quantum mechanics - Uncertainty Principle for a Totally Localized Particle

If a particle is totally localized at $x=0$, its wave function $\Psi(x,t)$ should be a Dirac delta function $\delta(x)$. Accordingly, its Fourier transform $\Phi(p,t)$ would be a constant for all $p$, thus the particle's momentum is totally uncertain. I guess this is what the Uncertainty Principle told us.

But in another hand, since the particle is totally localized at $x=0$, it is not moving and therefore static. Its velocity should be 0, so is its momentum, which contradicts the conclusion above.

So, what's wrong with my thinking?

Is it that there shouldn't be a TOTALLY STATIC particle, or it's in a non-normalizable state, so the uncertainty principle is not applicable?

Answer

The origin of your problem was already explained in the previous answers, let me just do so in a bit more detail. It is better to think of some normalizable wave function rather than the $\delta$-function itself. As you probably know, you can get arbitrarily close to a $\delta$-function by making a wave packet narrow and taking a suitable limit (see below for a concrete example).

Now you are right that you can localize your particle to an arbitrarily narrow region around $x=0$. You can even make it "static" in the sense that the average (quantum expectation value) of its momentum will be, and stay, zero. However, the uncertainty principle tells you that once the spread (dispersion) of coordinate $\Delta x$ is very small, the spread of momentum $\Delta p$ will be very large. Therefore, even if you initially localize the particle to a very narrow wave packet, it will broaden quickly with time. In fact, this broadening will be the faster, the sharper was the initial localization of the particle. If the wave function is initially Gaussian, then for a free particle you can solve the Schroedinger equation exactly and see that the wave function remains Gaussian, just its width grows (asymptotically as $\sqrt t$). So after some time, the particle is no longer localized, as a consequence of the very same uncertainty principle.

Even if the particle is not free but moves in some potential, making the initial wave packet at $t=0$ very narrow will result in the same spreading since the average kinetic energy (which is proportional to $\Delta p^2$ if the average momentum is zero) is much higher than the variation of the potential energy within the size of the wave packet.

There is one caveat in my above argument though. If you, simultaneously with narrowing down the wave packet, confine the particle by an increasingly strong potential, then you can keep it localized. Consider as a model the harmonic oscillator, defined by the Hamiltonian $$ H=\frac{p^2}{2m}+\frac12m\omega^2x^2. $$ The coherent states are Gaussian wave packets which thanks to the special form of the potential remain localized: both $\Delta x$ and $\Delta p$ are time-independent and given explicitly by $$ \Delta x=\sqrt{\frac{\hbar}{2m\omega}},\quad \Delta p=\sqrt{\frac{m\hbar\omega}2}. $$ If you now take the limit $\omega\to\infty$, the Gaussian packet goes asymptotically to the $\delta$-function. This corresponds to your particle localized infinitely close to $x=0$. The prize for this is that you have to make the potential infinitely strong, which simultaneously makes the particle oscillate with infinite frequency around the origin. So you cannot quite say that its momentum is zero.

gravity - A change in the gravitational law

What would happen if the force of gravitation suddenly starts varying as $1$$/$$r^3$ instead of $1/r^2$ ? Would the symmetry of universe now seen be disrupted?

wordplay - What is the name of this puzzle?

The English dictionary, as you may know, is a highly mutable entity.

While removing words such as irregardless might be considered a good thing, others, such as
snollygoster (n.): a shrewd, unprincipled person
are considered obsolete and will be missed, at least by me.

Other words have actually changed due to constant misuse. The one that disappoints me most is the fact that "literally" does not necessarily have to relate to its primary root definition:
literal (adj.) in accordance with, involving, or being the primary or strict meaning of the word or words; not figurative or metaphorical
It used to be fun to ask someone "Did you literally or metaphorically have your head in the toilet 'all night?'" Now, the correct answer might be "Both."

With this time-tested tradition of inventing new words and changing existing words "willy nilly," I propose the following twelve new definitions of existing words.

Your goal is to

• 
  

  determine the words being defined, and

  
  


• 
  

  discover the word that literally (I swear!) presents itself as a moniker for this type of word (the name of this puzzle).

  
  

---

---

1. Uninspired mobile phone photograph (6)
   _ _ _ _ _ _ ☑


2. Background checks on the key people in a company (8)
   _ _ _ _ _ _ _ _ ☑


3. Adorable (condescending?) term for an inexperienced user of the World Wide Web (8)
   _ _ _ _ _ _ _ _ ☑


4. An athletic protective cup (8)
   _ _ _ _ _ _ _ _ ☑


5. Blog about electric, wrought, and farm items (9)
   _ _ _ _ _ _ _ _ _ ☑



6. Spasmodic contraction caused by imprecise puzzle postings(8)
   _ _ _ _ _ _ _ _ ☑


7. Marijuana prescribed for an STD* (9)
   _ _ _ _ _ _ _ _ _ ☑


8. A Preggo's painful ped part (9)
   _ _ _ _ _ _ _ _ _ ☑


9. Donation by a celebrity look-a-like "baby daddy" (9)
   _ _ _ _ _ _ _ _ _ ☑


10. A trail to avoid in NYC's Central Park (10)
    _ _ _ _ _ _ _ _ _ _ ☑



11. What I'm doing when someone throws poop at me** (9)
    _ _ _ _ _ _ _ _ _ ☑


12. Discounted feminine product for everyone. (7)
    _ _ _ _ _ _ _ ☑

---

*Possibly not technically an STD
**Happens more often than you'd think
☑ indicates a correct answer has been found

Note: Please create a chatroom and offer invitations if you want to discuss the concept, as opposed to discussing the solution to the puzzle.

***Hint Regarding Title:

substr(0, 1) || substr(-1, 1)

Answer

So, all the words have been discovered. Another answer already lists them all, so I won't repeat them. What is missing is the name of the puzzle. We have this clue:

substr(0, 1) || substr(-1, 1)

which refers to a function used in some programming languages. Translated in English, it means:

Take either the first character, or the last one.

So, what it means for us is that we have to take the list of all the answers and

Choose, from each word, whether to take the first or the last letter (nitpicking: it could also mean that we have to take both, but it turns out that it is not the case).

We also know, from a comment that the OP left, that

The Questions are in random order.

So, this is the list that we have to work on:

M C
C E
N G
T D
F T
P C
C S

A Y
S E
P H
T N
G N

Now... I tried to figure it out, and some partial words that do fit are:

mend, amend, def

Which I would combine into:

Amendef (=Amend definition)

Which I find fitting, because it is formed in the way that this game works, and it even describes what it does! Unfortunately, this word doesn't exist. So, please bear with me if I say "amen" to that! ;-)

But this can't be the solution, because it doesn't use all the answers. So, I've written a program to try a brute-force with all the 12-letters words that I've found, and the result is...

Enchantments

For completeness, here is how it can be formed (there are actually 24 different ways to form it, assuming my program has no bugs, but I'll show just this one):

Letter 'e' taken from pair 1 (ce)
Letter 'n' taken from pair 2 (ng)
Letter 'c' taken from pair 5 (pc)
Letter 'h' taken from pair 9 (ph)
Letter 'a' taken from pair 7 (ay)
Letter 'n' taken from pair 10 (tn)
Letter 't' taken from pair 3 (td)
Letter 'm' taken from pair 0 (mc)
Letter 'e' taken from pair 8 (se)

Letter 'n' taken from pair 11 (gn)
Letter 't' taken from pair 4 (ft)
Letter 's' taken from pair 6 (cs)

lateral thinking - Disaster in Madrid!

A couple were in a car racing madly through Madrid in the middle of the night. Suddenly the engine juddered, whirred, and died. The broken-down car rolled to a halt, leaving them stranded with nobody else in sight. The husband had to go and get help. He was afraid for his wife left alone in the car, so he rolled up the windows and locked the doors before leaving the car, taking the keys with him.

It took him a long time to find aid, since he was a foreigner in the city and his Spanish wasn't too good. He returned more than half an hour after leaving, to find that disaster had struck! The exterior of the car was in the same state as when he left it, but his wife was dead and there was blood on the floor and a stranger in the car.

What happened?

Edit for clarity: all the events that night were down-to-earth and believable. There were no supernatural occurrences, extraterrestrial activity, or anything outside of what is commonly taken to be "the real world".

Answer

They were

On their way to the hospital at high speed because the woman was pregnant and the baby was on his way. Then their car broke down. While the husband was looking for someone to repair their car, the woman died while giving birth to her child (the stranger).

general relativity - Does a non-spherical black hole have distribution of mass like an empty body, solid body or pointed object?

Suppose a non-spherical (say, rotating or under distortion of another gravity source) black hole.

Does it have its mass distributed as if all the mass was on its surface, or as if the mass were distributed over its volume as some density or it would behave like a body with all its mass in the center?

For a spherical BH it would be all indistinguishable, but what about a non-spherical case?

