of course, an interferometer is an ingenious device to measure extremely small changes in speed - assuming the changes are orthogonal. Of course, an interferometer, per se, can't at all measure the expansion of the universe since that is uniform in both directions. The sense of my question is, can something that measures distance changes > as accurately as < the LIGO does, measure the expansion of the universe? How big or small is the ongoing expansion of the universe compared to the wibble from the black hole system in question?



Answer



Well, CuriousOne gives the most direct answer, that the universe is not expanding on scales which the gravitational attraction between objects dominates, like on the Earth (indeed, the entire Milky Way). But, let's pretend that we take LIGO, stick it out in space (even away from our local group, to be sure it's in an isolated region), and ask it to measure the expansion of the universe. I'll call this XLIGO to make sure we don't confuse it with reality.


Answer A: I guess you're saying 0.01 s because the frequency of that particular observation was around 100 Hz, but in any case, the maximum strain detected by LIGO for this specific event was around $1\times 10^{-21}$ (see here, look at Figure 2). So, over 4 km that's simply $1\times 10^{-18}~m$.


Answer B: I'm going to do this using the Hubble constant, with tells us how fast two objects are moving away from each other depending on their distance. The approximate value of the Hubble Constant is 75 km/s/Mpc. Read that "For every megaparsec difference, the objects receed at a speed of 75 km/s."



So, over 4 km, $$\rm (75~km/s/Mpc)(1~Mpc/3 \times10^{19}~km)(4~km)\approx 100\times10^{-19}~km/s$$


Or an expansion speed along a single arm of XLIGO of $1\times 10^{-14}~m/s$. Over a time period of 0.01 s that means an absolute shift of $$ (1\times 10^{-14}~m/s)(0.01~s)=1\times 10^{-16}~m$$ So the sensitivity of XLIGO, modeled by the only positive observation, is around 100 times larger then the rough expansion of the universe.


But there is a bigger problem here, and that is that the expansion of the universe is isotropic, identical in every direction. So in order to see the effect, we would have to try to measure the aniotropies (actually, we would need to see the higher-order pole, specifically the quadrapole term). I found a paper (here) discussing that $\Delta H/H\approx 3\%$ is not excluded, so that actually brings us down to the same level of approximate scales.


standard model - chiral symmetry condensate and 2SC, CFL breaking C, P and T symmetry?



Because we know that chiral symmetry condensate causes the chiral symmetry breaking, and it produces Goldstone modes of pseudo-scalars, so I believe that chiral symmetry breaking also breaks the T symmetry.



question: Do we have similar statements for C, P and T symmetry? For other QCD phases? Say for chiral symmetry breaking phases, 2SC, CFL breaking of color superconductor?



Namely,




  1. Does chiral symmetry breaking break C, P and T symmetry?





  2. Does 2SC two quark color/flavor break C, P and T symmetry?




  3. Does CFL three color/flavor locking break C, P and T symmetry?




By playing around, I know some partial answers, like chiral symmetry breaking breaks T, but the CFL preserves all C,P,T. But I like to hear from the experts, just to confirm the correct answer.




pattern - Fill the grid - 2.0



Can you fill this grid ? (Version 2.0 )




Grid


Text version



714523
191303
165088
213443
573211
91?12?


Answer



The missing numbers are



6 and 7.



To find them, observe that



on row $k$, the two digit number in the middle of the row is obtained by adding the two digit numbers to the left and right, adding $4\times(-1)^k$ to their sum and then dividing by 2.



probability - Monty Hall all over again?


Often, when people ask the Monty Hall problem, they omit some important details, such that the problem becomes ambiguous, or they add more details, such that the problem changes drastically. For example, what is the answer for the following problem?


You are participating in a show. There are three doors in front of you. There’s a car behind one of them; there’s a goat behind another; and there’s a sheep behind the third. To the best of your knowledge, every assignment of the three prizes to the three doors is equally probable. You do want the car, badly, and you don’t care about either of the animals at all. You can choose one door, and whatever is behind that door will be awarded to you.


So, without loss of generality, you choose the leftmost door.


After that, to your complete surprise, the host opens the middle door, and you observe a sheep behind it.


The host then offers you a chance to change your choice. Based solely on the information you have, without trying to guess the motives of the host, is it beneficial for you to change your choice?


EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are.



Answer




Here's the formal expression of what we seek. Let $C$ be the event "chose the door with the car" and $R$ be the event "revealed the door with the car". Now, assuming that the door opened is random (but not the one the player chose), we have


$$ Pr(C) = \frac13\\ Pr(R|C) = 0\\ Pr(R|\bar C) = \frac12\\ Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C)Pr(\bar C)} = \frac{1/3}{2/3}=1/2 $$ And therefore the probability of winning by switching or by staying are equal.


This differs from the Monty Hall problem through the value of $Pr(R|\bar C)$, which is zero for the Monty Hall problem. In that case, we have $$ Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C)Pr(\bar C)} = \frac{1/3}{1}=1/3 $$ and thus, switching doubles the chance of winning. This is not true if the host can open the door with the car, where the probability of winning is equal in either direction.


It's tempting to think that the situation is the same (the door that is opened is one without the car) between this situation and the Monty Hall one, but the balance of probabilities leading to the situation is different.


It's a mistake involving the confusion between counting the number of arrangements and determining the actual chance of each event. In the Monty Hall problem, if the player chooses a wrong door, which has a probability of 2/3, the host will necessarily open the other wrong door, and thus the chance of winning by switching is 1. In this problem, if the player chooses a wrong door, there's a 50% chance that the car will be opened. This halves the number of times that the player will choose the wrong door and have an opportunity to switch.


