lateral thinking - The Cryptic Lab ~3~

SPOILER ALERT PLEASE READ The Cryptic Lab * 1 * AND The Cryptic Lab +2+ BEFORE READING PART 3!

The series is getting less and less attention, so I might close it sooner than I think; after part 5 is out. If you guys enjoyed this, I'm glad you did. Still, enjoy part 3.

LAST SPOILER ALERT!!! SERIOUSLY DON'T READ FURTHER IF YOU HAVEN'T LOOKED AT THE PREVIOUS TITLES IS THE SERIES.

You open the door quietly. As the planks of wood fall onto the floor, with a loud, echoing sound, you jump at the loud noise. As soon as you got out of range from the door, it shut behind you. The mysterious man started to talk on the speaker again.

I'm impressed, test subject 137! You really are special, aren't you? Now don't get too comfortable, You won't like what's coming next. Now as I said before, comfort is earned. Here are some pants and a jacket, both have a wool interior to keep you extra cozy. Now, you know the drill. Another room, another furnace, another set of rules, and another door to unlock. Good luck, don't die.

You put the jacket and the pants on. Nice and cozy.

You walk towards the new locked door. Seems like it has a keyhole. You know you're gonna have to pick the lock again. One problem: What do you pick the lock with?

You look around the room, laying eyes on a box, with a glass cover. You look through the cover, and you see a wood skewer. Looks like it could be used to pick the lock, but it's bent in the middle. You attempted to open the glass cover, but it was screwed tight. Seems like whoever this mystery man is likes flat screwdrivers...

You see a fireplace in the corner of the room. A note on it says,

Warm up, I don't want my "guest" to die of cold.

So, so housewarming..., you thought.

Oh? There's a pot in the corner of the room. You walked up to it. It's stainless steel, premium quality. Nice. Maybe you can use this to hit "Mr. Housewarming", or whoever he is on the head once you finally lay eyes on him.

You suddenly realize you're thirsty. Oh look, a water tap! You run towards it as fast as you can, stand on your knees, and turn the tap. You immediately start drinking, but the water just doesn't taste right... It's salty and makes you want to puke... You spit the water out, leaving your thirst not quenched.

Finally, you look at the rules:

Basically, don't break anything, especially the fourth wall. Don't damage the pot, the box, the glass, the door, the lock, the walls, the ceiling, and the floor. Don't burn anything either. You can't use the pot to unscrew the screws. The only way you're allowed to get the glass off is by unscrewing the screws. P.S. Warm up by the fire for as long as you need. Good luck, don't die.

You sit by the fireplace, burning wood in all its glory. You look to see if there's anything in the fire to help you, but nothing is in there. You sit down and think. What will you do to get past this door?

OBJECTIVES: {PICK THE LOCK ON THE DOOR} (QUENCH YOUR THIRST)

Answer

Step 1:

Let's get the wood skewer. Since the screws are flat heads, I'll assume that the rim of the pot just fits into the head of the screws, and thus you can unscrew them using the pot as a makeshift driver.
Okay, well I guess we still have a few options:
Keep in mind, we technically walked through the door with the makeshift saw from the previous puzzle. (Nothing noted that we dropped it, and I don't see why we, as a puzzle solving survivor, would).
So let's try a different couple of things and see what works. If the first attempt doesn't work/isn't allowed, move on to the next.
Attempt 1: Use one of the edges of the glass blade to unscrew the screws.
Attempt 2: The now cut wood that was previously locking the door we just came through could be sharpened or shaved using the glass saw we have. Refine the wood down to a point that will fit the screws, and then unscrew them.
Attempt 3: I guess if all else fails, @Yessoan came up with a pretty interesting idea we could try. I'm willing to use this mainly because I have other ideas and because their water idea and mine differ.

Step 2:

We need water. For drinking and other reasons... So, over to the tap we go. Put the pot under the tap, along with you coat ontop of the pot (wool side facing up towards the faucet.) Turn the tap on low and let water go through the coat into the pot. (Or if it can't, wring the coat out into the pot every now and then.) This will reduce the salt content to help give you drinkable water. Also, we'll need some for the next step.

Step 3:

Boil the bent wood skewer in the pot of water using the fire. Eventually you can remove the pot and straighten the wood back out. (Use your wool clothes as oven mits if needed).

Step 4:

Pick the lock with the now straight wood skewer. (Or, if needed, bent into whatever shape helps)

word square - A Trickier Wordsquare Riddle

My favorite Wordsquare Riddles,
Inventing them, I can't resist.
But this, with four stanzas,
There might be a twist!

My first is perfectly happy,
Featured in a song of Coldplay's.
My second is roughly a hertz,

Back in Mozart days.

My third does not empty graves.
National, personal, why not?
My fourth determines your destiny.
Also, a task of the pilot.

A bit tricky? I apologize.
The clues are obvious as can be.
For the geniuses on this site
I bet they solve this easily!

Hint 1:

This is a Wordsquare, which is a crossword shaped like a square, and having same across and down clues. No other specification.

Hint 2:

For the first word, it is just a word in the lyrics, not the title. To make it easier: the song has the same birth date and month as Robert Baden Powell.

Hint 3:

I must admit that you must think laterally a bit, but I'm afraid that if I add the tag, the answers will go wild.

Hint 4:

Underestimating the last stanza will make you fail this puzzle.

Answer

The answer is a Wordsquare, but ...

... it is not a 4×4 square, despite there being four clues on the face of it. It turns out that the square is 5×5. Only four of the words are defined explicitly in the puzzle and all words are five letters long.

The first:

Bliss is a word in the lyrics of Something Just Like This: "Some superhero. Some fairytale bliss ..."

The second:

Lento is a musical instruction to play slowly and could be more or less one beat per minute. Thanks to MMAdams for the useful link and also for having the right idea about the Hertz clue.

The third:

To inter is to place a corpse in a grave at a funeral. And there are references to international and interpersonal.

The fourth:

Steer: a pilot steers an aircraft or it steers other ships into the harbour. And one is steered by one's destiny.

The final Wordsquare:

 B L I S S
 L E N T O

 I N T E R
 S T E E R
 S O R R •

But there's a twist:

It is clear by now that there is a fifth word:
Sorry: A bit tricky? I apologize.

Well, it's never too late to say that. :)

friction - Maximum acceleration for wheel-driven vehicle from standstill

Once I discussed cars and there was a question about rate at which the car gains its speed ($100\:\mathrm{km/h}$) from standstill (i.e. acceleration). At some point of conversation it came up that cars cannot gain their $100\:\mathrm{km/h}$ speed for such tiny time interval I proposed (less than $3$ seconds), because it can't go with acceleration more than $g$ i.e. free fall accel on Earth (we talked about wheel-driven vehicles - not a rocket-driven - and all we know that at certain acceleration value the wheels lose their grip with ground and vehicle is not moving forward it's burning tyres istead).

I was disagree, I feel intuitively that the critical acceleration for the wheel traction exists but it couldn't be equal to $g$ or approaching to, because it should look like a some deep law of the nature (between friction and acceleration) and this phenomenon would be widely known (but I couldn't find anything).

We started searching for contra-evidence. I found Bugatti Veyron which achieves $100\:\mathrm{km/h}$ speed roughly for $2.5\:s$ and some Formula 1 cars. But the sources of acceleration values were considered as non-reliable and not proven.

I know that the friction and cohesion proccesses are very untrivial, it brings electromagnetic interactions between atoms and molecules on microscopic level as well as surface geometry on macrolevel - that was another my argument against the statement that acceleration can't be more than $g$ without traction loss - because such complex and heterogeneous process as wheel traction can't have such a simple solution and equal to $g$.

So, could someone prove it or provide a formula or methods to show that critical acceleration of wheel-driven body without traction loss could be more than $g$ (or less)?

Answer

There is an approximate law that states that the frictional force, $F$, is given by:

$$ F = \mu W $$

where $W$ is the normal force and $\mu$ is the coefficient of friction; $\mu$ is normal taken to lie in the range zero to one. For a car the weight $W$ is given by $mg$, and the acceleration would be $F/m$, so dividing through by $m$ we get th acceleration to be:

$$ a = \mu g $$

If $\mu$ has the maximum value of one then the acceleration cannot exceed $g$ because trying to accelerate faster would just make the tyres spin.

But ...

