calculation puzzle - Create all numbers from 1-100 by using 1,3,3,7

Create all the numbers from $1$ to $100$ using the numbers $1,$$3,$$3,$ and $7.$

You can only use each number once, except for the $3,$ of which you have two.
You can use addition $(x+y),$ subtraction $(x−y),$ division $(\frac{x}{y}),$ multiplication $(x\times y),$ exponentiation $(x^y),$ roots $(y\sqrt x)$, factorials $(x!)$ and ceiling and floor $(\lceil x\rceil,\lfloor y\rfloor)$.

You can combine numbers like $1$ and $7$ to $17$ etc.
Use of any types of brackets are also allowed.
Good luck!

Answer

1 number missing. I give up

$0 = 7-3-3-1$
$1 = 7 \times 1 - 3 - 3$
$2 = 7 + 1 - 3 - 3$
$3 = \frac{7 - 1 + 3}{3}$

$4 = \sqrt{17-\frac{3}{3}}$
$5 = 3 \times (3 + 1) - 7$
$6 = 7 - 1 + 3 - 3$
$7 = 7 \times 1 - 3 + 3$
$8 = 7 + 1 - 3 + 3$
$9 = 13 + 3 - 7$
$10 = 3^3 - 17$
$11 = 17 - 3 - 3$
$12 = 3 \times (7 - 3) \times 1$
$13 = 3 \times (7 - 3) + 1$

$14 = 7 \times (1 + \frac{3}{3})$
$15 = 3 \times (7 - 3 + 1)$
$16 = 17 - \frac{3}{3}$
$17 = 17 - 3 + 3$
$18 = 17 + \frac{3}{3}$
$19 = 7 \times 3 + 1 - 3$
$20 = 3^3 - 7 \times 1$
$21 = 3^3 - 7 + 1$
$22 = (7-3)! - 3 + 1$
$23 = 7 \times 3 + 3 - 1$

$24 = \frac{73 - 1}{3}$
$25 = 33 - 7 - 1$
$26 = 33 - 7 \times 1$
$27 = 33 - 7 + 1$
$28 = 3 \times 7 + 3! + 1$
$29 = 3! \times 3! - 7 \times 1$
$30 = 37 - 3! - 1$
$31 = 37 - 3! \times 1$
$32 = 37 - 3! + 1$
$33 = 37 - 3 - 1$

$34 = 37 - 3 \times 1$
$35 = 37 - 3 + 1$
$36 = (7-1) \times (3+3)$
$37 = 3! \times (3! - 1) + 7$
$38 = 3! \times 7 - 3 - 1$
$39 = 3! \times 7 - 3 \times 1$
$40 = 3! \times 7 - 3 + 1$
$41 = 37 + 3 + 1$
$42 = 37 + 3! - 1$
$43 = 37 + 3! \times 1$

$44 = 37 + 3! + 1$
$45 = 3! \times 7 + 3 \times 1$
$46 = 3! \times 7 + 3 + 1$
$47 = 3! \times 7 + 3! - 1$
$48 = 3! \times 7 + 3! \times 1$
$49 = 7 \times (3+3+1)$
$50 = (7+3) \times (3!-1)$
$51 = 17 \times (3! - 3)$
$52 = 13 \times (7-3)$
$53 = \lfloor\sqrt{7^3}\rfloor \times 3 - 1$

$54 = 17 + 3^3$
$55 = 7 \times 3 + (3+1)!$
$56 = (3\times 3 - 1) \times 7$
$57 = 17 \times 3 + 3!$
$58 = \lfloor\frac{7^3}{3!}\rfloor + 1$
$59 = 3! \times (7+3) - 1$
$60 = 3! \times (7+3) \times 1$
$61 = 3! \times (7+3) + 1$
$62 = 7\times 3^3 - 1$
$63 = 7\times 3^3 \times 1$

$64 = 7\times 3^3 + 1$
$65 = (7+3!) \times (3!-1) $
$66 = 7\times 3! + (3+1)!$
$67 = (3+1)^3 + \lceil\sqrt{7}\rceil = 64 + 3$
$68 = \lfloor\frac{7^3}{3!-1}\rfloor$
$69 = \lceil\frac{7^3}{3!-1}\rceil$
$70 = (7 + 3) \times (3! + 1)$
$71 = 71 -3 + 3$
$72 = (3! + 3!) \times (7-1)$
$73 = 73 + \lfloor\frac{1}{3}\rfloor$

$74 = 3^{3+1} - 7$
$75 = 73 + 3 - 1$
$76 = 73 + 3 \times 1$
$77 = 73 + 3 + 1$
$78 = 3! \times (7 + 3!) \times 1$
$79 = 3! \times (7 + 3!) + 1$
$80 = 3^{7-3} - 1$
$81 = 3^{7-3} \times 1$
$82 = 3^{7-3} + 1$
$83 = 7 \times (3! + 3!) - 1$

$84 = 7 \times (3! + 3!) \times 1$
$85 = 7 \times (3! + 3!) + 1$
$86 = 73 + 13$
$87 = \lceil{\sqrt{\sqrt{13^7}}}\rceil - 3 = \lceil{89.00222..}\rceil - 3$
$88 = 3^{3+1} + 7$
$89 = 13 \times 7 - \lceil\sqrt{3}\rceil$
$90 = 13 \times 7 - \lfloor\sqrt{3}\rfloor$
$91 = (3!+3!+1) \times 7$
$92 = 13 \times 7 + \lfloor\sqrt{3}\lfloor$
$93 = 13 \times 7 + \lceil\sqrt{3}\lceil$

$94 = 13 \times 7 + 3$
$95 = $
$96 = 17 \times 3! - 3!$
$97 = 73 + (3+1)!$
$98 = 71 + 3^3$
$99 = (3! - 1)! - 3\times 7$
$100 = 31 \times 3 + 7$

Bonus:

If we allow log functions we can generate every number like this:
$x = \log_{\frac{1}{\lfloor\sqrt7\rfloor}}\left({\log_3\underbrace{\sqrt{\sqrt{\dots\sqrt{3\,}\,}\,}}_\text{x square roots}}\right)$
This is equivalent to
$x = \log_{\frac12}\left({\log_3{3^{\frac{1}{2^x}}}}\right)$
Going further:
$x = \log_{\frac12}\left({\frac{1}{2^x}}\right)$

riddle - A firm farewell - Rest in piece

---

Here is what I believe is the last masterpiece of Avigrail. May God rest his tiny soul.

gauge theory - primary constraints for constrained Hamiltonian systems

I would be most thankful if you could help me clarify the setting of primary constraints for constrained Hamiltonian systems. I am reading "Classical and quantum dynamics of constrained Hamiltonian systems" by H. Rothe and K. Rothe, and "Quantization of gauge system" by Henneaux and Teitelboim.

Consider a system with Lagrangian $L(q,\dot{q})$ and define the momenta $p_j = \partial L/\partial{\dot{q}_j}$, with $j$ from 1 to $n$ (for $n$ degrees of freedom), and the Hessian $W_{ij}(q,\dot{q}) = \partial^2 L/\partial{\dot{q}_i}\partial{\dot{q}_j}$. Let $R_W$ be the rank of the Hessian $W_{ij}$.

Assume that $L$ is singular and $R_W < n$. In order to ask the question, I'm now quoting how the Rothes describe the setting of primary constraints on pages 26 and 27 of their book.

Let $W_{ab}$, ($a,b = 1,...,R_W$) be the largest invertible submatrix of $W_{ij}$, where a suitable rearrangement of the components has been carried out. We can then solve the eqs. $p_j = \partial L(q,\dot{q})/\partial{\dot{q}_j}$ (1) for $R_W$ velocities $\dot{q}_a$ in terms of the coordinates $q_i$, the momenta $\{p_a\}$ and the remaining velocities $\{\dot{q}_\alpha\}$: $\dot{q}_a = f_a(q,\{p_b\},\{\dot{q}_\beta\})$, with $a,b = 1,...,R_W$ and $\beta = R_W + 1,...,n$.

Inserting this expression into the definition of $p_j$, one arrives at a relation of the form $p_j = h_j(q,\{p_a\},\{\dot{q}_\alpha\})$. For $j = a$ ($a = 1,...,R_W$) this relation must reduce to an identity. The remaining equations read $p_\alpha = h_\alpha (q,\{p_a\},\{\dot{q}_\beta\})$. But the rhs cannot depend on the velocities $\dot{q}_\beta$, since otherwise we could express more velocities from the set $\{\dot{q}_\alpha\}$ in terms of the coordinates, the momenta, and the remaining velocities.

This is where the presentation from Rothe stops, and my concern is that equations of the form $p_\alpha = h_\alpha (q, \{p_a\},\{\dot{q}_\beta\})$ (2) with all $\{\dot{q}_\beta\}$ present can still be possible, and yet one cannot solve for more velocities from the set $\{\dot{q}_\alpha\}$ in terms of the coordinates, the momenta and the remaining velocities if the conditions stipulated in the Implicit Function Theorem are not met, for not all equations of the type (2) can be solved implicitly for $\{\dot{q}_\alpha\}$. Therefore it is not proven that there are $(n - R_W)$ primary constraints of the form $\phi_\alpha (q,p) = 0$.