Answer

Caveat reader: the current community voting on this answer suggests it doesn't have enough disclaimers about how mixing classical and relativistic calculations is a recipe for saying things that don't make sense in an erudite-sounding way. There should be one such disclaimer after about every sentence. See also the comments. Having said that:

You can probe the mass distribution of a spherically symmetrical object by spinning it and measuring its moment of inertia, $I = J/\Omega$. A solid classical sphere with mass $M$ and radius $R$ has $I_\text{solid}=\frac25MR^2$, while a thin spherical shell with the same mass and radius has the larger $I_\text{shell}=\frac23MR^2$ because more of the mass is further from the rotation axis.

A rotating black hole has a nonlinear relationship between $J$ and $\Omega$. Using this notation (see also), including the gravitational radius $R_G = GM/c^2$, there's a parameter $-\pi/2 \leq \Phi \leq \pi/2$ which characterizes the rotation by

$$ a = \frac{J}{Mc} = R_G\cos\Phi $$

In this case $\Phi=0$ (or $a=R_G$) corresponds to a maximally-rotating Kerr black hole and $|\Phi|=\pi/2$ collapses to the non-rotating case. The rotating black hole has "outer" and "inner" event horizons, with radii

$$ r_\pm = R_G\cdot(1\pm\sin\Phi) = R_G \pm \sqrt{R_G^2 - a^2} $$

The outer radius, $r_+\to2R_G$, is the Schwartzchild radius, the size of the event horizon in the non-rotating limit. There are also angular frequencies associated with these horizons,

\begin{align} \Omega_\pm &= \frac{c\cos\Phi}{2r_\pm} = \frac{ca}{2R_G^2 \pm 2R_G\sqrt{R_G^2 - a^2}}, \end{align}

although interpreting $\Omega$ as the angular frequency of a rigid classical object raises some thorny questions. The definition can be solved for the specific angular momentum $a$:

\begin{align} a = \frac{J}{Mc} &= \frac\Omega{c} \frac{4R^2}{1 + (2R\Omega/c)^2} \quad(\text{both of }\Omega_\pm) \\ J &= \frac{4}{1 + (2R\Omega/c)^2}\times MR^2\Omega \end{align}

This suggests you might consider a "moment of inertia" $I=J/\Omega$ of

\begin{align} I_\text{slow} &\approx 4MR_G^2 \approx Mr_+^2 \\ I_\text{max} &= 2MR_G^2 = 2Mr_+^2 \end{align}

That's interesting. When finding moments of inertia in classical physics one always finds $I=fMR^2$ by dimensional analysis, and if the radius $R$ is the maximum size of the rotating object one always finds $f\leq1$. For both limits of the moment of inertia here, the coefficient is $f\gt1$, which (combined with the default assumption of spherical-ish symmetry) suggests that $R_G$ is an underestimate of the classical size of the rotating mass distribution. If you wanted a sphere of mass $M$ to have the same classically-computed moment of inertia as a non- or slowly-rotating black hole with that mass, you'd make a spherical shell or a uniform-density sphere with radius larger than the Schwartzchild radius of the event horizon. (A thin hollow shell would go at $\sqrt6 R_G = 1.23 r_+$.) The maximally-spinning black hole, which has an outer event horizon of size $r_+=R_G$, likewise has "too large" of a moment of inertia.

Conclusion: Using classical moment-of-inertia considerations to analyze data on $J/\Omega$ for rotating black holes would lead you to require that some or all of their mass distribution were outside of their event horizons.

I personally find that kind of satisfying in a hand-waving, non-mathy kind of a way. After all, we no longer interact with matter that has crossed inside of the event horizon. If the mass-energy distribution of a black hole were actually found inside of its event horizon, wouldn't we be unable to interact with it? In electromagnetism, energy is stored in the fields, and the gravitational field of a black hole certainly extends outside of its horizon, so perhaps it's not nutty to locate some of the energy density near but outside of the event horizon. But this hand-wavy interpretation probably would not survive contact with a careful relativist.

quantum mechanics - Infinite halving of a distance

If an object is, say, 100 cm. from a wall, and I move the object halfway to the wall and stop, then the distance is reduced to 50 cm. If I continually move the object by one half of the remaining distance and stop, and keep repeating this procedure, why is it that the object eventually makes contact with the wall (on a macroscopic level at least), considering that no matter how infinitely small the remaining distance is, I can theoretically cut that distance by half, and thus never reach the wall. Does the Planck distance, the theoretically smallest unit of distance, have anything to do with this? And if so, how does any object get "around" that infinitely small Planck distance in order to move at all?

strategy - Sink the Submarine

There is submarine somewhere on the number line. Initially (say, at midnight), it is at position $A$, and is moving $B$ units per hour to the right, where $A$ and $B$ are both integers; both $A$ and $B$ are unknown to you. You have an unlimited number of torpedoes. Starting at midnight, and at the top of every hour, you can fire a torpedo somewhere on the number line, sinking the sub if it is there. What strategy can you use to guarantee you eventually sink the sub?

Answer

We are given the following information
P(0) = A

V(t) = B
P(t) = A + Bt

Since we know our current time, we can cycle through all the possibilities of P(t) by cycling through all the combinations of A and B at the current time and we will eventually land on the correct combination of A and B.
NOTE: There are actually an infinite amount of combinations because the set of integers are unbounded. But since we know that A and B are finite numbers, this algorithm will eventually find the answer.

Here is a picture which shows a method for computing all the possible combinations. I'm sure there's probably an actual function out there that is able to compute the unique pair of values given an integer.

NOTE: The image depicted here is not the exact same as the function used in the solution below (it's simply been rotated). To view the actual geometry of the function used, visit the link.

Note that we have to increment t by 1 each time because we are firing the torpedo with respect to the current time (1 hour intervals).

I'll try to come up with a better explanation and maybe an animation if I have the time.

EDIT: (Sketchy) Algorithm for enumerating 2 rational numbers:

Since enumerating 2 rational numbers would essentially require enumerating 4 integers (i.e. integers a,b,c,d correspond to rational numbers a/b and c/d), this is what this algorithm essentially does.

First add 0 (special case that only needs to be considered once)

Then find all permutations where 1 is the biggest

1 1 1 1

Then find all permutations where 2 is the biggest

2 1 1 1
...

2 2 2 2

and so on...

Now for each line we need to enumerate all possible permutations of +/- (2^4 permutations).

terminology - In layman's terms, what is a quantum fluctuation?

What causes it and how does it occur? If you do post some mathematics, please explain what each term means too please.

electromagnetism - Getting incorrect results applying Ampere's law

Let’s say we have a current wire with a current $I$ flowing. We know there is a field of $B=\frac{\mu_0I}{2\pi r}$ by using Ampère's law, and a simple integration path which goes circularly around the wire. Now if we take the path of integration as so the surface spans doesn’t intercept the wire we trivially get a $B=0$ which is obviously incorrect.

I see that I have essentially treated it as if there is no current even present. But a similar argument is used in other situations without fault.

Take for example a conducting cylinder with a hollow, cylindrical shaped space inside. By the same argument there is no field inside.

To further illustrate my point, the derivation of the B field inside of a solenoid requires you to intercept the currents. You can’t simply do the loop inside of the air gap.

This, at least to me, seems like the same thing, and I can’t justify why one is incorrect and the other is incorrect. Please point out why I am stupid.

particle physics - If atoms never "physically" touch each other, then how does matter-antimatter annihilation happen?

It is known that matter and antimatter annihilate each other when they "touch" each other. And as far as I know, the concept of "touching" as our brain gets it is not true on the atomic level since atoms never really touch each other but only get affected by different forces.

If this is true, then when does the annihilation really happen ? Does it happen when two atoms are affected by each others repulsion force for example ?

Answer

You should not think a particle as point like. Classically, the probability of two point like particles colliding with random location and velocity is 0, that is why you said it never happens.

However, at quantum mechanical level, these particle are described by wavefunction. It means that there is spreading in its spatial location, say 0.1nm (the minimum spreading is guaranteed by Heisenberg's uncertainty principle). The probability of annihilation can therefore be calculated by the overlap of these wavefunction and their interaction. This overlap is the meaning of "touching" in some sense.

particle physics - The way light travels

Light travels in the form of a wave.It has masless photons travelling at the speed of light.Does it mean that tha photons travels in the trajectory same as that of a wave.

newtonian mechanics - Time derivative of angular velocity in rotating reference frame

I am going through a section in a textbook regarding the Newton Euler equations for a system of rigid bodies (robotics text). There is a particular line in the derivation I don't understand, I've attached the image to make it clear:

I think the derivative of the second term on the right-hand side of 3.28 is using the product rule to get to 3.32, but I don't understand where the cross-product comes from. Could someone explain this particular part of the derivation? Thanks.

Answer

It is worth it for you reading about differentiating vectors on rotating frames.

Appendix

In addition, to prove equation (3.28) follow this answer https://physics.stackexchange.com/a/65768/392 or https://physics.stackexchange.com/a/105099/392

english - Word Tank-chain-wheel Challenge

My niece gave me this word wheels challenges the other day and I comes up with an alternative. Here's how this challenge works:

Find a word that can form another word by moving its first letter to the end of the word, or vice versa.

Example:

EMIT
MITE
ITEM

Until you can no longer form another word. Here's the rule:

• Must be a word found in dictionary, or a common name.


