homework and exercises - Is it possible to produce gravitational waves artificially?

1. 
   

   Why don't they make a ball with irregularities, say the size of a tennis ball, then spin it very rapidly, so it would produce gravitational waves like a spinning star with irregularities on it? Is that not possible with our current technology?

   
   


2. 
   

   Also since gravitational waves can cause time dilation, wouldn't we be able to make some sort of a time machine with that concept?

   
   

quantum mechanics - What is the spin state of a spin-1/2 particle when it comes out of a Stern-Gerlach apparatus?

Having a particle entering the apparatus with spin state $|+\rangle$, for which $\hat S_x|+\rangle=+\frac\hbar 2|+\rangle$, I have a question about how to express the spin state when it comes out. I mean: will the spin state be an eigenstate of the spin $z$ component? or would it be a superposition of the $z$ component's eigenstate ?

Answer

A Stern-Gerlach apparatus oriented along the $z$ axis acts as a measurement on the basis of the $z$ component of the particle's spin. What that means is that the particle will always come out in an eigenstate of $\hat S_z$. You don't know which one, of course, as that is decided probabilistically.

Even if you don't actually observe the outcome of the measurement (in which case the system would be in the corresponding eigenstate), it is not accurate to say that the particle comes out in a superposition of $z$ eigenstates. While what comes out is a probabilistic mixture, it is a mixed state in which the coherence between the two contributions has been lost.

nucleosynthesis - Does all hydrogen originate from the Big Bang?

I was wondering, if every single hydrogen in the universe originate from the time about ~3 min after the Big Bang.

I know there are nuclear fusion processes going on in stars like the pp-chain reaction which 'emit' hydrogen, too, but those reactions needed several hydrogen atoms to begin with.

So are there really processes which create hydrogen 'from scratch'?

Now, assuming that the human body consists of ~60% water and a taking a body weight of ~75kg => 45kg water => 5kg hydrogen (taking m(H) = 1u & m(O) = 16u => m(H2) / m(H2O) = 1/9) [I'm no biochemist so I don't really know if there are significant amounts of hydrogen in other compounds within the human body], does this mean that roughly 95% (70kg/75kg) of me is - like Carl Sagan said - made of stardust, but even more mesmerizing, ~5% of the stuff that makes up my body made shortly after the Big Bang, nearly 14 billion years ago?

Answer

Free neutrons can be generated by spallation reactions between cosmic rays and heavy metal targets. Such processes do occur in the cosmos.

The free neutrons will then beta decay to yield protons (hydrogen nuclei) and electrons. This is a weak reaction, so the protons produced here were not present shortly after the big bang.

So not all hydrogen atoms originate from protons made in the big bang, though other sources are negligible, so your calculation looks OK to me.

mathematics - Dominos on a checkerboard

What's the maximal number of dominos (2x1 tiles) that can be placed on a checkerboard (8x8 square) so that every domino covers exactly 2 squares of the checkerboard and no two dominos form a 2x2 square?

Answer

Looks like:

  31 dominos can be placed  

Thanks to @Gamow's comment, this number's maximality can be proved by self-contradiction of the assumption that it is not maximal. Any more dominos would cover all 64 squares.

Assumption to be disproved: All squares can be covered with dominos.

   A. As the top left corner must be covered, start with a horizontal domino there. (Every other possible corner domino is equivalent to this by rotation and reflection.) From here on:

Successive placements of a diagonal series of dominos are forced to form a descending herringbone staircase in order to prevent a 2x2 square from being formed in combination with each previous domino.
B. The domino that neighbors A along the left edge must be vertical.
C. The inside corner formed by A and B must be covered by a horizontal domino.
D −  M. Likewise, until a horizontal domino-shaped hole at the bottom right corner, if filled, would form a 2x2 square with M.
   

Therefore the top left corner cannot be covered, which negates the assumption and proves that a 32-domino solution is impossible.

Further reading, courtesy of @Fimpellizieri's comment:
Conway's Tiling Groups _PDF [and height functions] – W. P. Thurston
Tiling with Polyominoes and Combinatorial Group Theory _PDF – J. H. Conway & J. C. Lagarias
Domino Tilings of the Torus _PDF [and the plane] Abstract – F. de Souza Lima Impellizieri

quantum mechanics - Coupled Harmonic Oscillator - Solve by diagonalization

I have a problem where I have two massive Particles $M$ and one particle with mass $m

This probably means to switch to new coordinates \begin{equation} y_1 = -q_1+q_2\\ y_2 = -\frac{1}{\sqrt{2}}q_1-\frac{1}{\sqrt{2}}q_2+q_3\\ y_3 = \frac{1}{\sqrt{2}}q_1+\frac{1}{\sqrt{2}}q_2+q_3 \end{equation}

But what do I do with the momentum operators. They should change accordingly but I am confused as to how exactly

general relativity - Does antimatter curve spacetime in the opposite direction as matter?

According to the Dirac equation, antimatter is the negative energy solution to the following relation:

$$E^2 = p^2 c^2 + m^2 c^4.$$

And according to general relativity, the Einstein tensor (which roughly represents the curvature of spacetime) is linearly dependent on (and I assume would then have the same mathematical sign as) the stress-energy tensor:

$$G_{\mu \nu} = \frac{8 \pi G}{c^4}T_{\mu \nu}.$$

For antimatter, the sign of the stress-energy tensor would change, as the sign of the energy changes. Would this change the sign of the Einstein tensor, causing spacetime to be curved in the opposite direction as it would be curved if normal matter with positive energy were in its place? Or does adding in the cosmological constant change things here?

Answer

Antimatter has the same mass as normal matter, and its interaction with gravity should be the same according to GR and QM.

That said, antimatter has only been created in tiny amounts so far and only few experiments have been performed to confirm there is no new physics involved.

The gravitational interaction of antimatter with matter or antimatter has not been conclusively observed by physicists. While the overwhelming consensus among physicists is that antimatter will attract both matter and antimatter at the same rate that matter attracts matter, there is a strong desire to confirm this experimentally, since the hypothesis is still open to falsification.

https://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter

quantum field theory - Calculation of Feynman invariant amplitude with internal global symmetry indices: trace overand isospin

This is a complete rewriting of the older post, making more clear the problem.

The issue here is to compute the $$|M|^2 =4a^2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) ^2 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'd}^{\nu}\bar{u}_{2'\nu c}u_{2d}^{\nu}\bar{u}_{2\nu c}) $$ invariant Feynman amplitude, and in fact, the search for a possible, if it exists, way for writing this as a trace both on $SU(2)$ and Lorentz indices.

One of terms of the above amplitude is: $$T_1 = \delta_{ad}\delta_{bc}\delta_{ad}\delta_{bc} \omega = 4 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})(u_{2'a}^{\nu}\bar{u}_{2'\nu b}u_{2a}^{\nu}\bar{u}_{2\nu b}),$$where $\omega $ is the series of spinors before contracting the deltas and the 4 coefficient comes from the contraction of deltas.

Question: I'm having a problem seeing how could I simplify this expression. Can I simultaneously reform this term to a trace? The possibilities are a trace over:

1. Just spin indices; but what happens then with the internal indices?


2. Internal isospin trace; but what then happens with spin indices?


3. A trace over both indices?

Thank you.

---

Details

Assume a Lagrangian with an interaction term of the form $L_{int}=\bar \psi \phi \cdot \tau \psi$ , globally symmetric under the action of $SU(2)$ group. Think that the field $\psi$ describes a nucleon as an isospin SU(2) doublet with entries the proton and neutron and the field $\phi$ an isospin triplet (in the adjoint representation) of SU(2) formed by the three scalar pions.

The problem is to calculate the invariant Feynman amplitude. We have, writing all the indices explicitly:

$$M=a(q^2) (\bar{u}_{1'\mu a}\tau^{ab}_k u^{\mu}_{1b})( u_{2'\nu c}\tau^{cd}_k\bar{u}^{\nu}_{2d}) ,$$ where $\tau_k$ are the $SU(2)$ generators; Pauli matrices and q is the interaction momentum of the propagator. The complex conjugate then is: $$M^{*}=a^{*}(q^2) (\bar{u}_{1\mu a}\tau^{ab}_k u^{\mu}_{1' b}) (u_{2\nu c}\tau^{cd}_k\bar{u}^{\nu}_{2' d}) .$$

This way one should recognise the scalar nature over both $\mu, \nu $ Lorentz indices and over $a,b,c,d, k $ isospin indices. A similar calculation can be found in chapter 46 of Srednicki, without the extra internal symmetry and it's indices. There, it's simple then to recognise the trace over spin indices of the $|M|^2 =MM^{*} $.

For the generators there is the following completeness relation: $\tau_{ab}^k \tau_{cd}^k = 2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) .$

So finally, the amplitude is:

$$|M|^2 =4a^2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) ^2 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'd}^{\nu}\bar{u}_{2'\nu c}u_{2d}^{\nu}\bar{u}_{2\nu c}) $$

The delta identity gives three terms; the first, let's say $T_1$, is: $$T_1 = \delta_{ad}\delta_{bc}\delta_{ad}\delta_{bc} \omega = 4 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'a}^{\nu}\bar{u}_{2'\nu b}u_{2a}^{\nu}\bar{u}_{2\nu b}),$$ where $\omega $ is the series of spinors before contracting the deltas and the 4 coefficient comes from the contraction of deltas.

thermodynamics - Radiation– white vs black house, hot or cool?