Henneaux and Teteilboim even state that these $(n - R_W)$ constraints of the form $\phi_\alpha (q,p) = 0$ are functionally independent, but give no justification to this statement.

I would be most thankful if you could help clarify my above concern and also if you could clarify the statement by Henneaux and Teitelboin as to the fact that the constraints are functionally independent. Thank you!

Answer

OP wrote

This is where the presentation from Rothe stops, and my concern is that equations of the form $p_\alpha = h_\alpha (q, \{p_a\},\{\dot{q}_\beta\})$ (2) with all $\{\dot{q}_\beta\}$ present can still be possible, and yet one cannot solve for more velocities from the set $\{\dot{q}_\alpha\}$ in terms of the coordinates, the momenta and the remaining velocities if the conditions stipulated in the Implicit Function Theorem are not met, for not all equations of the type (2) can be solved implicitly for $\{\dot{q}_\alpha\}$. Therefore it is not proven that there are $(n - R_W)$ primary constraints of the form $\phi_\alpha (q,p) = 0$

I sensed a misunderstanding of Rothe's statements, ignore me if I'm not understanding OP correctly:

Rothe is arguing at least one of the $\dot{q}_\beta$'s can be expressed as a function of $p_a,p_\alpha$ and remaining $\dot{q}_\alpha$'s. For any particular $\alpha$ in your equation (2), by implicit function theorem applied to one-variable($\dot{q}_\beta$) function, it's always doable unless $\frac{\partial h_\alpha}{\partial \dot{q}_\beta}=0$ for our chosen $\alpha$ and $\beta$, but the latter case simply means $h_\alpha$ does not depend on $\dot{q}_\beta$, so either case Rothe's statement is correct.

Updates:

As for the functional independence part, if I'm not mistaken, is quite trivial. It's because in your equation (2), all the $p_\alpha$'s are on LHS and RHS doesn't contain any $p_\alpha$ thus it's impossible to find a inter-relation among these equations(plural in the sense that $\alpha$ can take many values, from $1$ to $M'$). Or in differential calculus language, the constraint functions from (2) will be $\phi_\alpha(q,p)=p_\alpha-h_\alpha(q,\{p_a\})$, thus the Jacobian of these functions will simply be

$\frac{\partial \phi_\beta}{\partial \{p_\alpha,p_a, q\}}= \begin{bmatrix} 1 & 0 & \cdots &0&\cdots&\frac{\partial h_1}{\partial p_a}&\cdots&\frac{\partial h_1}{\partial q_i}&\cdots\\ 0 & 1 & \cdots&0&\cdots&\frac{\partial h_2}{\partial p_a}&\cdots&\frac{\partial h_2}{\partial q_i}&\cdots\\ \vdots & \vdots & \ddots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0 &0 &\cdots&1&\cdots&\frac{\partial h_{M'}}{\partial p_a}&\cdots&\frac{\partial h_{M'}}{\partial q_i}&\cdots\end{bmatrix}$

And this Jacobian is of maximal rank because of the identity submatrix on the left, and this is the same as saying these contraint functions are functionally independent.

P.S.: Now I'm not quite sure what situation H&T were referring to when they said $M

Why can a particle decay into two photons but not one?

I recently read an old physics news about the Higgs boson where it was observed to decay into 2 photons and I was wondering why it wouldn't have decayed into a single photon with the combined energy of 2 photons?

Answer

No massive particle can decay into a single photon.

In its rest frame, a particle with mass $M$ has momentum $p=0$. If it decayed to a single photon, conservation of energy would require the photon energy to be $E=Mc^2$, while conservation of momentum would require the photon to maintain $p=0$. However, photons obey $E=pc$ (which is the special case of $E^2 = (pc)^2 + (mc^2)^2$ for massless particles). It's not possible to satisfy all these constraints at once. Composite particles may emit single photons, but no massive particle may decay to a photon.

astrophysics - What elements can be created in the fusion process of different types of stars?

As I understand it fusion inside a sun can produce heavier and heavier elements until some sort of "nucleus size limit" is reached. As far as I understand, the limit is thought to be reached with the element iron.

What accounts for this "fusion element limit"? Is it the battle between the nuclear strong force and the electromagnetic force which limits the size of the nucleus that can exist in a fusion environment?

Is the limit of what elements can be created the same for all types of suns?

astronomy - Distance away from earth to see it as a full disk

This question is more space-related than physics-related, but here goes...

How far away the earth would I have to be in order to see the earth as a full disk? What I'm looking for is a distance in kilometers or miles. For example, when I fly in an airplane at 40,000 feet (about 12000 m), I can begin to see the curvature of the earth, but the view I have of the earth is a tiny piece of the total. Also, I know if I'm on the moon I can see the earth as a full disk. But what is the minimum distance away the earth I'd have to be to see the full disk?

The main reason I ask this question is because I am interested to know how much of the earth you can see from the International Space Station (ISS). I've seen various photo collections that supposedly show views of the earth from the ISS and some indicate that you can see the earth as a full disk. However, I am skeptical that you can see the earth as a full disk from the ISS. Below I've done some calculations to try and determine how far away from the earth you would have to be to see the earth as a full disk.

The human visual field of view is approximately 120 degrees in both the horizontal and vertical directions. If we construct a right triangle where one angle is 60 degrees (half of 120 degrees), "d" is the distance to the earth, and "r" is the radius of the earth, then d = r/tan(60) = 6371 km/1.732 = 3678 km = 2285 miles. This says that you would have to be 3678 km (2285 miles) away from the earth to see it as a full disk. Since the ISS is orbiting at an altitude of 347 km (216 miles) perigee and 360 km (224 miles) apogee (the mean is about 353 km (219 miles)), I believe that you will not be able to see the earth as a full disk from the space station.

Is this analysis correct? If not, what is the correct analysis?

Mike

newtonian mechanics - Violating Newtons First Law!

Suppose you are inside a very large empty box in deep space , floating ( i.e not touching the box from anywhere initially).The box is at complete rest. Now you push the box forward from inside. Now you would go backwards but the box will move forward to conserve momentum. However since you were inside the box your force is an internal force but the box would have moved forward. So doesnt this violate newtons 1st law as an internal force made a body move from state of rest?

mazes - Haisu: Pixel Perfect

HAISU is a portmanteau of three Japanese words - 'hairu', to enter, 'su', number, and 'hausu', an English borrow word meaning house, of course.
Together, we get a meaning of 'enter number house', which I have roughly translated to English as 'Room Count'.

The rules are simple - draw a path from the O to the X, passing through every cell in the grid exactly once. The grid is divided into several rooms. When your path passes over a cell with the big number N, it must be the Nth time you have entered the room. If a room has a small number m in the top left corner, you must enter that room a total of m times. An example Haisu puzzle and its unique solution are shown below.

Hopefully this example puzzle clarifies the rules.

My previous Haisus have been over-complicated and/or difficult, so here is a puzzle which is MUCH more moderate in difficulty!

Answer

Here is the unique solution:

and here's one way to get there, all steps being forced so that the outcome is known to be unique. Let's begin with the 3-visit region near the top left. There's only one way to arrange that.

And now the regions to its left are straightforward.

Let's add the obvious things in the corners, which we should really have done first on autopilot.

Now we can fill in the centre-left region:

The path can't cross the north and west boundaries of the central region for fear of giving too many visits to the L-shape above it.

Thinking about that region at top centre, we have to do this for fear of too many visits to adjacent regions:

Count entrances and exits for the 2-visit L-shape near the middle:

If we join the upper two bits of path in that region near the bottom left, they'll end up being part of a loop that doesn't visit every square. So don't do that:

We can't enter the single-visit region near top right from the left, because that would make the top-right region impossible, so:

There's very little flexibility at top right and we fairly quickly find we have to do this:

Counting entrances and exits of the bottom area, we must have this:

leading quickly here:

There's only one way to handle the cell to the right of the start, and then we can count entrances and exits of the 3-visit L-shaped region:

Now there's only one way to fill in that narrow corridor:

and all that remains is the earliest portion of the path:

calculation puzzle - MC Frontalot 80085

MC Frontalot is a musician, and one of his songs is a word problem that seems to me like it belongs here. If you want to listen to it instead of just reading the lyrics, a quick video search of "MC Frontalot 80085" will bring it up on YouTube. Here is the full song, copied (with permission) directly from his website:

80085

Last time I had a math class, there wasn't any internet

invented yet. That isn't on the level but I'll try to pique your interest

with half-truths and lies.

As ever, MC Frontalot feigns innocence and denies.

I won't admit it. You can't make me say it:

that I dropped Calc B more than a year before Mosaic.

Oh no now it's out, now it's shouted from the balconies:

that Frontalot's about to be engaging in some alchemy.

I'll turn a string of operands into some smut.

If that sort of thing's offensive to you keep eyes shut.

Or better yet, don't even enter, into calculator, song.

But if you're ready to be titillated, follow along.