• Must be at least 3 letters long.


• Must have at least 1 rotation.


• Plural form (end with s) counts as 1 rotation only if you can form another word afterward.


• Please don't be like my niece and use the example as an answer...

Scoring:

• 1 point for each letter used ("emit" will has 4 points)


• 1.5 point for each 'rotation' it can have ("emit" has 2 rotations, thus 3 points)



• 1 point if the word can be loop infinitely.

Highest score wins.

This challenge will remains open for at least 2 days.

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Time's up!

Highest score:
Dr Xorile - 11.5
Engineer Toast - 11.5

Base on the time the answer is submitted, Dr. Xorile will be the winner of this challenge! There are some creative answers which I like much! Thanks for everyone's participation.

Answer

So the obvious "trick" for getting around the plural thing is to use a past-tense:

denunciate -> enunciated gives 11.5

is the best I could find. In a similar vein:

devaluate -> evaluated gives 10.5

If you avoid this trick, then there are other options:

trusties -> rustiest gives 9.5

Although this all works with the scoring system, I think the prettiest answer so far was given by @gannolloy

stripe -> tripes -> ripest (9.0)

The only other one I can think of that's similar is:

stable -> tables -> ablest (9.0)

I'm struggling to find anything that has more rotations apart from 3 letter words (ate->tea->eat; asp->spa->pas (which is a bit weak)). I can't find anything longer than the two above.

I quite like this one which is one letter shorter:

route-> outer->utero (8.0 points)

because it doesn't use any "s" or "d" at all! Similar, but not as good (imho) are:

amass->massa->assam (if that second word is not too politically incorrect), and saver->avers->versa

Why is the "complete metric space" property of Hilbert spaces needed in quantum mechanics?

I have been learning more about Hilbert spaces in an effort to better understand quantum mechanics. Most of the properties of Hilbert spaces seem useful (e.g. vector space, inner product, complex numbers, etc), but I don't understand why we need the inner product space to form a complete metric space.

My Questions:

1. Why do we need the inner product space to form a metric space?


2. Why does it have to be a complete metric space?

electrical resistance - Is this circuit a series or parallel circuit?

I saw a definition of a parallel circuit as a circuit with more than one path for current to flow. Is that the definition that's accepted? It seems like a good definition to me, but does that mean this circuit:

is series because the current doesn't flow through the voltmeter (in theory)?

The definition is quite good because this circuit:


I would say is a series circuit and the definition works for that, but I don't see much of a difference between the two circuits, so if the second one is a series circuit is the first one also series?

Edit, for clarity - The question:
What is the definition of a parallel circuit and does that make the first circuit (with the voltmeter) a series or parallel circuit?

electricity - Why don't electric workers get electrocuted when only touching one wire?

I know that when electricians work on the poles on the streets, if they only touch one wire at a time they will be fine. However, from my understanding, the negative wire is connected to a large negative terminal, and the positive wire to a large positive terminal. Now for example, if you touch the positive wire, wouldn't all of the electrons get stripped away from your body? Another way to look at it is that current always flows from a higher to a lower potential. If the potential in the wire is 100V (I have no idea what it is) and your potential is 0, wouldn't current flow through you?

Answer

AC or DC, you only get electrocuted if current passes through your body. (Current passing through any part of your body can be dangerous, and possibly cause an electrical burn, but current passing across your heart is the one that's really dangerous.) Touching just one wire at a time gives the current nowhere much to go. You are right to think that some electrons can get stripped from your body when you touch a bare wire. But not many. Once they've gone, unless your body gets new electrons from somewhere else, the current stops. If you're standing in a pool of water, or touching a metal pole, or another wire that can conduct lots of electrons from somewhere else, you're fried.

So how many electrons get conducted away from the human if it has no other source of electrons? In this case, the human acts as a capacitor. Now, Wikipedia tells me that one standard for this approximates a human as having capacitance $C=100\,\mathrm{pF}$. This is pretty tiny. (For comparison, the capacitors you might see if you open up a computer or other electronics can easily have capacitance billions of times larger.) If the voltage is $V=100\,000\,\mathrm{V}$ (which is really quite high), the charge that would be transferred is $Q = C\, V = 10^{-5}\,\mathrm{C} \approx 10^{14}$ electrons. You probably have $10\,000$ times more electrons in a single eyelash, so that's not many. The danger there is that at such high voltages, a lot of things become conductive that aren't normally.

P.S. Most electrical wires that you see in a city (though not those for trams, or the really big high-voltage lines) are actually insulated. (I worked as an electrician through college, and have touched many such wires.) So the birds aren't frequently touching anything dangerous to begin with. But even if they were, they'd only be in danger if they had somewhere to get new electrons. There are plenty of places (especially around those big transformers you see) where you can find exposed power, though, so be careful.

homework and exercises - Why should a solution to the wave equation be finite?

A function which represents a wave must satisfy the following differential equation:

$$\frac{\partial^2 y}{\partial t^2} = k\frac{\partial^2 y}{\partial x^2}$$

Any function that satisfies the wave differential equation represents a wave provided that it is finite everywhere at all times.

What does "it is finite everywhere at all times" mean?

Question:Which of the following functions represent a wave?

a) $(x - vt)^2$

b) $\ln(x + vt)$

c) $e^{-(x - vt)^2}$

d) $(x + vt)^{-1}$

Only option (c) is given as the answer though all 4 satisfy the differential equation.

I believe I did not understand the significance "function should be finite everywhere at all times" which is why I am unable to answer the aforementioned question.

Answer

It's semantics. Whoever wrote the problem prefers to refer to a wave as "A function which satisfies the wave equation and which is bounded" instead of "a function which satisfies the wave equation".

Unfortunately there are bound to be conventions which you disagree with, but in academics (undergrad and lower) the only way to deal with it is to figure out which conventions the professor (or problem writer) is working with before you read the problems. It's too easy for conversations on convention to turn into, "technically, it is a wave even though it's not physical" countered with "technically, it's not a wave because it's not bounded." The best you can do is recognize an issue in terminology ASAP and deal with it in a constructive way.

A better statement, which is more objectively true, would be: "Functions like $(x-vt)^2$ solve the wave equation, but generally don't come up and are not useful in physical solutions."

newtonian mechanics - What is the direction of gravitational constant?

Our book says:

$$\vec F=\dfrac{GMm}{r^2}$$

If Force is a vector quantity, $G$ must also be vector quantity so what is the direction of $G$.

I am a school kid please explain simply and kindly

Answer

$G$ is not a vector quantity, but just a number - a so-called scalar quantity - just like mass $m$. True, force is a vector quantity, but the formula you show here is a formula for the "size" of the force only!

The "size" or "strength" of a vector is called the magnitude. Some formulas give magnitudes only, other formulas give directions and yet other formulas give the full vector quantity including both its magnitude and direction. The formula you showed here does not give the direction, but only the magnitude of the force.

Compare these two formulas:

$$F=G\frac{Mm}{d^2}\qquad,\qquad \vec F=G\frac{Mm}{d^2}\hat r$$

The first formula gives the force magnitude $F$. That is the one you showed. The second gives the full force vector $\vec F$. In the second one, there is included a unit direction vector that points in the direction of the force, which I here call $\hat r$. (A unit vector is a vector with a magnitude of 1.) Without this, the direction is not involved at all because - as you rightfully question - none of the other parameters are vectors.

_{(You might here and there see other versions of this formula depending on how $\hat r$ is defined. For instance, on this Wikipedia page as well as in an answer below they have flipped $\hat r$ to mean the opposite of how I have used it, and then they add a minus sign so the formula still fits. And in another answer below, a different symbol $\mathbf e$ is used, which isn't a unit vector, so it must be divided by its length $d$ so it still fits (so that $\hat r=\frac{\mathbf e}d$). It is thus important that it is clear each time what such parameters exactly mean.)}

thermodynamics - Is the Boltzmann constant really that important?

I read a book in which one chapter gave a speech about the fundamental constants of the Universe, and I remember it stated this:

If the mass of an electron, the Planck constant, the speed of light, or the mass of a proton were even just slightly different (smaller or bigger) than what they actually are, then the whole Universe would not exist as we know it. Maybe we all wouldn't exist.

This speech works for all the fundamental known constants of the Universe but one: the Boltzmann constant. Its value is well known but even if its value were $10$ times bigger or if it were exactly $1$, or $45.90$ or $10^6$ well... the Universe would remain the same as it is now. The Boltzmann constant is not really fundamental to the existence of the Universe.

Maybe they weren't the exact words, but the concept is correct.

Now I ask: is that true, and why?

Answer

We can understand all of this business if we visit the statistical mechanics notion of temperature, and then connect it to experimental realities.

First we consider the statistical mechanics way of defining temperature. Given a physical system with some degree of freedom $X$, denote the number of possible different states of that system when $X$ takes the value $x$ by the symbol $\Omega(x)$. From statistical considerations we can show that modestly large systems strongly tend to sit in states such that $\Omega(x)$ is maximized. In other words, to find the equilibrium state $x_\text{eq}$ of the system you would write $$ \left. \left( \frac{d\Omega}{dx} \right) \right|_{x_\text{eq}} = 0$$ and solve for $x_\text{eq}$. It's actually more convenient to work with $\ln \Omega$ so we'll do that from now on.