In my book's thermodynamics chapter, it says that an

"object that radiates heat faster also absorbs heat faster. This means that an object that is a more efficient radiator comes to equilibrium with its environment more quickly. With this in mind, is it better to paint your house black or white?"

I am confused which it would be. The book says white because

"in summer, you house is cooler than the environment and white reflects away the heat. In winter your house is warmer than the environment and white radiates away the heat"

general relativity - Friedmann vacuum solution linear rather than exponential?

Does the Friedmann vacuum equation have a linear solution rather than an exponential one?

Using natural units one can write Friedmann's equation for the vacuum as $$ \begin{eqnarray} \left(\frac{\dot a}{a}\right)^2 &=& \frac{8\pi G}{3}\rho_{vac}\\\tag{1} &=& L^2 \left(\frac{\rho_0}{L^4}\right) \end{eqnarray} $$

where I define the Planck length $L=(8\pi G \hbar / 3 c^3)^{1/2}$, $\hbar = c = 1$, and $\rho_0$ is a dimenionless constant.

Now let us interpret the Planck length $L$ to be the size of the smallest volume of space that can be described by general relativity.

But the Weyl postulate, together with cosmological observations, also imply that space is expanding.

Therefore we must have

$$L = a(t) L_0\tag{2}$$

where $L_0$ is the Planck length measured at the reference time $t_0$ where $a(t_0)=1$.

Inserting Eq.(2) into Eq.(1) we find

$$\left(\frac{\dot a}{a}\right)^2 = L_0^2 \left(\frac{\rho_0}{a^2L_0^4}\right)\tag{3}$$

where the Friedmann equation (3) has been rescaled in terms of the Planck length $L_0$ measured at the reference time $t_0$.

Eq.(3) has a linear solution

$$a(t) = \frac{t}{t_0}.$$

The scaled mass density $\rho(t)$ of the vacuum is not constant but rather given by

$$\rho(t) = \frac{\rho_0}{a^2 L_0^4} = \frac{1}{t^2 L_0^2}.$$

terminology - Planets and Pluto? Neptune?

If one of the rules to be a planet is that it needs to clear ALL objects from their orbit, does this also make Neptune a non-planet? Since it has thus far failed to clear Pluto from it's orbit. Or does this rule not really apply to planets and we should welcome Pluto and a few other dwarf planets into our family of planets?

quantum mechanics - How to write "postselection" operator?

Suppose, I wish to know an operator, which eigenvalue is 1 if state is exactly F and 0 otherwise. I.e. this operator should select some desired state.

Which is this operator?

One of it's eigenfunctions is F, but what one is another?

Answer

All other functions not containing a non-zero factor of $F$ are eigenfunctions with eigenvalue $0$ of this operator, the eigenspace with that value is degenerate with dimension $\mathrm{dim}(\mathcal{H}) - 1$, where $\mathcal{H}$ is the whole space.

Obviously, the operator is simply the projection on $\lvert F \rangle$, so it is $P_F = \lvert F \rangle \langle F \rvert$.

logical deduction - Where did you end up? [Part 1]

Follow the clues and tell me where you are in the end.

1. PSE


2. To the opposite of the opposite of 2 + 2 ~= 4


3. Go to the way you round .5


4. If up is down and down is up go down



5. You're on the right track


6. You've made it, where are you?

This is part 1. I may not make a part 2, but others in the community may.

Answer

1. PSE very likely stands for puzzles.stackexchange, hence I'm starting right here.


2. The statement 2 + 2 ≠ 4 is wrong. The opposite of wrong is right. The opposite of right is left.


3. .5 is rounded up (usually, on pen and paper).



4. Up (proper) is down (1), and down (1) is up (2). Ergo up (2) is up (proper), and we can deduce that down (2) is therefore down (proper). Hence by going down (2), I go down.


5. You're on the right track.


6. I've gone up, down, right, and left, thus I'm right back where I started: on puzzles.stackexchange! :D

general relativity - What is the meaning of the "expansion of space"?

When we say that "the space between galaxies is expanding," what do we really mean? For instance, if I think of space as being a Cartesian grid, then when space expands should I think of it as adding more grid-points or as making the distance between the grid-points larger? Or is this a flawed picture of space-time even when far from significant mass densities?

It seems to me that it must be the case that we are "adding more grid-points" because otherwise we would not observe the expansion since anything that occupied that part of the grid (e.g. light, my hand, etc.) would also expand by a corresponding amount (otherwise, after a sufficient amount of expansion, we would be able to observe sub-atomic scale processes as being macroscopic in scale). But if it is the case that we are adding more grid-points, then the "expansion" of space seems like a misnomer: shouldn't it be called the "creation" of space?

That being said, either case (adding grid-points or stretching distances) could conceivably be the same process depending on the nature of space-time (i.e. if it were discrete but dense - like the Rationals - rather than either strictly discrete (Integers) or strictly continuous (Reals)... I hope that analogy makes sense).

So what is really going on in intergalactic space? What is the meaning of its expansion? Does it correspond to creating more space, stretching the space that's already there, or something more subtle?

mathematics - Coin weighing problem

You are given ten stacks of golden coins, each stack consisting of ten coins and a digital scale with arbitrary precision.

You know that all stacks of coins are made from gold, weighing 10 grams per coin except one stack, which is made from silver but is painted golden. A silver coin weighs 1 gram less than a golden one, which makes its weight 9 grams.

With only being able to weigh exactly once, how can you identify the stack containing the silver ones?

Answer

1. You can number stacks from 1 to 10.


2. From each stack take as many coins as the stack number.


3. Weigh them all.


4. Subtract the weight from 10+20+...+100=550. The result is the number of the silver stack.

P.S. You can do the same even if you have 11 stacks of 10: just number them from 0 to 10.

energy conservation - Collision and impulsive forces: a formal approach

Consider two bodies $m$ and $M$. Suppose that $m$ is moving with constant velocity $v_0 > 0$ along a certain axis (e.g., it is moving on the right on the $x$-axis), and at a certain time, it collides with $M$ at $t=0$. Suppose that after the collision, $m$ has velocity $v_1 \in \mathbb{R}$ and $M$ has velocity $w_1>0$.

During the collision, the two bodies are subject to impulsive forces. Given $f>0$, these forces are $-f\delta(t)$ on body $m$ and $f\delta(t)$ of body $M$ due to the third Newton's law. Notice that $\delta(t)$ is the Dirac delta. Moreover, $f$ is $[N\cdot s]$ and $\delta(t)$ is $[s^{-1}]$.

In this setup, I can say that the velocities of two bodies over time are the followings:

$$\begin{cases} v(t) & = & v_0 + (v_1 - v_0)H(t)\\ w(t) & = & w_1 H(t) \end{cases}, $$

where $H(t)$ is the Heaviside step. According to these, we can write that:

$$\begin{cases} \dot{v} & = & (v_1 - v_0)\delta(t)\\ \dot{w} & = & w_1 \delta(t) \end{cases}, $$

since $H'(t) = \delta(t)$.

The laws of motion for the bodies are:

$$\begin{cases} m\dot{v} & = & -f\delta(t)\\ M\dot{w} & = & f \delta(t) \end{cases}, $$

and hence:

$$\begin{cases} m(v_1-v_0)\delta(t) & = & -f\delta(t)\\ Mw_1\delta(t) & = & f \delta(t) \end{cases} \Rightarrow \\ \begin{cases} m(v_1-v_0) & = & -f\\ Mw_1 & = & f \end{cases} \Rightarrow\\ f = Mw_1 = m(v_0-v_1). $$

As a side note, $f$ and $-f$ are the variations of the momenta of $M$ and $m$, respectively. Thus, I have obtained the equation of the conservation of the momentum of the whole system $m+M$.

Suppose that we don't know if the collision is elastic or not. How can we evaluate the variation of kinetic energy without using formulas like $\frac{1}{2}mv^2$, but just looking to the laws of motion and the velocity expression of the two bodies?

Answer

I would have written this as a comment, not an answer, but I don't have the reputation!

Usually, this type of problem is rewritten in the centre of mass frame of the colliding pair. This automatically takes care of conservation of overall momentum. The problem then just involves the relative motion of the two bodies, not their absolute motion. Then, an inelastic collision is specified as a relation of some kind between the pre-collisional and post-collisional relative velocities, perhaps involving a coefficient of restitution.

You can't avoid inputting some kind of equation like this, so as to specify the physics. The collision is still impulsive, but you deduce the value of $f$ from that relation. The consequences for the change in kinetic energy follow from those equations. If the collision is elastic, of course, energy is conserved.

knowledge - Traversing the US and Canada: A Cryptic Journey

Each answer is a place, landmark or attraction in the United States or Canada. The answers traverse all Canadian provinces, Canadian territories, U.S. states and the District of Columbia. This is done by crossing land borders where possible, although in a couple of cases crossing the sea is necessary as there are no land connections.