Ready? Go. Eighty women went to the podiatrist.

Arrive: simultaneous. Soon the scene's riotous.

Nine just leave. Those in the difference persevere,

packing up the lobby very tightly, domineered

by one Sally Gorey (that's her given name)

(though her title is Reception) (and professional acclaim

is due her) ('cause she did what needed doing). And it's done:

she opened up the schedule, slotted every single one.

But, um... not many on a Friday afternoon!

All but an eighteenth of the women in the room

had to vrooooom. For each remaining patient

x-rays were taken. Then the doctor took vacation.

Why was that vacation germane to the math?

'Cause of good data policy in the office and a vast

abundance of caution on the part of our Sally:

eight backups nightly, automated, and the tally

only ever shrinking when manually deleted.

All of this occurring in the box behind reception so she needed

a full backup of that box, noons.

These weren't incremental, so her server needs ballooned.

Who deserved to flee Duluth? The doctor was in Rio

for three work weeks and another Monday just to be so

thoroughly relaxed upon return.

Have you gathered all the facts that you needed to discern?

Morning in the office, after vacatings:

out of those belonging to the original 80 ladies.

How many digital toes were in images grand total?

Your evidence so far is largely anecdotal.

And you're keen to know if any had deformity. So icky!

Ten toes per customer; this puzzle: not that tricky.

Key in your calc. Check your seven-segment indicator.

Now add my eliteness. Notice that the sum is greater

than expected. You still have to subtract

two for a pair of things Sally has that I lack.

I warned you it was kind of immature; I wasn't skirting the issue.

Still you snicker at the calculator. "Dirty! I need a tissue."

I hope this is assumed, but I will state anyway that how you reached your answer is as important as the answer itself, and just a number (even if it's the right number) is not an acceptable answer.

Answer

The answer is

already in the title: 80085.

Here are the steps to getting there:

80 women went to the podiatrist, of which 9 just leave. There are 71 left, plus Sally, so 72. All but an eighteenth had to leave, so 4 stayed (Sally and 3 patients). 3 patients times 2 feet, so we start with 6 pictures. Then, during each of the following 25 nights (Friday to Tuesday, 3 weeks), the number of images at the box behind the reception increases by 48 (8*6) and the complete reception box gets a backup in the 'backup box' at the noon afterwards. Now, the nightly increase happens 25 times, and the daily backup happens 24 times. That results in a total of 15750 images. The amount of toes is 15750*5, so 78750. Now add my eliteness, so 1337. This results in 80087. We have to subtract 2, so we end up with 80085.

And now in plain calculations:

women = 80-9 = 71
women = 71+1 = 72
women = 72/18 = 4
patients = 4-1 = 3
pictures = 3*2 = 6
reception(1) = 54, backup(1) = 54
reception(2) = 102, backup(2) = 156
reception(3) = 150, backup(3) = 306
reception(4) = 198, backup(4) = 504

reception(5) = 246, backup(5) = 750
reception(6) = 294, backup(6) = 1044
reception(7) = 342, backup(7) = 1386
reception(8) = 390, backup(8) = 1776
reception(9) = 438, backup(9) = 2214
reception(10) = 486, backup(10) = 2700
reception(11) = 534, backup(11) = 3234
reception(12) = 582, backup(12) = 3816
reception(13) = 630, backup(13) = 4446
reception(14) = 678, backup(14) = 5124

reception(15) = 726, backup(15) = 5850
reception(16) = 774, backup(16) = 6624
reception(17) = 822, backup(17) = 7446
reception(18) = 870, backup(18) = 8316
reception(19) = 918, backup(19) = 9234
reception(20) = 966, backup(20) = 10200
reception(21) = 1014, backup(21) = 11214
reception(22) = 1062, backup(22) = 12276
reception(23) = 1110, backup(23) = 13386
reception(24) = 1158, backup(24) = 14544

reception(25) = 1206
totalimages = 1206+14544 = 15750
toes = 15750*5 = 78750
elitetoes = 78750+1337 = 80087
answer = 80087-2 = 80085

quantum mechanics - Position Operator Eigenvectors change with Space Displacement

I am working through Ch. 3 of Ballentine where he finds the commutator relationships between various operators.

He begins on p.78 with a space displacement

$$\mathbf{x'} = \mathbf{x} + \mathbf{a}$$

Which involves a corresponding displacement of position eigenvectors

$$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|\mathbf{x}\rangle = |\mathbf{x} + \mathbf{a}\rangle .$$

However, I believe it should be instead

$$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|\mathbf{x}\rangle = |\mathbf{x} - \mathbf{a}\rangle .$$

Because, he mentions that we are taking the "active" point of view, where we keep the coordinates the same but instead shift our vectors and operators. In this case, $|\mathbf{x'}\rangle$ is shifted $+\mathbf{a}$ which in turns means that it is equal to unprimed position eigenvector $-\mathbf{a}$.

Answer

After going through the text again, I wanted to give an answer more motivated by the exposition in the text.

From, p.63, since the laws of nature are invariant under certain space-time transformations, if $A | \phi_n \rangle = a_n | \phi_n$ with $A$ representing an observable, and $\phi_n$ an eigenvector, then after the transformation $A' | \phi_n' \rangle = a_n | \phi_n' \rangle$ because they represent the same observable.

And from p.64 and p.65, given a space-time transformation $U$

$$|\phi_n' \rangle = U|\phi_n \rangle$$

$$A' = UAU^{-1}$$

Now, for a space displacement of the position operator we have $$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}}|\mathbf{x}\rangle = |\mathbf{x}+\mathbf{a}\rangle$$ $$\mathbf{Q'} = e^{-i\mathbf{a}\cdot \mathbf{P}}\mathbf{Q}e^{i\mathbf{a}\cdot \mathbf{P}}$$ And for a single direction of Q, $$Q_\alpha'|\mathbf{x'}\rangle = x_\alpha |\mathbf{x'}\rangle $$

Which suggests, $$\mathbf{Q'} = \mathbf{Q} - \mathbf{a}I$$ because for a single direction $$Q_\alpha'|\mathbf{x'}\rangle = x_\alpha |\mathbf{x'}\rangle = (Q_\alpha -a_\alpha I)|\mathbf{x}+\mathbf{a}\rangle$$ $$ = Q_\alpha |\mathbf{x}+\mathbf{a}\rangle - a_\alpha |\mathbf{x}+\mathbf{a}\rangle $$ $$ = x_\alpha + a_\alpha |\mathbf{x}+\mathbf{a}\rangle - a_\alpha |\mathbf{x}+\mathbf{a}\rangle $$ $$ = x_\alpha|\mathbf{x}+\mathbf{a}\rangle$$

So, thinking back to the active/passive view of transformations where I was confused, what is happening is that we are shifting the position operator $Q$ forward by $\mathbf{a}$ which corresponds to an active point of view.

This results in the position eigenvectors for a specific eigenvector to be shifted forward or in other words for the eigenvalues of a given eigenvector to be shifted backwards! All very confusing, but I believe this clarifies my original question more in the spirit of the text exposition.

quantum mechanics - Shankar's Active/Passive Change of Basis

I'm working my way through Shankar's Quantum Mechanics (7th printing, and I'm doing it alone, so I apologize if I have core concepts completely wrong).

He has a section on Active and Passive Transformations (which seems to be slightly different from what I know as Active & Passive Transformations, but that would be a different question :) ), and it reads like this:

Suppose we subject all the vectors $|V\rangle$ in a space to a unitary transformation $$ |V\rangle \rightarrow U|V\rangle $$ Under this transformation, the matrix elements of any operator $\Omega$ are modified as follows: $$ \langle V'|\Omega|V\rangle \rightarrow \langle UV'|\Omega|UV\rangle = \langle V'|U^\dagger\Omega U|V\rangle $$ It is clear that the same change would be effected if we left the vectors alone and subjected all operators to the change $$ \Omega\rightarrow U^\dagger\Omega U$$

I was trying to wrap my head around this, so I was working through some examples I made up and realized that the formula $\Omega\rightarrow U^\dagger\Omega U$ only works when $U$ is unitary. In my own examples I worked it out for an arbitrary (viz. non-unitary) change of basis T to be $$\Omega\rightarrow T^{-1}\Omega T$$ Like I said, I worked this out myself, so it may be wrong, but it checks out for the unitary case because when $U$ is unitary, $U^{-1} = U^\dagger$.

My question is: what part of Shankar's logic above makes use of the fact that U is unitary? It seems like his logic holds fine even without that constraint, but then it gives us a formula that isn't true for non-unitary U.

Answer

By definition the matrix elements of an operator $\Omega$ are the expressions $\langle V|\Omega|V'\rangle$, and they transform as DanielSank described. However, their interpretation as actual coefficients of the matrix of $\Omega$ in the basis $V,V',\ldots$ only is correct when the $V,V',\ldots$ form an orthonormal basis. If $A$ is the matrix of $\Omega$ in any basis, and $T$ is the matrix mapping it to a different basis, the matrix of $\Omega$ in the new basis becomes $T^{-1}AT$.