As a result, while the probability rules around the actual situation are the same, the underlying probabilities are shifted - there's only a 1/3 chance that the player will choose a wrong door and the host will open the other wrong door, and a 1/3 chance that the player will choose the right door. In the Monty Hall problem, the probability of the former is 2/3. So even though the situation itself looks the same, the chance of being in that situation while having chosen the wrong door is lower, with the chance of each of them being 1/2 (given that the situation arose), rather than 1/3 and 2/3.


Note that the probability that the player chose the car given that the car wasn't behind the door the host revealed depends on only two parameters. If we express our equation for $Pr(C|\bar R)$ slightly differently, we have $$ Pr(C|\bar R) = \frac{Pr(C)}{Pr(C)+\frac{1-Pr(R|C)}{1-Pr(R|\bar C)}(1-Pr(C))} = \frac{A}{A+B(1-A)} $$ where $A=Pr(C)$ and $B=\frac{1-Pr(R|C)}{1-Pr(R|\bar C)}$ - as such, these are the only two values that are relevant. In the Monty Hall case, $A=1/3$ and $B=2$. In our case, $A=1/3$ and $B=1$. If the host is incredibly malicious, almost certainly set to reveal the car if the player doesn't choose it, then $A=1/3$ and $B\approx 0$, meaning that the player is practically guaranteed to win by staying.


If we were to allow the host to open other doors, then we have to handle those distinctly. As such, it is better to consider $\hat R$, which is the event "the host opened one of the other two doors, and it contained an animal" (or specifically "the sheep" - it's equivalent). For the three-door-only case, $\hat R \equiv \bar R$. So the general expression would be $$ Pr((C|\hat R) = \frac{Pr(C)}{Pr(C)+\frac{Pr(\hat R|C)}{Pr(\hat R|\bar C)}(1-Pr(C))} = \frac{A}{A+B(1-A)} $$ where $A=Pr(C)$ again and $B=\frac{Pr(\hat R|C)}{Pr(\hat R|\bar C)}$. In this expression, it is easily seen that adding extra doors that aren't among the three we care about doesn't change the result. With an extra door, we have $Pr(\hat R|C) = 2/3$ and $Pr(\hat R|\bar C) = 1/3$, and we again have $B=2$.


Note - the assumption was described above as "assuming that the door opened is random (but not the one the player chose)". This is for the purposes of the calculations. The true assumption is actually that the host could have any possible motive with equal probability, and looks at the expected probability. It just happens that the two assumptions are equivalent.


Update: As I noted, taking the host's motive as "random" is the same as randomly choosing the host's motive. It's worth explaining properly how this works, and explain an easy pitfall in determining it.



Suppose that $X$ is a uniformly-distributed random variable (between 0 and 1) representing the probability that the host reveals the car given that the player didn't choose the car's door. That is, $$ Pr(R|\bar C\land X=x)=x $$ Because $C$ and the value of $X$ are independent of each other, we can deal with these separately - that is, we can treat it as $(R|\bar C)|X=x$ or $(R|X=x)|\bar C$, and nothing will be wrong.


And here's where the pitfall comes in. It can be tempting to calculate $$ Pr(C|\bar R\land X=x) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+Pr(\bar R|\bar C\land X=x)Pr(\bar C)} = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+(1-x)Pr(\bar C)} $$ and then integrate from there. However, there's a problem - $\bar R$ and $X$ are not independent, and thus we cannot treat $C|\bar R\land X=x$ as $(C|\bar R)|X=x$, which would be necessary to determine $C|\bar R$ by integration.


Instead, we need to calculate it through the full process. That is, we must get $$ Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R)} $$ and to calculate the denominator, we must integrate. That is, $$ Pr(\bar R) = \int_0^1 Pr(\bar R|X=x) dx $$ where $$ Pr(\bar R|X=x) = Pr(\bar R|C)Pr(C) + Pr(\bar R|\bar C\land X=x)Pr(\bar C)\\ = Pr(\bar R|C)Pr(C) + (1-x)(1-Pr(C)) $$ and this quickly gives us $$ Pr(\bar R) = Pr(\bar R|C)Pr(C) + \frac12(1-Pr(C)) $$ at which point the equivalence with assuming the host chooses at random becomes clear.


geophysics - Does the Earth's core have tides?


Does the moon produce a measurable tidal-effect on the Earth's (liquid) core? If so, how strong is it? Would it play a factor in other geological effects like earthquakes, volcanoes, etc?



Answer



The gravitational fields of the Sun and the Moon do produce measurable effects on the shape of the Earth. The tidal distortion of the solid Earth (and the liquid outer core) is often referred to as the Earth Tide to distinguish it from ocean tides.


The gravitational effects of the Earth tide do cause small strains in the Earth which might influence geophysical processes such as volcanic eruptions and earthquakes - however, it is hard to prove specific cause and effect instances.


This paper: Earth Tides provides a detailed explanation of Earth tide physics and this figure provides an example of measurements.


Agnew, D. C. (2007). Earth Tides, pp. 163-195 in Treatise on Geophysics: Geodesy, T. A. Herring, ed., Elsevier, New York.



mathematics - Picking up stones


Here is puzzle that was lifted from an old essay of mine, that was adapted from a puzzle in an old puzzle book.



The great shah has invited a young boy into his palace as a student. While he is not busy managing his kingdom, he is a scholar of mathematics and invites the brightest learners from all over the country to his palace to study.


He sits the boy down at a small table, on which is a bag of stones. He then gently pours out the stones and lays them out in rows on the table.


"Here are 42 stones, my son," he tells the boy. "We will play a game with them. First, I will take one, three, or five stones from the pile. Then you will do the same, choosing between one, three, or five stones. We will continue until all the stones are gone. And neither of us may take more stones than there are remaining in the pile. The winner of the game will be the person who takes the last stone."


The boy nods silently at the shah, and turns back to looking at the pile of stones.