Now suppose we add the constraint that the system has a certain amount of energy $E_0$. Denote the energy of the system when $X$ has value $x$ by $E(x)$. In order to find the equilibrium value $x_\text{eq}$, we now have to maximize $\ln \Omega$ with respect to $x$, but keeping the constraint $E(x)=E_0$. The method of Lagrange multipliers is the famous mathematical tool used to solve such problems. One constructs the function $$\mathcal{L}(x) \equiv \ln \Omega(x) + t (E_0 - E(x))$$ and minimizes $\mathcal{L}$ with respect to $x$ and $t$. The parameter $t$ is the Lagrange multiplier; note that it has dimensions of inverse energy. The condition $\partial \mathcal{L} / \partial x = 0$ leads to $$t \equiv \frac{\partial \ln \Omega}{\partial x} \frac{\partial x}{\partial E} \implies t = \frac{\partial \ln \Omega}{\partial E} \, .$$ Now remember the thermodynamic relation $$\frac{1}{T} = \frac{\partial S}{\partial E} \, .$$ Since the entropy $S$ is defined as $S \equiv k_b \ln \Omega$ we see that the temperature is actually $$T = \frac{1}{k_b t} \, .$$ In other words, the thing we call temperature is just the (reciprocal of the) Lagrange multiplier which comes from having fixed energy when you try to maximize the entropy of a system, but multiplied by a constant $k_b$.

If not for the $k_b$ then temperature would have dimensions of energy! You can see from the discussion above that $k_b$ is very much just an extra random constant that doesn't need to be there. Entropy could have been defined as a dimensionless quantity, i.e. $S \equiv \ln \Omega$ without the $k_b$ and everything would be fine. You'll notice in calculations that $k_b$ and $T$ almost always shows up together; it's no accident and it's basically because, as we said, $k_b$ is just a dummy factor which converts energy to temperature.

Folks figured out thermodynamics before statistical mechanics. In particular, we had thermometers. People measured the "hotness" of stuff by looking at the height of a liquid in a thermometer. The height of a thermometer reading was the definition of temperature; no relation to energy. Entropy was defined as heat transfer divided by temperature. Therefore, entropy has dimensions of $[\text{energy}] / [\text{temperature}]$.$^{[a]}$

We measured the temperatures $T$, pressures $P$, volumes $V$, and number of particles $N$ of some gasses and found that they always obeyed the ideal gas law$^{[b,c]}$

$$P V = N k_b T \, .$$

This law was known from experiment for a long time before Boltzmann realized that entropy is actually proportional to the logarithm of the number of available microstates, a dimensionless quantity. However, since entropy was already defined and had this funny temperature dimensions, he had to inject a dimensioned quantity for "backwards compatibility". He was the first to write $$ S = k_b \ln \Omega$$ and this equation is so important that it's on his tomb.

In practice, it is actually rather difficult to measure temperature and energy in the same system over many orders of magnitude. I think that it's for this reason that we still have independent temperature and energy standards and units.

• 
  

  Boltzmann's constant is just a conversion between energy and a made-up dimension we call "temperature". Logically, temperature should have dimensions of energy and Boltzmann's constant is just a dummy that converts between the two for historical reasons. Boltzmann's constant contains no physical meaning whatsoever. Note that the value of $k_b$ isn't the real issue; values of constants depend on the units system you use. The important point is that, unlike the speed of light or the mass of the proton, $k_b$ doesn't refer to any unit-independent physical thing in Nature.

  
  


• 
  

  Temperature is the Langrange multiplier that comes from imposing fixed energy on the problem of maximizing entropy. As such, it logically has dimensions of energy.

  
  



• 
  

  Boltzmann's constant $k_b$ only exists because people defined temperature and entropy before they understood statistical mechanics.

  
  


• 
  

  You will always see $k_b$ and $T$ together because the only logically relevant parameter is $k_b T$, which has dimensions of energy.

  
  

$[a]$: Note that if temperature had dimensions of energy then under this definition entropy would have been dimensionless (as it "should" be).

$[b]$: Actually, this law was originally written as $PV = n R T$ where $n$ is the number of moles of a substance and $R$ is the ideal gas constant. That's not really important though because you can group Avogadro's number in with $R$ to get $k_b$. $R$ and $k_b$ have equivalent "status".

$[c]$: Note again how $k_b$ and $T$ show up together.

mathematical physics - Tensor Operators

Motivation.

I was recently reviewing the section 3.10 in Sakurai's quantum mechanics in which he discusses tensor operators, and I was left desiring a more mathematically general/precise discussion. I then skimmed the Wikipedia page on tensor operators, and felt similarly dissatisfied. Here's why

In these discussions, one essentially defines an indexed set of operators $T_{i_1\cdots i_k}$ to be a "cartesian" tensor operator of rank $k$ provided $$ U(R)\, T_{i_1\cdots i_k}\, U^\dagger(R) = R_{i_1}^{\phantom{i_1}j_1}\cdots R_{i_1}^{\phantom{i_1}j_1}T_{j_1\cdots j_k} $$ for each rotation $R\in\mathrm{SO}(3)$ where $U$ is some unitary representation of $\mathrm{SO}(3)$ acting on a Hilbert space (usually that of some physical system whose behavior under rotations we with to study). Similarly one defines a "spherical" tensor operator of rank $n$ as an indexed set of operators $T^{(n)}_{q}$ with $-n

Based on these standard definitions, I would think that one could define something less "coordinate-dependent" and extended to representations of any group, not just $\mathrm{SO}(3)$, as follows.

Candidate Definition. Let a group $G$ be given. Let $U$ be a unitary representation of $G$ on a Hilbert space $\mathcal H$, and let $\rho$ be a representation of $G$ on a finite-dimensional, real or complex vector space $V$. A $k$-multilinear, linear operator-valued function $T:V^k\to \mathrm{Lin}(\mathcal H)$ is called a tensor operator relative to the pair of representations $U$ and $\rho$ provided \begin{align} U(g) T(v_1, \dots, v_k) U(g)^\dagger = T(\rho(g)v_1, \dots, \rho(g)v_k) \end{align} for all $g\in G$ and for all $v_1, \dots, v_k\in V$.

Notice that if a basis $u_1, \dots, u_N$ for $V$ is given, and if we define the components $T_{i_1,\dots i_k}$ of $T$ in this basis by \begin{align} T_{i_1 \dots i_k} = T(u_{i_1}, \dots, u_{i_k}) \end{align} and if $\rho(g)_i^{\phantom ij}$ denotes the matrix representation of $\rho(g)$ in this basis, then by using multilinearity the defining property of a tensor operator can be written as follows \begin{align} U(g) T_{i_1\cdots i_k} U^\dagger(g) = \rho(g)_{i_1}^{\phantom {i_1}j_1}\cdots \rho(g)_{i_k}^{\phantom {i_k}j_k} T_{j_1\cdots j_k} \end{align} So this definition immediately reproduces the cartesian tensor definition above if we take, $V =\mathbb R^3$, $G=\mathrm{SO}(3)$, and $\rho(R) = R$, and similarly for the spherical tensor definition if we take $V=\mathbb C^{2n+1}$, $G=\mathrm{SO}(3)$, $\rho = D^{(n)}$ and $k=1$.

Question.

Is the sort of object I just defined the "proper" formalization/generalization of the notion of tensor operators used in physics; it seems to contain the notion of tensor operator used in the physics literature? Is there any literature on the sort of object I define here? I would think that the answer would be yes since this sort of thing seems to me like a natural generalization a mathematically-minded physicist might like to study.

Answer

OP's candidate definition is a direct transcription of the tensor operator notion used in physics (and e.g. in Sakurai section 3.10) into a manifestly coordinate-independent mathematical construction. Tensor operators are e.g. used in the Wigner-Eckart theorem.

In this answer we suggest the following slight generalization of OP's candidate definition. Let the following five items be given:

1. 
   

   Let $G$ be a group.

   
   


2. 
   

   Let $H$ be a complex Hilbert space.

   
   


3. 
   
   

   Let $\rho: G \to GL(V,\mathbb{F})$ be a group representation.

   
   


4. 
   

   Let $R:G \to B(H)$ be a group representation.

   
   


5. 
   

   Let $T:V\to L(H;H)$ be a linear map.

   
   

Definition. Let us call $T$ for a $G$-equivariant map if $$ \forall g\in G, v\in V :\quad T(\rho(g)v)~=~ {\rm Ad}(R(g))T(v)~:=~R(g)\circ T(v)\circ R(g)^{-1}. \tag{*} $$

OP's candidate definition may be viewed as a special case of definition (*). For instance, if $\rho_0: G \to GL(V_0,\mathbb{F})$ is a group representation, then one may let $\rho: G \to GL(V,\mathbb{F})$ in point 3 be the tensor product representation $\rho=\rho_0^{\otimes m}$ with vector space

$$V~=~V_0^{\otimes m}~=~\underbrace{V_0\otimes \ldots \otimes V_0}_{m \text{ factors}}.$$

cosmology - The age of the universe

Many times I have read statements like, "the age of the universe is 14 billion years" . For example this wikipedia page Big Bang.

