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Saturday, August 24, 2019

acoustics - Why does water make a sound when it is disturbed?

When I disturb a body of water, what causes the familiar "water moving" sound?

entropy - Can anyone prove this overstated-but-almost-never-justified fact from thermodynamics?

Clausius inequality states that $\oint {\delta Q\over T}$ equals zero for a system undergoing a reversible cycle, whereas it can’t be greater than zero for an irreversible cycle.

But everywhere, I see people and resources mentioning that “it follows” that for irreversible, it is strictly negative. HOW??

Mostly, they mention that since entropy changes for irreversible processes are positive, the above follows, except that this follows from the above once the above is justified! Else, give me an independent justification of this argument!

Otherwise, it is mentioned that irreversible processes involve dissipation, so entropy generation. But wait! An irreversible process doesn’t have to be dissipative. Say, irreversibility is due to non-quasi-staticity. Now what?

Similar is the case with the efficiencies of irreversible engines working between two temperatures.

Carnot’s theorem merely says that an engine working between two temperatures can’t be more efficient than a reversible engine working between the same temperatures.

I know that this implies that all reversible engines working between these temperatures have the same efficiency. But HOW does this imply that an irreversible engine between the same two temperatures will have a strictly less efficiency? Still people and resources blatantly mention this without any justification.

Someone please resolve this!

Friday, August 23, 2019

newtonian mechanics - Changing Rotation Direction in Mid Air

Not sure whether this is an appropriate question for this site, but could anyone explain the physics behind how this skier is able to change his direction of rotation mid-air? https://www.youtube.com/watch?v=iCKNid-ZkIk

Answer

There are two and possibly three factors at work here:

One which I don't think has been mentioned is that unless a body is rotating about a so-called principal axis, in general the angular momentum and the angular velocity vectors do not point in the same direction: they are related by a nontrivial matrix (the inertia tensor) in Euler's second law of (rotational) motion. The best you can do is diagonalise the inertia tensor: this diagonalisation discovers three orthogonal principal axes and one rewrites Euler's second law in a frame that rotates with the principal axes in the so called Euler equations. The inertia tensor is still not proportional to the identity hence the difference in direction. The skier probably begins to "fly" (i.e. freefall) with an angular momentum not aligned with the angular velocity: it is the latter that catches your eye. So the direction of "apparent spin" can be quite different from that of the angular momentum.

To add to David Hammen's answer: The skier with his skis is not a rigid body but a deformable one, and by, smoothly cycling his shape through a shape configuration space, can change his orientation even though his angular momentum cannot change. As David says this is exactly how a cat flips over when falling or how an astronaut can shift his or her orientation in space.

I say a great deal more about astronauts in my answer here (I give a simple model system illustrating cyclic shape shifting) and about falling cats in my article "Of Cats and Their Most Wonderful Righting Reflex" my my website here. The Kane reference in David Hammen's answer is heavy going, but uses only elementary dynamical concepts. So it should be accessible a bright freshman willing to do a bit of work. If, however, you have a knowledge of fibre bundles, the Montgomery reference is a much clearer overview: we begin with the shape configuration space: this can be any of several reasonable assumptions, but in Kane's and Montgomery's simplified analysis, the cat's shape is defined by two angles and leads to a pinched cylinder, pinched torus or 2-sphere, depending on how you exactly model things. The exact topology is not important: what is important is that now we (or rather, Montgomery) kit the shape space with a fibration: at each point the fibre we add is a space of orientations in 3-space (i.e. $SO(3)$) for the cat and show that the conservation of angular momentum defines a nontrivial fibre connexion. Depending on the cat's symmetry assumed, the fibre bundle that results either has a structure group ("gauge group" if you will) of $SO(2)$ (the tailess, symmetrical cat can flip about one axis only) or the full $SO(3)$ acting on the fibre $SO(3)$. The computation of the structure group is the messy bit and where the conservation of angular momentum comes in to compute the connexion ("gauge covariant derivative") for the bundle - this is what is effectively done in the Kane paper to find the Kane dynamical equations. Of course, one can also think of these ideas in the language of anholonomy (curvature of the fibre bundle).