The shah starts by taking three stones. "We shall play seven games. If you win exactly six out of seven games against me, my student, you will have a place in my study."


The boy nods again, and looks hard at the stones. Suddenly, a look on his face changes as he realizes that he has a winning strategy.





  1. What is the boy's strategy to win the first six games?


The next part of the story was written in the essay after the first question had already been answered; try to answer the first question before looking at the problem statement of the second.



The boy easily wins all six of the first six games following his strategy. As they prepare for the seventh game, the shah tells the boy, "I can already tell you, boy, that you will win all seven of these games here, not six."


Surely enough, the shah starts by taking one stone, and the boy plays wildly, trying his hardest not to follow his winning strategy from earlier. But no matter how he plays, he cannot find a way to lose this last game and earn his place in the shah's study. As he removes the last stone from the pile, he hangs his head in shame.


The shah looks pitifully at the boy. "Poor child," he says. "If you can tell me why you could not lose this seventh game, you will still have a place in my study."


The boy thinks long and hard again, and suddenly his face lights up again as he turns to the shah and explains why he could not lose the game.





  1. Could the boy have lost the seventh game? Why or why not?



Answer



Regardless of what one player takes, the other can adjust accordingly so that 6 in total are removed. (1+5, 3+3, 5+1). So as long as the number of remaining stones before the shah's turn is dividable by 6, the student can force a win.


The 'trick' lies in the first turn. Given 42 is a multiple of 6, no matter what opening move the shah makes, he can make the numbers 'work' to a multiple of 6 remaining after his turn.


Now, why can he not lose on the 7th game?


Simple, the game starts with an even number of 42, and the shah always gets to play after an even number of turns. Given that two odd numbers have been subtracted from the pile, the remaining number has to be even. So when there are 0 stones left in the pile, it's the shah's turn, and it was the boy who removed the last stone.


Cloud chamber video showing large particles - What are they?


Watching a video of a cloud chamber on wikipedia (http://en.wikipedia.org/wiki/File:Cloud_chamber.ogg), I cannot help noticing the large collisions that take place at 00:12 and 00:24.


What are they? Alpha particles? They are huge compared to the small ones (i guess they are electrons)



Answer



I tutor a cloud chamber workshop at CERN weekly, so I have some regular experience with recognising cloud chamber tracks.


There are only four particles (plus their antiparticles) we can observe in a cloud chamber:



  • $\alpha$ (He nucleus),

  • p (proton),


  • $\mu$ (muon),

  • $\beta$ (electron).


All other particles are either uncharged (and hence don't ionise the cloud) or decay too fast to survive at least a few millimeters (the minimal amount to be able to see them in a cloud chamber).


The width of a track depends on the ionising power of a particle, which in turn directly depends on its charge and its mass. From the above four particles, the $\alpha$ particle has charge +2 while the others have charge $\pm 1$. Their masses are roughly $m_\alpha = 4$GeV, $m_p=1$GeV, $m_\mu=0.1$GeV, and $m_e=0.0005$GeV. From this we estimate an $\alpha$ particle to be 8 times as ionising as a proton (with the effect of the double charge included), which in turn is 10 times as ionising as a muon, which finally is 200 times as ionising as an electron. This is of course merely a rule of thumb while the complete calculation is a bit more intricate. But the main result is correct:


The thick tracks are $\alpha$ particles or protons, while the thin tracks are muons or electrons.


But how do we distinguish between an $\alpha$ particle and a proton? Well, as you maybe remember from your high school nuclear physics class, $\alpha$ radiation can be stopped with a thin sheet of paper. In other words, there is no way that an external $\alpha$ particle could penetrate the casing of a cloud chamber. Hence, $\alpha$ particles that are observed in a cloud chamber are created inside. This can be due to an $\alpha$ source which has been positioned inside, but can also be due to natural background radiation: air contains a certain fraction of Radon, which is a natural $\alpha$ source. In our cloud chambers at CERN (which have no internal $\alpha$ source, so they only show the effect of Radon decay) this accounts to more or less 3 $\alpha$ particle tracks per minute (but the Radon concentration varies geographically, with the weather, and has some other factors, so it could be much more or less). Protons on the other hand cannot be created from radioactive decay, so they come from cosmic radiation.


Why does this matter? Because it is easy to distinguish radioactivity from cosmic rays based on their energy: while particles generated from radioactive decay have kinetic energies in the range of a few MeV's, cosmic rays have kinetic energies in the range of TeV - EeV (yup that's Exa-electronvolt, or $10^{18}$eV). This means that $\alpha$ particles, due to their huge mass relative to their kinetic energy, have very low penetration potential (that's why they are stopped by a sheet of paper) and can travel only a few centimetres in air. Protons are of much higher energy and can travel through the cloud unimpededly, and will hence form long straight thick tracks. Even though the length of a track also depends on the angle at which the particle crosses the cloud (implying that short thick tracks could be protons at a steep angle), it is known that very few protons are generated in a cosmic air shower (in one year of weekly tutoring workshops of 3 hours each, I have only seen 6 protons up to today). Furthermore, a last discriminating parameter is the particle's speed: cosmic particles move at speeds close to the speed of light, while (radioactive) $\alpha$ particles are so low in energy that they have speeds in the order of a few centimeters per second.


To conclude, the tracks from the video you refer to are thick, relatively short, and slow, so they are undoubtedly $\alpha$ particles.


Ps: the same reasoning can be applied to distinguish between muons and electrons. Muons can only be cosmic, while most observed electrons will be radioactive. Even though comparing the speeds won't be easy (even radioactive electrons move too fast for the eye to be called slow), we can make the difference based on the form of the track: long and straight implies high energy and hence a muon, curved and full of cusps implies lots of interaction so low energy, and hence an electron.