Now, my question is, which observers' are these time intervals? According to whom 14 billion years?

Answer

An observer with zero comoving velocity (i.e. zero peculiar velocity). Such an observer can be defined at every point in space. They will all see the same Universe, and the Universe will look the same in all directions ("isotropic").

Note that here I'm talking about an "idealized" Universe described by the FLRW metric:

$$\mathrm{d}s^2 = a^2(\tau)\left[\mathrm{d}\tau^2-\mathrm{d}\chi^2-f_K^2(\chi)(\mathrm{d}\theta^2 + \sin^2\theta\;\mathrm{d}\phi^2)\right]$$

where $a(\tau)$ is the "scale factor" and:

$$f_K(\chi) = \sin\chi\;\mathrm{if}\;(K=+1)$$ $$f_K(\chi) = \chi\;\mathrm{if}\;(K=0)$$ $$f_K(\chi) = \sinh\chi\;\mathrm{if}\;(K=-1)$$

and $\tau$ is the conformal time:

$$\tau(t)=\int_0^t \frac{cdt'}{a(t')}$$

The peculiar velocity is defined:

$$v_\mathrm{pec} = a(t)\dot{\chi}(t)$$

so the condition of zero peculiar velocity can be expressed:

$$\dot{\chi}(t) = 0\;\forall\; t$$

The "age of the Universe" of about $14\;\mathrm{Gyr}$ you frequently hear about is a good approximation for any observer whose peculiar velocity is non-relativistic at all times. In practice these are the only observers we're interested in, since peculiar velocities for any bulk object (like galaxies) tend to be non-relativistic. If you happened to be interested in the time experienced by a relativistic particle since the beginning of the Universe, it wouldn't be terribly hard to calculate.

homework and exercises - Poisson brackets and angular momentum

I'm trying to find $[M_i, M_j]$ Poisson brackets.

$$\{M_i, M_j\}=\sum_l \left(\frac{\partial M_i}{\partial q_l}\frac{\partial M_j}{\partial p_l}-\frac{\partial M_i}{\partial p_l}\frac{\partial M_j}{\partial q_l}\right)$$

I know that:

$$M_i=\epsilon _{ijk} q_j p_k$$

$$M_j=\epsilon _{jnm} q_n p_m$$

and so:

$$[M_i, M_j]=\sum_l \left(\frac{\partial \epsilon _{ijk} q_j p_k}{\partial q_l}\frac{\partial \epsilon _{jnm} q_n p_m}{\partial p_l}-\frac{\partial \epsilon _{ijk} q_j p_k}{\partial p_l}\frac{\partial \epsilon _{jnm} q_n p_m}{\partial q_l}\right)$$

$$= \sum_l \epsilon _{ijk} p_k \delta_{jl} \cdot \epsilon_{jnm} q_n \delta_{ml}- \sum_l \epsilon_{ijk}q_j \delta_{kl} \cdot \epsilon_{jnm} p_m \delta_{nl}$$

Then I have thought that values that nullify deltas don't add any informations in the summations. And so, $m=l, j=l$ but so I obtain $m=j$. But if $m=l$, the second Levi-Civita symbol in the first summation is zero... And if I go on, I obtain $\{M_i, M_j\}=-p_iq_j$ instead of $\{M_i, M_j\}=q_ip_j-p_iq_j$

Where am I wrong? Could you give me some hints to continue?

Answer

You are confusing in the index, such calculations must be carried out very carefully. I would start with your difention. $$M_i=\epsilon _{ijk} q_j p_k$$

$$M_p=\epsilon _{pnm} q_n p_m$$ $$\{M_i, M_p\}=\sum_l \left(\frac{\partial M_i}{\partial q_l}\frac{\partial M_p}{\partial p_l}-\frac{\partial M_i}{\partial p_l}\frac{\partial M_p}{\partial q_l}\right)$$

First term

$=\epsilon _{ijk}p_k\delta_{jl}\epsilon _{pnm}q_n\delta_{ml}=\epsilon _{ilk}p_k\epsilon _{pnl}q_n=(-1)\epsilon _{lik}p_k(-1)^2\epsilon _{lpn}q_n=-\epsilon _{lik}p_k\epsilon _{lpn}q_n=-\left(\delta_{ip}\delta_{kn}-\delta_{in}\delta_{kp}\right)p_kq_n$

Here I used the antisymmetry of $\epsilon _{lik}$ and equation $\epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$

Second term

Absolutely the same calculations. $=\epsilon _{ijk}q_j\delta_{kl}\epsilon _{pnm}p_m\delta_{nl}=\epsilon _{ijl}q_j\epsilon _{plm}p_m=\epsilon _{plm}p_m\epsilon _{ijl}q_j=-\epsilon _{lpm}p_m\epsilon _{lij}q_j=-\left(\delta_{pi}\delta_{mj}-\delta_{pj}\delta_{mi}\right)p_mq_j=$

Make the change $m=k,j=n$. Then

$=-\left(\delta_{pi}\delta_{kn}-\delta_{pn}\delta_{ki}\right)p_kq_n$

All together

$\{M_i, M_p\}=-\left(\delta_{ip}\delta_{kn}-\delta_{in}\delta_{kp}\right)p_kq_n+\left(\delta_{pi}\delta_{kn}-\delta_{pn}\delta_{ki}\right)p_kq_n=\delta_{in}\delta_{kp}p_kq_n-\delta_{pn}\delta_{ki}p_kq_n=p_pq_i-p_iq_p=q_ip_p-p_iq_p$

logical deduction - Can you solve this logic puzzle in disguise? The color code behind the tomb

Another themed puzzle for you to enjoy. As with all my puzzles, please provided feedback about pro's and cons of the puzzle. The puzzle is self-contained (the link is just to provide a theme) and I believe it contains all necessary information to find the unique solution. If you don't think so or if you discover any problems, please let me know.

---

After you have made it through the door of the ancient tomb (puzzle here), you find yourself in an odd cavern which partly looks natural, partly artificial. From the ceiling hang several stalactites but of a strange, crystalline nature with facets. They also seem to shimmer in different colours.

You walk a bit deeper and find a very technical looking panel on top of a short column. The panel has the shape of truncated pyramid and on its top-facet are 36 wheels with a marker which can be set to one of 6 positions. The four truncated sides of the pyramid each contain what appears to be a socket of some sort with 6 finger-thick holes in a row.

It takes not long, and you find a very modern looking device nearby which has a cable and plug, obviously fitting the sockets. The device has a single button in form of an arrow, and a liquid-crystal-like display without any visible reading. You push the button, but nothing seems to happen. Next you plug the thing into one of the four sockets and press the button again. Suddenly you get a reading of some tree-like symbols on the display - as long as you keep the button pressed.

You turn a few of the wheels on the panel (while and also in between using the strange device at the same time), but that does not seem to change any of the readings.

Knowing that the ancient people who've built this place seem to be keen on puzzles and testing the skills of archaeologists and grave-robbers alike, you soon decide that this has to be another logic test to be solved. But what is to do? Surely, there have to be more hints somewhere. Sure enough you find three smaller drawing on the right wall of the cavern, and a single, larger one on the other side of the cavern. Will they tell you how you can solve the puzzle?

---

The goal of the puzzle

..is to find a combination of the 36 dials on top of the panel, i.e. a single combination of 36 colours. The only clues you have are the pictures given below and 4 device-readings when you plug in the device on each of the 4 sockets.

In your answer, you have to explain why you've chosen a particular combination. Please give your reasoning as detailed as possible.

---

The socket & the device:

The 4 display readings on the device when plugged in as indicated, respectively:

The 3 drawings on the left wall of the cavern:

The two, large drawings on the right wall of the cavern:

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Spoilers & Hints

If you feel really stuck (and only then) you may look at the following spoiler hints. They are meant to guide your thoughts in the right direction.

How do I even start thinking about this?

While this is an abstract riddle, it's themed in the real world. It can sometimes be helpful to think with some physics in mind. On the other hand, the crystals in this cave are of a strange, crystalline nature as mentioned in the introduction. Don't expect them to act like simple coloured glass.

So many arrows - what could they mean?

Did you notice that there are 3 different arrow shapes? This may not be accidental.

All those dial combinations in the image, what should they mean?

Have you read the story? You've already tried turning those dials and seeing if it makes a difference - it doesn't. But your pretty convinced that a final, correct setting will be what you're after...

---

Okay, I have found the solution - what now?

Please post the solution and how you derived at it as an answer. Please use spoiler tags to hide the essential parts, but do not use one big single spoiler. There should be readable text interrupted by spoiler sections, similar to what you see just below here:

The solution

which consists of 36 dial positions on the socket

could potentially be noted down as

a 6x6 grid of dial positions I to VI using the coordinate notation as in the following image:
Then produce the 6x6 grid with the dial-positions in place.

---

---

References

WARNING looking at the references will be a strong and fundamental spoiler, as a large part of the puzzle is based on finding out how this represents a puzzle.