Each clue is missing a state/province name from another clue. (Not the answer to that clue, just the name of the administrative division.)

Note that some long names have been shortened. For example, if the full place name is "Xyz Royal National Park", the answer will be just "Xyz" (often with "royal national park" or similar as the definition).

1. Islamic extremists leave chaotic {43} USA salad to historic settlement (1'4 3 7)
2. {34} capital's borders hold extravagant Hopis regarding museum (9)
3. Anne's farm actor Clark and family coloured like {15} bay (5 6)

4. Familiar Margaret's top {11} vehicle duos produce fishing village (5'1 4)
5. Rapids turning around cliff's bottom, Cool J in author {24}'s birth initials (9 5)
6. National park's academic beginners with {4} finalists (6)
7. Actor Phoenix and second striker George follow series filmed in {64} canyon (4 5 5)
8. Region's east becomes west for {52}'s starters next to monarchy (9 7)
9. Primate and heartless CEO in briefly backwater {23} peninsula (4 3)
10. AC/DC bassist is after top-three {7} wine to stroll passageway (7 5 4)
11. Former {42} governor Eleanor familiarly leads to public park (8 4)
12. Felines unwell in short {57} back mountains (9)
13. Short {10} strangely cued a retired Wachowski waterway (6 5)

14. Fur post finally keeps top cards in {5} top two by the fifth street in hold'em (5 5)
15. Biblical figure's top-three {35} water park (4'1 3)
16. "The Knife" in a capital conundrum next to {3}'s last resort area (8 6)
17. Short {49} between pillars by haphazard sauna den sand formations (7 5)
18. Resort town president with top billing outside {12} capital's borders (3-2-3)
19. TV's Gomer east of {17} neighbour resort town (8)
20. Strike east twice from excited {21} in half-hearted book by ocean namesake town's pathway (8 4 9)
21. In front of seashore, bathe vigorously with {53} outsiders returning to coastal city (7 5)
22. Ford follows half of {26} capital with event host to stronghold (4 7)
23. Obelisk is {60} memorial structure (10 8)

24. Plantation east of {8} with one string instrument (10)
25. Historic town's sharper leader goes right, actress Tina embraces {48} capital couple (7 5)
26. Bullets and finally brought rich cognac within the edges of {50} govern central cavern system (7 4)
27. Elegance with {22}'s ending mansion (9)
28. Aged {61} capital is historic district (3 5)
29. Cay of friends having {33} loner is barrier island (4 2 5)
30. Forest's peculiar echo in {63} lake follows discussion (13)
31. National park's island near {51} mysteriously loses poles, takes five before finishing classes (10)
32. Coasts by body of water west of {31} resort city (4 6)
33. Without question, Vicky with rest of {19}'s capital leads to siege site (9)

34. Fifteen minutes behind {2}'s language neighbourhood (6 7)
35. Greek exclamations about the western part of {37} capital city (6 7)
36. Mountain city's {9} capital gets incomplete ransom for body (7)
37. Skyscraper or peach-shaped {29} building losing fluid for actor Bruce (6 5)
38. Settlements from {32}'s end with strange anise around colon (5 8)
39. Dale leaves {44} suburb, with deceit makes natural landmark (6 5)
40. Old cattle town is {16}-made car (5)
41. Lake's broken head following broken heart of {38} (6 3)
42. Battle site's returning {39} city not high end, keeps head low (5)
43. In the case of {27}, a soup brewed bloating heads leading to native community (4 6)

44. Lucky charm to twist a {46} River meander (9 4)
45. Half-fleecy westward ravine following {55}'s first and second national park (5 6)
46. Hold speech alongside {20} road with no turn to mountain (5 4)
47. Geyser not adulterous after initiating {41}'s long distance (3 8)
48. {13} band with additional mountain (8)
49. {6} loses foot, access lobster cooked within cottage (7 5 5)
50. Meeting place for short {40} (5)
51. Even {45} or steep mountain (4)
52. Okinawa city and {14} twins inspire top national park (7)
53. Gold rush town or {62} capital has endless dawn for a start (6 4)

54. Mountain's odd diesis holds {1} initials (6)
55. Last twos of East Asian capital and {59} make national park (4)
56. Mountain town to bar a duo from {36} capital (5)
57. Flexible EU skis alongside the shores of {25} lake (8)
58. Battle site is small, born holding {28} capital finishers (6 7)
59. National preserve to half-heartedly create soft heroin in designer {56}'s heart by beginning moonlight (7 2 3 4)
60. Volcano street next to {58} capital making a short head (2. 6)
61. Body of water's confusingly rare talk inside {47} capital's borders (6 4)
62. Left insane next to {18} company's man-made barrier (6 3)
63. {30} capital scorcher held in dodgy wash riff waterfront area (9'1 5)

64. Mountain top monitors a nuke blown up within {54}'s borders (5 3)

Answer

1. Islamic extremists leave chaotic [43] USA salad to historic settlement (1'4 3 7)

L'ANSE AUX MEADOWS in NEWFOUNDLAND AND LABRADOR; 43 is NEW MEXICO

2. [34] capital's borders hold extravagant Hopis regarding museum (9)

BIOSPHERE in QUEBEC; 34's capital is BATON ROUGE

3. Anne's farm actor Clark and family coloured like [15] bay (5 6)

GREEN GABLES in PRINCE EDWARD ISLAND; 15 is WISCONSIN but I don't fully understand the wordplay

4. Familiar Margaret's top [11] vehicle duos produce fishing village (5'1 4)

PEGGY'S COVE in NOVA SCOTIA; 11 is CONNECTICUT

5. Rapids turning around cliff's bottom, Cool J in author [24]'s birth initials (9 5)

REVERSING FALLS in NEW BRUNSWICK; Virginia Woolf was born Adeline Stephen

6. National park's academic beginners with [4] finalists (6)

ACADIA in MAINE; 4 is NOVA SCOTIA

7. Actor Phoenix and second striker George follow series filmed in [64] canyon (4 5 5)

LOST RIVER GORGE in NEW HAMPSHIRE; 64 is HAWAII

8. Region's east becomes west for [52]'s starters next to monarchy (9 7)

NORTHEAST KINGDOM in VERMONT; 52 is NORTHWEST TERRITORIES

9. Primate and heartless CEO in briefly backwater [23] peninsula (4 3)

CAPE COD in MASSACHUSETTS; 23 is DC and apparently "backwater" means "backwards" now

10. AC/DC bassist is after top-three [7] wine to stroll passageway (7 5 4)

NEWPORT CLIFF WALK in RHODE ISLAND; 7 is NEW HAMPSHIRE

11. Former [42] governor Eleanor familiarly leads to public park (8 4)

BUSHNELL PARK in CONNECTICUT; 42 is Texas (not sure the "PARK" in the answer is actually clued here)

12. Felines unwell in short [57] back mountains (9)

CATSKILLS in NEW YORK; 57 is SASKATCHEWAN

13. Short [10] strangely cued a retired Wachowski waterway (6 5)

RIDEAU CANAL in ONTARIO; 10 is RHODE ISLAND

14. Fur post finally keeps top cards in [5] top two by the fifth street in hold'em (5 5)

SNAKE RIVER in MINNESOTA; 5 is NE(w brunswick).

15. Biblical figure's top-three [35] water park (4'1 3)

NOAH'S ARK in WISCONSIN; 35 is ARKANSAS

16. "The Knife" in a capital conundrum next to [3]'s last resort area (8 6)

MACKINAC ISLAND in MICHIGAN; 3's last is ISLAND

17. Short [49] between pillars by haphazard sauna den sand formations (7 5)

INDIANA DUNES in INDIANA; 49 is NORTH DAKOTA

18. Resort town president with top billing outside [12] capital's borders (3-2-3)  

PUT-IN-BAY in OHIO; 12 is NY, capital ALBANY

19. TV's Gomer east of [17] neighbour resort town (8)

OHIOPYLE in PENNSYLVANIA; 17 is Indiana, its neighbour Ohio

20. Strike east twice from excited [21] in half-hearted book by ocean namesake town's pathway (8 4 9)

ATLANTIC CITY BOARDWALK in NEW JERSEY (21 is DELAWARE)

21. In front of seashore, bathe vigorously with [53] outsiders returning to coastal city (7 5)

BETHANY BEACH in DELAWARE; 53 is YukoN

22. Ford follows half of [26] capital with event host to stronghold (4 7)

FORT MCHENRY in MARYLAND (26 is KY, capital Frankfort)

23. Obelisk is [60] memorial structure (10 8)

WASHINGTON MONUMENT in DC; 60 is WASHINGTON

24. Plantation east of [8] with one string instrument (10)

MONTICELLO in VIRGINIA; 8 is VERMONT

25. Historic town's sharper leader goes right, actress Tina embraces [48] capital couple (7 5)

HARPERS FERRY in WEST VIRGINIA; "48 capital couple" is RR (Pierre, SD)

26. Bullets and finally brought rich cognac within the edges of [50] govern central cavern system (7 4)

MAMMOTH CAVE in KENTUCKY; 50 is ManitobA

27. Elegance with [22]'s ending mansion (9)

GRACELAND in TENNESSEE (22 is Maryland)

28. Aged [61] capital is historic district (3 5)

OLD SALEM in NORTH CAROLINA; 61 is OREGON whose capital is SALEM

29. Cay of friends having [33] loner is barrier island (4 2 5)

ISLE OF PALMS in SOuTH CAROLINA; 33 is MISSISSIPPI and I assume "loner" here means "single first letter of".