A wonderful quick summary is given by this video (NOT mine: I found it on Youtube)

https://www.youtube.com/embed/yGusK69XVlk

The analysis in this video is correct for a cat which is symmetrical about its transverse plane (in the anatomists Coronal/Sagital/Transverse convention), in particular for a tailess cat. Most domestic cats don't use their tail a great deal in the righing reflex and consequently their flip tends to be confined to one axis, but small tree-dwelling cats of South East Asia like the Marbled Cat and Clouded Leopard have a huge tail (more like a club) in relation to their body and use it for re-orientation in all axes as they divebomb their prey.

At the speed the skier is flying, air resistance means that the flight will not be torque free. However, this hopefully is not a major factor, and indeed one of the skills needed by the skier is to hold his or her body so that it "tumbles" (i.e. moves essentially torque free, but buffetted by perturbational torques) in a stable way and keeps the perturbative torques from becoming a problem. Tiny buffetting torques can otherwise make the apparent spin of the body flail about wildly indeed.

As discussed in this exposition here J. Peraire, S. Widnall, "Lecture L28 - 3D Rigid Body Dynamics: Equations of Motion; Euler’s Equations", we suppose that a body is at first rotating about a principal axis $z$ and is then buffetted by a tiny angular impulse. The linearised Euler equations for the torque free motion that follows the impulse described by the angular velocities $\omega_x,\,\omega_y,\,\omega_z$ relative to the rotating principal axis frame are:

$$\dot{\omega}_x - \frac{(I_{yy}-I_{zz})(I_{zz}-I_{xx})}{I_{yy}\,I_{xx}} \omega\,\omega_x = 0$$

with a like equation for $\omega_y$ (here $\omega$ is the initial angular speed before the impulse) The solutions to this DE are $e^{\pm \sqrt{a}\,t}$ where

$$a=\frac{(I_{yy}-I_{zz})(I_{zz}-I_{xx})}{I_{yy}\,I_{xx}} \omega$$

If $I_{zz}$ is either less than both or greater than both $I_{xx}$ and $I_{yy}$, $a$ is negative and both the terms $e^{\pm \sqrt{a}\,t}$ are oscillatory. The body wobbles a little bit, but is otherwise unaffected by the perturbation. If, however, our hapless skieer gets into a situation where $I_{zz}$ lies between $I_{xx}$ and $I_{yy}$, then $a$ is positive and one of the solutions grows exponentially. A large, torque free change in angular velocity follows. Such a body's attitude is essentially out of control, even in the presence of tiny, buffeting winds.

general relativity - Is there a maximum possible acceleration?

I'm thinking equivalence principle, possibilities of unbounded space-time curvature, quantum gravity.

Answer

The spit horizon in a Rindler wedge occurs at a distance $d~=~c^2/g$ for the acceleration $g$. In spatial coordinates this particle horizon occurs at the distance $d$ behind the accelerated frame. Clearly if $d~=~0$ the acceleration is infinite, or better put indefinite or divergent. However, we can think of this as approximating the near horizon frame of an accelerated observer above a black hole. The closest one can get without hitting the horizon is within a Planck unit of length. So the acceleration required for $d~=~\ell_p$ $=~\sqrt{G\hbar/c^2}$ is $g~=~c^2/\ell_p$ which gives $g~=~5.6\times 10^{53}cm/s^2$. That is absolutely enormous. The general rule is that Unruh radiation has about $1K$ for each $10^{21}cm/s^2$ of acceleration. So this accelerated frame would detect an Unruh radiation at $\sim~10^{31}K$. This is about an order of magnitude larger than the Hagedorn temperature. We should then use the string length instead of the Planck length $4\pi\sqrt{\alpha’}$ and the maximum acceleration will correspond to the Hagedorn temperature.