This puzzle was inspired by

the "Towers" puzzle one can find elsewhere.

In my case, I've seen such puzzles in the app

Time Killers by Andrea Sabbatini.

However, the (accepted) solution by Misch also provides a link to

this site.

I do not know about any other "original" copyrights, but if you alert me to any, I'll put them here.

Answer

I'll answer this riddle with my own notation, not the one provided by the author. This notation makes more sense, but maybe the author didn't chose it because it could possibly spoil the riddle?

My solution is

541326
132564
423615
256431
614253
365142

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The socket

The 36 wheels on the socket are connected to crystals, as shown on a drawing on the wall. First interpretation could be that the wheel can change the aspect of the crystal which is below it. However, as changing the wheels doesn't seem to have any effect, it is more likely, that the crystals have a given color, and the solution is to find out which color, and to turn the wheels accordingly.

The device

The device has six sensors, which as shown in a drawing on the wall send light (the colored array) into the socket. Each sensor sends light to and "reads" information back from exactly one of the rows of crystals. As seen in the drawings, the light emitted from the device consists of 6 different colors. Additionally, there is one additional color (I shall call it the reflection color). The device measures how many reflection rays are coming back from the crystals.

Crystals modifying the light

First, it is shown, that all colors let the reflection ray pass. Additionally, it is shown for each crystal, which colors it absorbs, and that it reflects a reflection ray in the opposite direction, if if is illuminated by all color rays. What is missing here, is the information, if the crystal also emits a reflection ray if it is only lit by several instead of by all color rays. There is no information provided to find out, which crystal needs which color to let it emit a reflection ray. Here, I used the hint, that a little physical thinking would be useful, and just guessed, that a crystal emits a reflection ray, exactly then, if it is hit by a colored ray, which it doesn't let pass.

The solution

At this stage, this puzzle reminded me of an other puzzle, which, as it turned out is the same puzzle, just presented differently. The game Towers works the following way: You have a grid, which you have to fill with towers, with different sizes. On each border, there are numbers, which indicate, how many towers you see from there. You see a tower only if it isn't hidden by a bigger tower which is in front of it. When you think a little about it, you'll see that both puzzles have the exact same concept. So I converted the given indications into a tower game, and solved it (as that's easier than playing with strange colors. The color code, I gave above symbolizes the tower heights.

The last constraint:

After solving still one thing was not clear to me: in the Towers game, there is the constraint, that in each row and column, each tower size can only be placed once, but I was missing that constraint in this riddle (otherwise, more solutions would surely have been possible. Only now, I understand that exactly that is symbolized by the last drawing.

The analogy:

From Simon Tatham's puzzle collection

general relativity - Does a non-spherical black hole have distribution of mass like an empty body, solid body or pointed object?

Suppose a non-spherical (say, rotating or under distortion of another gravity source) black hole.

Does it have its mass distributed as if all the mass was on its surface, or as if the mass were distributed over its volume as some density or it would behave like a body with all its mass in the center?

For a spherical BH it would be all indistinguishable, but what about a non-spherical case?

Answer

Caveat reader: the current community voting on this answer suggests it doesn't have enough disclaimers about how mixing classical and relativistic calculations is a recipe for saying things that don't make sense in an erudite-sounding way. There should be one such disclaimer after about every sentence. See also the comments. Having said that:

You can probe the mass distribution of a spherically symmetrical object by spinning it and measuring its moment of inertia, $I = J/\Omega$. A solid classical sphere with mass $M$ and radius $R$ has $I_\text{solid}=\frac25MR^2$, while a thin spherical shell with the same mass and radius has the larger $I_\text{shell}=\frac23MR^2$ because more of the mass is further from the rotation axis.

A rotating black hole has a nonlinear relationship between $J$ and $\Omega$. Using this notation (see also), including the gravitational radius $R_G = GM/c^2$, there's a parameter $-\pi/2 \leq \Phi \leq \pi/2$ which characterizes the rotation by

$$ a = \frac{J}{Mc} = R_G\cos\Phi $$

In this case $\Phi=0$ (or $a=R_G$) corresponds to a maximally-rotating Kerr black hole and $|\Phi|=\pi/2$ collapses to the non-rotating case. The rotating black hole has "outer" and "inner" event horizons, with radii

$$ r_\pm = R_G\cdot(1\pm\sin\Phi) = R_G \pm \sqrt{R_G^2 - a^2} $$

The outer radius, $r_+\to2R_G$, is the Schwartzchild radius, the size of the event horizon in the non-rotating limit. There are also angular frequencies associated with these horizons,

\begin{align} \Omega_\pm &= \frac{c\cos\Phi}{2r_\pm} = \frac{ca}{2R_G^2 \pm 2R_G\sqrt{R_G^2 - a^2}}, \end{align}

although interpreting $\Omega$ as the angular frequency of a rigid classical object raises some thorny questions. The definition can be solved for the specific angular momentum $a$:

\begin{align} a = \frac{J}{Mc} &= \frac\Omega{c} \frac{4R^2}{1 + (2R\Omega/c)^2} \quad(\text{both of }\Omega_\pm) \\ J &= \frac{4}{1 + (2R\Omega/c)^2}\times MR^2\Omega \end{align}

This suggests you might consider a "moment of inertia" $I=J/\Omega$ of

\begin{align} I_\text{slow} &\approx 4MR_G^2 \approx Mr_+^2 \\ I_\text{max} &= 2MR_G^2 = 2Mr_+^2 \end{align}

That's interesting. When finding moments of inertia in classical physics one always finds $I=fMR^2$ by dimensional analysis, and if the radius $R$ is the maximum size of the rotating object one always finds $f\leq1$. For both limits of the moment of inertia here, the coefficient is $f\gt1$, which (combined with the default assumption of spherical-ish symmetry) suggests that $R_G$ is an underestimate of the classical size of the rotating mass distribution. If you wanted a sphere of mass $M$ to have the same classically-computed moment of inertia as a non- or slowly-rotating black hole with that mass, you'd make a spherical shell or a uniform-density sphere with radius larger than the Schwartzchild radius of the event horizon. (A thin hollow shell would go at $\sqrt6 R_G = 1.23 r_+$.) The maximally-spinning black hole, which has an outer event horizon of size $r_+=R_G$, likewise has "too large" of a moment of inertia.

Conclusion: Using classical moment-of-inertia considerations to analyze data on $J/\Omega$ for rotating black holes would lead you to require that some or all of their mass distribution were outside of their event horizons.

I personally find that kind of satisfying in a hand-waving, non-mathy kind of a way. After all, we no longer interact with matter that has crossed inside of the event horizon. If the mass-energy distribution of a black hole were actually found inside of its event horizon, wouldn't we be unable to interact with it? In electromagnetism, energy is stored in the fields, and the gravitational field of a black hole certainly extends outside of its horizon, so perhaps it's not nutty to locate some of the energy density near but outside of the event horizon. But this hand-wavy interpretation probably would not survive contact with a careful relativist.

general relativity - How does light behave within a black hole's event horizon?

If the event horizon of a black hole is the distance from the center from within which light cannot escape, imagine a person with a flashlight falls into the black hole.

He points his flashlight in a precisely radial direction and turns it on. Now there is a light ray moving outward at the speed of light. If it now cannot move in an arc, but rather is constrained to radial motion, it must, at some point before the horizon, switch directions and fall back into the black hole.

If the speed of light is constant, how does it suddenly change directions, without either decelerating, or requiring an infinite amount of energy?

quantum mechanics - What is the experiment used to actually observe the position of the electron in the H atom?

Prior to observation, the electron can be found anywhere (from inside the nucleus to the ends of the universe), but once its position is determined the answer is precise (albeit its momentum is not due to the uncertainty principle).

I have several questions related to this idea

First, how do you actually determine the position of the electron without "kicking" it out of the atom?

Second, if you were able to determine its position very precisely, wouldn't its momentum be so high that it would exceed the speed of light? (or does it just become more massive? Either way, it doesn't seem like it could remain bound to the nucleus.

Third, if you were able to determine its position, how does your knowledge of its position degrade with time? It would appear that to get back to its original probability distribution (over all space) it would need a great deal of time, again so as not to violate the speed of light (unless it can pop in and out of existence far, far away).

Answer

First, how do you actually determine the position of the electron without "kicking" it out of the atom?

When talking of quantum mechanical entities, as the atom and the electron are, one has to keep clearly in mind that our well validated models that allow us to probe their behavior are probabilistic, the probability given by the square of the wave function.

The wavefunction is a function of (x,y,z,t) . The way it has been validated is by making probability distributions and checking them against the data. The only way of measuring positions for an electron in the atom is by the electron interacting. This might be by its being kicked off and measured, giving one point eventually in the probability distribution under measrurement, or in fitting weak scattering data, for example, like light through a crystal, or x-rays , the interferences of light giving information of position . Again a statistical distribution. In this case the end result is a probing of the atom's position as a whole, as the electron orbitals define the size of the atoms.

Second, if you were able to determine its position very precisely, wouldn't its momentum be so high that it would exceed the speed of light? (or does it just become more massive? Either way, it doesn't seem like it could remain bound to the nucleus.