Wednesday, August 28, 2019

general relativity - Communication near a black hole


One ship is in a stable orbit outside of the event horizon of a very large black hole so that there is significant time dilation.


Another ship is in its own orbit around the same black hole, but at a much greater distance, so that time dilation is minimal.


If someone on the ship much farther away turns a very bright light on and off again every 10 seconds, what will the observer on the ship near the black hole see?




Why is gravity such a unique force?


My knowledge on this particular field of physics is very sketchy, but I frequently hear of a theoretical "graviton", the quantum of the gravitational field. So I guess most physicists' assumption is that gravity can be described by a QFT?


But I find this weird, because gravity seems so incredibly different from the other forces (yes, I know "weirdness" isn't any sort of scientific deduction principle).


For relative strengths:



  • Strong force: $10^{38}$

  • Electromagnetic force: $10^{36}$


  • Weak force: $10^{25}$

  • Gravity: $1$


Not only does gravity have a vastly weaker magnitude, it also has a very strange interaction with everything else. Consider the Standard Model interactions:


Standard model interactions


No particle (or field) interacts directly with all other fields. Heck, gluons only barely interact with the rest of them. So why is it then that anything that has energy (e.g. everything that exists) also has a gravitational interaction? Gravity seems unique in that all particles interact through it.


Then there's the whole issue of affecting spacetime. As far as I'm aware, properties such as charge, spin, color, etc. don't affect spacetime (only the energy related to these properties).



Answer



The short answer for why gravity is unique is that it is the theory of a massless, spin-2 field. To contrast with the other forces, the strong, weak and electromagnetic forces are all theories of spin-1 particles.


Although it's not immediately obvious, this property alone basically fixes all of the essential features of gravity. To begin with, the fact that gravity is mediated by massless particles means that it can give rise to long-range forces. "Long-range" here means that gravitational potential between distant masses goes like $\dfrac{1}{r}$, whereas local interactions most commonly fall of exponentially, something like $\dfrac{e^{-mr}}{r}$, where $m$ is the mass of the force particle (this is known as a Yukawa potential).



Another important feature of massless particles is they must have a gauge symmetry associated with them. Gauge symmetry is important because it leads to conserved quantities. In the case of electromagnetism (a theory of a massless, spin-1 particle), there is also gauge symmetry, and it is known that the conservation of electric charge is a consequence of this symmetry.


For gravity, the gauge symmetry puts even stronger constraints on the way gravity interacts: not only does it lead to a conserved "charge" (the stress energy tensor of matter), it actually requires that the gravitational field couple in the same way to all types of matter. So, as you correctly noted, gravity is very unique in that it is required to couple to all other particles and fields. Not only that, but gravity also doesn't care about the electric charge, color charge, spin, or any other property of the things it is interacting with, it only couples to the stress-energy of the field. For people who are familiar with general relativity, this universal coupling of gravity to the stress energy of matter, independent of internal structure, is known as the equivalence principle. A more technical discussion of the fact that massless, spin-2 implies the equivalence principle (which was first derived by Weinberg) is given in the lecture notes found at the bottom of this page.


Another consequence of this universal coupling of gravity is that there can only by one type of graviton, i.e. only one massless, spin-2 field that interacts with matter. This is much different from the spin-1 particles, for example the strong force has eight different types of gluons. So since gravity is described by massless, spin-2 particles, it is necessarily the unique force containing massless spin-2 particles.


In regards to the geometric viewpoint of gravity, i.e. how gravity can be seen as causing curvature in spacetime, that property also follows directly (although not obviously) from the massless spin-2 nature of the gravitons. One of the standard books treating this idea is Feynman's Lectures on Gravitation (I think at least the first couple of chapters are available on google books). The viewpoint that Feynman takes is that gravity must couple universally to the stress tensor of all matter, including the stress tensor of the gravitons themselves. This sort of self-interaction basically gives rise to the nonlinearities that one finds in general relativity. Also, the gauge symmetry that was mentioned before gets modified by the self-interactions, and turns into diffeomorphism symmetry found in general relativity (also known as general covariance).


All of this analysis comes from assuming that there is a quantum field theoretic description of gravity. It may be concerning that people generally say we don't have a consistent quantum theory of gravity. This is true, however, it can more accurately be stated that we don't have an ultraviolet complete theory of quantum gravity (string theory, loop quantum gravity, asymptotically safe gravity are all proposed candidates for a full theory, among many others). That means that we don't believe that this theory of massless spin-2 particles is valid at very high energies. The cutoff where we think it should break down is around the Planck mass, $M_p \approx 10^{19}$ GeV. These energies would be reached, for example, at the singularity of a black hole, or near the big bang. But in most regions of the universe where such high energies are not present, perturbative quantum general relativity, described in terms of gravitons, is perfectly valid as a low energy effective field theory.


Finally, you noted that the extremely weak coupling of gravity compared to the other forces also sets it apart. This is known as the hierarchy problem, and to the best of my knowledge it is a major open problem in physics.


Regardless, I hope this shows that even hierarchy problem aside, gravity plays a very special role among the forces of nature.


solid state physics - Relation between density and refractive index of medium


Is there any relation between Refractive index and density of a material? It is not found to be proportional in my experimental results. Is there any equation to relate these parameters?




material science - If aerographite is lighter than air, why doesn't it float?


Air is 6 times denser than aerographite but looking at pictures or videos presenting the material, I see it resting on tables rather than raising to the ceiling.
Also, since the material is made of carbon nanotubes, I assume there are empty spaces between those tubes. Why are those spaces not occupied by air?



Answer



Your two questions are connected. There is a huge amount of empty space in aerographene (and other aerogels). However this space is filled with air, and precisely because it is filled with air it doesn't float.