The equation we started with is an approximation, and friction is actually a far more complicated phenomenon that that simple law suggests. For example car tyres deform and can key into irregularities on the road to increase the friction. Racing car tyres can have effective values of $\mu$ far greater than one, so they can accelerate at more than 1g. In fact a drag racer can accelerate at around 4g.

symmetry - Understanding units and the units of the derivative operator

Suppose that $f$ is a function from unit $A$ to $B$, then what is the unit of $f'(x)$?. We can do $f'(x)\Delta x$ to get an estimate of $f(x + \Delta x)$. Since the latter has unit $B$, so has the former. $\Delta x$ has unit $A$, so $f'(x)$ has unit $B/A$. So far so good, but what if $A$ is the unit of temperature? Intuitively it does not make sense to give $\Delta x$ the same unit as temperature, because a temperature difference cannot be used interchangeably with an absolute temperature. In particular it does not make sense to do $f'(x)\cdot y$ where $y$ is an absolute temperature. Are there extensions to the system of units that let you recognize that as an error just by looking at the units, i.e. a system of units where $\Delta x$ has a different unit than $x$? Similarly, adding two absolute temperatures is invalid. Is there a system of units that lets you only add two relative temperatures to get a new relative temperature or a relative plus an absolute tempurature to get a new absolute temperature?

Now for a more general and more vague question:

I was reading the page on extensions to dimensional analysis on Wikipedia. It describes a system of units that seems to capture the rotational invariance of physical laws. All physical laws should be rotationally invariant, and Siano's system of units makes sure that if a law is not rotationally invariant then it has a unit error (i.e. it does not matter what we define as our zero angle). Is this a correct intuition? Are there similar extensions to the unit system that capture other physical invariances? It seems like the standard units capture scale invariance (i.e. it does not matter what mass we call 1 kg as long as we do it consistently), and the system that I'm after in the first paragraph should capture translation invariance (it does not matter which point we define as zero). Of course there are other symmetries. Is there a system of units that lets you easily check from the units whether a law satisfies these symmetries?

Another limitation of our current unit system is that some mathematical laws are not expressible. For example $e^{ab}=(e^a)^b$ is not expressible if $a$ and $b$ have units. Is there a way to solve this problem?

Answer

As you correctly stated at the beginning, the units of $f'(x)$ are easily seen from writing the derivative as $$ f'(x) = \frac{df}{dx} $$ so the units are the same as units of $f/x$. However, your concerns about the temperature can't be justified. The units of absolute temperature and the temperature difference are the same. In particular, the international system of units, SI, uses 1 kelvin for both.

One may also work with temperature scales that are not absolute, e.g. the Celsius degrees. The absolute zero doesn't correspond to 0 °C; in this sense, these scales are "nonlinear". However, a Celsius degree is still a unit of temperature as well as temperature difference. As a unit of temperature difference, 1 °C and 1 K are the very same thing. You can surely never forget or omit units such as degrees (of temperature) from physical quantities, whether they are computed as derivatives or not.

Similarly, adding two absolute temperatures is invalid.

It may be unnatural or useless in most physical situations (see Feynman's "Judging Books By Their Cover") but it is a valid procedure when it comes to the units. After all, absolute temperatures are just energies per degree of freedom so adding absolute temperatures isn't much different from adding energies which is clearly OK.

In particular it does not make sense to do $f'(x)\cdot y$ where $y$ is an absolute temperature.

It makes perfect sense. Thermodynamics is full of such expressions. For example consider $f(x)=S(t)$, the entropy as a function of time. Then $S'(t)\cdot T$ is a term that appears in the rate of change of some energy according to the first law of thermodynamics.

Quite generally, it is not sensible to single out temperature in these discussions. The same comments hold for distances, times, or pretty much any other physical quantities. Take time. One may consider the "current year". It's some quantity whose unit is 1 year. (Similarly, the position of something in meters.) And one may consider durations of some events whose units may also be years. It is the same unit. It is obvious that the difference $A-B$ i.e. any difference has the same units as $A$ as well as $B$. In my "current year" analogy, $Y=0$ corresponds to the birth of Jesus Christ, a random moment in the history of the Universe. That's totally analogous to $t=0$ °C, the melting point of ice. But in both cases, the time differences or temperature differences have the same units as the quantities from which the differences were calculated – such as temperature (whether they're absolute or not) or dates.

It wouldn't make any sense to have different units for quantities and their differences because dimensional analysis would cease to hold: one could no longer say, among other things, that the units of $A-B$ are the same as units of $A$ or $B$ separately.

It is very correct that one cannot calculate a sensible value of $\exp(a)$ if $a$ is dimensionful i.e. if it has some nontrivial units. Such an exponential would be adding apples and oranges, literally. Express $\exp(a)$ as the Taylor expansion, $1+a+a^2/2+a^3/6+\dots$. If $a$ fails to be dimensionless, each term has different units so it's not dimensionally correct to add them. For this reason, all exponentials (and, with a somewhat greater tolerance, logarithms) in physics are exponentials of dimensionless quantities (which have no units). The desire to avoid physically (and mathematically) meaningless quantities such as exponentials of dimensionful quantities is one of the very reasons why we use units and dimensional analysis at all. It is not a "problem"; it is a virtue and the very point of these methods.

general relativity - How is the approximate gravitational wave stress energy momentum tensor not 0?

In Section 35.7 of Misner, Thorne, and Wheeler, p. 955, an "effective" stress energy momentum tensor for gravitational waves is defined:

$$T^{\text{GW}}_{\mu \nu} = \frac{1}{32 \pi} \left< \bar{h}_{\alpha \beta, \mu} \bar{h}^{\alpha \beta}_{\space\space,\nu} - \frac{1}{2}\bar{h}_{,\mu}\bar{h}_{\nu} - \bar{h}^{\alpha \beta}_{\space \space ,\beta} \bar{h}_{\alpha \mu, \nu} - \bar{h}^{\alpha \beta}_{\space \space, \beta} \bar{h}_{\alpha \nu,\mu} \right>.$$

The brackets indicate an averaging over a region of space much larger than a wavelength of the wave. The text then says that on a background space time with Einstein tensor $G^{\text{B}}_{\mu \nu}$,

$$G^{\text{B}}_{\mu \nu} = 8 \pi \left(T^{\text{GW}}_{\mu \nu} + T^{\text{other fields}}_{\mu \nu} \right).$$

But for a gravitational wave propagating through empty space, with $R^{\text{B}}_{\mu \nu} = 0$, wouldn't this then imply that $T^{\text{GW}}$ is $0$? What am I missing?

Answer

The treatment of this in section 35.7 of MTW says they're only presenting the result, whereas the actual derivation is given in section 35.13, p. 964. The basic idea is as follows. First you take the vacuum field equation $R_{\mu\nu}=0$, and expand it in powers of the wave's amplitude $A$. They argue that the term linear in $A$ must vanish, because any effect of gravitational waves on the background curvature must be a second-order effect. Turning next to the $A^2$ term, they break it up into a fluctuating part and a smooth background. Setting the sum of these equal to zero, we get something that looks like the ordinary Einstein field equations, for the background, with the wave term interpretable as a source, i.e., as an effective stress-energy.

[EDIT] In a comment, the OP asks:

Thank you! But how does that address my question, exactly? Wouldn't this still mean that a gravitational wave propagating in vacuum carries no energy/momentum?

Locally (i.e., in any experiment on scales smaller than a wavelength), there is no energy-momentum, because the stress-energy tensor is zero. But in GR it is not true that you can simply integrate the stress-energy tensor over some volume and find the total energy-momentum inside. This is why, for example, we say that a Schwarzschild black hole has mass, even though the stress-energy tensor is zero everywhere. Integration of a tensor with rank>0 fails because of the ambiguities introduced by the path-dependence of parallel transport.

optics - Why does the intensity of a LED change as the temperature increases

I am working with an array of LEDs and I am looking into how there properties change as they heat up. One of the things I found out is that the intensity of these the light decreases as the LEDs heat up. I knew that this would happen from the many reports I read on the internet about this, but I cant find a clear explaination on the internet for the theoretical background, one site talked about a reducion in the forward voltage, the other about an increase in the resistance of the P-N junction.

So I was wondering of anybody here could give me a clear explaination as to why the intensity of my LEDs decreases.

Thanks for taking the time to read my problem and I hope to see your awnser soon.
G.Wesseling

P.S. I am sorry for my english, its not my first language

logical deduction - Infinite precision

The man tells you, "To enter the door of infinite precision, you must complete all of these tasks perfectly:

• Hold in front of me a baby that is exactly two days old.


• Bring me a metal object that weighs exactly one kilogram.*


• Bring me a stone object that weighs exactly one pound.*


• Bring me a glass containing something other than water that is exactly the same density as water.



• Write all the digits of pi."

"I'm sorry", you say, "but that is impossible. I can complete four, but there is one I cannot do."

Which task can't you accomplish, and why?

(The explanation should state how you can do the other four tasks, as well).

Hint:

Keep in mind that this puzzle was spoken, not written down. This may help with one of the tasks.

*If you want to get technical (and I know some of you do), consider these objects with a mass of one kilogram and one pound.

riddle - The sands of life are running out

The one I obey, my dear commander
Keep on running I will not slander
For I know you will follow my lead
Call me in if you are in need

If one commands the other will follow

That's the deal, one has to swallow
Either way, I rely on you
We kiss goodnight and start anew

Mighty Dictator of all places
followed by all human races
Watch over all the new born lives
earn the fruit of fertile strifes

Endless love, you never disclaim
nor would you repel and never blame
Unbribable Protector, punish me

and I will follow tacitly

---

This is an easy riddle. For all the others:

Every verse contains the same solution. The whole riddle is metaphorically describing one solution.