In other words, the inner products $\langle TV|\Omega|TV'\rangle$, called matrix elements, are equal to $\langle V|T^\dagger\Omega T|V'\rangle$, but the elements of the matrix of $\Omega$ in the basis consisting of the columns of $T$ are equal to $\langle V|T^{-1}\Omega T|V'\rangle$, if the $V,V',\ldots$ form an orthonormal basis. These are equal only if the $TV, TV',\ldots$ still form an orthonormal basis, which is equivalent to $T$ being unitary.

everyday life - Does plucking a guitar string create a standing wave?

About two weeks ago there was a mock test in Korea, and a physics question asked if a plucked guitar (it was actually a gayageum, a traditional instrument, but I'll just call it a guitar for convenience) string creates a standing wave.

I've learned in school that this is true, and the answer was true as well. But today my physics teacher said that this is actually false. Because a standing wave is caused by two identical waves traveling in opposite directions, a guitar string cannot create a standing wave. So a plucked guitar string only makes a vibration, not a standing wave.

But this is also mentioned in school textbooks. On the page explaining standing waves, there's a picture of a vibrating string and the caption says, "A string tied at both ends makes a standing wave, causing resonance."

I am confused. Does plucking a guitar string make a standing wave on the string? Or is this just a vibration?

Answer

Yes, plucking a guitar string does create standing waves, but...

No, plucking a guitar string does not create a standing wave, as the sum of standing waves is in general not a standing wave (thanks for Ben Crowell for pointing this out), since a standing wave must have a stationary spatial dependence and a well-defined frequency:

$$ y(x,t) \propto \sin(2\pi x/\lambda)\cos(\omega t).$$

The initial perturbation is not sinusoidal, but instead contains a plethora of frequencies, of which only remain, after a transient, the resonant ones - which correspond to some of the possible standing waves. It's the sum of those that compose the vibration you'll observe.

The counter-propagating waves, if you want to model each of the standing waves this way, you get from the reflections at the cord's ends.

For more details see this answer and, especially, the answers to the question Why do harmonics occur when you pluck a string?.

group theory - Integrating the generator of the infinitesimal special conformal transformation

(c.f Di Francesco, Conformal Field Theory chapters 2 and 4).

The expression for the full generator, $G_a$, of a transformation is $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta \omega_{a}} \partial_{\mu} \Phi - \frac{\delta F}{\delta \omega_a}$$ For an infinitesimal special conformal transformation (SCT), the coordinates transform like $$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$$

If we now suppose the field transforms trivially under a SCT across the entire space, then $\delta F/\delta \omega_a = 0$.

Geometrically, a SCT comprises of a inversion, translation and then a further inversion. An inversion of a point in space just looks like a translation of the point. So the constant vector $b^{\mu}$ parametrises the SCT. Then $$\frac{\delta x^{\mu}}{\delta b^{\nu}} = \frac{\delta x^{\mu}}{\delta (x^{\rho}b_{\rho})} \frac{\delta (x^{\gamma}b_{\gamma})}{\delta b^{\nu}} = 2 x^{\mu}x_{\nu} - x^2 \delta_{\nu}^{\mu}.$$ Now moving on to my question: Di Francesco makes a point of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$ I was wondering if somebody could point me to a link or explain the derivation. Is the reason for its non appearance due to complication or by being tedious?

I am also wondering how, from either of the infinitesimal or finite forms, we may express the SCT as $$\frac{x'^{\mu}}{x'^2} = \frac{x^{\mu}}{x^2} - b^{\mu},$$ which is to say the SCT is an inversion $(1/x^2)$ a translation $-b^{\mu}$ and then a further inversion $(1/x'^2)$ which then gives $x'^{\mu}$, i.e the transformed coordinate.

Answer

In order to determine the finite SCT from its infinitesimal version, we need to solve for the integral curves of the special conformal killing field $X$ defined by \begin{align} X(x) = 2(b\cdot x) x - x^2 b. \end{align} I explain why this is equivalent to "integrating" the infinitesimal transformation below. This means we need to solve the differential equation $X(x(t)) = \dot x(t)$ for the function $x$. Explicitly, this differential equation is \begin{align} \dot x = 2(b\cdot x) x - x^2 b. \end{align} This can be done with a trick, namely a certain change of variables. Define \begin{align} y = \frac{x}{x^2}, \end{align} then the resulting differential equation satisfied by $y$ becomes simple; \begin{align} \dot y = -b. \end{align} I urge you to perform the algebra yourself to confirm this. It's kind of magic that it works if you ask me, and the change of variables is precisely an inversion, so I think there's something deeper going on here, but I'm not sure what it is. The solution to this equation is simply $y = y_0 -tb$, so we find that the original function $x$ satisifes \begin{align} \frac{x}{x^2} = \frac{x_0}{x_0^2} - tb. \end{align} In other words, we've turned a monstrous nonlinear system of ODEs into a simple algebraic equation. In fact, one can show that the algebraic eqution $x/x^2 = A$ has the solution $x = A/A^2$, from which it follows that the solution to our original problem is \begin{align} x(t) = \frac{x_0 - x_0^2(tb)}{1-2x_0\cdot(tb) + x_0^2(tb)^2}, \end{align} as desired, since this is precisely the form of the "finite" SCT. Note that these only are local integral curves; the solution hits a singularity when $t$ is such that the denominator vanishes.

Why solve for integral curves?

If you're wondering what your original question has anything to do with solving the integral curves of the special conformal vector field I wrote down, then read on.

It helps to start with the concept of a flow.

Transforming points via flows.

Let a point $x\in\mathbb R^d$ be given, then for each $b\in\mathbb R^d$, we assume, at least in a neighborhood of that point, that there is a $\epsilon$-parameter family of transformations $\Phi_b(\epsilon):\mathbb R^d \to \mathbb R^d$ with $\epsilon\in [0,\bar\epsilon)$ such that $\Phi_b(\epsilon)(x)$ tells you what an SCT corresponding to the vector $b$ does to the point $x$ is you ``flow" in $\epsilon$. At $\epsilon = 0$, this flow just maps the point to itself; \begin{align} \Phi_b(0)(x) = x, \end{align} namely it starts at the identity. For $\epsilon >0$, the flow translates the point along a curve in $\mathbb R^d$. If you change $b$, then this corresponds to moving way from $x$ in a different initial direction under the flow.

Infinitesimal generator of a flow.

When we talk of an infinitesimal generator of such a transformation, we are talking about the term that generates the linear approximation for the flow in the parameter $\epsilon$. In other words, we expand \begin{align} \Phi_b(\epsilon)(x) = x + \epsilon G_b(x) + O(\epsilon^2), \end{align} and the vector field $G_b$ is called the infinitesimal generator of the flow. What you have pointed out in your question is that we know this infinitesimal generator; \begin{align} G_b(x) = 2(b\cdot x)x - x^2 b, \end{align} and we now want to reconstruct the whole flow simply by knowing this information corresponding to its linear approximation at every point. This is equivalent to solving some first order ordinary differential equations, which is why people often say we want to "integrate" the infinitesimal transformation to determine the finite one; integration is a perhaps somewhat archaic way of solving the corresponding differential equation.

Finding the whole flow given its generator.

Ok, so what differential equation do we solve? Well, note that the vector field $G_b$ is tangent to the curves generated by the flow by its very construction (we took a derivative with respect to $\epsilon$ with is the "velocity" of the curve generated by the flow), so the differential equation we want to solve is \begin{align} \dot x(\epsilon) = G_b(x(\epsilon)), \end{align} and we want to solve for $x(\epsilon)$. The solutions to this differential equation are referred to as integral curves of the vector field $G_b$.

Acknowledgements.

I didn't figure out the first part of this answer completely on my own. The idea for making the substitution $y=x/x^2$, which is really the crux of everything, came from here http://www.physicsforums.com/showthread.php?t=518316, namely from user Bill_K.

The idea for how to solve the algebraic equation $x/x^2 = A$ came from math.SE user @HansLundmark after I posted essentially your question in mathy language on math.SE here.

I should another math.SE user @Kirill solved for the integral curves in a totally different way in his answer to the question I posted.

Addendum.

How does one get $\dot y = -b$ from the change of variable $y=x/x^2$ as claimed in the first section? Well, let's calculate: \begin{align} \dot y &= \frac{x^2\dot x - x(2x\cdot \dot x)}{(x^2)^2} \\ &= \frac{x^2(2(b\cdot x) x - x^2 b) - x(2x\cdot (2(b\cdot x) x - x^2 b))}{(x^2)^2} \\ &= \frac{2x^2(b\cdot x) x - (x^2)^2b - 4x^2(b\cdot x) x + 2x^2(b\cdot x)x}{(x^2)^2} \\ &= -\frac{(x^2 )^2b}{(x^2)^2} \\ &= -b \end{align} Magic!

fluid dynamics - Why do ice cubes stick together or to the edges of a drinking glass?

I was drinking iced-water from a drinking glass (made of glass) at a restaurant yesterday when I was taking a drink, I noticed that there is very little ice water coming out and then suddenly, the ice water mixture comes crashing down. With my open mouth overfilled with ice water, my shirt gets wet – I hate when this happens. After drying myself off, I started thinking why this does happen:

1. 
   