This is because the density reported is the density the material would have if the air was sucked out (i.e. in vacuum), and it is so low because the material is extremely porous. But in the atmosphere, the air fills the immense empty space. The effective volume of air displaced by aerographite now takes up only the volume of the constituent nanotubes of aerographite, which is extremely small. The tiny weight of this displaced air presents the buoyant force, which is not sufficient to counter the weight of the structure. Effectively because it is so porous the aerographite's density increases when not in vacuum.


On the other hand, given that graphene is known to be non-permeable for atoms, if you sucked the air out of aerographene and encased it in graphene, and if outside air didn't squish the whole thing so that it's density surpassed that of air, than the resulting balloon might float.



energy - What is Work? What does the quantity suggest intuitively?




The mathematical formula for work says that work is force into displacement, but what is the philosophy behind it? I mean what does the quantity suggest?




Tuesday, August 27, 2019

riddle - The Hangman Game Backwards Again


It's been a while since one of these puzzles (created by Alex, see here for the first) has shown up, and I thought some might enjoy another one.


fill in an "n" - You don't need to throw me out into the bin.

fill in an "f" - You wish you could see my insides.
fill in an "m" - I may seem very familiar.
fill in an "v" - Your soul now belongs to me.


As always, your answer can be in form of:



The question was: _ p p l e



followed by explanation.



Answer





The question was sa_e



fill in an "n" - You don't need to throw me out into the bin.



sane - don't need to go to the looney bin



fill in an "f" - You wish you could see my insides.



safe - I would love to see the insides of many fancy safes




fill in an "m" - I may seem very familiar.



same - things that are the same are quite familiar.



fill in an "v" - Your soul now belongs to me.



save - If I save your soul, it may belong to me at that point.



geometry - Klotski Puzzle 1


Klotski is a sliding-blocks puzzle game very similar in nature to rush-hour/unblock me. In a given puzzle, a certain number of blocks labeled Z, A, B, C, .... are given. The goal is to move the Z block into the goal area in the fewest number of possible moves.


Each move consists of choosing a certain letter, and then moving the block corresponding to that letter in one of the four main directions (up, down, left, right) as many times as you like. At every point in a move, the block must be moving into adjacent empty spaces only.


In some Klotski puzzles, there are several parts to some blocks which are not connected. Imagine that they are joined by an invisible rigid bar; that is, if one A moves up, all A's must move up as well. For a move to be valid, all such parts of a block must be able to simultaneously make the move.


As an example, consider the following puzzle: Example



The Z pieces will always be marked in red, while the goal area will always be marked in gold. One solution to this puzzle would be BrZruBlAdCdZur(using d=down, u=up, l=left, r=right):


Br (moving both B blocks to the right once)


Zru (moving the Z block right once and up once)


Bl (moving the B block left once)


Ad (moving the A block down once)


Cd (moving the C block down once)


Zur (moving the Z block up once, and right once into the goal)


The puzzles I post here definitely have solutions with the given # of moves, but that won't necessarily be optimal.


Anyway, here's a puzzle! Note that the invisible-bar analogy still holds over multiple grids. enter image description here



Answer




15 moves:



Gr, Fr, Cu, Zu, Br, Au, Zl,Dl, Eu, Brd, Ed, Fd, Cr, Ar, Zu



Monday, August 26, 2019

Using Lagrangian mechanics instead of Newtonian mechanics



When studying advanced classical mechanics, we all study about Lagrangians and the Euler-Lagrange equations and their importance. Of course, the Lagrangian is calculated based on the potential and kinetic energy. But usually, this is calculated using the position function and some constants. For example, potential energy can be defined as:


$$U(x) = - \int F(x)dx$$


and


$$F = ma \, .$$



This means that potential energy can be defined with mass and acceleration, which is the second derivative of the position function. Of course kinetic energy is defined as


$$(1/2) mv^2 \, .$$


Again, kinetic energy is defined with mass and velocity, the derivative of the position function. As you can see, it is kind of repetition to use the position function to derive the Lagrangian to figure out the dynamics of a system because that is what the position function tells you. So, what is the point of the Lagrangian? I heard from @Sofia that it may be used when only the potential and kinetic energy functions are known. But in what practical experiments it that done? It id much easier to find the position function. So what uses of the Lagrangian are there that can't be done with the position function?




homework and exercises - How does symmetry allow a rapid determination of the current between $A$ and $B$?


The following was originally given to me as a homework question at my physics 2 course:



Consider the following circuit



enter image description here


The difference of potentials between the point $V_{1}$and the point $V_{2}$ is $4.4$ volts, the resistance of all the resistors is the same $R=1\Omega$.


Find the current between point $A$ and point $B$.



The answer given is simply $0$ and the argument was just the pair of words ``using symmetry''.


I don't really understand the answer:


First, it is not completely symmetric: There is a difference of potentials so the potential at the point $V_{1}$ is not the same as the potential at the point $V_{2}$.


Secondly: How can I see that the symmetric structure will give me that the current between $A$ and $B$ is $0$ ?


Also, I would appreciate to see a calculation of this current to get a better feel for whats going on, I know the rule $V=IR$ (which seems the most useful here, but I also know other rules that can be used), but I don't understand how to use this rule to find the current.