30. Forest's peculiar echo in [63] lake follows discussion (13)

CHATTAHOOCHIE in GEORGIA; 63 is California (the lake is TAHOE)

31. National park's island near [51] mysteriously loses poles, takes five before finishing classes (10)

EVERGLADES in FLORIDA; 51 is NUNAVUT, the island is GREENLAND

32. Coasts by body of water west of [31] resort city (4 6)

GULF SHORES in ALABAMA; 31 is Florida, west of which is the GULF of Mexico

33. Without question, Vicky with rest of [19]'s capital leads to siege site (9)

VICKSBURG in MISSISSIPPI; 19 is PENNSYLVANIA whose capital is (Harri)SBURG

34. Fifteen minutes behind [2]'s language neighbourhood (6 7)

FRENCH QUARTER in LOUISIANA; 2 is QUEBEC

35. Greek exclamations about the western part of [37] capital city (6 7)

EUREKA SPRINGS in ARKANSAS; 37 is ILLINOIS, capital SPRINGFIELD

36. Mountain city's [9] capital gets incomplete ransom for body (7)

BRANSON in MISSOURI; 9 is Massachusetts, capital BOSTON

37. Skyscraper or peach-shaped [29] building losing fluid for actor Bruce (6 5)

WILLIS TOWER in ILLINOIS; the Gaffney Peachoid is a water tower

38. Settlements from [32]'s end with strange anise around colon (5 8)

AMANA COLONIES in IOWA (alabAMA + COLON in NAIES)

39. Dale leaves [44] suburb, with deceit makes natural landmark (6 5)  

SCOTTS BLUFF in NEBRASKA; 44 is Arizona; Scottsdale is a suburb of Phoenix

40. Old cattle town is [16]-made car (5)

DODGE in KANSAS; 16 is MICHIGAN

41. Lake's broken head following broken heart of [38] (6 3)

BROKEN BOW in OKLAHOMA; 38 is iOWa

42. Battle site's returning [39] city not high end, keeps head low (5)

ALAMO in TEXAS; 39 is NEBRASKA, city is OMAHA

43. In the case of [27], a soup brewed bloating heads leading to native community (4 6)

TAOS PUEBLO in NEW MEXICO (27 is Tennessee so TE)

44. Lucky charm to twist a [46] River meander (9 4)  

HORSESHOE BEND in ARIZONA; 46 is COLORADO

45. Half-fleecy westward ravine following [55]'s first and second national park (5 6)

BRYCE CANYON in UTAH; 55 is BRITISH COLUMBIA

46. Hold speech alongside [20] road with no turn to mountain (5 4)

PIKES PEAK in COLORADO; 20 is NEW JERSEY and the road is its TURNPIKE

47. Geyser not adulterous after initiating [41]'s long distance (3 8)

OLD FAITHFUL in WYOMING; 41 is _O_KLAHOMA

48. [13] band with additional mountain (8)

RUSHMORE in SOUTH DAKOTA; 13 is ONTARIO

49. [6] loses foot, access lobster cooked within cottage (7 5 5)

MALTESE CROSS in NORTH DAKOTA; 6 is MAIN(e)

50. Meeting place for short [40] (5)

(The) FORKS in MANITOBA; 40 is KS

51. Even [45] or steep mountain (4)

THOR in NUNAVUT; 45 is UTAH

52. Okinawa city and [14] twins inspire top national park (7)

NAHANNI in NORTHWEST TERRITORIES; 14 is miNNesota

53. Gold rush town or [62] capital has endless dawn for a start (6 4)

DAWSON CITY in YUKON; 62 is NEVADA, capital CARSON CITY

54. Mountain's odd diesis holds [1] initials (6)

DENALI in ALASKA; 1 is NEWFOUNDLAND AND LABRADOR

55. Last twos of East Asian capital and [59] make national park (4)

YOHO in BRITISH COLUMBIA (tokYO + idaHO)

56. Mountain town to bar a duo from [36] capital (5)

BANFF in ALBERTA; 36 is MISSOURI, capital is Jefferson City

57. Flexible EU skis alongside the shores of [25] lake (8)

WASKESIU in SASKATCHEWAN; 25 must be WEST VIRGINIA

58. Battle site is small, born holding [28] capital finishers (6 7)

LITTLE BIGHORN in MONTANA; so 28 must be NORTH CAROLINA with capital RALEIGH

59. National preserve to half-heartedly create soft heroin in designer [56]'s heart by beginning moonlight (7 2 3 4)

CRATERS OF THE MOON in IDAHO; the designer is Alberta Ferretti with RE in the middle

60. Volcano street next to [58] capital making a short head (2. 6)

ST. HELENS in WASHINGTON; 58 is MONTANA

61. Body of water's confusingly rare talk inside [47] capital's borders (6 4)

CRATER LAKE in OREGON (47 is WYOMING, capital CHEYENNE)

62. Left insane next to [18] company's man-made barrier (6 3)

HOOVER DAM in NEVADA; 18 is OHIO where the Hoover company was founded

63. [30] capital scorcher held in dodgy wash riff waterfront area (9'1 5)

FISHERMAN'S WHARF in CALIFORNIA; 30 is GEORGIA (Sherman)

64. Mountain top monitors a nuke blown up within [54]'s borders (5 3)

MAUNA KEA in HAWAII; 54 is AlaskA

quantum field theory - Some questions about Ward-Takahashi Identity

I'm a learner of Peskin and Schroeder's textbook of quantum field theory.

I have proceeded to Ward-Takahashi identity and have one question when I look for Wikipedia for reference.

The following is the Ward-Takahashi identity, where $M^{\mu}$ is the correlation function for $n$ inserting electrons and $n$ out-going electrons. $$k_{\mu}M^{\mu}(k;p_1...p_n;q_1...q_n)=-e\sum_i[M_0(k;p_1...p_n;q_1...(q_i-k)...q_n)-M_0(k;p_1...(p_i+k)...p_n;q_1...q_n)].$$

The wiki says that

Note that if ($M^{\mu}$) has its external electrons on-shell, then the amplitudes on the right-hand side of this identity each have one external particle off-shell, and therefore they do not contribute to S-matrix elements.

Does on-shell means a divergent contribution according to LSZ reduction formula?

Besides, can you tell me why the S matrix is zero if all the external electrons in the left hand side are on shell?

experimental physics - 2002 research: speed of light slowing down?

Back in 2002 there was some research published hinting that $c$ may have been faster at some distant point. It was based on measurements of the fine-structure constant, $$ \alpha = \frac1{4\pi\epsilon_0} \frac{e^2}{\hbar c} \approx \frac 1{137}, $$ in light from distant (and thus ancient) quasars.

Has there been any recent developments on this? I know that at the time there was considerable doubt as to whether $c$ was inconstant. Have there been further measurements? Is it accepted now that alpha is changing? What's the current thinking on whether that means $c$ has changed?

http://www.theage.com.au/articles/2002/08/07/1028157961167.html

Answer

That result has been controversial since the beginning. A comparable survey looking at a different part of the sky saw no effect, but the original authors and some new collaborators combined data from a most-of-the-sky survey and found hints that the fine-structure constant might be large in one direction of space and small in another.

One of the strengths of the quasar observation was that was based on spectroscopic observations of atomic transitions. Since a slight change to the fine-structure constant pushes some energy levels up and others down, there were transitions from the same sources which were both redder and bluer than predicted. This was the main argument against the effect being some sort of redshift miscalibration.

If the fine-structure constant is changing over time, or if Earth is moving through regions of space where the fine-structure constant has different values, those same sorts of energy-level shifts would occur on Earth. A long-running experiment has compared the atomic-clock transition in cesium, which should be relatively insensitive to changes in α, to a particular transition in dysprosium which should have enhanced sensitivity to changes in α. So far, no earthbound effect has been seen.

Conclusion: still an open question. Stay tuned.

mathematics - Teacup geometry

Inspired by the three utilities puzzle from prog_SAHIL I'm now posting a similar puzzle that makes use of the topology of a cup with a handle:


The question is:

How many distinct points can you draw on the surface of this cup, such that it is possible to connect each point with all other points pairwise without any connections crossing each other (or one of the points)?

For example if you have 4 points the connections could look like this:

Answer

This is topologically equivalent to a torus, and you can go up to 7 points:

as shown by this Math.SE answer.