   How does ice that starts off loose and separated in water fuse together to form one large clump?

   
   


2. 
   

   What causes the ice-water mixture to stop flowing momentarily and then suddenly starting flowing again? Is it that the ice fuses together in the glass and forms one large clump, and then the friction with the edges builds a "dam-like" situation? Or are there individual pieces of ice fusing with the edges of the drinking glass damming-up the ice?

   
   

calculation puzzle - What is a Surpassing Phrase™?

If a phrase conforms to a certain rule, I call it a Surpassing Phrase™.

Use the examples below to find the rule.

Some details to save you time:
1. Font doesn't matter.
2. The number of words in the phrase can vary, but I chose short 2-word phrases to keep the list concise.

3. Case doesn't matter.
4. Latest EDIT: calculation-puzzle tag added

What is a Surpassing Phrase™?

Answer

A Surpassing Phrase™ is one where consecutive pairs of letters in each word get farther apart in the alphabet. For example:

su up pe er rb / su ub bw wa ay
 2  5 11 13 16 /  2 19 21 22 24

but not:

ex xc ce el ll le en nt / tr ra ai in
19 21  2  7  0  7  9  6 /  2 17  8  5

The name comes from each successive distance between letters "surpassing" the previous one.

research level - Proof for the Mass gap of sine-Gordon action with $g cos(beta Phi)$

This is the sine-Gordon action: $$ \frac{1}{4\pi} \int_{ \mathcal{M}^2} dt \; dx \; k\, \partial_t \Phi \partial_x \Phi - v \,\partial_x \Phi \partial_x \Phi + g \cos(\beta_{}^{} \cdot\Phi_{}) $$ Here $\mathcal{M}^2$ is a 1+1 dimensional spacetime manifold, where 1D space is a $S^1$ circle of length $L$.

At $g=0$ : it is a chiral boson theory with zero mass, gapless scalar boson $\Phi$.

At large $g$ : It seems to be well-known that at large coupling $g$ of the sine-Gordon equation, the scalar boson $\Phi$ will have a mass gap.

Q1: What is the original Ref which states and proves this statement about the nonzero (or large) mass gap for large $g$?

-

Q2: What does the mass gap $m$ scale like in terms of other quantities (like $L$, $g$, etc)?

-

NOTE: I find S Coleman has discussion in

(1)"Aspects of Symmetry: Selected Erice Lectures" by Sidney Coleman

and this paper

(2)Quantum sine-Gordon equation as the massive Thirring model - Phys. Rev. D 11, 2088 by Sidney Coleman

But I am not convinced that Coleman shows it explicitly. I read these, but could someone point out explicitly and explain it, how does he(or someone else) rigorously prove this mass gap?

Here Eq.(17) of this reference does a quadratic expansion to show the mass gap $m \simeq \sqrt{\Delta^2+\#(\frac{g}{L})^2}$ with $\Delta \simeq \sqrt{ \# g k^2 v}/(\# k)$, perhaps there are even more mathematical rigorous way to prove the mass gap with a full cosine term?

standard model - A pedagogical exposition of the hadron physics?

I am looking for a textbook/lecture notes/etc. on the basics of hadron physics.

I wish to understand how to construct the effective Lagrangian for pions and nucleons starting from the QCD Lagrangian. In other words, from $$\mathcal{L}_{QCD}=i\bar{Q}\hat{D}Q+(\text{mass terms})+ (\text{gauge fields})$$

derive

$$\mathcal{L}_{hadrons}=-\frac{f_{\pi}^2}4Tr(\partial^\mu U\partial_\mu U^{\dagger})+i\bar{N}\hat{D}N+\text{(gauge interactions)+(higher-order terms)}$$

I've read Peskin and Schroeder and Srednicki's textbooks and sort of understand the general idea but still missing a lot of important points. Among the unclear things are the following

• 
  

  What is the exact quantitative relation between the quark fields and the hadron fields? To my understanding, this is not a direct one. However, in order to make computations something more then "hadrons are composed out of appropriate combinations of quarks" is needed.

  
  



• 
  

  How do we construct terms in the hadron Lagrangian? It is relatively clear to me how pion terms appear, but the nucleon ones are confusing. For example, in Srednicki's book there are two different bases for nucleons non-trivially related to each other and it is not clear to me why do we interpret one of them and not the other as "real" protons and neutrons.

  
  

In brief, I would like a text on the subject which is i) as pedagogical and ii) as self-contained as possible. I'm not concerned too much with generality, detailed treatment of the lowest-terms only would suffice. Also, I would strongly favour more modern expositions since some old-fashioned terms and approaches are by themselves a great source of confusion for me.

Answer

What you are looking for are explanations of effective field theory (for example see this review by Burgess http://arxiv.org/abs/hep-th/0701053) and chiral perturbation theory in particular (for example see this review by Scherer http://arxiv.org/abs/hep-ph/0210398, and here are some slides by Tiburzi that look good at first glance: http://www.int.washington.edu/PROGRAMS/12-2c/week2/tiburzi_01.pdf).

To briefly answer your questions--I think you will end up being disappointed! The exact quantitative relationship between the quarks and hadrons is extremely complicated and not fully understood. Indeed fully understanding this is more or less what the Millenium Prize Problem on "Yang Mills Existence and Mass Gap" is about.

The relationship between quarks and hadrons is mainly that the lagrangian for the hadrons should reflect the symmetries of the underlying QCD lagrangian (and also including the effects of chiral symmetry breaking). This may seem like an uncomfortably indirect mapping, and it is true that it would be better to have a firm, direct map that showed in detail how to go from one picture to the other. Unfortunately, we have to make do with what we have: the spirit of effective field theory is that we don't have to know exactly how the low energy behavior emerges from the high energy one, it is enough to know what degrees of freedom we want to describe in the low energy theory and what symmetries the lagrangian should have. This effective lagrangian, with all possible operators consistent with all the symmetries, will produce the most general S-matrix consistent with the symmetries, and combined with power counting we can turn this into a useful scheme where we can determine a finite number of parameters from experiment and from those parameters predict the results of other experiments.

The method that is usually taken for Chiral Perturbation Theory is:

1. 
   

   Write down the fields corresponding to the degrees of freedom you want to describe--for example the pions.

   
   


2. 
   

   Write down all operators consistent with the symmetries of the theory (gauge invariance, lorentz invariance, spontaneously broken chiral symmetry, etc). At this stage you have an infinite number of operators.

   
   


3. 
   

   Order these by they scaling dimension, which tells you at what order in an expansion in energy that they will appear. Decide what order in energy you want to work to. This will leave you with a finite number of operators.

   
   



4. 
   

   Fix the coefficients of these operators by experiment.

   
   

In principle one could derive the coefficients from QCD. However there is much too hard to be done analytically. People also compute these coefficients using lattice QCD.

As to your question about Srednicki, I'm not exactly sure what you mean but typically the notion of an asymptotic state / particle that propagates to infinity is clearest when the terms in the lagrangian are quadratic in fields (the free part) has the mass and kinetic terms diagonal. So it might be that the field redefinitions you are talking about are needed to put the lagrangian in that form.

astrophysics - Why more Fe-56 than Ni-62 as fusion product in heavy stars?

Suppose we create an Fe-56 nucleus and an Ni-62 nucleus, each from individual protons and neutrons. In the case of Ni-62, more mass per nucleon is converted to binding energy. Thus we could argue the Ni-62 nucleus to be more strongly bound than the Fe-56 nucleus, if I'm correct so far.
1. Why is Fe-56 is mentioned in many astrophysics texts as the most strongly bound of all nuclei?

Fe-56 is commonly mentioned as the dominant end product of fusion reactions in the core of massive stars. If I'm correct, fusion reactions beyond Si-28 are accompanied by partial disintegrations, resulting in a cocktail of fragments, not exclusively multiples of He-4 (nuclear statistical equilibrium).
2. Why is much more Fe-56 than Ni-62 produced in the core of a massive star, although Ni-62 is more tightly bound than Fe-56? What determines the share of each nuclide in the resulting iron group?

Answer

The final stage of nucleosynthesis at the core of a massive star involves the production of iron-peak elements, mostly determined by competition between alpha capture and photodisintegration. The starting material is mostly Si28 and weak processes are unable to significantly alter the n/p ratio from unity on short enough timescales. Thus the expected outcome of these quasi-equilibrium reactions should be nuclei with $Z \simeq N$. Subject to that constraint, then the most stable nucleus resulting from alpha captures onto Si is Ni56.

To produce heavier nuclei (e.g. Zn60) by alpha capture requires higher temperatures (because of the higher Coulomb barrier) and at these higher temperatures photodisintegration drives the equilibrium back towards smaller nuclei.

So where does all the confusion arise? Most of the iron-peak material ejected in a supernova is formed slightly further out from the core in explosive Si burning. The major product is Ni56, as above, and this then undergoes weak decays to Co56 and then Fe56 with half lives of 6 days and 77 days respectively. Thus the most common iron-peak product that ends up in the interstellar medium is Fe56 (also from alpha capture in type Ia supernovae).

general relativity - Finding geodesics: Lagrangian vs Hamiltonian

I have a question referring to how to compute geodesics of a given spacetime (say, Kerr).