Thursday, August 22, 2019

mathematics - Restoring order in a deck of playing cards (I)

Michelle has a deck of 52 playing cards, stacked in a pile with their backs facing up. She takes the first 2 cards in the pile, turns them over, and places them at the bottom of the pile. She now takes the next 3 cards in the pile and, once again, turns them over, and places them at the bottom of the pile. She proceeds like this, taking each time the next prime number of cards from the top, turning them over, and placing them at the bottom of the deck. Once she has done this for all all primes up to 47 (the largest prime less than 52), she continues in the same fashion counting in turn 2, 3, 5, etc. cards from the top and placing them at the bottom of the deck.

Will the deck of cards ever have all their backs facing up again?

Answer

The cards will again be all face down . . .

. . . after 11700 operations. I'll just show the first 20


 0  0000000000000000000000000000000000000000000000000000
 2  0000000000000000000000000000000000000000000000000011
 3  0000000000000000000000000000000000000000000000011111
 5  0000000000000000000000000000000000000000001111111111
 7  0000000000000000000000000000000000011111111111111111
 11 0000000000000000000000001111111111111111111111111111
 13 0000000000011111111111111111111111111111111111111111

 17 1111111111111111111111111111111111100000011111111111
 19 1111111111111111000000111111111110000000000000000000
 23 1111111111000000000000000000001111110000000000000000
 29 0111111000000000000000011111111111111111110000000000
 31 1111111111100000000000000000011111111111111110000001
 37 1111111100000010000000011111111111111111100000000000
 41 0000000000000000000000000000011111111011111100000000
 43 1000000000000010000000011111111111111111111111111111
 47 1111100000000000000000000000011111111011111111111110
 2  1110000000000000000000000001111111101111111111111000

 3  0000000000000000000000001111111101111111111111000000
 5  0000000000000000000111111110111111111111100000011111
 7  0000000000001111111101111111111111000000111111111111
 11 0111111110111111111111100000011111111111111111111111

But if the rules are changed slightly and the cards are moved from top to bottom in the same (not reversed) sequence, it only takes $56$ operations.


 0  0000000000000000000000000000000000000000000000000000
 2  0000000000000000000000000000000000000000000000000011
 3  0000000000000000000000000000000000000000000000011111

 5  0000000000000000000000000000000000000000001111111111
 7  0000000000000000000000000000000000011111111111111111
 11 0000000000000000000000001111111111111111111111111111
 13 0000000000011111111111111111111111111111111111111111
 17 1111111111111111111111111111111111111111111111000000
 19 1111111111111111111111111110000000000000000000000000
 23 1111000000000000000000000000000000000000000000000000
 29 0000000000000000000000000001111111111111111111111111
 31 1111111111111111111111111111111111111111111111110000
 37 1111111111100000000000000000000000000000000000000000

 41 0000000000000000000000111111111111111111111111111111
 43 1111111111111111111111111111111000000000000000000000
 47 0000000000000000000000000000000000001111111111111111
 2  0000000000000000000000000000000000111111111111111111
 3  0000000000000000000000000000000111111111111111111111
 5  0000000000000000000000000011111111111111111111111111
 7  0000000000000000000111111111111111111111111111111111
 11 0000000011111111111111111111111111111111111111111111
 13 1111111111111111111111111111111111111111111111100000
 17 1111111111111111111111111111110000000000000000000000

 19 1111111111100000000000000000000000000000000000000000
 23 0000000000000000000000000000000000000000111111111111
 29 0000000000011111111111111111111111111111111111111111
 31 1111111111111111111111111111111100000000000000000000
 37 0000000000000000000000000000000000000000000000011111
 41 0000001111111111111111111111111111111111111111111111
 43 1111111111111110000000000000000000000000000000000000
 47 0000000000000000000011111111111111111111111111111111
 2  0000000000000000001111111111111111111111111111111111
 3  0000000000000001111111111111111111111111111111111111

 5  0000000000111111111111111111111111111111111111111111
 7  0001111111111111111111111111111111111111111111111111
 11 1111111111111111111111111111111111111111111100000000
 13 1111111111111111111111111111111000000000000000000000
 17 1111111111111100000000000000000000000000000000000000
 19 0000000000000000000000000000000000000000000000011111
 23 0000000000000000000000001111111111111111111111111111
 29 1111111111111111111111111111111111111111111111100000
 31 1111111111111111000000000000000000000000000000000000
 37 0000000000000000000000000000000111111111111111111111