Answer

After reading this I am under the impression that this has to be the answer.

Time

The one I obey, my dear commander Keep on running I will not slander For I know you will follow my lead Call me in if you are in need

We all know that we humans can command time, which simply means to use the time wisely (or simply changing the time on a watch). As we carry on with our daily activities, time follows us regardless whether we use it wisely or not, which then explains why the time will not 'slander' as we 'follow (its) lead'(and we will not slander when time is running). And of course, when we are in need of time, we can call him in, which just means that we would grasp time more efficiently and thus, have 'more time'(plus, using the watch again: it will tell/call us in us when batteries are empty).

If one commands the other will follow That's the deal, one has to swallow Either way, I rely on you We kiss goodnight and start anew

We both can command each other. When we command time, we are using it wisely whereas when time commands us, we will follow it as well. That's the case as one has to swallow, and swallow itself does not mean the physical act of swallowing, it could also mean 'put up with or meekly accept something', which in this case, the one which is commanding. When the clock hits twelve, a new day starts, hence 'we kiss goodnight and start anew'.

Mighty Dictator of all places followed by all human races Watch over all the new born lives earn the fruit of fertile strifes

Time is indeed the mighty dictator of all places, and we all humans follow it. Time watches over new born lives as when time passes, the newborns gradually age to become an adult, which then gives the metaphorical meaning that time is 'watching over' them. In the case of the fruit of fertile strife, it would mean that the humans themselves are displeased and hurt because of an argument, and to have the fruit of the fertile strife would be the solution to the argument, something that can dissipate the hurt between both of them, and as we have all heard before, 'Time heals all wounds.'
Double meaning of races!

Endless love, you never disclaim nor would you repel and never blame Unbribable Protector, punish me and I will follow tacitly

We all know that time is unbiased towards anyone, and that it will still pass regardless of situation, thus explaining why it would never disclaim, repel and blame. We all know that time can never be bought no matter how much money you have, hence giving it the title, Unbribable Protector. When time 'punishes us', it means that we have not used it wisely and we realise the fact that we have wasted precious time, something that can not be bought by money and we learn the importance of time. After being punished, we would learn to appreciate and use time efficiently, and hence 'follow tacitly'.

Title Interpretation

The sands could mean hourglasses itself as hourglasses use sand to embody the meaning of time. And when turned upside down and there is no more time left, hence 'The sands of life are running out'.

mathematics - Gambling in a rigged casino

A well-known crooked casino is offering the following bet with a pair of (identically looking and indistinguishable) dice that both show the numbers $1,2,3,4,5,6$ on their faces.

• The first die is fair and perfectly standard: throwing it yields each of the outcomes $1,2,3,4,5,6$ with equal probability $1/6$.


• The second die is loaded: throwing it yields each of the five outcomes $1,2,3,4,5$ with equal probability $1/5$ (while $6$ is impossible to throw).

We stress that there is no way whatsoever for an honest gambler to tell the two dice from each other. Now, in the bet that we are considering, the gambler is allowed to throw these dice as often as he likes, subject to the following casino rules:

• The gambler has to pay $1$ Euro for every single throw of a single die.


• Whenever he switches from throwing one die to throwing the other die, he must pay $5$ Euro.
  For his very first throw, he may arbitrarily choose either of the dice (without paying anything).


• As soon as he throws a $6$, the game is over. The gambler wins $20$ Euro.


• At any moment in time, the gambler may decide to quit the game and leave the table at no further cost.

Question: What is the best strategy for this game and what is the expected profit?

---

Answer

What follows is a strategy for earning an expected profit of about $3.63$ Euro, and an argument that this strategy is optimal.

First we'll pick some positive integer $k$. We will switch dice after every roll number that is an odd multiple of $k$, and continue to roll until we get a 6. So we will roll the first die $k$ times, then the second die $2k$ times, then the first die $2k$ times, and so on.

With probability $1$ we will eventually roll a 6 and win the 20 Euro. To compute the expected cost, we consider two cases, depending on which die we pick up first.

First die is fair. We expect to roll the fair die $6$ times. We expect to roll the loaded die $$ \sum_{n\geq 0}\left(\frac{5}{6}\right)^{2nk}2k = \frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}} $$ times. The expected number of switches is $$ \left(\frac{5}{6}\right)^k+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{(2n+1) k} = \frac{2(5/6)^k}{1-(5/6)^{2k}}. $$ So if we start with the fair die, our expected profit is $$ 20-6-\frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}}-5\cdot\frac{2(5/6)^k}{1-(5/6)^{2k}}. $$

First die is loaded. We again expect to roll the fair die $6$ times. We expect to roll the loaded die $$ k+\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk} 2k=\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}} $$ times. The expected number of switches is $$ 1+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk}=\frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}. $$ So if we start with the loaded die, our expected profit is $$ 20-6-\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}}-5\cdot \frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}. $$

Expected profit Our overall expected profit is the mean of the expected profit if the first die is fair and the expected profit if the first die is loaded. The expression simplifies to $$ 14-\frac{1}{2}\frac{(5+k)\left(1+(5/6)^{k}\right)^2}{1-(5/6)^{2k}}. $$ Numerically, I found that profit is maximized taking $k=9$. The expected profit using this strategy is $$ \frac{4218321}{1160653}\approx 3.63\,\,\,\text{Euro}. $$

Is this strategy optimal? Here's a sketch of a proof that this strategy is optimal.

Recall that the odds a given die is loaded is the ratio $$ \frac{\text{probability that die is fair}}{\text{probability that die is loaded}}. $$ The product of the odds that each die is fair is $1$.

During the course of the game, we only get information of one kind, namely that the die we just rolled did not turn up 6 (if it turns up 6, the game ends). Bayes' Theorem says that whenever a die turns up something other than 6, the odds that it is fair get multiplied by $\frac{5}{6}$ (and so the odds the other die is fair get multiplied by $\frac{6}{5}$). This means that assuming no 6's have been rolled, the odds that a die is fair depend only on the different between the number of times that die has been rolled, and the number of times the other die has been rolled.

When we are deciding whether to roll our current die, switch dice, or walk away, we need only consider the odds our current die is fair. Any costs already paid (sunk costs) should not be taken into account, as future payments and earnings do not depend on them. This means that possible strategies fall into two possible kinds:

• If the odds the current die is fair are below a certain threshold, leave the game. Otherwise roll.


• If the odds the current die is fair are below a certain threshold, switch dice. Otherwise roll.

The strategy described above is of the second kind, where the threshold is $\left(\frac{5}{6}\right)^k$. We choose $k$ to optimize profit among strategies of this kind. The only other possibility is a strategy where we roll the current die $n$ times (for some fixed $n$), then leave the game if we haven't rolled a 6. It's not too hard to write down the expected profit here as a function of $n$, find the $n$ that maximizes profit (if I computed correctly it is $n=5$), and see that this doesn't do better.

quantum mechanics - Orthonormality of Radial Wave Function

Is the radial component $R_{n\ell}$ of the hydrogen wavefunction orthonormal? Doing out one of the integrals, I find that

$$\int_0^{\infty} R_{10}R_{21}~r^2dr ~\neq~0$$

However, the link below says that these wave functions should be orthonormal (go to the top of page 3):

http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-05.pdf

Am I doing something wrong? Are the radial components orthogonal, or aren't they? Are there some kind of special condition on $n$ and $\ell$ that make $R_{n\ell}$ orthogonal? Any help on this problem would be appreciated.

Answer

No, the radial parts of the wavefunctions are not orthogonal, at least not quite to that extent.

The radial components are built out of Laguerre polynomials, whose orthogonality only holds when leaving the secondary index fixed (the $\ell$ or $2\ell+1$ or whatever depending on your convention). That is, $$ \langle R_{n'\ell} \vert R_{n\ell} \rangle \equiv \int_0^\infty R_{n'\ell}^*(r) R_{n\ell}(r) r^2 \, \mathrm{d}r = \delta_{nn'}. $$ You can check this yourself, using some of the lower-order functions, e.g. \begin{align} R_{10}(r) & = \frac{2}{a_0^{3/2}} \mathrm{e}^{-r/a_0}, \\ R_{21}(r) & = \frac{1}{\sqrt{3} (2a_0)^{3/2}} \left(\frac{r}{a_0}\right) \mathrm{e}^{-r/2a_0}, \\ R_{31}(r) & = \frac{4\sqrt{2}}{9 (3a_0)^{3/2}} \left(\frac{r}{a_0}\right) \left(1 - \frac{r}{6a_0}\right) \mathrm{e}^{-r/3a_0}. \end{align} (Note that $R_{10}$ and $R_{21}$ are in fact both strictly positive, so they can't integrate to $0$.) You should find $$ \langle R_{10} \vert R_{10} \rangle = \langle R_{21} \vert R_{21} \rangle = \langle R_{31} \vert R_{31} \rangle = 1 $$ and $$ \langle R_{21} \vert R_{31} \rangle = \langle R_{31} \vert R_{21} \rangle = 0, $$ as expected. However, $\langle R_{10} \vert R_{21} \rangle = \langle R_{21} \vert R_{10} \rangle$ and $\langle R_{10} \vert R_{31} \rangle = \langle R_{31} \vert R_{10} \rangle$ are very much neither $0$ nor $1$.