I know that the direct way is via the geodesic equation

$$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\nu\kappa}\dot{x}^{\mu}\dot{x}^{\kappa}~=~0.$$

But also read that one can write down the geodesic equations using the Lagrangian formalism. From what I have seen so far, there are two approaches: either to write down the Euler-Lagrange equations with Lagrangian

$$L~=~\frac{1}{2}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}.$$

and then Euler-Lagrange equations are exactly the geodesic equations.

OR: write down the Hamiltonian equations using the Lagrangian above, and then use those equations as the geodesic equations.

My question: are Euler-Lagrange and Hamiltonian approaches fully equivalent when it comes to writing down the geodesic equations? Does one have advantage over the other?

general relativity - How to define the proper time of a photon?

I'm writing a paper about the motion of photons near a Schwarzschild black hole. At some point there's a derivative of the Hamiltonian of the system with respect to time $\tau$. I need to explain what the proper time is $\tau$, but it's quite odd because photons don't have any proper time.

The Hamiltonian that I have is

$$H = - \left( 1-\frac{2M}{r} \right)^{-1} \frac{p_{t}^2}{2}+\left( 1-\frac{2M}{r} \right) \frac{p_{r}^2}{2}+\left( \frac{p_{ \theta}^2}{2r^2}+\frac{p_{\phi}^2}{2r^2sin^2\theta} \right).$$

• So what would be the definition in this case?



• "the proper time is the time for the photon although he doesn't have one?"

Does anyone know?

quantum mechanics - Do electric and magnetic lines of force physically exist?

As per my imagination any thing can't impose force on the other by not giving even a touch(i,e action at a distance). So I thought there must be some physical existence of lines of force. Although virtual particles hypothesis has been proposed. Up to my knowledge virtual particles are just said to be disturbance (field) and not the one what we thought as materialistic "particle". Anyway I have heard about Faraday speaking about such similar view about there physical existence. But I do have a naive idea about it. I would greatly appreciate if any one can input something about your view on the physical existence of force lines. It would be more appreciable if any one can speak about Faraday's and Maxwell's approach on this concept.

I am providing Faraday's and Maxwell's basic views about the concept, so that anyone can speak about it.

FARADAY'S VIEWS: At first Faraday considered the physical reality of the lines of force as a possibility, yet several scholars agree that for Faraday their physical reality became a conviction. One scholar dates this change in the year 1838. Another scholar dates this final strengthening of his belief in 1852. Faraday experimentally studied lines of magnetic force and lines of electrostatic force, showing them not to fit action at a distance models. In 1852 Faraday wrote the paper "On the Physical Character of the Lines of Magnetic Force" which examined gravity, radiation, and electricity, and their possible relationships with the transmission medium, transmission propagation, and the receiving entity.

MAXWELL'S VIEW: Initially, Maxwell took an agnostic approach in his mathematization of Faraday's theories. This is seen in Maxwell's 1855 and 1856 papers: "On Faraday's Lines of Force" and "On Faraday's Electrotontic State". In the 1864 paper "A Dynamical Theory of the Electromagnetic Field" Maxwell gives scientific priority of the electromagnetic theory of light to Faraday and his 1846 paper "Thoughts on Ray Vibrations". Maxwell wrote: Faraday discovered that when a plane polarized ray traverses a transparent diamagnetic medium in the direction of the lines of magnetic force produced by magnets or currents in the neighborhood, the plane of polarization is caused to rotate. The conception of the propagation of transverse magnetic disturbances to the exclusion of normal ones is distinctly set forth by Professor Faraday in his "Thoughts on Ray Vibrations." The electromagnetic theory of light, as proposed by him, is the same in substance as that which I have begun to develop in this paper, except that in 1846 there were no data to calculate the velocity of propagation.

I hope anyone from quantum physics can speak about virtual particles.
Here is a opinion of Matt Strassler (click on this) about Virtual particles hypothesis

physical chemistry - How would steel degrade in space

If there is a steel plate floating in space, for ever, within the solar system, how is it going to degrade? Of course without oxygen it is not going to oxidize (rust), but how are cosmic rays, UV light, alpha particles and the general space environment going to affect the structure of steel on the long term?

EDIT: Let's imagine a plane of steel on orbit around the earth.

Just curious.

Answer

As previously mentioned, micrometeorite pitting will occur over sufficiently long time scales.

Steel is composed of microcrystalline grains, and as such it is likely that cosmic rays will cause the formation of crystallographic defects, leading to embrittlement of at least the periphery of the steel in the long term. Inelastic collision between the steel and ions or molecules floating in space may result in interesting chemistry on the surface of the steel, however this absolutely depends on the particular species the steel impacts with. Note also that space is very empty and as such collisions will happen with a very low frequency.

Ultraviolet light will cause photoelectronic emission. As such, the steel will develop a positive charge until the electron work function exceeds the energy of the light that is hitting it. This shouldn't degrade the metal, however very high energy photons (such as gamma rays) can cause crystallographic defects as discussed in this report by N.P. Baumann.

electrostatics - How would charge be distributed in charged conductors if the Coulomb law was not ${1}/{r^2}$?

Would the excess charge on a conductor move to surface until the electric field inside become zero if the Coulomb law was for example $\frac{1}{r^3}$? If yes, would the distribution $\sigma(x,y)$ be different from when it is $\frac{1}{r^2}$?

Answer

Suggestion to the question (v3): Generalize the question to a $1/r^s$ potential law in $n$ spatial dimensions! Then according to Henry Cohn's mathoverflow answer here, the charges rush to the boundary iff $s\leq n-2$. So in OP's example $(s=2,n=3)$, the charges don't rush to the boundary, in contrast to the real world $(s=1,n=3)$.

nuclear physics - How does Positronium exist?

I've just recently heard of Positronium, an "element" with interesting properties formed by an electron and positron, and I was shocked to hear that physicists were actually working with this element, even if for a very short lifetime. I was always under the impression that matter and antimatter annihilated when they came even remotely close to each other, which is apparently not the case.

How do these two particles combine to form an element if they're oppositely charged and roughly the same mass? What kind of interactions could possibly take place before they're pulled together and annihilated?

Answer

As you've noticed, it's not automatically true that a particle and its antiparticle will annihilate each other when they get close to each other. In fact, no interaction between particles is really certain to happen. Quantum mechanics (and at a higher level, quantum field theory) tells you that all these interactions happen with certain probabilities. So for instance, when a particle and its antiparticle come into close proximity, there is only a chance that they will interact within any given amount of time.

However, the longer the particles remain together, the greater the probability that they will interact and annihilate each other. This is responsible for the 142 ns lifetime of positronium as reported in the Wikipedia article: the probability of annihilation increases with time in such a way that the average lifetime of an "atom" of positronium is 142 ns.

As Cedric said, as long as the positron and electron don't annihilate each other (and remember, there is only a limited chance of that happening in any given time), they can interact in much the same way as any other charged particles, such as the proton and electron. Being bound together by the electromagnetic interaction, as in a hydrogen atom or a positronium "atom," is just one example.

computational physics - Single slit diffraction simulation (mathematica)

I'm trying to simulate the Fraunhoffer diffraction at slits(single,double,triple) with Mathematica.

In the picture, the red one is analytical result and the green one is numerical result.

The question is, why does it oscilate on the numerical result? how can I remove it?

The mathematica source file is here.

source code_.pdf

The slit size is 50um.

I don't think it is the sampling problem, because when I changed the sampling number 500 to 5000, the oscillation was still exist.

Answer

I have used python for solving your problem but no oscillation

Here is my code

import pylab as py
lambda1=1000e-9 #wavelength is 1000 nm
pi=3.1416 #defining py


k=2*pi/lambda1
w=30e-6#slit width is 30 micron
D=0.1#screen distance
N=20# number of points inside a slit
slit=py.asarray(range(N,0,-1))*-w/N #points inside slit
screen=py.frange(-10e-3,10e-3,1e-5)
m=len(screen)
n=len(slit)
elfield=py.zeros(m).astype('complex')


r=(D**2+screen**2)**0.5

for i in range (m):
    r12=0   
    for j in range(n):
        r12=(D**2+(screen[i]-slit[j])**2)**0.5
        elfield[i]=elfield[i]+py.exp(1j*k*r12)
intensity=abs(elfield*py.conj(elfield))   
py.plot(abs(intensity))

I think 1. You might not using complex datatype 2. You are trying to plot the real part or abs of electric field. try to multiply by the complex conjugate to get the intensity.

quantum mechanics - Why doesn't the phase operator exist?

In many articles about quantum optics, the phase-number uncertainty relation $$\Delta \phi \Delta n \ge 1$$ has been mentioned and used as a heuristic argument, but they say that the phase-number uncertainty relation does not exist in a strict sense. Also many textbooks say there does not exist the phase operator.