If you only determine the position, it could be as precise as your measurement capabilities. The Heisenberg uncertainty constrains one only if both momentum and position are required together.

Third, if you were able to determine its position, how does your knowledge of its position degrade with time? It would appear that to get back to its original probability distribution (over all space) it would need a great deal of time, again so as not to violate the speed of light (unless it can pop in and out of existence far, far away).

Again, please note that experiments are one off for individual interactions. One photon goes through the crystal and interacts with the field of the electron and is registered as one point in a probability distribution. Or one electron is kicked off and its track is measured and projected back to its position , as in this recent expreriment, giving the distribution of the electron in the hydrogen orbitals.

hydrogen orbitals

In the case of photons probing non destructively the atom there is no way that one can know what an individual electron is doing in its orbital after that slight interaction. So there is no degradation detectable with time as the electron is still in its orbital.

In the case of scattering electrons off the hydrogen atom, the process is completely destructive of the atom, the electron flies off and is detected in an appropriate detector system, and the hydrogen becomes an ion, a proton seeking for an electron from the environment to return to neutrality.

homework and exercises - Question about canonical transformation

I was going through my professor's notes about Canonical transformations. He states that a canonical transformation from $(q, p)$ to $(Q, P)$ is one that if which the original coordinates obey Hamilton's canonical equations then so do the transformed coordinates, albeit for a different Hamiltonian. He then considers, as an example the Hamiltonian

$$H=\frac{1}{2}p^2,$$

with a transformation:

$$Q = q,$$ $$P = \sqrt{p} - \sqrt{q}.$$

The notes state that this transformation is locally canonical with respect to $H$, and that in the transformed coordinates the new Hamiltonian is:

$$K = \frac{1}{3} \left( P + \sqrt{Q} \right)^3.$$

I don't understand how we know that this is locally canonical, or what it really even means to be locally canonical. Also, where do we get K from? Considering that the inverse transformation would be:

$$q=Q,$$ $$p=\left( P + \sqrt{Q} \right)^2,$$

Why isn't the new Hamiltonian this:

$$K= \frac{1}{2} \left(P + \sqrt{Q} \right)^4,$$

where all I've done is plug the inverted transformation into the original Hamiltonian?

I'm a bit confused by all this. Would appreciate any help.

Answer

The original coordinates satisfy the equations of motion when the integral of $p\, \dot{q} - H(p,q)$ is minimized, and the new coordinates satisfy the equations of motion when the integral of $P\, \dot{Q} - K(P,Q)$ is minimized. There is no requirement that $H$ and $K$ be numerically equal.

The transformation is canonical if the Poisson bracket remains invariant.

The EOMs are

$\dot{p} = 0$

$\dot{q} = p$

and from the new Hamiltonian, we get

$\dot{P} = -(P+\sqrt{Q})^2 \frac{1}{2\sqrt{Q}} = - \frac{p}{2\sqrt{q}} = \frac{d}{dt} \left(\sqrt{p} - \sqrt{q} \right)$

$\dot{Q} = (P+\sqrt{Q})^2 = \dot{q}$

thus the equations of motion are numerically equal.

electric circuits - Electricity takes the path of least resistance?

Electricity takes the path of least resistance!

Is this statement correct?

If so, why is it the case? If there are two paths available, and one, for example, has a resistor, why would the current run through the other path only, and not both?

Answer

It's not true. To see this, you can try an experiment with some batteries and light bulbs. Hook up two bulbs of different wattages (that is, with different resistances) in parallel with a single battery:

 ------------------------------------------
|                     |                   |
Battery              Bulb 1              Bulb 2
|                     |                   |
 ------------------------------------------

Both bulbs will light up, although with different brightnesses. That is, current is flowing through the one with more resistance as well as through the one with less resistance.

differential geometry - How are the *constant vectors* different from *vector fields* in terms of their respective transfomation properties?

How does one distinguish between the transformation properties of a scalar field $\phi(\textbf{r})$ or vector field $\textbf{A}(\textbf{r})$ (more generally, the tensor fields) from the transformation of ordinary scalars or vectors (which are not fields in the sense that they are not defined as the function of space)?

EDIT: For simplicity, I want to understand the transformation rule of constant vector and and vector field under rotation. Under rotations, the components of a constant vector $\textbf{A}$ transforms as: $$A^\prime_i=R_{ij}A_j$$ Does the same rule apply for vector fields too? Will it not matter that the argument of $\textbf{A}(\textbf{r})$ also transforms under rotation?

Answer

Yes it does affect the argument as well. The full transformation rule is given by: $$ \mathbf{A}(\mathbf{x}) \rightarrow R\mathbf{A}(R^{-1}\mathbf{x}) , $$ where $\mathbf{A}(\mathbf{x})$ is the vector field as a fucntion of the position vector $\mathbf{x}$ and $R$ is the rotation matrix.

quantum mechanics - What would be the associated wavelength of the particle if its velocity is zero?

What will be the wavelength of a particle whose velocity is zero?

According to de Broglie's hypothesis, then the wavelength would become infinite as the momentum is zero. But, I think for a stand still particle, its particle nature should be more dominant, as at that moment it is highly localized.

Answer

To the contrary, the slower the particle moves, the more its wavelike properties show up. Compare e.g. electron in an atom, where its energy is at its lowest, with an electron flying out of a CRT. In the former case we need quantum mechanics to describe its motion (it's where QM originates), while in the latter case classical mechanics is sufficient.

So the wavelength becoming infinite for a resting electron is a completely consistent result. And it's also consistent with Heisenberg's uncertainty principle: momentum is exactly defined while position is completely undefined.

quantum mechanics - Does the wave nature of a particle refer to the wave function?

In quantum mechanics when we talk about the wave nature of particles are we referring in fact to the wave function? Does the wave function describes the probability of finding a particle (ex: photons) at some location? So do the "waves" describe probabilities just the way in classical physics the electromagnetic waves describe the perturbations of the electric and magnetic fields?

Answer

No, because the wavefunctions are not waves in space. They are waves in enormous high-dimensional spaces of possibilities. If you have two particles, the wavefunction is waving in 6 dimensions (the two positions of the two particles make a six dimensional space of possibilities), if you have three particles, the wavefunction is in 9 dimensions. So it is always wrong to think of it as a wave in space, like a field.

There is a field which obeys the Schrodinger equation, but this classical field is a classical wave, like E and B, which describes many coherent bosons in the same quantum state all moving together, like a superfluid or a Bose-Einstein condensate.

mathematics - Professor Halfbrain's first cutting theorem

Professor Halfbrain has recently made several fascinating discoveries on cutting convex polygons in the plane.

Halfbrain's first cutting theorem: Every convex polygon can be cut (by a perfectly straight cut) into two polygons $A$ and $B$ so that $A$ and $B$ have the same perimeter and so that the length of the longest side of $A$ equals the length of the longest side of $B$.

Question: Is this theorem indeed true, or has the professor once again made one of his notorious mathematical blunders?

Answer

Yes, it is always possible.

Pick a point $P$ on the original convex polygon, and then place the point $Q$ on $P$ and move it continuously clockwise along the perimeter of the polygon until it comes back to $P$. As $Q$ is moving, keep track of the clockwise distance along the perimeter from $P$ to $Q$. By the intermediate value theorem, at one point that distance was equal to half the perimeter. Thus for any point on the perimeter, there is a unique "opposite" point that splits the perimeter in half.

Now move $P$ and $Q$ along the perimeter of the polygon together. As they are moving, they cut the original polygon into $A$ and $B$, and we consider the longest edge of each of the two polygons. Clearly the length of the longest edge is continuous because it's unchanging when it is an edge of the original polygon and changing continuously when it is $PQ$. We move $P$ and $Q$ until they switch places, so $A$ and $B$ also switched places, so the lengths of the longest edges of $A$ and $B$ also swapped.

Thus by the intermediate value theorem there was a $PQ$ that split the original polygon into $A$ and $B$ with equal perimeters and equal longest edges.

mathematics - Curios observation about a special grid - Why?

[0,1,6,4,3]
[4,5,6,0,9]
[9,9,0,1,1]
[1,0,4,5,6]
[7,6,4,9,0]

This 5x5 table has unique properties.
Each number in a cell means :

The cell = last digit of (sum of its neighbour (including diagonals))

or

The cell = The remainder of the sum of its neighbour divided by 10

Here is another example

 [0,1,6,4,8]   [1,2,1,3,2]   [1,0,9,0,1]
 [4,5,6,0,4]   [4,5,1,0,9]   [6,5,9,5,6]
 [4,4,0,6,6]   [3,3,0,7,7]   [5,5,0,5,5]
 [1,0,4,5,1]   [1,0,9,5,6]   [9,0,1,0,9]

 [2,1,4,4,0]   [8,7,9,8,9]   [4,5,1,5,4]

What surprised me is, every table like this will follow this :

• The middle cell is always 0.
  


• Any other cell $(i,j)$ and $(6-i, 6-j)$ adds upto 0 modulo 5
  That means, $(i,j) + (6-i, 6-j)$ is completely divisible by 5.
  