You can recover the full orthogonality you expect, but only by adding on the angular dependence given by the spherical harmonics for the full wavefunction.

pattern - Find the missing result

On a scrap of paper you read the following results:

$$ \begin{array}{r c l c} NETHERLANDS & - &FINLAND & 3-1 \\ JAPAN & - &IRAQ & 0-0 \\ RUSSIA & - &INDONESIA & 1-0 \\ AUSTRALIA & - &CHINA & 0-1 \\ CANADA & - &CUBA & 2-1 \\ SWITZERLAND & - &FRANCE & 0-0 \\ MADAGASCAR & - &EGYPT & 2-0 \\ BRAZIL & - &BELGIUM & 5-1 \\ IRELAND & - &GERMANY & 2-3 \\ ALGERIA & - &COLOMBIA & 3-2 \\ GREECE & - &MEXICO & ... \end{array} $$

Can you figure out what the result of Greece - Mexico should be?

Hint:

I've written the countries' names in uppercase for a reason.

Hint 2:

Most of these "results" have been the same for quite a while. One that has changed relatively recently is BRAZIL-BELGIUM, which used to be 0-1.

Answer

The Greece - Mexico result is a complete blow out:

1-6

The hints gave it away for me:

"countries' names in uppercase" hinted at the country capitals, and
"BRAZIL-BELGIUM, ... used to be 0-1" reinforced that (Brazil's capital was Rio up until 1960, when it changed to Brasilia)

So, the trick is that the scores are based on:

the number of characters that are the same, and in the same location between a country and its capital. Meaning that the specific pairing is just a red herring, since each individual country is evaluated in isolation.

The full list explained:

NETHERLANDS     FINLAND
amstERdAm       helsiNki      3−1

JAPAN           IRAQ
tokyo           baghdad       0−0

RUSSIA          INDONESIA

moScow          jakarta       1−0

AUSTRALIA       CHINA
canberra        beIjing       0−1

CANADA          CUBA
ottAwA          havAna        2−1

SWITZERLAND     FRANCE
bern            paris         0−0


MADAGASCAR      EGYPT
antAnAnarivo    cairo         2−0

BRAZIL          BELGIUM
BRAsILia        Brussels      5−1
rio de janeiro  Brussels     (0-1)

IRELAND         GERMANY
dubLiN          bERliN        2−3


ALGERIA         COLOMBIA
ALGiers         bOgOta        3−2

GREECE          MEXICO
athEns          MEXICO city   1-6 

conventions - Can units be plural?

I was in a conversation with my senior engineer where he kept on insisting that we can use plural when we write down any unit. I argued that it is not the 'common' practice or even throughout my whole academic career (unfortunately) I haven't found any instance where there was any plural unit used in the text books. He argued that if I said that it was not correct then it should have a good reason for that.

When I searched for this topic I couldn't come to any conclusive decision. Such as this thread and the other links those have been referred there (some leads to English.SE). These answers gave me the impression that it is grammatically acceptable provided the right circumstances.

But I felt that it would be rather ambiguous to accept plurals on scientific and engineering notations.

For example we were talking about output rate of a boiler which is measured in $\mathrm{kg/hr}$. My senior said that it is okay if anyone writes $\mathrm{kgs/hr}$.

To me it looks ambiguous. If anyone writes $\mathrm{s}$ after $\mathrm{kg}$ it may give a plural sense but as well it may refer to second also. Moreover if anyone argues that this is acceptable in some cases (like $\mathrm{kgs/hr}$) then what would be the yard stick to find out accepted cases? For instance can we add $\mathrm{s}$ in $\mathrm{m/s}$ or $\mathrm{km/hr}$ like $\mathrm{ms/s}$ or $\mathrm{kms/hr}$?

There is The NIST Guide for the Use of the International System of Units, which has this example.

the length of the laser is $5\ \mathrm{m}$ but not: the length of the laser is five meters

But I want to have more conclusive answer to which one is acceptable i.e. $\mathrm{kg/hr}$ or $\mathrm{kgs/hr}$ (or other similar instances).

Answer

According to the International System of Units (SI)

Unit symbols are mathematical entities and not abbreviations. Therefore, they are not followed by a period except at the end of a sentence, and one must neither use the plural nor mix unit symbols and unit names within one expression, since names are not mathematical entities.

as well as to the international standard ISO/IEC 80000 Quantities and units

Symbols for units are always written in roman (upright) type, irrespective of the type used in the rest of the text. The unit symbol shall remain unaltered in the plural and is not followed by a full stop except for normal punctuation, e.g. at the end of a sentence.

it is not acceptable to use the plural of unit symbols.

By the way, it is also not permissible to use abbreviations such as “hr” for unit symbols (“h”) or unit names (“hour”).

dark matter - Relativity and Galaxy Rotation Speed

If time travels slower nearer gravity wells, why can't the galaxy rotation speeds being faster on the outer edges than the inner areas be explained by relativity? What necessitates dark matter?

Answer

See Why isn't the center of the galaxy "younger" than the outer parts? and Galactic Rotation Speeds - Ehrenfest Paradox, Gravitational time dilation, Dark Matter - all of the above?. The time dilation effects are tiny and far too small to explain the observed rotation curves.

quantum mechanics - Explanation of why this derivation of Schmidt decomposition works

I'm following Preskill's notes and he derives the Schmidt decomposition in the following way:

Let a bipartite state be $\psi_{AB} = \sum_{i,j}\lambda_{ij}\vert i\rangle\vert j\rangle = \sum_{i} \vert i\rangle\vert \tilde{i}\rangle$, where I simply choose $\sum_j \lambda_{ij}\vert j\rangle = \vert \tilde{i}\rangle$.

I choose a set of basis vectors $\vert i\rangle$ such that the partial state is diagonal, that is $\rho_A = \sum_i p_i\vert i\rangle\langle i\vert$. But I can also obtain $\rho_A = Tr_B(\rho_{AB}) = Tr_B\sum_{i,j} \vert i\rangle\langle j\vert \otimes \vert \tilde{i}\rangle\langle \tilde{j}\vert = \sum_{ij} \langle \tilde{j}\vert\tilde{i}\rangle \vert i\rangle\langle j\vert$. The last part can be computed by explicitly writing out the trace over $B$ and using the properties of an orthonormal basis.

Thus, we have $\rho_{A} = \sum_i p_i\vert i\rangle\langle i\vert = \sum_{ij} \langle \tilde{j}\vert\tilde{i}\rangle \vert i\rangle\langle j\vert$. That is $\langle \tilde{j}\vert \tilde{i}\rangle = p_i\delta_{ij}$. Suddenly, the $\vert\tilde{i}\rangle$ are all orthogonal to each other.

Why does choosing the basis where $\rho_A$ is diagonal also give you orthogonal vectors in $B$? This seemed to drop out of the sky for me although the math is clear. What is the physical meaning of this?

Answer

Let us start from the Schmidt decomposition $|\psi\rangle = \sum s_i |a_i\rangle |b_i\rangle$.

Now consider the reduced state of $A$: $\rho_A=\sum s_i^2 |a_i\rangle\langle a_i|$. This is, the eigenbasis of A is exactly the basis you need for the Schmidt decomposition!

Thus, if you write your state using that eigenbasis of Alice, $$ |\psi\rangle = \sum_i |a_i\rangle \Big(\sum_j \lambda_{ij}|j\rangle\Big)\ , $$ the part $|\tilde b_i\rangle=\sum_j \lambda_{ij}|j\rangle$ must be equal to $s_i|b_i\rangle$, since the Schmidt decomposition is unique (modulo degeneracies).

word - A connection between Riley Riddles and I

The following riddle was inspired by what is referred to as "Riley Puzzles / Riddles" (example).

---

Another Riley Riddle! This one keeps to the traditions, but it might be a little too hard. I will give out hints every $24$ hours.

My prefix is a question.

My suffix, mean but golden.

My infix has two heads,

But I have bars instead.

What am I?

The answer has $4$ letters, and the title is a clue.

---

Other inspirations include this riddle and this one, too.

Answer

I think you are

wifi.

My prefix is a question.

"Why?"

My suffix, mean but golden.

Phi.