Why isn't it possible to define the phase operator? If I define in such way $$\hat{\phi}\vert \phi \rangle=\phi \vert \phi \rangle$$ does this cause any problem? What is the major obstacle in defining the phase operator? Furthermore, how can I derive the phase-number uncertainty relation?

cipher - Decode the message enciphered in these symbols: ◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉

◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉

---

Administrative details (not part of the puzzle):

• 
  

  The answer is a clue to The Security to the Party [12] (now with party soundtrack!)

  
  


• 
  

  For the benefit of anyone without a Unicode-capable browser, the puzzle should look like this (image of the puzzle text):

  
  


• 
  

  The answer is written text in the English language. The puzzle is: "◳◰ ◓◨ ◨◧◕ ◎◌ ◱◯◱◯ ◍◌○ ◉◉ ◇◔◓◕ ◐►◓◒ ◒◑ ◈◑ ◆◆◓ ◉◉◉" is the cyphertext. What is the plaintext?

  
  



• 
  

  This type of puzzle is called a Cryptogram. The Google search "How to solve a cryptogram" leads to all kinds of useful resources, hints and tips.

  
  


• 
  

  The puzzle is not case-sensitive (see question below).

  
  


• 
  

  The plaintext does not contain any slang or acronyms (see question below).

  
  


• 
  
  

  Chat about this question is here: http://chat.stackexchange.com/rooms/18659/discussion-on-question-by-a-e------------- (contains spoilers!)

  
  


• 
  

  There are no proper nouns (names) in the plaintext. (question from chat)

  
  


• 
  

  The plaintext does not contain roman numerals. (question from chat)

  
  


• 
  

  There is no punctuation in the plaintext. (question from chat)

  
  
  

---

Thank you @Kevin for providing a list of names of the symbols, for anyone who's having trouble with fonts (also good from an accessibility point of view, I guess). So everyone's on a level playing field, I've copied it below:

WHITE SQUARE WITH UPPER RIGHT QUADRANT
WHITE SQUARE WITH UPPER LEFT QUADRANT
SPACE
CIRCLE WITH UPPER HALF BLACK
SQUARE WITH RIGHT HALF BLACK
SPACE

SQUARE WITH RIGHT HALF BLACK
SQUARE WITH LEFT HALF BLACK
CIRCLE WITH ALL BUT UPPER LEFT QUADRANT BLACK
SPACE
BULLSEYE
DOTTED CIRCLE
SPACE
WHITE SQUARE WITH LOWER LEFT QUADRANT
LARGE CIRCLE
WHITE SQUARE WITH LOWER LEFT QUADRANT

LARGE CIRCLE
SPACE
CIRCLE WITH VERTICAL FILL
DOTTED CIRCLE
WHITE CIRCLE
SPACE
FISHEYE
FISHEYE
SPACE
WHITE DIAMOND

CIRCLE WITH UPPER RIGHT QUADRANT BLACK
CIRCLE WITH UPPER HALF BLACK
CIRCLE WITH ALL BUT UPPER LEFT QUADRANT BLACK
SPACE
CIRCLE WITH LEFT HALF BLACK
BLACK RIGHT-POINTING POINTER
CIRCLE WITH UPPER HALF BLACK
CIRCLE WITH LOWER HALF BLACK
SPACE
CIRCLE WITH LOWER HALF BLACK

CIRCLE WITH RIGHT HALF BLACK
SPACE
WHITE DIAMOND CONTAINING BLACK SMALL DIAMOND
CIRCLE WITH RIGHT HALF BLACK
SPACE
BLACK DIAMOND
BLACK DIAMOND
CIRCLE WITH UPPER HALF BLACK
SPACE
FISHEYE

FISHEYE
FISHEYE

Answer

MAGICSIBLINGS
It's morse code, with every circular symbol representing a dot and every other symbol representing a dash. The content of the symbols is a red herring.

wordplay - Alternating Audio Waves

Try solving this cipher from these strange memos:

Memo 1

Welcome Comet, today's Thisbe plus Russell here!

Memo 2

Reading outlaws loudness.

Memo 3

Retro memo: Berlin today upheld voting thesis, "Putin Salutes". Thanks be to God. U

P.S. The same algorithm is used on all memos

Enjoy ;)

---

Hint:

1

This is the largest evidence

Answer

The gimmick

The title offers a fairly large hint: "Alternating Audio Waves" suggests that this could be a pronunciation puzzle. But what could alternating mean? Perhaps we should take every other syllable in each word.

Translation

Memo 1: Welcome Comet, today's Thisbe plus Russell here!
Wel-come Com-et, to-days, this-bee, plus, rus-sell, here
Taking every other syllable gives: "Wel-com-to-this-plus-sell" which approximately sounds like "Welcome to this puzzle".

Memo 2: Reading outlaws loudness.
Read-ing out-laws loud-ness
Every other syllable gives "Read out loud"

Memo 3: Retro memo: Berlin today upheld voting thesis, "Putin Salutes". Thanks be to God. U
Re-tro mem-oh Ber-lin to-day up-held vot-ing the-sis pu-tin sal-utes thanks be to god u
Every other syllable gives "Re-mem-ber to up-vot the pu-sal thanks to u" or "Remember to upvote the puzzle, thanks to you"

I'm not sure what the hint means though.

lateral thinking - Where do all the doors lead to?

You are walking in a hallway and find a door. You decide to enter the room. The door closes behind you, the room suddenly becomes dark and starts spinning. You fall on the floor in the middle of the room, totally disoriented. You have no idea which way you came from. The light is back on.

There are four doors - one at each side:

Where do all the doors lead to? Which one would you choose and why?

Note: Excuse my poor drawing skills.

PS. Due to my poor drawing skills, I decided to change the second image. Hopefully, it is less confusing now.

Answer

Since the 4 different puzzle parts were solved by 4 different people, I am adding this community wiki answer with links to their answers as they should be the ones receiving upvotes:

Door 1 was solved by March Ho: https://puzzling.stackexchange.com/a/41523/29050

1666 - The Great Fire of London

In March Ho's own words:

Since Door 1 is the only unsolved door, here is my guess:

The Devil refers to the Number of the Beast (usually quoted as 666).

Additionally,
The Great Fire of London occurred in the year 1666, which is linked to the red colour due to the fire.

Additional details:

Hell usually is associated with fire. In addition, the devil is smiling, so he probably feels at home wherever this place is. Thus, additional hints for the fire. Also, by pure coincidence, if you write 666 and fire in Google, one of the first suggestions is the Great fire in London, 1666.

Door 2 was solved by Yandrakus: https://puzzling.stackexchange.com/a/41503/29050

1812 - Napoleon's invasion of Russia

In Yandrakis' own words:

Could be that Door nº2 leads to...
Napoleonic invasion to Russia

Because:

The time on the clock is set at 18:12 (I think), and the symbol (asterisk) depicted on the door is related to a snow flake. Napoleon tried to invade Russia during winter and this was one of his biggest mistakes in his military career.

Additional details:

The light blue color of the door is intentional - to represent ice, snow, winter.

Door 3 was solved by Radoslav Hristov: https://puzzling.stackexchange.com/a/41497/29050

2016 - the present

In Radoslav Hristov's own words:

3rd door:

It is a whole number 5776 and refers to "The year of Light" a.k.a. "Beginning of the End of Time". Further on the same: according to https://en.wikipedia.org/wiki/Anno_Lucis the 2016 a.d. actually is anno lucis, or year of light 2016, i.e. the current year!

Additional details:

It was only natural that there would be a door leading to back where we came from (the hallway, presumably the present). The yellow color is not intentional and does not have hidden meaning.

Door 4 was solved by Gareth McCaughan: https://puzzling.stackexchange.com/a/41502/29050

1773 - The Boston Tea Party

In Gareth McCaughan's own words:

I think the fourth door might possibly be

pointing to tea, or perhaps more specifically to the Boston Tea Party.

The quotation

is from Psalm 73 verse 17, perhaps suggesting the year 1773

and the plant on the door

seems like it might plausibly be a tea plant.

Additional notes:

The orange color of the door and the brownish color of the text actually resemble the color of the drink itself (tea). The leaf is indeed a tea leaf.

electromagnetism - What is the answer to Feynman's Disc Paradox?

[This question is Certified Higgs Free!]

Richard Feynman in Lectures on Physics Vol. II Sec. 17-4, "A paradox," describes a problem in electromagnetic induction that did not originate with him, but which has nonetheless become known as "Feynman's Disc Paradox." It works like this: A disc (Feynman's spelling) that is free to rotate around its axis has set of bead-like static charges near its perimeter. The disc in it also has a strong magnetic field whose north-south axis is parallel to the rotation axis of the disc. The disc with its embedded static charges and magnetic field is initially at rest.

However, the magnetic field is was generated by a small superconducting current. The disc is permitted to warm up until the magnetic field collapses.

The paradox is this: Conservation of angular momentum says that after the field collapses, the disc must of course remain motionless. However, you could also argue that since the collapsing magnetic field will create a strong circular electric field that is tangential to the perimeter of the disc, the static charges will be pushed by that field and the disc will necessarily begin to rotate.