The diagram for this could look for example like this:

In this picture the lines going "under" the square represent connections going through the cup handle and the lines going "over" the square would go along the handle of the cup.

One can also just look up the answer if you know the question asks for the complete graph $K_n$ of degree $n$ with maximum $n$ such that the graph genus $\gamma (K_n)$ is at most $1$. Then

if you take the equation from Wolfram MathWorld $$ \gamma (K_n) = \left\lceil \frac{(n-3)(n-4)}{12} \right\rceil $$ you see that the genus $\gamma (K_n) \le 1$ as long as $n \le 7$.

energy conservation - Why can't we harness gravity?

We harness solar energy with constant supply of light but not with gravity. Why? What are the latest developments that can harness gravity? Why do Perpetual machines fail?

visible light - What's the intensity of a laser pointer?

The unit of intensity or the irradiance of the light is $\mathrm{W/m^2}$. So what's the intensity of a light pointer?

Searching on the web, the laser pointer's power is about $5\mathrm{mW}$, imagine the spot is $1\mathrm{mm^2}$, then the power is about $5\mathrm{kW}/m^2$, the same magnitude as sunlight on earth.

I read a paper on floquet topological phase, the light intensity required is about $10^{17}\mathrm{W/m^2}$, I can't imagine how strong this light intensity is. Please give some real physical situation where this strong light intensity can occur.

I would assume this strong light intensity would "destroy" any material, is it true? If just shed the light for a femtosecond, I believe the material will still be OK.

Answer

The system with the greatest intensity I know of is at the National Ignition Facility. This generates a peak power of around 500TW and I think the target area is around 10 mm$^2$ (I can't find a detailed description of the target). That means the power density is around $5 \times 10^{19}$ Watts per square metre, but allow a factor of ten either side due to the uncertainty in the target size.

The effect of this high power density is to compress and heat deuterium and tritium until they undergo nuclear fusion.

What are the physical explanations one can derive from determining the spectrum of cosmic ray nuclei of carbon, oxygen and boron

What are the physical explanations one can derive from determining the spectrum of cosmic Ray nuclei of carbon, oxygen and boron.

mathematics - Fair tournament for rotating teams of two

new to puzzles here. A real life situation led me to consider this. We have a small league of 6 people playing in teams of two. I would like to rotate the teams in such a way that everyone gets an equal chance to play with every other person and against every other person.

e.g. in the first game, it's A+B vs C+D and E+F sit out.

How do I arrange rotations in such a way that everyone gets an equal number of chances to play with every other player and against every other player?

Answer

Each person plays the same number of matches with each of the other 5, so the number of matches each person plays must be a multiple of 5.

If each person plays 5 matches, then there are a total of 30 players across the matches. But this is impossible because 4 players are in each match and 30 isn't divisible by 4. So each player must play at least 10 matches, with a total of 15 matches.

You can make five sets of three games each: for each set, divide the players into three pairs, and have each pair play the other two once. This is one possible way to divide the pairs:

1. AB/CD/EF


2. AC/BE/DF


3. AD/BF/CE


4. AE/BD/CF


5. AF/BC/DE

You can check that for every two players, they are together in two games (in the set where they are paired) and against each other in four games (once in each of the other sets).

quantum field theory - What is the physical implication(s) of the isomorphism between ${rm SO}(2)$ and $mathbb{R}/mathbb{Z}$?

In the book Mathematical physics by V. Balakrishnan, he says (on page 329) that the isomorphism between ${\rm SO}(2)$ and $\mathbb{R}/\mathbb{Z}$, and the fact that $\mathbb{R}$ is the universal covering group of ${\rm SO}(2)$ has deep implications in two-dimensional systems in condensed matter physics and quantum field theory without any further explanation. Can someone elaborate on what he might have in mind?

Answer

In $d\ge 3$, the first homotopy group of $\mathrm{SO}^+(1,d)$ is $\mathbb Z_2$, which essentially leads to spin quantisation. In $d=2$, and due to $\mathrm{SO}(2)\sim\mathbb R/2\pi\mathbb Z$, we have $\pi_1(\mathrm{SO}^+(1,d))=\mathbb Z$, and therefore we don't have spin quantisation anymore. Particles are no longer classified into bosons vs. fermions, but they may have any statistics. We may find anyons, which lead to a very rich phenomenology (think fractional quantum Hall effect, etc.).

Recall that spin comes from the projective representations of the little group, to wit, $\mathrm{SO}(d)$. Unlike in higher dimensions, in $d=2$ we have that $\mathrm{Spin}(d)$ is not the universal cover of $\mathrm{SO}(d)$; indeed, $\widehat{\mathrm{SO}}(2)=\mathbb R$, which is non-compact. We thus no longer require $U(4\pi)=1$, so that spin is no longer a half-integer. Fun!

quantum mechanics - In a Bell scenario, why can correlations be nonlocal only if there are at least two measurement settings to choose from?

In (Brunner et al. 2013), the authors mention (end of pag. 6) that a set of correlations $p(ab|xy)$ can be nonlocal only if $\Delta\ge2$ and $m\ge2$, where $\Delta$ is the number of measurement outcomes for the two parties (that is, the number of different values that $a$ and $b$ can assume), and $m$ the number of measurement settings that one can choose from (that is, the number of values of $x$ and $y$).

A probability distribution $p(ab|xy)$ is here said to be nonlocal if it can be written as $$p(ab|xy)=\sum_\lambda q(\lambda) p(a|x,\lambda)p(b|y,\lambda).\tag1$$ This means that if either there is only one possible measurement outcome, or only one possible measurement setting, then all probability distributions can be written as in (1).

The $\Delta=1$ (only one measurement outcome) case is trivial: if this is the case, denoting with $1$ the only possible measurement outcome, we have $p(11|xy)=1$ for all $x,y$. Without even needing the hidden variable, we can thus just take $p(1|x)=p(1|y)=1$ and we get the desired decomposition (1).

The $m=1$ case, however, is less trivial. In this case the question seems equivalent to asking whether an arbitrary probability distribution $p(a,b)$ can be written as $$p(a,b)=\sum_\lambda q(\lambda)p(a|\lambda)p(b|\lambda).$$ The paper does not mention any reference to support this fact. How can this be proven?

Answer

Make a $\lambda_{a,b}$ for every pair $(a,b)$.

Then make $q(\lambda_{a,b}) = p(a,b)\,$, and

$p(a|\lambda_{a,b}) = p(b|\lambda_{a,b}) = 1.$

optics - Virtual vs Real image

I'm doing magnification and lens in class currently, and I really don't get why virtual and real images are called what they are.

A virtual image occurs the object is less than the focal length of the lens from the lens, and a real image occurs when an object is further than focal length.

By why virtual and real? What's the difference? You can't touch an image no matter what it's called, because it's just light.

Answer

You can project a real image onto a screen or wall, and everybody in the room can look at it. A virtual image can only be seen by looking into the optics and can not be projected.

As a concrete example, you can project a view of the other side of the room using a convex lens, and can not do so with a concave lens.

---

I'll steal some image from Wikipedia to help here:

First consider the line optics of real images (from http://en.wikipedia.org/wiki/Real_image):

Notice that the lines that converge to form the image point are all drawn solid. This means that there are actual rays, composed of photon originating at the source objects. If you put a screen in the focal plane, light reflected from the object will converge on the screen and you'll get a luminous image (as in a cinema or a overhead projector).

Next examine the situation for virtual images (from http://en.wikipedia.org/wiki/Virtual_image):

Notice here that the image is formed by a one or more dashed lines (possibly with some solid lines). The dashed lines are draw off the back of solid lines and represent the apparent path of light rays from the image to the optical surface, but no light from the object ever moves along those paths. This light energy from the object is dispersed, not collected and can not be projected onto a screen. There is still a "image" there, because those dispersed rays all appear to be coming from the image. Thus, a suitable detector (like your eye) can "see" the image, but it can not be projected onto a screen.

milky way - Andromeda Collision vs Intergalactic Expansion

If the expansion of the universe is continuing unabated, which means intergalactic (perhaps even interstellar) space is continuingly expanding, then:

Why is the Andromeda galaxy on a collision course with the Milky Way?

Surely if the atoms/sub atomic particles of each galaxy were thrown in different directions at the Big Bang then surely they would never collide (especially given the 'existence' of dark energy).

Answer

The universe expands continuously but the gravitational pull between Andromeda and Milkyway is still enough to overcome this expansion. That's why there are still celestial formations and even atomic nuclei(the space expands everywhere even inside atoms). Five fundamental forces can overcome the expansion of the universe and hold particles together in an ordered formation. Keep in mind this answer is just an elementary theoretical explanation and there is complex mathematical data that comes with it.

string theory - "low-energy effective action" but in what sense?