 41 1111111111111111111111111111111111111111110000000000
 43 0000000000000000000000000000000000000000000000000001
 47 0000111111111111111111111111111111111111111111111111
 2  0011111111111111111111111111111111111111111111111111
 3  1111111111111111111111111111111111111111111111111110
 5  1111111111111111111111111111111111111111111111000000
 7  1111111111111111111111111111111111111110000000000000
 11 1111111111111111111111111111000000000000000000000000
 13 1111111111111110000000000000000000000000000000000000
 17 0000000000000000000000000000000000000000000000000011

 19 0000000000000000000000000000000111111111111111111111
 23 0000000011111111111111111111111111111111111111111111
 29 1111111111111111111111111111111000000000000000000000
 31 0000000000000000000000000000000000000000000000000000
 result = 56

Found by C program.

mathematics - The Determinant Sudoku

The determinants of $2\times 2$ matrices are calculated this way: $$\det\begin{bmatrix}a & b \\c & d \end{bmatrix}=ad-bc$$

And here's the sudoku-determinant puzzle:

What's the solution for this sudoku?

The original puzzle can be found here, Thanks for reddit user Wizediablo for sharing this puzzle!

Answer

Final solution

Step-by-step explanation

In each of the 2x2 boxes with determinant 0, we have four numbers $a,b,c,d$ between 1 and 9 such that $ad=bc,a\neq b\neq d\neq c\neq a$. This leaves surprisingly few possibilities:

5 and 7 can't be involved at all, since they're prime and have no multiples less than 10

The only possibilities for the pairs $\{a,d\},\{b,c\}$ (unordered in all ways; these are just two pairs with equal products) are:

{1,9},{3,3}

{2,9},{3,6}

{4,9},{6,6}

{1,8},{2,4}

{2,8},{4,4}

{3,8},{4,6}

{2,6},{3,4}

{1,6},{2,3}

{1,4},{2,2}

Making a start then: we know that any box/row/column with all but two of its cells contained in the determinant-zero boxes must have 5 and 7 in those two cells.

(Black numbers indicate definitely filled cells, small red numbers indicate all possible positions of the remaining 5's and 7's.)

Consider the two determinant-zero boxes which overlap in two adjacent cells. We now have three pairs of numbers whose ratios are equal, which means that ratio must be

either 1:2 or 1:3 or 2:3. So the three pairs of numbers must be either {1,3},{2,6},{3,9} or {2,3},{4,6},{6,9} or three of {1,2},{2,4},{3,6},{4,8}.

Consider the top middle 3x3 box. It contains the numbers 1,2,3,4,6,8,9 in two determinant-zero boxes with exactly one cell of overlap. Since 9 must be contained somewhere and {1,9},{3,3} and {4,9},{6,6} are clearly impossible, it must be part of a {2,9},{3,6} box. Similarly, 8 must be contained somewhere and {2,8},{4,4} is impossible, but also {3,8},{4,6} is impossible as there is only one cell of overlap, so it must be part of a {1,8},{2,4} box, with the overlap at 2.

We can follow exactly the same reasoning for the right middle 3x3 box, so the middle cell must be 2 in both cases and the cells diagonally next to it must be 4 and 9.

Consider the determinant-zero box straddling four 3x3 boxes. Its top left and bottom right cells must each be 4 or 9. Clearly they can't both be 9.

If there's one 4 and one 9, then this box must be {4,9},{6,6}, so the top right and bottom left cells must both be 6. But the 9 must be part of a {2,9},{3,6} box, and that 6 would share a row or column with one of the 6's we've just filled in. Contradiction, so both the top left and bottom right corner are 4's, making this box {2,8},{4,4}.