My infix has two heads,

"if". Not sure exactly why "two heads", though. Perhaps they refer to the antecedent and consequent. Perhaps they're the two letters. Perhaps something I haven't thought of. [EDITED to add:] OP explains in comments that this is a reference to the $\Leftrightarrow$ symbol for "iff", though personally I pronounce that "if and only if" which isn't exactly the infix here :-).

But I have bars instead.

Used to indicate strength of wifi signal.

The title ("A connection between Riley Riddles and I")

is presumably punning on the word connection: OP reads and posts Riley riddles over wifi.

general relativity - Why is Einstein gravity not renormalizable at two loops or more?

(I found this related Phys.SE post: Why is GR renormalizable to one loop?)

I want to know explicitly how it comes that Einstein-Hilbert action in 3+1 dimensions is not renormalizable at two loops or more from a QFT point of view, i.e., by counting the power of perturbation terms. I tried to find notes on this, but yet not anything constructive. Could anybody give an explanation with some details, or a link to some paper or notes on it?

Answer

you're quite right that Einstein gravity is not renormalizable by powercounting. Be careful though, this is not a rigorous proof, it's a mere estimation. In fact there is not proof to this date which once and for all proves that gravity is really not renormalizable. If you think in terms of Feynman diagrams (which are a nightmare for Einstein gravity), there might be non-trivial cancellations hidden within the sum of graph which tame divergences. It might also be that the potential counterterms are related by some non-obvious symmetry, so that in the end only a finite number of field redefinitions is necessary to get rid of the divergences -- or in other words that a sensible implementation of renormalization is possible. In fact, the question about UV finiteness is currently being addressed by Zvi Bern and friends who could show using sophisticated techniques that maximally supersymmetric quantum gravity is much less divergent than one would naively think. The buzzwords here are color-kinematics duality and the double copy construction which basically says that a gravity scattering amplitude is in some sense the square of a gauge theory amplitude. Check the arxiv, there's a plethora about this.

Now, regarding powercounting the reasoning is roughly as follows: the EH action is basically $$\mathcal{L} = \frac{1}{\kappa} \int d^4x \sqrt{-g}R $$ with $g$ the determinant of the spacetime metric $g^{\mu\nu}$. The mass dimension of the Ricci scalar $R$ is $[m^2]$, that of the integral measure $[m^{-4}]$, i.e.in order for the whole expression to be dimensionless $\kappa$ has to have mass dimension $[m^{-2}]$. If you now do a perturbative expansion around a flat background of the metric, you'll encounter at each step more and more powers of one over $\kappa$. Graphically, this expansion is an expansion in numbers of loops in Feynman diagrams. At each step, i.e. at each loop level the whole expression should be dimensionless, i.e. at each step you need more and more powers of loop momentum (at each loop level two more powers, to be precise), s.t. in the end your expressions become the more divergent the higher you go in the perturtabive expansion. In order to cancel these ever sickening divergences you'd have to introduce an infinite number of counterterms which -- in terms of renormalization -- makes no sense, hence this theory is by powercouting non-renormalizable.

There are nice lecture notes about this, cf http://arxiv.org/abs/1005.2703 (Notes by Lance Dixon about supergravity but the introductory bit is quite general).

geometry - Putting the pips on a d6

Using a blank cube and a bunch of circular stickers, an average person constructs a d6. A d6 is also known as a six-sided die, or sometimes, a dice.

For the purpose of this puzzle, the average person

• knows the shape of the correct pip pattern for each number (as in "a 3 is three pips on a diagonal line" and "a 5 is like a 4 but with a pip in the middle")


• knows that the sum of two opposing sides is always 7


• doesn't know any more specifics on dice manufacturers' conventions (and doesn't have a reference handy)


• given options, chooses one randomly, as long as the things he knows will apply in the resulting die.

What is the probability for the resulting pip pattern being the exact same one that official casino dice always use? (Seen in both of the dice in the image below)

(There's no need to account for any minor inaccuracies in the pip positions, we are only interested in the overall pattern being the same.)

Answer

I make it

1 in 16.
There are two possible orientations of the Six and the One - Six horizontal and Six vertical - the One is symmetrical about all axes.

There are also two possible orientations of the Five and the Two - Two Up left to right, and Two down left to right - the Five is symmetrical about both axes.
There are also two possible orientations of the Four and the Three - Three Up left to right, and Three down left to right - the Four is symmetrical about both axes.
Finally there are two possible positions for the numbers around the middle faces of the cube - 5 4 2 3 or 5 3 2 4 (its mirror).
2 x 2 x 2 x 2 = 16.

statistical mechanics - Probablistic interpretation of entropy

After taking a statistical mechanics course, I'm somewhat surprised that my intuitive highschool understanding of entropy doesn't match my current understanding.

When I was introduced to entropy, I was told that it (and the second law of thermodynamics) is just a statement that things move probabilistically toward their most likely state. I remember an example given with atoms, how entropy being defined (for atoms) as dq/T makes sense from this perspective.

But now in graduate school I'm told that entropy is the expected value of "information," that it's the log(Z), that it's always increasing, AND that it is dq/T.

I've had a really hard time trying to really connect all these facts mathematically. I've read a little bit on wikipedia about H-theorem, and tried to piece together an understanding of entropy but I'm still left with quite a few questions:

If entropy can be explained probabilistically, does that mean we can describe a system's entropy as a random variable? Can we find the probability a given system changes entropy as a function of time (microcanonically, canonically, or whatever)? And if we did this, could we show more explicitly how entropy is always increasing at large timescales?

Can we derive the first law of thermodynamics using microscopic arguments? In stat mech, we always calculate the partition function, and then just assume that the first law (or maxwell relations) holds - and then just plug-and-chug. But I always imagined a more statistical argument taking place.

I've seen entropy talked about so casually; whether it's the heat death of the universe, or violation of laws of thermodynamics - that I really wish I was more comfortable with the logic behind going from microstates to macrostates.

newtonian mechanics - How do we define what is "External" force or "Internal" force in the context of momentum conservation?

I know that without presence of any "External" force momentum is always conserved. But how do we distinguish between "External" force and "Internal" force where all are "Force"?

cipher - A married couple

Disclaimer: This puzzle is from a university in Israel, computer science. Its in a real document written in Hebrew so no point in posting it. There is no link to it on the internet.

There is a couple that has been married for 10 years. It all started when the husband wrote this cipher to his wife.

91
21
62
62
21

23
63
61
32
82
71
33
63
73
21

23
63
33
33
32
32
?

The answer is supposed to be in English.

optics - How to bend light?

As we all know that light travels in rectilinear motion. But can we bend light in parabolic path? If not practically then is it possible in paper? Has anyone succeeded in doing that practically ?

Answer

Light does not, in general circumstances, travel in straight lines (although it does do so in the ones we usually encounter).

For one, light is really a wave and can only approximately be thought of as consisting of independently-propagating rays. This happens when the wavelength of the light is much smaller than the distances it is propagating over, which is usually the case for light (whose wavelength in the visible range is $0.4$ to $0.7\,\mu\textrm{m}$) but is not necessarily the case e.g. for radio waves and when nanoparticles are involved.

In this short-wavelength limit, wave propagation gives way to ray propagation (which is a special, approximate case of the former), and specifically to Fermat's principle for the mathematical description of light. This principle states that light rays starting at $A$ and ending up at $B$ will follow the path that minimizes the travel time $$S=\int_A^B n(s)\textrm{d}s,$$ where $n(s)$ is the (possibly spatially dependant) refraction index along the path.

For a homogeneous medium, this does indeed give straight lines for propagation. For a planar interface between two different media it gives Snell's law for refraction and it also describes reflection. (However, because it does not account for the actual nature of light as an oscillating electric field, this description cannot predict transmission or reflection coefficients.

However, if the medium is not homogeneous, then light will not travel on a straight line, and for complicated inhomogeneities the path can be correspondingly difficult to calculate. For an example, see the formation of mirages or more generally atmospheric refraction. Conversely, if one has a path one wishes a given light ray to take, then it is possible to engineer a refractive index spatial dependence that will make light bend that way. (Of course, whether such a dependence is physically reasonable is another matter; if the path bends too sharply then it may not be possible to find materials with the correspondingly large index and index gradients necessary.)

mathematics - A classical combinatorial puzzle

It is a classical puzzle by Edsger Dijkstra. Not quoting the original problem but changing it into bag and balls, the puzzle is:

A bag contains some black and white balls. The following process is to be repeated as long as possible (assuming that we have infinite supply of black and white balls).

• Randomly select two balls from the bag. If they are the same color, throw them out, but put an extra black ball in.


• If they are different colors, place the white one back into the bag and throw the black one away.

As you can see that each iteration of the process reduces the number of balls in the bag by one. Also, repetition of the process must terminate with exactly one ball in the bag. The question is:

What, if anything, can be said about the color of the final ball based on the number of white balls and the number of black balls initially in the bag.