In string theory, consistency with Weyl invariance imposes dynamics on the background fields through the vanishing of the beta functions. Those dynamics can also be derived from the so-called low energy effective action: $$S = \frac{1}{2\kappa_0^2}\int d^{26} X\; \sqrt{-G}\; \mathrm{e}^{-2\Phi}\,(R-\frac{1}{12}H_{\mu\nu\lambda}H^{\mu\nu\lambda}+4 \partial_{\mu}\Phi\partial^{\mu}\Phi)$$ (at least in bosonic string theory)

Maybe I shouldn't worry over lexical denomination, but I find this naming of "low-energy" a bit obscure. In what sense is it used? Would it be because the background fields are supposed to emerge from the fundamental strings? Or because we forget the massive excited states of the string (with masses around the Planck scale and irrelevant for low energy phenomenology)?

Answer

It's a standard terminology – and set of insights – not only in string theory but in quantum field theories or anything that can be approximated by (other) quantum field theories at... low energies.

Such a low-energy action becomes very accurate for the calculation of interaction of particles (quanta of the fields) of low energies, in this case $E\ll m_{\rm string}$. Equivalently, the frequencies of the quanta must be much smaller than the characteristic frequency of string theory. The previous sentence may also be applied in the classical theory: the low-energy effective action becomes accurate for calculations of interactions of waves whose frequency is much lower than the stringy frequency or, equivalently, whose wavelength is much longer than the string scale, $\lambda\gg l_{\rm string}$.

Low-energy effective actions may completely neglect particles whose mass is (equal to or) higher than the characteristic energy scale, in this case $m_{\rm string}$, because such heavy particles can't be produced by the scattering of low-energy particles at all – so they may be consistently removed from the spectrum in this approximation.

The scattering of the light and massless particles that are kept may be approximately calculated from the low-energy effective action and this approximation only creates errors that are proportional to positive powers of $(E/m_{\rm string})$ so these errors may be ignored for $E\ll m_{\rm string}$. You may imagine that there are corrections in the action proportional to $\alpha'$ or its higher powers that would make the effective action more accurate at higher energies but become negligible for low-energy processes.

There are lots of insights – conceptual ones as well as calculations – surrounding similar approximations and they're a part of the "renormalization group" pioneered mainly by Ken Wilson in the 1970s. In particular, by "low-energy effective actions", we usually mean the Wilsonian effective actions. But they're pretty much interchangeable concepts to the 1PI (one-particle-irreducible) effective actions, up to a different treatment of massless particles.

It is impossible to teach everything about the renormalization group and effective theories in a single Stack Exchange answer. This is a topic for numerous chapters of quantum field theory textbooks – and for whole graduate courses. So I just conclude with a sentence relevant for your stringy example: string theory may be approximated by quantum field theories for all processes in which only particles much lighter than the string mass are participating and in which they have energies much smaller than the string scale, too. If that's the case, predictions of string theory for the amplitudes are equal to the predictions of a quantum field theory, the low-energy effective field theory, up to corrections proportional to powers $(E/E_{\rm string})$.

What does the discovery of the Higgs Boson mean for physics?

Will this unite some theories, or cause some other change in physics, and perhaps our undertanding of the universe?

quantum mechanics - What are the (philosophical) implications (if any) of this experiment in relation to the Bohr-Einstein debates and hidden variable theories?

What are the (philosophical) implications  (if any ) of this experimental result, in relation to the Bohr - Einstein debates and hidden variable theories of quantum mechanics? A related question can be found here

Answer

There seems to be a confusion in this experiment, mixing frameworks and models.

In the publication in the beginning of the description of the experiment they state:

"First, we developed a superconducting artificial atom with the necessary V-shape level structure " .

Super conductivity is a meta level , emergent from the underlying quantum mechanics, and modeled with quantum mechanics. That it can be modeled quantum mechanically does not subtract from the fact that it is emergent from zillions of particles. My opinion is supported by the statement in this tutorial on quantum trajectories , that everything is within the scope of standard quantum mechanics and its postulates.

"In these systems, the stochastic equations arise as effective evolution equations, and are in no sense anything other than standard quantum mechanics(except, perhaps, in the trivial sense of approaching the limit of continuous measurement)

One cannot argue for the basic postulates of quantum mechanics on emergent data from underlying quantum mechanics, imo. In a sense it is like taking the mechanics in a video game to deduce the mechanics in real three dimensional space.

Let me give an example from classical physics:

Thermodynamics emerges from classical statistical mechanics, can one deduce from thermodynamic states the underlying particle interactions and assert their continuity and change classical mechanics postulates?

It so happens that the emergent behavior of electron pairs in metals can be modeled quantum mechanically , that means that the theory fits the data with postulates appropriate to the data. If for this superconducting construct, "atom", there is a discrepancy with the basic quantum mechanical postulates, too bad, it means that the model's postulates have to change for superconductors, with whatever this implies.

In my opinion to really question the underlying quantum mechanical framework of elementary particles one needs experiments with elementary particles, not with collective phenomena emergent from zillions of such particles. Hidden variable theories for the basic quantum mechanics framework cannot be based on emergent states.

quantum mechanics - In double slit experiment how do we observe electrons near slit?

If we observe the electron near the slit, does it produces two bands instead of interference pattern. Also how do we observe electron. Is it by producing electric field. Also does double slit experiment only works with coherent sources of light or all sources? What do you think about the below explanation of wave. http://www.bioinsilico.blogspot.in/2015/03/intelligence-and-in-search-of.html In the figure, the length of up and down represents amplitude and the speed of up and down represents wavelength. If we are observing electrons by producing electric field, won't it disturb the wave nature of electrons.

cosmology - Is there no radioactive decay between nuclear fusion and solid material formation?

I'm aware my question might be considered a duplicate of this one: Radio-dating and the age of the earth

I read that one and I looked everywhere and I still can't find my piece of mind. I would really like to understand the following points:

• 
  

  does one uranium nucleus start the radioactive decaying process the instant it is born inside a supernova ? (I understand that for a single nucleus the chance of decaying which is an instantaneous event, grows from 0% to 100% in the time from the beginning to the half-life)

  
  



• 
  

  while talking about a collection of atoms, Wikipedia states here that "This predictability allows the relative abundances of related nuclides to be used as a clock to measure the time from the incorporation of the original nuclides into a material to the present."

  
  


• 
  

  @JohnRennie helped me understand that chemically speaking zircon and lead do not mix whereas zircon and uranium do. So basically macroscopic contamination via accretion with flying lead is an event too rare to take into consideration right ? Also... I understand that the uranium-lead decay series has very short half-lives for every nuclide except for the beginning and the end (uranium and lead) and so, from a cosmological perspective it's quite improbable that zircon would mix with anything other than uranium, should a chemical compatibility exist with another nuclide.

  
  

Please bare with me and excuse my "highly scientific" vocabulary.

I have an experiment I would like to imagine in order to discover what I'm doing wrong:

1. A star goes supernova


2. We concentrate our attention on one uranium nucleus which is fused together during the violent event


3. The uranium nucleus travels for millenia towards the an accretion cloud surrounding a newborn star


4. The nucleus collides with brand new molten zircon crystal and becomes part of it.

Let's say that event 4. was happening after 0.99 uranium half-lives and the uranium nucleus still hadn't decayed. Isn't it imminent that the nucleus will decay pretty soon after event 4. ?

I mean, I understand that if the nucleus would've become thorium or protactirium or lead it wouldn't've gotten stuck in the crystal for chemical reasons, but if it was still uranium what does that say about the internal clock of the nucleus which had been ticking for 99% of the half-life ?

Is that clock reset because of external electromagnetic reasons the moment the nucleus becomes part of the crystal ?

Thank you for your indulgence

Answer

Here's the key point:

If [the nucleus] was still uranium what does that say about the internal clock of the nucleus which had been ticking for 99% of the half-life ?

The answer is "nothing," because a single nucleus doesn't have an internal clock.

Here's a better model for what's happening in an unstable nucleus. Think of it as an enclosed container with a small hole, or a tunnel, somewhere near the top.

Somewhere down in the bottom of the container is a rubber ball. That's the alpha particle. But the whole thing isn't just sitting there, static, because oscillators in quantum mechanics don't sit still even in their lowest energy states. The container is jumping and jiggling around, like it's on a paint shaker, and the ball is constantly bouncing off the bottom and off the walls. Importantly, the hole in the container always stays up near the top.

What happens to this system over the long term? Most of the time the ball only bounces partway up the walls of the container and falls back down. Sometimes it goes higher than the opening — but because the opening is small, the ball usually misses and falls back to the floor again. Eventually, by chance, the ball will find just the right trajectory so that it hops right out the tiny exit hole and leaves the container for ever. But the vibrations that take the ball near the hole are chaotic, and there's no way for you know whether the ball is "about" to come out or how long it's been bouncing around. Every time the ball hits the bottom, it gets a new random trajectory. The dice roll again.

If you have an unstable system, there's 50% chance it'll still be intact after one half-life. If you check on it later, and it's still there, there's still a 50% chance that it'll be intact after another half-life. But there's no memory, and no "almost decayed," for a single nucleus.

riddle - WHAT?! AM I BOTHERING YOU?

So often forgotten, unless accidentally used,
How am I so rotten; why am I so often abused?

I'm loud and obnoxious, it's obvious when I'm there,
For I'm not for the cautious, you might use me when you swear.