Let's go back to those two determinant-zero boxes which share two adjacent cells. Now they're (in order, top to bottom, left to right) either {3,1},{9,3},{6,2} or {6,4},{9,6},{3,2}. The former possibility forces this (click for full-size image):

And now that upper right determinant-zero box must have a 1, 2, or 3 opposite the 8, none of which is possible. Contradiction, so we have {6,4},{9,6},{3,2}. This means that upper right determinant-zero box can't contain a 4 (since there are already 4's in the second and third rows of the whole thing) and therefore can't contain an 8. That tells us which way round the 2 and 8 are in the determinant-zero box which straddles four 3x3 boxes.

In the second row, the only possibilities for the seventh and eighth cells are 3,8,9. We already know that determinant-zero box can't contain an 8, so it must be a {2,9},{3,6} box. That enables us to complete this whole mess of interlinked determinant-zero boxes.

In the seventh column, the only cells unfilled are 1,6,9. We know 1 can't be adjacent to either 6 or 9 in a determinant-zero box, so we must have 1 in the top cell and a {2,9},{3,6} box down at the bottom.

In the eighth column, the only cells unfilled are 4 and 8, and clearly a {2,8},{4,4} box is impossible within a single 3x3 box. So the bottom right determinant-zero box must be {1,8},{2,4} or {3,8},{4,6}, and the latter is impossible because there's already a 6 in the ninth column. Thus the bottom right cell is 3.

Now we can use pure Sudoku techniques to fill in the top chunk of the ninth column, some more 5's and 7's, and most of the sixth column:

By considering the possibilities for the middle left determinant-zero box, we can show that it must be either {2,9},{3,6} or {1,6},{2,3}, with the 6 in the top left cell and the 2 in the top right cell. See Mystery's answer for details.

We can then use pure Sudoku arguments to fill in the 2 and 3 in the middle 3x3 box, and to know which two of the remaining cells are 5 and 9, meaning the other two must be 1 and 4.

We know that the right-hand ends of the seventh and eighth rows must be (in order, left to right) 2 6 4 1 and 3 9 8 2. Disregarding the 5 and 7 in each row, whose locations we (almost) know, the remaining cells must be {3,8,9} and {1,4,6} respectively.

If the 2 6 4 1 is in the seventh row, then the 6 in the first column must be in the bottom left determinant-zero box, but this box can't contain 3 or 9 anywhere. Contradiction, so the 3 9 8 2 is in the seventh row. Now the bottom left determinant-zero box can't contain any of 1,2,4,6 opposite each other, so it must be {2,9},{3,6}. The 6 must be in the bottom left cell, the 3 in the top right, the 9 in the top left, and the 2 in the bottom right.

Now we can use pure Sudoku techniques to fill in the eighth row, the rest of the 7's, the bottom left 3x3 box, the ninth row, the seventh row, and the 1 and 4 in the middle 3x3 box.

Finally we can finish the very last determinant-zero box: its bottom row must be 9 and 3, not 3 and 1. And then Sudoku techniques enable us to quickly fill in the rest:

_{Thanks to @Goinghamateur for help in comments!}

story - Forget the freehand red circles - it's a freehand rebus! - Clue Twenty Eight

_{<<---First clue
<--Previous clue}

_{This is an entry in the 26th Fortnightly Topic Challenge.}

You speak the password. The entire space shudders, and the wall that you were leaning on open, spilling you headfirst into... an art museum? There are all of the pictures from the previous Clues there, but you ignore those. You focus on the new ones, which appear to be freehand drawn on the computer:

You look around for a place to enter the answer. Finally, you see a door on the other side of the room, with the now-familiar keypad.

You cross the room and type in...

_{Next clue--->}

Answer

The answer is

Tungsten.

First

This looks like a smiley. So, may be it;s STICKING TONGUE OUT.

Second

Could be Parseltongue (Snakes language)

Third

This could be either number 10 or Tally Marks.

So the answer is

Tungsten. i.e. Two tongues and a ten.

OP's intended solving

The first image

This is indeed sticking it's tongue out. So that's tongue, or tung.

Second image

This is a picture of a snake, with a question mark for what it says. SO that's s.

Third

These are 10 tally marks, so ten.

All together, that's

tungsten.