Needless to say, you can't have it both ways!

Feynman, bless his heart, seemed to have an extraordinarily optimistic view of the ability of others to decipher some of his more cryptic physics puzzles. As a result, I was one of many people who years ago discovered to my chagrin that he never bothered to answer his own question, at least not in any source I've ever seen.

In the decades since then, that lack of resolution has produced a surprisingly large number of published attempts to solve the Feynman Disc Paradox. Many of these are summarized in a paper that was written and updated a decade ago by John Belcher (MIT) and Kirk T. McDonald (Princeton) (Warning: I can see the paper, but it may have access restrictions for others.)

My problem is this: I more-or-less accidentally came up with what seems to be a pretty good resolution of the paradox, and it ain't the one described in any of the papers I've seen on it. But I can't easily back off, because the solution is a bit too straightforward once you look at it the right way. I think!

I also think that Feynman's solution was very likely to have been relatively simple, and not some kind of tremendously detailed exercise in relativistic corrections. He was after all trying to teach freshmen, and he honestly seemed to think they would all figure it out with a bit of thought!

So, help me out here folks: Does anyone know for sure what Feynman's solution to this little puppy was? Along those lines, is Laurie M Brown from Northwestern by any chance linked into this group? I can't imagine anyone who knows more about Feynman's published works!

I will of course explain why I think there's a simple solution, but only after seeing if there's something simple (but apparently hard to find) already out there.

Addendum: The Answer!

I am always delighted when a question can be answered so specifically and exactly! @JohnMcVirgo uncovered the answer, right there in Volume II of the Feynman Lectures... only 10, count 'em 10, chapters later, in the very last paragraph of FLoP II 27, in Section 27-6 ("Field Momentum"), p 27-11:

Do you remember the paradox we described in Section 17-4 about a solenoid and some charges mounted on a disc? It seemed when the current turned off, the whole disk should start to turn. The puzzle was: Where did the angular momentum come from? The answer is that if you have a magnetic field and some charges, there will be some angular momentum in the field. It must have been put there when the field was built up. When the field is turned off, the angular momentum is given back. So the disc in the paradox would start rotating. This mystic circulating flow of energy, which at first seemed so ridiculous, is absolutely necessary. There is really a momentum flow. It is needed to maintain the conservation of angular momentum in the whole world.

Feynman hints at the above answer in earlier chapters, but never comes right out with a direct reference back to his original question.

John McVirgo, again, thanks. I'll review FLoP II 27 in detail before deciding whether to post that "other viewpoint" I mentioned. If Feynman already covers it, I'll add another addendum on why I think it's important. If the viewpoint is not clear, I'll need to do some simple graphics to explain how it may add some clarity to how the angular momentum conservation part works.

Addendum 2012-07-08: Not The Answer!

In the comments, @JohnMcVirgo has very graciously noted that I read more into his answer than he had intended, and for that reason he did not feel he should receive the answer mark. By finding that bit of text at the very end of the chapter John mentioned, I may in fact have answered my own question, at least in the literal sense of "what did Feynman say about it?" But John also points out his own surprise on how Feynman answered, which is different from points made by both him and @RonMaimon. So for now I'm leaving this question open. I will assign an answer eventually, but only after I've read up on FLoP II 27 to the point where I feel I know it inside out.

Addendum 2012-07-08: New Answer!

Well that was a short several weeks! @RonMaimon's additions to his initial answer, combined with his latest comment clarifying the difference between field momentum and "mechanical" momentum, demonstrate a deep understanding of the issues. Since @JohnMcVirgo already suggested the updated Ron Maimon text as the answer, I agree and have so designated it. I am still deeply grateful to John for pointing me to FLoP II 27, since without that clue I never would have found Feynman's answer in his own words.

I will at some point bring up my "other view" of Poynting problems as a new question. I now have two of those pending, since I am also still planning an updated Dual Cloud Chamber problem at some point.

standard model - The Higgs field a new Luminiferous aether?

As of this writing it has been made clear to me that classical physics' Luminiferous aether was a terriblly poor discriptor of space. With the advent of Special Relativity and General Relativity, that aether ceased to be applicable. Today I began reading on the Higgs boson, and the Higgs field. How are those theories different from or similar to the seemingly irrational Luminiferous aether?

P.S. I am a first year student at Bakersfield Community College. If this question seems naive, you now have an explanation.

(EDIT) To me the similarity rests upon their fundamental influences regarding static positions in space. I currently only accept models in which static spacial positions are impossible. Contrastingly relative positions seem quite reasonable. In my humble opinion my naivatisms inhibit me from absorbing and understanding the Higgs boson. If they actually are similar, public protest would probably be more apparant. However, I have put forth my best effort to pose an direct question here.

P.S. The questions Where does space go when it falls? and What happens to time lost when matter accelerates? come much earlier in my studies, and they seem relevant.

Answer

Here comes a simple minded answer by an experimentalist.

The luminiferous ether was discarded because it violated special relativity. It presupposed a fixed reference frame of the ether against which everything moved. In special relativity there exists no absolute frame of reference, and special relativity has been vindicated many times experimentally.

The Higgs field, as also the vacuum sea in general, comes from quantum field theory formulations of the interactions of elementary particles. All quantum field theories are consistent with special relativity , and thus the Higgs field is also consistent with special relativity. It therefore cannot play the role of the luminiferous ether, and also the same holds for the vacuum sea, which is seething with virtual pairs of particle/antiparticle.

At the level of your studies this should suffice.

general relativity - How do we know that gravity is spacetime and not a field on spacetime?

How do we know that gravity is the curvature of spacetime as opposed to a field, which couples equally to all objects, on spacetime?

wordplay - Why Alexa Shouldn't Teach Non-Native English Speakers About Idioms

1. 
   

   They cost ≈ $0.00833333333333333333333333333333 each.
   (9) -_----_-_-----

   
   


2. 
   

   Go! Intentionally fracture your fibula!
   (8) -----_-_---

   
   


3. 
   

   You are the joints of the lower appendages of a eusocial animal.
   
   (4) --- --_---_--- -_-----*

   
   


4. 
   

   I bequeath unto you the frigid glenohumeral joint.
   (10) - -_------_---_---_----_--------*

   
   


5. 
   

   You and the curvy peg are now parting ways, thanks to me.
   (13) - -_-------_---_---_---_----*

   
   



6. 
   

   A conglomeration of wind, precipitation, humidity and temperature is above me as I randomly touch various things with my hands.
   (11) - -_-------_-----_---_-------*

   
   


7. 
   

   Prevent blood-sucking parasites from using their mandibles, and slumber securely.
   (23) -----_-----_---_--- -_---_---_---_----_----*

   
   


8. 
   

   Refrain from enumerating avian live stock prenatally.
   
   (21) --- -_-----_----_--------_------_---- --_-------*

   
   


9. 
   

   The pump which both delivers and retrieves molecules from every component of my construct has stopped its function.
   (4) --_-----_--_------

   
   


10. 
    

    Visible collections of gaseous vapors are plated without exception.
    (2) -----_-----_---_-_------_------

    
    



11. 
    

    Retrieve and refasten your various body parts at once!
    (10) ----_--------_--------

    
    


12. 
    

    While you are entertained, the fourth dimension is airborne.
    (13) ----_-----_----_---_---_------_---

    
    

ANSWER: ----__---__-----

---

*Punctuation is indicated, but doesn t count.

(A-la Rand al'Thor, the answer is an unsolicited hint (and shameless plug) for this question.

Answer

They cost ≈ $0.00833333333333333333333333333333 each.

A DIME A DOZEN (9th letter = Z)

Go! Intentionally fracture your fibula!

BREAK A LEG! (8th letter = E)

You are the joints of the lower appendages of a eusocial animal.

YOU'RE THE BEE'S KNEES (4th letter = R)

I bequeath unto you the frigid glenohumeral joint.

I'M GIVING YOU THE COLD SHOULDER (10th letter = O)

You and the curvy peg are now parting ways, thanks to me.

I'M LETTING YOU OFF THE HOOK (13th letter = O)

A conglomeration of wind, precipitation, humidity and temperature is above me as I randomly touch various things with my hands.

I'M FEELING UNDER THE WEATHER (11th letter = N)

Prevent blood-sucking parasites from using their mandibles, and slumber securely.

SLEEP TIGHT AND DON'T LET THE BED BUGS BITE (23rd letter = E)

Refrain from enumerating avian live stock prenatally.

DON'T COUNT YOUR CHICKENS BEFORE THEY'VE HATCHED (21st letter = S)

The pump which both delivers and retrieves molecules from every component of my construct has stopped its function.

MY HEART IS BROKEN (4th letter = E)

Visible collections of gaseous vapors are plated without exception.

EVERY CLOUD HAS A SILVER LINING (2nd letter = V)

Retrieve and refasten your various body parts at once!

PULL YOURSELF TOGETHER (10th letter = E)

While you are entertained, the fourth dimension is airborne.

TIME FLIES WHEN YOU ARE HAVING FUN (13th letter = N)

ANSWER: ----__---__-----

ZERO__ONE__SEVEN