Take away all restraint, and I'm what you're left with.

What am I?

Answer

You are:

Caps Lock

So often forgotten, unless accidentally used,

How am I so rotten; why am I so often abused?

People often forget they have caps lock on, or accidentally use caps lock, especially when entering passwords

I'm loud and obnoxious, it's obvious when I'm there,

For I'm not for the cautious, you might use me when you swear.

Caps lock enable "loud" capital letters, and it's obvious when people are using it. When people are swearing, I'm sure they use caps lock

Take away all restraint, and I'm what you're left with.

People go wild with caps lock

Additionally

The first letter of each line spells SHIFT, which some people might opt for rather than caps lock

And the title...

Is written with caps lock (or shift, who knows)

relativity - Wormholes, time travel, and time dilation

I believe I understand the "wormhole time machine" now, but still have one question regarding what happens if you enter the stationary mouth.

A wormhole is created and (of course) has two mouths, $x$ and $y$.

$x$ remains stationary on Earth while $y$ is sent away at relativistic speeds (.99c) for (example) five years as measured on Earth (observer at the stationary mouth $x$).

The spaceship with mouth $y$ returns to Earth. Mouth $y$ and any occupants of the spaceship have obviously aged less than five years, though observers on Earth and wormhole mouth $x$ have aged 5.

What now happens if someone enters into wormhole mouth $y$?

As explained below by Ben Crowell, you would exit mouth $x$ at a time in the past. This happens because the time (as measured by clocks, if you wish) inside the mouths of the wormholes remains syncronized. So, when you enter $y$ (the younger wormhole mouth), it takes you to the mouth of $x$ when it was the same age. You end up in the past.

What happens if someone enters into wormhole mouth $x$?

I can't find a definite answer on this, so I'll try to work through it. I think you would emerge from mouth $y$ - the one in the present time (5 years old). Because the time inside remains synchronized between the wormholes regardless of the outside age, then entering $y$ takes us to $x$ in the past (sometime between <5 and 5 years ago); however, entering $x$ does not take us to the future because mouth $y$ (outside) has not yet aged beyond 5 years since we started our experiment.

Can someone confirm if my reasoning is correct or not?

big list - What are the next generation physics experiments?

The LHC and LIGO are two recent examples of hugely ambitious experiments in fundamental physics, both of which took decades to develop.

What are the next major experiments currently being planned and developed? What will they measure? What impact are they expected to have? And when are they expected to go live?

One example:

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  eLISA due 2034

  
  

  Developed by the ESA, eLISA will be the first dedicated space-based gravitational wave detector. Consisting of three probes spanning millions of kilometres, it will provide a hugely more accurate window to gravitational waves.

  
  

  Possible signal sources: the usual GW stuff, the early phase of the big bang, and even speculative objects like cosmic strings.

  
  

I suggest, at some point, we collate all of the answers into a single community post.

Can Impossible Vacuum balloon be possible with this idea?

As it is well known, a vacuum balloon using the materials we have at our disposal is not possible, because of the sheer force they have to resist from the air outside.

I have an idea and I want to know if there is anything that makes it impossible.

1) Build a balloon out of graphene which is very light and it is conductive too. 2) Since graphene is not airtight, cover graphene (that is not yet inflated) with a good electrical insulator which is also airtight. 3) Now make the graphene store electric charge. 4) This would exert force from inside as the electric charges on the graphene repel each other. 5) This causes the balloon to inflate while having a vacuum inside.

Now my question is there any theoretical problems (not engineering problems) with this model?

Answer

Interesting idea! Let's check the numbers.

For this to work at Earth's surface, the electric force has to cancel $P = 10^5 N/m^2$.

The electric field $E = \sigma / \epsilon_0$ in the case (no field inside, all the field goes outside).

For an area $A$, the force on a conductor due to a field $E$ is $F = QE/2 = \sigma A E / 2$ so the force per unit area is

$P' = F/A = \sigma E / 2 = \epsilon_0 E^2$

where I've expressed it in terms of $E$ so we can solve for the electric field $E$ needed to set the electric "pressure" equal to atmospheric pressure (ideally, you'd want a bit more to keep things stable):

$ \epsilon_0 E^2 = 10^5$

$ E^2 = 10^5 / 9\times10^{-12} \sim 10^{16}$

So the electric field you need is about $10^8$ volts/meter. That's a lot. It's 3,000 times more than the 30kV/m breakdown voltage of air.

What does that mean? It means that the charge on your balloon is going to instantly spark off into the air and the balloon won't be held up against the atmosphere any more.

Note that insulating the balloon won't help. The electric field has to go off to infinity to hold up the balloon against pressure. If you add insulation against sparking, several things can happen:

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  That insulation will still have the electric field outside it, so where the insulation ends, the sparking starts, and the balloon ends up neutralized.

  
  
  


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  Or you make the insulation so thick, without adding weight, that the surface area of the ballon is much, much, much (3000X) times bigger than the balloon itself, spreading the field out and reducing it.

  
  

So it looks like this isn't going to work at Earth's surface. Bottom line, the atmosphere pushes too hard, electrostatic forces are too weak, and the atmosphere isn't a good enough insulator.

But it's still a cool idea.

mathematics - The Ebbozonian coin weighing puzzle

In the country of Ebbozonia, there are only two type of coins: light coins and heavy coins. The weights of these coins satisfy the following properties:

• All light coins have the same weight $L$.


• All heavy coins have the same weight $H$.


• Heavy coins are heavier than light coins: $H>L$

The precise values of $L$ and $H$ are not known to the public. The difference between $L$ and $H$ should be fairly small, as there is no way of distinguishing the two coin types without a balance. His Highness, the Honourable Minister of Treasury, has recently announced the following important property of the Ebbozonian coin system:

• If a balance is in perfect equilibrium with $h_1$ heavy and $\ell_1$ light coins on the right pan and with $h_2$ heavy and $\ell_2$ light coins on the left pan, then $h_1=h_2$ and $\ell_1=\ell_2$ must necessarily hold true.

Cosmo puts 10 Ebbozonian coins on the table and asks Fredo to determine quickly whether these 10 coins all have the same weight. On the table, there is a balance with two pans (but there are no weights).

Question: Can Fredo solve Cosmo's problem by using the balance at most three times?

Answer

The answer is

Yes

I ran a test with 6 coins, leaving 2 off each time. I got that to work. This made me think I could scale it up to 10 coins by doing 3 on each side. I tried that and got all the cases where 2 coins were different, but 3 coins would stop me (and I thought only the even sets would be a problem). That made me try a different number of coins in each weighing. That's when I found:

Weighing 1: 1 2 3 vs 4 5 6
Weighing 2: 1 2 3 6 vs 7 8 9 10
Weighing 3: 1 2 3 4 5 vs 6 7 8 9 10

quantum information - Can 2D graph state be represented by branching MERA?

A 2D graph state is a highly entangled state to support general measurement based quantum computation. But its state complexity is relative low. Branching MERA represents also a set of low complexity but high entanglement states.

Question: Can 2D graph state (for example on a regular 2D lattice) be represented by branching MERA? Why or why not?

A little bit confused about this problem. I guess the answer is no, otherwise 2D graph state can be classically approximated. But is there a conclusion somewhere?

wordplay - Silly Sally likes wearing glasses but not spectacles

Silly Sally likes balloons, but not party hats.

Silly Sally likes eating pizza, but not burgers.

Silly Sally likes using alliterations, but she doesn't like Metaphors.

Silly Sally likes browsing Reddit, but not Stack Overflow.

What does Silly Sally like?

Answer

I think the idea here is that:

Silly Sally likes doubled letters. From the limited examples we have, this makes sense.

BONUS:

Silly Sally likes Puzzles! But not Questions!

general relativity - Are all black holes singularities identical?

There seems to be a problem between a singularity and the event horizons size. My logic is this if u have two collapsing stars with different masses there horizons will be different in diameter yet both singularities are said to be the same infinite density points. This would make them to be identical structures since theres no difference between two infinite dense points. So how does one have a larger horizon. And another problem is that theory says that light coming from the collapsed star inside is curved back in what is converging light rays but if this is the case then why does the horizon become larger the more matter the black hole consumes. Shouldnt the light within curve even more since more mass is introduced and shouldnt the horizon shrink instead of grow?

general relativity - The vacuum light speed: Is it really constant, i.e., independent of location in space-time?

I am by no means an expert in this field, however something puzzles me about the speed of light and the relativity of time and space (space-time).

Is is universally acknowledged that the speed of light (299,792,458 m/s) is the universal speed limit, and that nothing can travel faster than light? That is a measurement based on a man-made interpretation of time (hours, minutes and seconds etc. are man made...there is nothing natural dictating how long a second should be).

For instance, according to Einstein, time and space bend around the physical matter of the universe, so for example, time near or on the surface of a "super massive black hole" should be drastically slower, relative to that of earth. Lets say for example that for every second that passes on the black hole, 10 seconds pass on earth, so essentially time on the surface of the black hole is 10 times slower than the time on earth.