Answer




This is your circuit:


enter image description here


The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, $I_1=I_2$ , So the potential difference across yellow resistors is the same. It means that the potential of point $\mathbf{A}$ is equal to potential of point $\mathbf{B}$ :


$$I_1=I_2\to V_A=4.4-I_1R \text{ , }V_B=4.4-I_2R\to V_A=V_B$$


So there isn't any potential difference across the blue resistor, and the current through it is 0, and it can be omitted from the circuit without any change in the behavior of the circuit.


quantum field theory - Change of variables in path integral measure


In fermion's path integral we have a measure that you can write, in terms of the Grassmann variables $\psi, \bar{\psi}$ as


$$ D\bar{\psi}D\psi, \quad \psi(x) = \sum_n a_n\phi_n(x), \quad \bar{\psi}(x) = \sum_n \bar{a}_n \bar{\phi}_n(x) $$


Where $a_n, \bar{a}_n$ are Grassmann variables and $\phi_n(x)$ a set of orthonormal functions such that


$$ \int d^3x\ \phi^\dagger_n(x)\phi_m(y) = \delta_{nm} $$


Now if you perform a change of variables in, for instance, axial group $U(1)_A$ with an small parameter $\alpha(x)$, this renders


$$ a'_m = \sum_n(\delta_{mn} + i\int d^3x\ \alpha(x)\phi^\dagger_m(x)\gamma^5\phi_n(x))a_n = \sum_n(1 + C)_{mn}a_n $$ $$ \bar{a}'_m = \sum_n(1 + C)_{mn}\bar{a}_n $$ $$ C_{mn} = i\int d^3x\ \alpha(x)\phi^\dagger_m(x)\gamma^5\phi_n(x)), \quad 1\ {\rm is\ the\ identity} $$


Now, following Peskin (chapter 19.2, Eq. (19.69)), the path integral meassure should change as


$$ D\bar{\psi}'D\psi' = D\bar{\psi}D\psi·(det[1 + C])^{-2}. $$


I don't understand where the -2 power for the Jacobian ($det[1 + C]$) came up since if we were talking about a usual integral with usual variables we would end up with +2 power.



What am I missing?




EDIT


Thinking about the problem I found a possible explanation. Grassmann variables, let's call it $\eta$, are forced to satisfy


$$ \int d\eta\ \eta = 1 $$


Therefore, a change of variables such that $\eta$ changes to $$\eta' = A\eta\tag{A}$$ and $\eta'$ is still a Grassmann variable should fulfill


$$ \int d\eta'\ \eta' = 1 \tag{B} $$


But if we follow the change of variables given by Eq. (A) and we want Eq. (B) satisfied,


$$ \int d\eta\ \eta = A^{-2}\int d\eta'\ \eta' = 1 \Leftrightarrow \int d\eta'\ \eta' = A^2 $$


Then, we are violating Eq. (B) and $\eta'$ isn't a Grassmann variable. So, if $\eta, \eta'$ are Grassmann variables then the Jacobian ($j = A^{-1}$) among themselves must be introduced in the measure with the opposite power sign, so:



$$ \int d\eta'\ \eta' = \int j^{-1}·d\eta\ j·\eta \equiv 1 $$


Fine or something to complain about?



Answer





  1. OP is right. Grassmann-odd integrals are the same thing as Grassmann-odd derivatives $\int \!d\theta^j=\frac{\partial}{\partial\theta^j}$, cf. e.g. this Phys.SE post and above comments by user knzhou.




  2. For this reason, if $\theta^{\prime k} = M^k{}_j ~\theta^j$ is a linear change of Grassmann-odd variables [where the matrix elements $M^k{}_j$ are Grassmann-even], then $$\begin{align}\int \!d\theta^1 \ldots \int \!d\theta^n &~=~\frac{\partial}{\partial\theta^1} \ldots \frac{\partial}{\partial\theta^n} ~=~\sum_{k_1=1}^n\frac{\partial\theta^{\prime k_1}}{\partial\theta^1} \frac{\partial}{\partial\theta^{\prime k_1}} \ldots \sum_{k_n=1}^n\frac{\partial\theta^{\prime k_n}}{\partial\theta^1} \frac{\partial}{\partial\theta^{\prime k_n}} \cr &~=~\sum_{\pi\in S_n} M^{\pi(1)}{}_1 \frac{\partial}{\partial\theta^{\prime \pi(1)}} \ldots M^{\pi(n)}{}_n \frac{\partial}{\partial\theta^{\prime \pi(n)}}\cr &~=~\sum_{\pi\in S_n} M^{\pi(1)}{}_1 \ldots M^{\pi(n)}{}_n (-1)^{{\rm sgn}(\pi)} \frac{\partial}{\partial\theta^{\prime 1}} \ldots \frac{\partial}{\partial\theta^{\prime n}}\cr &~=~\det M \frac{\partial}{\partial\theta^{\prime 1}} \ldots \frac{\partial}{\partial\theta^{\prime n}}~=~\det M \int \!d\theta^{\prime 1} \ldots \int \!d\theta^{\prime n}. \end{align}$$





  3. More generally, change of super-integration variables transform with the superdeterminant/Berezinian.




mathematics - How Many Undefined Magic Constants are there?


Magic Square: An nxn square where every horizontal,vertical, and diagonal line all add up to the same number.


Magic Constant: The number which every line with a magic square adds up to


So most magic squares, have a magic constant. But some do not. A fairly trivial answer is 0x0. Where no numbers are present, therefore it cannot sum up into any number.



A less than trivial answer but not complicated answer is 2x2. No possible configuration of 1,2,3,4 within a 2x2 block can have all lines add up to the same number. If it were possible it would theoretically be 5


So besides 0, and 2 are there any other positive dimensions for a magic square that do not have a magic constant? Technically making all possible configurations invalid magic squares?


Mathematical Proof is Appreciated!



Answer



From Wikipedia (my emphasis):



Normal magic squares of all sizes except 2 × 2 (that is, where n = 2) can be constructed. The 1 × 1 magic square, with only one cell containing the number 1, is trivial.



For a mathematical proof of this, see the Nrich article here. The idea is to turn an n x n square into a diamond (an n x n square standing on one corner), fill in that diamond with the numbers 1 to n^2 in the natural way, and then fold everything into the original square. Some pictures to demonstrate (also taken from the Nrich article):


3x3 case



5x5 case


mathematics - How can 12 teams rotate through 6 games without overlaps?