I have checked this with my computer.

Why this happens ?

Answer

Here is an almost-insight-free but perfectly straightforward proof. It involves computer calculation, but there's nothing here that couldn't be done unaided by a human with a reasonably high boredom threshold.

First of all, we're just working modulo 5 here. Your condition on each cell's relation to its neighbours is a mod-10 condition, which is equivalent to one mod 2 and one mod 5. The mod 2 one is irrelevant to the puzzle.

Now, think of possible grids as 25-element vectors. Our condition on each cell gives a linear constraint on these vectors. We have 25 constraints and a 25-dimensional space of vectors.

"In general" this setup will leave us with a 0-dimensional space of solutions, in other words no solutions but the trivial one where (when we're working mod 5) all cells' entries are multiples of 5. But in this case it happens that the conditions are not quite independent, and in fact there are two linearly independent solutions; every solution is a linear combination of these (mod 5), so there are $5^2=25$ solutions mod 5. (One of them is the trivial one.)

How do we determine that there are two linear solutions and find what they are? We ask a computer. In my case, I asked Mathematica. The code is boring and ugly:

m = ConstantArray[0, {25, 25}]
For[i = 0, i <= 4, i++,

 For[j = 0, j <= 4, j++,
  For[di = -1, di <= 1, di++,
   For[dj = -1, dj <= 1, dj++,
    ii = i + di; jj = j + dj;
    If[0 <= ii <= 4 && 0 <= jj <= 4,
     m[[5 i + j + 1, 5 ii + jj + 1]] = 
      If[di == 0 && dj == 0, 4, 1]]
    ]]]]
NullSpace[m, Modulus -> 5]

(here the cleverness, such as there is, lies inside the invocation of NullSpace; as mentioned above, doing by hand what this does by machine is boring but not difficult) and the answer is

{{4, 0, 1, 0, 4, 4, 0, 1, 0, 4, 0, 0, 0, 0, 0, 1, 0, 4, 0, 1, 1, 0, 4, 0, 1}, {0, 4, 4, 1, 2, 1, 0, 4, 0, 1, 1, 1, 0, 4, 4, 4, 0, 1, 0, 4, 3, 4, 1, 1, 0}}

meaning that the solutions mod 5 are the linear combinations of these grids

4 0 1 0 4
4 0 1 0 4
0 0 0 0 0
1 0 4 0 1
1 0 4 0 1

and

0 4 4 1 2
1 0 4 0 1
1 1 0 4 4
4 0 1 0 4
3 4 1 1 0

and since each of these satisfies the given anti-symmetry condition, so does any linear combination of them, QED.

newtonian mechanics - Elongation in bar with unequal applied forces

How is elongation in a uniform rod with unequal forces acting on opposite sides calculated? If applied forces are equal and opposite, the elongation is defined by the formula ($\delta = \frac{FL}{AE}$). How does the solution change for the case when forces are unequal (as shown)?

Answer

As you correctly note, the solution is different when the applied forces are not equal. The bar is not in static equilibrium: Both static and dynamic forces deform the bar in motion. These concepts are illustrated by superposition. $$\delta = \delta_\text{static} + \delta_\text{dynamic} \qquad = \frac {F_\text{less}L}{EA} + \frac{(F_\text{more} - F_\text{less})L}{2AE}$$

---

During changes in acceleration (when $\frac{\mathrm{d\;a(t)}}{\mathrm{dt}} \neq 0$), forces and accelerations within the damped solid body are transient, where $a(x,t)$, until they reach steady-state, where $\frac{\partial{\;a(x,t)}}{\partial{x}} = 0$.

^{Transient deformations in solid bodies are illustrated by a mass/spring system, where each mass element can be thought to represent differential mass element.}

Newton's Second Law requires that the bar (of mass $M$) accelerate in the direction of $F_{net}$. $$\sum F \;\text{on bar:} \qquad F_\text{more}-F_\text{less} = Ma \qquad \Rightarrow \;\therefore a = \frac{F_\text{more} - F_\text{less}}{M}$$

The derivation of deformation is shown when a single force acts on the bar.

Dynamic Deformation:

A Free Body Diagram is taken at an arbitrary cross-section of the bar, where the mass of the split body is $m = (\frac{M}{L})x$. Summation of forces acting on $m$ is solved for $T(x)$.
$$\sum F \;\text{on split body:} \qquad F_{o} - T = ma$$ $$F_{o} - T = \overbrace{\left(\frac{M}{L}x\right)}^\text{m} \overbrace{\left(\frac{F_{o}}{M}\right)}^\text{a} = \frac{F_{o}}{L}x \qquad \Rightarrow \qquad T = F_{o} - \frac{F_{o}x}{L}$$ $$\therefore T = F_{o}\left(1-\frac{x}{L}\right)$$ The static axial deformation ($\delta = \frac{FL}{AE}$) written in differential form:
$$\mathrm{d \delta} = \frac{T \mathrm{dx}}{AE} = \frac{[F_{o}(1-\frac{x}{L})]\mathrm{dx}}{AE}$$ Integrate differential deformation over the length of the bar to determine total deformation: $$\delta = \int_0^L \mathrm{d \delta} \; = \frac{F_{o}}{AE} \int_0^L 1-\frac{x}{L} \mathrm{dx} \implies \; \delta = \frac{F_{o}L}{2AE}$$

The derivation can be generalized to include both forces, where integration of $T(x) = F_\text{more} - \dfrac{F_\text{more}-F_\text{less}}{L}x$ results in the same solution given by superposition.

$$\therefore \delta = \; \overbrace{\frac{F_\text{more}L}{2AE} + \frac{F_\text{less}L}{2AE}}^\text{Integration} \;=\; \overbrace{\frac{F_\text{less}L}{AE} + \frac{(F_\text{more}-F_\text{less})L}{2AE}}^\text{Superposition}$$

---

References:

mathematics - Important days in history... and to come!

The other day I decided I wanted to come up with another simple puzzle before releasing another toughy on puzzling.SE. I came across an interesting series of dates that I wanted to share and see if people could figure out what the next date in the series was.

The series of dates is:

     March 30, 1973

      April 1, 1975
      June 28, 1978
 September 27, 1983
     March 24, 1992
  December 18, 2005
............., .....

What is the next date in the series? Explain.

Answer

I would guess that the answer is

March 10, 2028

Argument:

Whenever I look at four consecutive dates $d_1,d_2,d_3,d_4$ in this sequence, then the third date is exactly in the middle between the first date and the fourth date; mathematically, this means $d_3=(d_1+d_4)/2$.

(Thanks to Z Dailey): The Unix timestamps of these dates are the Fibonacci numbers.

friction - What is the relationship between rolling resistance and velocity?

I'm a games programmer, trying to write a simple car physics simulation. I'm aware that a car travelling in a straight line will exert a traction force that drives it forwards (by turning the wheels and pushing back against the ground), and that the main forces that act against this acceleration are aerodynamic drag and rolling resistance. But I can't seem to find a sensible way of calculating the rolling resistance for a given vehicle.

This page here claims that

$F_{rr} = C_{rr} * v$

That is, the force is proportional to a rolling resistance coefficient multiplied by the velocity. In my tests, this seems wrong, and the page itself (and the sources it cites) admit that they're on shaky ground in terms of explaining or proving that formula.

This page can't make up its mind. For most of the page it says that

$F_{rr} = C_{rr} * W$

That is, the force is equal to a coefficient multiplied by the weight ($mg$) of the vehicle - i.e. the force is the same regardless of velocity. It even provides a table of coefficients for different circumstances. But if the force is constant, won't a car in neutral with the engine off be accelerated backwards by this force? What is rolling resistance at velocity 0?

Then, for a bit of that page it claims that velocity is a factor in calculating the coefficient:

The rolling coefficients for pneumatic tyres on dry roads can be calculated as

$c = 0.005 + 1/p (0.01 + 0.0095(v/100)^2)$

where $c$ = rolling coefficient

$p$ = tyre pressure (bar)

$v$ = velocity (km/h)

This makes no attempt to explain what all those "magic numbers" mean, and still produces a coefficient of ~0.0088, which in a 1500 kg car would yield a force of 129 N whilst the car was standing still. That can't be right...

So, which is right? Given basic information about a vehicle (mass, velocity, information about the wheels and the surface they're rolling over), and ignoring aerodynamic drag, what's a sensible way to come up with a broadly plausible rolling resistance?

general relativity - How does gravity truly work?

I Am only 12 years old and I'm constantly wondering and trying understand how gravity really works. On YouTube everyone always talks about objects wrapping space time around themselves and uses the analogy of a trampoline. I still don't understand gravity because if space were like a trampoline, then earth would be spiralling in towards the sun along with all the other planets, right? So could someone explain to me how gravity really works without the trampoline analogy.

Answer

I personally prefer this visual explanation of Einstein's general theory of relativy from Brian Greene, a professor at Columbia University. https://www.youtube.com/watch?v=0jjFjC30-4A

A follow up video answering why don't planets fall toward the sun. https://www.youtube.com/watch?v=uRijc-AN-F0