Given the example above, is the speed of light at the surface of the black hole still 299,792,458 m/s, or is it 299,792,458,0 m/s?

Answer

There are more ways to understand this, and the answer is the same: yes, the speed of light is constant. I will try to explain this in the way I consider simpler, but I am sure that others have their preferred explanation. Anyway, to receive a fair explanation, I suggest you to read something more detailed about special relativity, and then about general relativity. My explanation is geometric. Being mathematical, this means that you try first to understand what I mean in an imagined world of mathematics, to see that the ideas are consistent. Think at it as a sci-fi movie, in which you are only concerned with logical possibility. I suggest you that only after you are happy with the self-consistency of the model, you try to judge this image and compare it with what you know about physical world.

Spacetime is a space with four dimensions. Near each point, the spacetime is almost flat, but as we depart from the point, it becomes curved. On very short (infinitesimal) distances, the spacetime being almost flat, we can write a Pythagoras' theorem. In four dimension it is like

$$d s^2= - c^2d t^2+d x^2 + d y^2 + d z^2.$$

We are interested to make it work also in frames which are not normalized, and have different scales (denoted here by $g_{aa}$), and hence different measurement units:

$$d s^2= g_{00}d x_0^2+g_{11}d x_1^2 + g_{22}d x_2^2 + g_{33}d x_3^2.$$

Here I replaced $ct,x,y,z$ with $x_0,x_1,x_2,x_3$. But we also want to write this in coordinates whose axes are not necessarily orthogonal, so we have to add some cosines between the axies $a$ and $b$, which are written as $g_{ab}$:

$$d s^2=\sum_{a,b}g_{ab}d x_a d x_b.$$

This also works for curvilinear coordinates (we allow the metric coefficients $g_{ab}$ to vary from point to point), which are the suitable ones for curved spacetime.

The length is given by Pythagoras' theorem. Some infinitesimal distances are $d s^2>0$, and they separate points which can be in the same space. Some are $d s^2<0$, and they measure time intervals. Some are $d s^2=0$, and such directions are called light-like direction. So, if you measure the length of a curve described by a photon in spacetime, you measure it with this theorem, and you always obtain it to be $=0$. If you choose the reference frame so that $d x_2=d x_3=0$, and go back to $x,y,z,t$ notation, you obtain that

$$g_{11}d x^2+g_{00}c^2d t^2=0,$$

and the light speed is apparently

$$\frac{d x}{d t}=c\sqrt{-\frac{g_{00}}{g_{11}}},$$

which is not necessarily $=c$. Can we conclude that it is not constant? Well, not, because this formula doesn't show the speed of light in tehe units in which $c$ is expressed, but in some other units, which are scaled. To find the correct answer, we either choose the frame to be orthonormal, which gives $g_{11}=-g_{00}=1$, or we make sure to divide each infinitesimal distance with the unit "conversion factor" along that direction (they are just $\sqrt{-g_{00}}$ and $\sqrt{g_{11}}$). Hence, the speed of light is always $c$, although in rescaled coordinates may appear not to be. You cannot make it different, no matter how you try, unless you rescale it (i.e change the units).

Now, please see that this is not a proof that the speed of light is constant. It is constant by the very construction of the spacetime. It would be circular to claim that this shows the speed of light is constant. I showed you this construction to explain how it is consistent to have constant speed of light, even if time and space intervals change in different frames. Now you can compare this model with the physical data.

Now, the "true" speed of light is that present in the wave equation describing the light, in vacuum. And this is still $c$. Can it vary from point to point? It may be possible, in principle, but it is consistent with the observations that it remains constant. If it would vary, Maxwell's equations would not be covariant. This would not be a big deal, one can imagine worlds in which they are not covariant. But the theory of relativity originated from the study of their invariance.

statistical mechanics - How Non-abelian anyons arise in solid-state systems?

Recently it has been studied non-abelian anyons in some solid-state systems. These states are being studied for the creation and manipulation of qubits in quantum computing.

But, how these non-abelian anyons can arise if all particles involved are fermions?

In other words, how electronic states can have different statistics than the fermionic if all electrons are fermions?

Answer

The realization of non-Abelian statistics in condensed matter systems was first proposed in the following two papers. G. Moore and N. Read, Nucl. Phys. B 360, 362 (1991) X.-G. Wen, Phys. Rev. Lett. 66, 802 (1991)

Zhenghan Wang and I wrote a review article to explain FQH state (include non-Abelian FQH state) to mathematicians, which include the explanations of some basic but important concepts, such as gapped state, phase of matter, universality, etc. It also explains topological quasiparticle, quantum dimension, non-Abelian statistics, topological order etc.

The key point is the following: consider a non-Abelian FQH state that contains quasi-particles (which are topological defects in the FQH state), even when all the positions of the quasi-particles are fixed, the FQH state still has nearly degenerate ground states. The energy splitting between those nearly degenerate ground states approaches zero as the quasi-particle separation approach infinity. The degeneracy is topological as there is no local perturbation near or away from the quasi-particles that can lift the degeneracy. The appearance of such quasi-particle induced topological degeneracy is the key for non-Abelian statistics. (for more details, see direct sum of anyons? )

When there is the quasi-particle induced topological degeneracy, as we exchange the quasi-particles, a non-Abelian geometric phase will be induced which describes how those topologically degenerate ground states rotated into each other. People usually refer such a non-Abelian geometric phase as non-Abelian statistics. But the appearance of quasi-particle induced topological degeneracy is more important, and is the precondition that a non-Abelian geometric phase can even exist.

How quasi-particle-induced-topological-degeneracy arise in solid-state systems? To make a long story short, in "X.-G. Wen, Phys. Rev. Lett. 66, 802 (1991)", a particular FQH state $$ \Psi(z_i) = [\chi_k(z_i)]^n $$ was constructed, where $\chi_k(z_i)$ is the IQH wave function with $k$ filled Landau levels. Such a state has a low energy effective theory which is the $SU(n)$ level $k$ non-Abelian Chern-Simons theory. When $k >1,\ n>1$, it leads to quasi-particle-induced-topological-degeneracy and non-Abelian statistics. In "G. Moore and N. Read, Nucl. Phys. B 360, 362 (1991)", the FQH wave function is constructed as a correlation in a CFT. The conformal blocks correspond to quasi-particle-induced-topological-degeneracy.

newtonian mechanics - Train crash: are these situations alike?

I was just wondering... I believe that if a car travelling 50 miles per hour crashes into a wall, the result should be the same as crashing to another car also travelling 50 miles per hour (but in the other direction of course)

Is this true? Why is that?

Answer

This is true if the wall is so strong that it does not move and is not damaged in the course of the crash, so this can be only approximately true. So why is this (approximately) true? Because, on the one hand, the consequences of a crash are determined by the accelerations in the course of the crash, on the other hand, due to the symmetry, accelerations will be approximately the same in the crash of two identical vehicles as in a crash of a vehicle against a very strong wall (in both cases, no parts of the first vehicle can travel beyond the plane of symmetry of the two vehicles or beyond the nearest surface of the wall). However, some differences are still possible, as parts of the vehicle can rebound in a different way in these two cases (the collision can be (in)elastic to a different extent in these two cases).

special relativity - How does a photon experience space and time?

To an an external observer it appears that time has stopped for photon. But this relation is reflexive, so for an observer travelling with the photon it appears the universe has stopped everywhere.

Is this right?

Space also gets distorted parallel to the direction of motion, but not perpendicular to it.

Does this mean that for an observer travelling with a photon sees spacetime as a flat plane?

Note 1: I'm using language vividly not literally when I say a photon experiences space and time. Not that I'm against idealist or panpyschist interpretations of matter or energy come to that.

Note 2: Since it's been pointed out that an inertial frame can't be boosted into the frame of a photon and moreover there is no limiting behaviour here - no matter what frame you are boosted into, light will always appear to travel at c. Nevertheless, when an external observer watches a train chasing a photon by gradually increasing its speed, he will note that the train speed is approaching the speed of the photon (but will never match it) and when he looks at the clock inside the train he will not too that it is gradually slowing down (but never actually stops). Using this picture, we can attempt the thought experiment - if not one that can be carried out - of what a world looks like to a photon.

Answer

There is a more precise sense in which the question is ill-posed (at least mathematically); namely, it is a fundamental assertion of relativity (special and general) that the time 'measured' (counted, experienced, observed...) by an observer between two events occurring on her worldline is the length of her worldline-segment joining the two events (that's how we connect the physical notion of (personal) time with the mathematics of the theory). The way she determines motion depends on this notion of time. Equivalently, proper time is measured by the arc-length parameter of the observer. Now, since null curves have zero length (hence no arc-length parameter) the concept of proper time is not defined for null observers. Hence neither is (proper) relative motion (i.e `from the photon's perspective').

Also, the relation you describe between timelike and null (instantaneous) observers isn't reflexive at all (whereas it is for the timelike ones, via the `Lorentz boosts'): no isometry of Minkowski space can take a timelike vector to a null one.

Although the question doesn't make sense, in this strict sense, mathematically, perhaps there are other physical or mathematical tricks for interpreting it?