Given the following:



  • Six Number Teams (1 - 6)

  • Six Letter Teams (A - F)

  • Six Games (Basketball, Football, Baseball, Volleyball, Hockey, Rugby)

  • Six Time Slots (1pm - 6pm)


Set up a game schedule that follows these rules:



  • Each team must play each game once.


  • Each letter team must play against each number team exactly once.

  • Every game must be played once (by one letter team and one number team) during each time slot


Please provide either a solution or a mathematical prove of why a solution is impossible.



Answer



In order to do this puzzle, you'd need to create Mutually Orthogonal Latin Squares of order 6.


For example, say that instead, you had 6 teams (1-3 and A-C), 3 sports (baseball, football, hockey), 3 timeslots. Then, you could make the following schedule:


    A   B   C
1 b@1 f@2 h@3
2 h@2 b@3 f@1

3 f@3 h@1 b@2

So in this example, team A plays baseball against team 1 at 1:00pm.


This uses 2 Mutually Orthogonal Latin Squares of order 3.


1 2 3    b f h
2 3 1 h b f
3 1 2 f h b

This allows us to conform to the following rules:




  • Every lettered team plays every numbered team exactly once - simply by design of the table

  • Every team plays every sport exactly once

  • Every team plays in every timeslot exactly once


However, it is a known impossibility to create two MOLS of order 6, so the original question is not possible.


Sunday, August 25, 2019

gravity - Has the gravitational interaction of antimatter ever been examined experimentally?


I know that the gravitational interaction of antimatter is expected to be the same as normal matter.


But my question is, has it ever been experimentally validated?


I think it would not be a trivial experiment, because electromagnetic effects have to be eliminated, so neutral particles would be needed. Maybe diamagnetically trapped antihidrogen atoms could be examined as to which direction they fall?



Answer



The only experiment I know of was done by the ALPHA team at CERN. The results are published in this paper. The error bounds are huge - all the team were able to say is that the upper limit for the gravitational mass of antihydrogen is no greater than 75 times its inertial mass! However I believe an updated version of the experiment, ALPHA2, is in progress and will hopefully be able to do a bit better.


Other planned experiments are AEGIS and GBAR, both also at CERN. However neither have made any measurements yet.


This may seem like slow progress, but antihydrogen is extraordinarily difficult stuff to handle as contact with any normal matter will annihilate the antihydrogen.


newtonian mechanics - Internal forces in an isolated system


I did a weird activity and now I am trying to figure out the physics behind it:


While sitting on a chair I kept my feet above the ground and tried to move the chair forward. I was able to. Initially, I thought that perhaps the center of mass won't be changing as I was pushed backwards. But I moved the chair for a while and saw that it had moved a considerable distance and I was still sitting on the chair. Hence the center of mass did change for sure. According to Newton's second law, there must be an external force acting on the system(me and the chair). But I am not able to figure out what force is it. I have a feeling that it is somewhere related to the fact that the system is not completely isolated but I am not sure how.


Note:
I am not in contact with the ground or any other object close to me. I have lifted my feet and have my hands on my lap.



Answer



The movement you are performing is similar to hopping or jumping - see diagram below. It is much more difficult when you sit on the chair than when you stand, because you do not have as much flexibility, but it is essentially the same kind of movement. You could do even better by standing on the chair and tying it to your feet, then jumping on the chair as you would on the ground.



enter image description here


Using your arms (instead of your legs) as springs, you squat in the chair (b) then suddenly push yourself up and forward, dragging the chair with you when your arms are extended. It is slightly more effective if you hold yourself upright first (a) then swoop down before pushing all in the same movement. Swinging your legs also helps - when jumping you swing your arms and push with your legs; in the chair you swing your legs and push with your arms.


You could think of the chair as shoes - you push the ground through them, and because they are tied to your feet they leave the ground with you when you jump.


There are external forces acting here - you and the chair are not an isolated system. Contact forces with the ground - friction and normal reaction - as well as inertia/momentum, are essential - you cannot perform this kind of movement in space or when floating in water. You have to push down and back on the ground in order to get the ground to push upward and forward on you.


Why a type of motion is relative and another is not?



Two scientists are completely isolated in two different boxes:





  1. First box moves uniformly straight (in a perfect space).




  2. Second box rotates uniformly (also in a perfect space).




The scientists have any instrument to detect his movement. Who makes? The second.


Why a type of motion is relative and another is not? Why not are both relative? Why not are both absolute?



(I do NOT mean the theory of relativity by Albert Einstein)



Answer



This discussion is only in the context of newtonian mechnaics(since this is what the OP asked for).


The galilean principle of relativity states:


-The laws of physics hold in thier simplest form with respect to inertial frames of reference.


-Time intervals and distances between events are the same for all frames of reference.


The relativity of constant velocity motion is a direct consequence of the galilean relativiy:


Say you have two inertial frames of reference, $S$ and $S'$, with coordinates $(x,t)$ and $(x',t)$, where $S$ is at rest and $S'$ moves with velocity $v$ with respect to $S$.Then using galilean transformation which relates the coordinates of $S$ and $S'$, with the given information, we can deduce what $S'$ says about the motion of its frame and the motion of $S$ with respect to it:


$x'=x-vt$ .


The equation of motion of $S'$ with respect to $S$ is simply given by



$x=vt$.


Plugging in the transformation one gets: $x'=vt-vt=0$ .


Differentiating by $\dfrac{d}{dt}$ to obtain the velocity of $S'$ in his frame one gets:


$\dfrac{dx'}{dt}=0$.


So $S'$ claims that he's at rest and not moving.


In addition to that consider rearranging terms of the galilean transformation to look like this:


$x=x'+vt$.