Answer

What can be said is that

the parity of the number of white balls never changes. Therefore, if there is an odd number of white balls initially then the last ball in the bag must be white; if an even number, the last ball must be black.

The way the puzzle is stated is, I think, deliberately unclear. You could equivalently (and more transparently) say: at each step you either remove one black ball, or remove two white and add one black. This also makes it a little more explicit that

at each step the parity of the number of black balls changes, which of course it must since you're removing one ball each time and the white parity is invariant.

(Of course the two statements aren't quite equivalent if for some reason you care about probabilities, since in the original version you remove two balls at random and let that determine which of the two things you do; but the puzzle itself is interested only in the worst case.)

mathematics - An Hourglass Puzzle

This puzzle follows from the well known sand timer puzzles: A Sand Timer Challenge http://www.crazyforcode.com/sand-timer-puzzle/
http://www.cut-the-knot.org/hg_solution.shtml

Warmup:
You are given a timer of 4 minutes, and a timer of 7 minutes. The time starts when the first timer is flipped. Is there a time for which all larger integer values of time can be measured? Prove why or why not.

Easy:
Same rules as above, but now your timers are 5 minutes and 8 minutes.

Medium:
Same rules as above, but now your timers are 6 minutes and 11 minutes.

Hard:

Same rules as above, but now your timers are $m$ minutes and $n$ minutes, where $m>n>0$ and $m,n \in\mathbb{Z}$. (Basically find a generalization. If you want to solve this first be my guest...)

EDIT:

If you couldn't understand what I was asking.
Let $T$ be the set of all times that can be timed using timers $m$ and $n$.
Find $s\in T:s-1\notin T,\forall x>s:x\in\mathbb{Z},x\in T$.

Answer

Taking a huge risk, but hey...

W/U:

Times in the form of $4n + 7m$ can be measured. As long as a number is large enough, you can add 2*4-7 = 1 if the 7-minute hourglass is to be used at least once to measure the previous time, or 3*7 - 5*4 = 1 if the 4-minute hourglass is to be used at least four times (for at least 20 mins!). At our most conservative guess, it's 27, but since values greater than 19 will either have an $m$ that's at least 1 or $n$ that's at least 5, 20 will do as as a slightly less-conservative one. Also, 7+(7-4) = 10 and 4*2 + 4*2 (mod 7) = 9 minutes can be measured. 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7 are reachable while 6 isn't, so $T$=6.

E:

If it's $5n + 8m$, it can be measured. As long as a number is large enough, you can add 5*5 - 3*8 = 1 if the 8-minute hourglass is to be used at least thrice to measure the previous time, or 2*8 - 3*5 = 1 if the 5-minute hourglass is to be used at least thrice. At our most conservative guess, the first $5n + 8m$ number whose $n$ and $m$ equal 3 (39) will do. However, 8+(8-5) = 11, mins can also be measured. 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18 are reachable while 17 isn't. 5*2 + 5*2 (mod 8) = 12 mins can also be measured, so that makes 17 also reachable. 16, 15 are measurable, and so is 14 (see the image in the comments section), 13, 12, 11 and 10. 9 isn't, so $T$=9.

M:

For $6n + 11m$, it can be measured. As long as a number is large enough, you can add 6*2 - 11*1 = 1 if the 11-minute hourglass is to be used at least once to measure the previous time, or 11*5 - 6*9 = 1 if the 6-minute hourglass is to be used at least nine times. At our most conservative guess, it's 54+11 = 65, but since values greater than 53 will either have an $m$ that's at least 1 or $n$ that's at least 9, 54 will do as a slightly less-conservative one. However, 11+(11-6) = 16 mins can also be measured. Then again, if we start out flipping any hourglass as soon as it empties, only 1 minute's worth of sand will remain in the upper half of the 6-min hourglass after 11 mins, while the other empties. To measure any time longer than that, we can also flip the bigger one, then flip both when the smaller one empties, then do it again when the bigger one does, and so on. 11 can be measured while 10 can't, so $T$=10.

H (incomplete):

As long as $n$ and $m$ are co-primes, $k$ and $j$ values to satisfy $nk$ ≡ 1 (mod $m$) and $mj$ ≡ 1 (mod $n$) can be found. In addition, $mx$ + $mx$ (mod $n$) and $ny$ + $ny$ (mod $m$) values can be used to measure time. Also, if there's some amount of sand left in only one of the hourglasses in the middle of a measurement, the time this sand corresponds to can be continuously added by flipping the other at the time, then flipping both whenever one empties, ... (rinse and repeat), making $mx$ + $rmx$ (mod $n$) if the latter without r is less than $m$ and $ny$ + $sny$ (mod $m$) if the latter without s is less than $n$ also measurable. That means the smaller one of the aforementioned $nk$ and $mj$ ($(n,m)max*[(n,m)min-1]$ at most) is enough for the conservative limit.

---

---

VERY LATE EDIT:

Just wrote a Java program to find the generalized solution:

package hourglass;

import java.util.ArrayList;

import java.util.List;

public class Unreachables {

public static void main(String[] args) {
    // x = n, y = m
int x = 7;
int y = 9;

int MaxUnreachable = 0;


int ConsLim1 = 0;
int ConsLim2 = 0;

int ub1 = 1;
int ub2 = 1;
int SecGen = 0;

List FirstGenList = new ArrayList();
List SecGenList = new ArrayList();

List ReachablesList = new ArrayList();

SecGenList.add(0);

 for (int i=0; i    if ((i*x)%y == 1) {
        ConsLim1 = i*x;
    }
    if ((i*y)%x == 1 && i        ConsLim2 = i*y;

    }               
 }

 int LesserConsLim = Math.min(ConsLim1, ConsLim2);

int MaxXFac = LesserConsLim/x;
int MaxYFac = LesserConsLim/y;  

     ///////////////////////////////////////////////////////////
  for (int j=0; j<=MaxXFac; j++) {

    for (int k=0; k<=MaxYFac; k++) { 
        if (j*x + k*y < LesserConsLim) {
            FirstGenList.add(j*x + k*y);
        }
    }
  }

    ///////////////////////////////////////////////////////////
  for (int l=0; l    int base = l*x;

    int r = (l*x)%y;

     if (r      for (int m=0; m        SecGen = base + r*m;
        if (SecGen 0) {
            SecGenList.add(SecGen);
            ub1++;
        }
      }

     }              
   }

  ub2 = 1;
  for (int n=0; n    int base = n*y;
    int t = (n*y)%x;

     if (t      for (int o=0; o
        SecGen = base + t*o;
        if (SecGen 0) {
            SecGenList.add(SecGen);
            ub2++;
        }
      }
     }              
   }

  for (int p=0; p
      for (int r=0; r          int PotRes = FirstGenList.get(p)+SecGenList.get(r);
           if (PotRes               ReachablesList.add(PotRes);  
           } 
      }
  }

  for (int z=LesserConsLim-1; z>0; z--){
      if (ReachablesList.indexOf(z)<0){

          MaxUnreachable = z;
          break;
          }       
      }


  System.out.println(MaxUnreachable); // our solution

    }
}

pattern - Mensa IQ Test puzzle

Just played a Mensa IQ test online (http://test.mensa.no/)

Answers to these two questions elude me.

Could anyone find out the logic of them? I was puzzled & curious for few days.

Answer

First one:

Square + square gives opposite square.

Same for dots - only dot+dot gives opposite dot, otherwise nothing. You could also say it is multiplication for 1 and reverse solution.
So correct answer is D

Second one:

Look at diagonals: 1x3, 2x2, 3x1
1x1, 2x3, 3x2
and 1x2, 2x1, 3x3 = ?
Only one that could fit is E, since there is always only one change. Like one change from 1x3 to 2x2, one change to 3x1 and so on.

quantum chemistry - Why are do neutral atoms shrink as their valence shells approach 8 electrons?

Why do neutral, unbonded atoms shrink in size as they approach having 8 electrons in their valence shells? A good example is elements 3 through 10 in this table, that is, lithium (1 valence electron) through neon (8 valence electrons).

---

For posterity, below is my earlier incorrect version of my question. It's incorrect because it based on a table that used the atomic radii of bonded atoms to describe their sizes. That is a very different measurement from the sizes of neutral atoms, since for bonded atoms you are tossing in additional electrons via the bonds. Very small atomic radii for halogens are easy to explain in that case, since the incomplete shells must fight very hard over the shared electrons to complete their own octets, making their bonding radii anomalously short. This question had an interesting back-and-forth, and I thank the two contributors both who both caught my chart interpretation error and answered the real and interesting question that lurked beneath my incorrect one.

---

If you look here at the relative sizes of charge-neutral atoms, you can see something I never noticed or even thought about until I tried to think through this question in the new Chemistry beta.

Why are nearly-complete atomic shells, such as the seven valence electrons of fluorine, so compact in comparison to the bloated-but-very-stable situation created by adding one more electron to make neon?

I thought I could answer that in terms of Pauli exclusion and wound up completely baffled. Can anyone help on this one? Why does "almost complete" equal "very small" in a charge-neutral atom?

Answer

I'm not sure how easy it would be to be to give a rigorous explanation of this. Here's an explanation based on the atomic orbital approach to the electronic structure of atoms. This is only an approximation, but I think it gives a good flavour of what is going on.

To clarify Terry's question: if you take the atomic orbital approximation to electronic structure you can ask what happens when e.g. the $p$, $d$, or $f$ orbitals are filling up. For example, along the first row of the periodic table from Boron to Neon the $2p$ orbital is filling up, or along the first row of the transition metals from Scandium to Zinc the $3d$ orbital is filling up, and in both cases the radius of the atom decreases.

The $p$, $d$ and $f$ (and higher if we ever make atoms that heavy) orbitals are generally expressed as a product of a radial and angular part, and for this question it's the radial part we have to consider. The atoms are shrinking because the radial part shrinks towards the nucleus, and as you've probably guessed it's because of the increasing nuclear charge attracting the electrons.

Consider the first row from Boron to Neon, i.e. the $2p$ shell is filling. You can get six electrons into this shell. They all share the same radial part of the atomic orbital and they're distributed between the three angular orbitals, $2p_x$, $2p_y$ and $2p_z$. As you add more electrons to the $2p$ orbital (and increase the nuclear charge) you get some mutual screening of the nuclear charge. However the electrons avoid each other as much as possible, so although they do screen each other from the increased nuclear charge the increase in screening is less than the increase in the nuclear charge. The result is that the effective potential felt by the electrons deepens and the radial part of the $2p$ orbital shrinks towards the nucleus. That's why Neon is the smallest atom in this row. When you add another electron it has to go into a new atomic orbital, the $3s$, and the radial part of the $3s$ extends further than the radial part of the $2p$ so the atom suddenly grows.

The reason all this is an approximation is because the splitting of the electron configuration into separate orbitals only works if there are no electron correlation terms i.e. if all the electrons feel a central averaged field, and of course this isn't true. However it's a pretty good approximation for light atoms and I think gives a reasonable explanation of Terry's question.

newtonian mechanics - Why is momentum conserved when a ball hits a vertical wall?

Almost in every book on physics, there's an example of conservation of momentum when the ball that is moving horizontally in the air, hits some massive wall. They claim that the return speed of the ball when it bounces off is the same as it was before the hit. If there were no external forces acting on the system (or their net force was zero) that would be fine. But in this case, there is a gravitational force acting on the ball, and because there is no surface underneath it, there's no normal force and therefore it doesn't "cancel out" the gravitational force. So my question is, why they say that the momentum conserved? Do they neglect the gravitational force or what? I'm quite confused.

Answer

The assumption in these problems is that the collision takes place instantaneously so that gravity has no time to change the momentum of the ball during the collision.

To see why this is makes sense, let $y$ denote the vertical direction, and notice that if the collision took some small amount of time $\delta t>0$ then the change in vertical momentum of the ball would be (by integrating both sides of Newton's second law) $$ \delta p_y = \int_{t_0}^{t_0+\delta t}dt \,F(t) = F(t_0)\delta t + \mathcal O(\delta t^2) $$ so we see that as the collision time goes to zero, so does the change in momentum in the vertical direction.

electromagnetism - Why bar magnet can produce magnetic field?

Say I wrapped a piece of wire around an iron bar in a closed circuit connected to a DC power supply, the electrons starts flowing and moving charge produce magnetic field. Yet a bar magnet can produce a magnetic field without any charged particles moving around, are the magnetic fields produced by the two objects different types?

Answer

All macroscopic objects we observe are emergent from the underlying particle/atomic/molecular nature, which by the way follows quantum mechanical equations.

a) The reason we see electric fields is fundamentally because electrons and protons have electric charge, which generates their electric field. There are many ways that electric fields can be macroscopically induced. Analogously, the reason we see magnetic fields is because particles/atoms/molecules have magnetic dipole moments inherent in their nature, which have elementary magnetic fields.

b) A second source of electric and magnetic fields comes from the behavior of elementary and fields under motion: they obey Maxwell's equations, which tell us that changing electric fields generate magnetic fields and changing magnetic fields generate electric fields.

Say I wrapped a piece of wire around an iron bar in a closed circuit connected to a DC power supply, the electrons starts flowing and moving charge produce magnetic field.

A magnetic field is generated from the moving charges due to the laws of b), and the small dipole ferromagnetic domains described in a) orient themselves according to the external field and magnetism is retained even when the current is off. The current enhances the total magnetic field after first orientations of the dipoles.

Yet a bar magnet can produce a magnetic field without any charged particles moving around, are the magnetic fields produced by the two objects different types?

In a permanent magnet the domains after being oriented in a direction retain it because they are in the lowest energy state of the solid when oriented.

The magnetic field is the same type of field for all cases.

electromagnetism - How to read Maxwell's Equations?

I was talking with a colleague about Maxwell's Equation, in particular Faraday's Law of Induction, and I realised that I did not understand the following. Usually, this equation is expressed as:

The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux enclosed by the circuit. - Wikipedia

In other words, a magnetic field that varies in time generates a circular electric field. Now I tried to express the same idea but backwards, that is, a circular electric field generates a magnetic field that varies in time proportional to the curl of the electric field. My colleague did not agree with this idea and said that the only interpretation is the one equivalent with the statement from Wikipedia.

This got me thinking and the reasons to back up my point of view are:

1. Since the relation in discussion is $\nabla \times \vec E=-\partial_t \vec B$ one should be able to read it both ways, since the $"="$ sign follows the properties of an equivalence relation


2. If I consider the relations $\vec D=\epsilon_0 \vec E$ and $\vec B=\mu_0 \vec H$, we have the induced fields $\vec D$ and $\vec B$ as functions of the field $\vec E$ and $\vec H$. The way I understand the term "to induce", it would much rather make sense to say that Field A generates The induced Field B, than The induced field B generates Field A. Going back to Faraday law, this would mean that the electric field generated the induced magnetic field.

Although the second point is based more on my understanding of the terminology, which might be subjective, I do not find the first one to have the same issue.

Which is the correct way of looking at these equations (same problem arises on Ampere's law too) or is there a book/written material that addresses this topic, that I can find?

---

Edit: So far the question received two answers that are based on the causality of the problem, both of them suggesting that the electric and magnetic fields should be considered as a connected thing (thus the electromagnetic field). I did not search for material suggested by AlbertB, and will do in the following hours, but I have a follow up for both answers.

If there is no delay when it comes to "generating" one field from another, because they are intertwined, wouldn't that mean that an electromagnetic wave should propagate with an infinite velocity? This question neglects relativity, and I consider having a flawed image on how an EM wave propagates.

Answer

It appears that there are two phenomena intermixed, and that might be the source of the confusion. Both, Rob and Albert, are talking about the properties of electromagnetic waves. While you are talking about the properties of current and charges, independent of each other.

If you have a wire loop(s) being "crossed" by a magnetic field, the magnetic field will induce a current in the wire, which creates a voltage difference at its end points.
If you have a voltage source at the end point of a wire loop, the electric field will induce a magnetic field perpendicular to the loop.

From these examples, one can see that the processes (therefore the equations) are reversible.

optics - Why do objects appear smaller when farther away?

Why does the apparent size of an object change when it is farther away from me compared to when it is closer to me?

Can someone perhaps explain this through an optics perspective?

Answer

Apparent size is not measured as an ordinary size, in meters. It is actually an angle, so it is measured in degrees or radians.

See this picture:

The object on the left is the eye. Looks like as the object moves further, the angle becomes smaller. That is what is called perspective.

Sometimes people try to compare apparent size (solid angle) and real size, but that makes no sense because they have different dimensions. For example, I've been asked:

Is the Moon bigger or smaller than a 1€ coin?

The answer is that it is much, much bigger: about 3000 km vs 2 cm. What the question is trying to ask is compare the apparent size of the Moon with the real size of a coin, and that makes no sense. You should compare the apparent size of the Moon with the apparent size of the coin, but then you should say what distance the coin is.

For reference, the Moon apparent size is about half a degree. That is about the size of your thumbnail, with the arm extended. It does not matter if your hand is big or small because a big hand will also mean a big arm!

mathematics - Sorting a single number

What is the only number that, when written in English, gives a sequence of letters which is sorted alphabetically?

(The original phrasing of the question was meant to be a bit less straightforward to answer, as one had to come up with a way in which a single number can be 'sorted', but you didn't like it...)

Answer

I guess that the answer you are looking for is ...

... forty.

It is the only number whose letters are sorted alphabetically when written in English.

river crossing - Strategy to solve the Missionaries and Cannibals problem

In the Missionaries and Cannibals problem:

Three missionaries and three cannibals must cross a river using a boat which can carry at most two people, under the constraint that, for both banks and the boat, if there are missionaries present on the bank (or the boat), they cannot be outnumbered by cannibals (if they were, the cannibals would eat the missionaries). The boat cannot cross the river by itself with no people on board.