$S'$ can claim that actually he is at rest and it's $S$ that is moving with velocity $-v$.


But the ultimate reason why constant velocity motion is relative is the following argument :


In addition to the existence of $S$ and $S'$, consider a particle that is being acted upon by a force $F=m\dfrac{d^2x}{dt^2}$ in $S$.



how the laws of physics look like in $S'$?


Since we know that


$x'=x-vt$


Differentiating twice with time one gets:


$\dfrac{d^2x'}{dt^2}=\dfrac{d^2x}{dt^2}$.


So that we get $F=m\dfrac{d^2x'}{dt^2}$.


So the laws of physics look exactly the same in both $S$ and $S'$, satisfying the principle of relativity.


What implication does this result have on constant velocity motion? Well it implies that constant velocity is relative since:


-every observer can claim he's at rest and it's the other observer who's moving.


-since the laws of physics are the same in both the rest and the moving(with constant velocity) frame. If you locked both $S$ and $S'$ in a closed box, they won't be able to make an experiment to tell apart if they're moving or not. The world just behaves the same if you're at rest or moving with constant velocity.



What about motion that's not at constant velocity? It turns out that such frames are non-inertial frames of reference, in that the laws of physics don't hold true in them.


(To give you a fun task to do, consider two frames, $S$ at rest on the ground and $S''$ that is accelerating with $a$ with respect to $S$, thier coordinates are related by $x"=x-\dfrac{1}{2}at^2$, and say you have a particle that is acted upon by $F=m\dfrac{d^2x}{dt^2}$ in $S$, if the principle of relativity holds we should expect that in $S"$ $F=m\dfrac{d^2x"}{dt^2}$, check if this is true or not(it's not!) and since it's not true, $S"$ cannot claim rightly he's at rest and it's $S$ who's accelerating.)


Accelerated/non-inertial frames of reference are marked by the fact that Newton's laws like his second law and the law of conservation of momentum are violated. Or in other words, they're marked by the exsistence of ficticious forces, that are proprtional to mass in thier frames.


This brings us to your second question: is rotation relative or not?


This is the reason why rotation(and other forms of acceleration) are not relative:


-the laws of physics break down in such frames(that is, they don't hold in their simplest form as they do in inertial frames), so that even one is confined in a box, one can detect his absolute motion(rotation) through the existence of fictitious forces like the coriolis and centrifugal forces.


Now let's turn to the totally empty space case that you asked for:


There's an important point to tell. Newton regarded accelerated motion absolute in the sense that it moves with respect to absolute space, a structure whose exsistence he postulated to explain why acceleration is absolute and all observers agree on whether a body is accelerating or not. In modern formulations however, absolute space is not necessary and so it's discarded , it's said a frame's acceleration is absolute in the sense that it's accelerating to all inertial frames of reference.


Now let's answer your question(this is my understanding, and I'm ready to be corrected if I got something wrong):


-if you have totally empty space, and one observer in it, It makes no sense to tell if he's moving with constant velocity at all(that is if he's at rest or moving uniformly) or if he's accelerating or not. Since the concept of constant velocity motion makes sense only between two inertial frames of reference. Since there's only one frame, and the concept of acceleration in modern formulation only makes sense when one speaks of acceleration with respect to other inertial frames, It makes no sense to speak of any motion(be it uniform or rotation or whatever) in otherwise totally empty universe.



-Newton would agree upon my answer on uniform motion. But amazingly, he'd disagree regarding acceleration. For Newton who defines acceleration relative to absolute space, it makes total sense to speak of rotation of an object in an empty space, since it's rotating with respect to space itself.


So who is right? We don't know for these reasons: -the modern definition of acceleration(with respect to all inertial frames) vs newton's notion of acceleration against absolute space give the same experimental results in a universe that is populated by matter. To tell apart which is right, one has to do the experiment of rotating an object in a totally empty universe(which is impossible).


-Newtonian mechanics does not even hold in our universe,they're GR and QM that do(or more precisely, the to be found theory of quantum gravity).


I will leave others to talk about mach's principle.


Saturday, August 24, 2019

english - Two Bowls of Alphabet Soup


These are three different puzzles with a similar format - feel free to solve them separately!


1.


H, J, K, M, N, U, V, W, X, and Y are.


A, B, C, D, E, F, G, I, L, O, P, Q, R, S, T, and Z aren’t.


2.



B, D, E, F, H, I, K, L, M, N, P, R, T, U, V, W, X, Y, and Z are.


A, C, G, J, O, Q, and S aren’t.


3.


E, F, P, T, and Y are.


A, B, C, D, G, H, I, J, K, L, M, N, O, Q, R, S, U, V, W, X, and Z aren’t.




antimatter - Particle antiparticle annihilation-do they have to be of the same type?


I read that a particle will meet its antiparticle and annihilate to generate a photon. Is it important for the pairs to be of the same type? What will happen when for example a neutron meets an antiproton or a proton meets a positron? Are there any rules to determine what happens when such particles meet?



Answer



Yes, there are rules that depend on the quantum numbers carried by the particles under question and the energy available for the interaction.


In general we label as annihilation when particle meets antiparticle because all the characterising quantum numbers are equal and opposite in sign and add and become 0, allowing for the decay into two photons, two because you need momentum conservation.



A positron meeting a proton will be repulsed by the electromagnetic interaction, unless it has very high energy and can interact with the quarks inside the proton, according to the rules of the standard model interactions.


When a neutron meets an antiproton the only quantum number that is not equal and opposite is the charge, so we cannot have annihilation to just photons, but the constituent antiquarks of the antiproton will annihilate with some of the quarks in the neutron there will no longer be any baryons, just mesons and photons, and all these interactions are given by the rules and crossections of the standard model.


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 \...