The shortest solution for this puzzle has 11 one-way trips.

What should be the strategy to solve this puzzle for M missionaries and C cannibals, given that M is not less than C, or else the puzzle wouldn't be solvable.

Answer

Let's define a b * c d to represent the state of the river at any given time: a missionaries and b cannibals on the left, c missionaries and d cannibals on the right, and * being < if the boat is on the left and > if the boat is on the right.

The problem starts out in the state M C < 0 0, and we want to get 0 0 > M C.

For the case of M being more than C, here's an algorithm to transfer 1 missionary and 1 cannibal at a time:

• Bring 1 missionary and 1 cannibal over. (M-1 C-1 > 1 1)


• Bring the cannibal back. (M-1 C < 1 0; since M > C, M-1 >= C, as required.)


• Bring 1 missionary and 1 cannibal over again. (M-2 C-1 > 2 1)


• Bring a missionary back. (M-1 C-1 < 1 1).

And then you're left with the case of M-1 missionaries and C-1 cannibals on one side of the river. Just keep repeating this procedure (it works by induction) until you have only missionaries on the left side of the river. Then you're home free.

For M equal to C, I don't think you'll have a solution for M > 3, because the solution for M = 3 already depends on the fact that you can have only cannibals on one side and they won't be able to do anything to the missionaries no matter how many there are on one side.

The part of the solution where this applies looks something like this:

• 3 1 < 0 2


• 1 1 > 2 2


• 2 2 < 1 1


• 0 2 > 3 1

There's no way to get a similar arrangement for anything greater than three, because you can't send enough missionaries at once to balance out the rest of the cannibals.

quantum mechanics - Is there experimental verification of the s, p, d, f orbital shapes?

Have there been any experiments performed (or proposed) to prove that the shapes of the s,p,d,f orbitals correspond to our spatial reality as opposed to just being a figment of the mathematics that give us something to visualize?

Answer

A few years ago the XUV physics group at the AMOLF Institute in Amsterdam were (to my knowledge the first to be) able to directly image the orbitals of excited hydrogen atoms using photoionization microscopy. For more details see the paper,

Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States. A.S. Stolodna et al. Phys. Rev. Lett. 110 213001 (2013).

This was actually featured as one of Physics World Top 10 Breakthroughs of the year 2013. There is a nice open access Viewpoint on this if you want to read more

Viewpoint: A New Look at the Hydrogen Wave Function, C.T.L. Smeenk, Physics 6, 58 (2013)

For a more in-depth look, see

Taking snapshots of atomic wave functions with a photoionization microscope. A.S. Stodolna. PhD thesis, Radboud Universiteit Nijmegen, 2014.

riddle - See me once, See me Twice #7

Here I am again, with another See me once riddle. Have fun!

See me once, at the start of the week
See me twice, the men becoming bleak

See me once, when gambling on chinese ground
See me twice, saved but lost time I found

First Hint:

See me once, my dice are divine my friend

See me twice, my brother's story would never end

_{Here are the previous riddles in this series (the solutions there have nothing to do with this one, only the process of getting there).
#1, #2, #3, #4, #5, #6}

Answer

See me once, at the start of the week, when gambling on Chinese ground, my dice are divine.

Mo - may be short for Monday.
mo - the ISO code and Internet country code top-level domain for Macau.
Mo - Tibetan divination which uses dice.

See me twice, the men becoming bleak, saved lost time I found, my brother's story would never end.

This suggests the novel Momo, also know as The Grey Gentlemen, by Michael Ende. Momo is a girl who saves her society from the sinsister Men in Grey who are stealing the time of humans.
Michael Ende also wrote The Neverending Story, centred on the boy Bastian Balthazar Bux.

newtonian mechanics - Deriving a Schwarzschild radius using relativistic mass

Introduction

I have shown below two different approaches to deriving the Schwarzschild radius. I know these are less rigorous than the derivation of the Schwarzschild solution however the $\frac{2GM}{c^{2}}$ term still shows up in the metric anyways which in a sense validates the classical method shown below.

Classical method

\begin{equation} \frac{GMm}{r} = \frac{1}{2}mv^{2} \:\:\:\text{and}\:\:\:v =c \:\:\rightarrow \:\: R_{S} = \frac{2GM}{c^{2}} \end{equation}

However the classical method seems less general because it seems to ignore the Lorenz transformations. I offer no judgments rather I am looking to understand why this derivation reconciles with the Schwarzschild metric better.

This next approach is analogous except we set the relativistic potential energy equal to the relativistic kinetic energy. Since we are working in a frame where the gravitating mass $M$ is centered at the origin we only apply the gamma factor to the small mass $m$.

Relativistic method

\begin{equation} \frac{GMm}{r\sqrt{1-\frac{v^{2}}{c^{2}}}} = \frac{mc^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}-mc^{2} \:\:\:\text{and}\:\:\:v =c \rightarrow R_{S} = \frac{GM}{c^{2}} \end{equation}

I feel that the second derivation is more natural but I am somewhat unconvinced it is correct. I was hoping one of you fine stack exchange users could shed some words of wisdom.

Answer

The Newtonian derivation is not a derivation at all. It is a coincidental consequence of the way the Schwarzschild radial coordinate is defined. There is no physical insight to be gained from attempting to derive the Schwarzschild radius this way.

If we start with flat spacetime then the metric is:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

If we now introduce a weak gravitational field, where weak means that the gravitational potential per unit mass $\phi \ll c^2$, then we can use an approximation called the weak field limit to describe the curvature that corresponds to the weak gravitational field. In this approximation the metric becomes:

$$ ds^2 \approx -\left( 1 + \frac{2\phi}{c^2}\right) c^2dt^2 + \frac{1}{1 + 2\phi/c^2}\left(dx^2 + dy^2 + dz^2\right) \tag{1} $$

Remember that this approximation is only valid when $\phi \ll c^2$, but if we ignore this and blunder on regardless we would conclude that there is a coordinate singularity when:

$$ 1 + \frac{2\phi}{c^2} = 0 $$

or:

$$ \phi = -\tfrac{1}{2}c^2 $$

Both sides of this equation are an energy per unit mass, and putting the mass back in produces a possibly more familiar result:

$$ \phi m = -\tfrac{1}{2}mc^2 $$

which is exactly the argument used in the classical approach of calculating when the escape velocity reaches the speed of light.

If we rewrite the weak field equation (1) using polar coordinates:

$$ ds^2 \approx -\left( 1 + \frac{2\phi}{c^2}\right) c^2dt^2 + \frac{dr^2}{1 + 2\phi/c^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$

then substitute the Newtonian expression for the gravitational potential:

$$ \phi = -\frac{GM}{R} $$

we get something that looks like the Schwarzschild metric:

$$ ds^2 \approx -\left( 1 - \frac{2GM}{c^2R}\right) c^2dt^2 + \frac{dr^2}{1 - 2GM/c^2R} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$

But the Newtonian radial coordinate $R$ is not the same as the Schwarzschild radial coordinate $r$. The former is the distance measured from the central point to the position labelled by $R$ while the latter is the circumference of a circle passing through the position labelled by $r$ divided by $2\pi$. However it just so happens that the way the Schwarzschild radial coordinate is defined means that if we replace $R$ by $r$ we get an exact result:

$$ ds^2 = -\left( 1 - \frac{2GM}{c^2r}\right) c^2dt^2 + \frac{dr^2}{1 - 2GM/c^2r} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$

And this is why the Newtonian derivation gives the correct result for $r_s$. It is just a coincidence and shouldn't be regarded as a derivation at all.

time - Why do electrons, according to my textbook, exist forever?

Does that mean that electrons are infinitely stable? The neutrinos of the three leptons are also listed as having a mean lifespan of infinity.

Answer

Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that.

We'll go to your own rest frame--any decay you can do has to occur in all reference frames, and it's easiest/most limiting to talk about the electron's rest frame. In this frame, you have no kinetic energy, only rest mass energy equal to about 511 keV. So whatever you decay to has to have less rest mass than that--you might decay to a 300 keV particle, and give it 100 keV of kinetic energy, but you can't decay to a 600 keV particle. (There's no way to offset this with kinetic energy--no negative kinetic energy.) Unfortunately, every other charged lepton and every quark is heavier than that. So what options does that leave us? Well, there are massless particles (photon, gluon, graviton). There are also the neutrinos, which are all so close to massless that it took until very recently for anyone to tell that this was not the case. So you can decay to neutrinos and force carriers, maybe. Except then you run into a problem: none of these have any electric charge, and your decay has to conserve charge. You're stuck.

tl;dr: Electrons are the lightest negatively charged particle and therefore cannot decay into lighter particles without violating charge conservation.