special relativity - Could any object have zero mass?

Energy and mass are interrelated. As everything has energy could any object be massless? For example a photon is a packet of energy but still it is considered to be a massless particle. Why is it so?

Answer

There is only one mass. Lets make this clear. The concept of "relativistic mass" is not really a useful concept in my opinion. The invariant mass, or simply the mass, is defined as (in natural units, so $c = 1$):

$$E^2 - p^2 = m^2$$

The reason this is a much more useful definition for a mass, is because this quantity is Lorentz invariant, meaning it has the same value in every reference frame. If you define mass in any other way you are going to run into unnecessary trouble.

For the photon, this invariant mass is assumed to be 0, so its energy, $E$, gets a contribution only from the momentum of the photon, hence $E = p$. There are justifications for why we assume the photon has zero mass. The photon only has 2 degrees of freedom; the longitudinal polarisation does not exist precisely because the photon is massless. We also have other reasons to believe the photon is massless. Some laws of electromagnetism would have to be modified as well if the photon isn't massless, an example of which would be Coulomb's law. Hence Coulomb's law provides a good test of the photon mass (refer to this paper) which has been assigned the upper limit of $m ≲ 10^{−14}$ eV/$c^2$.

For other particles this is not the case; since they also possess this intrinsic mass they get contributions to the energy from that quantity as well and therefore $E^2 = p^2 + m^2$.

temperature - Intuitive reason for the $T^4$ term in Stefan Boltzmann law

The Stefan Boltzmann Law gives a relation between the total energy radiated per unit area and the temperature of a blackbody. Specifically it states that, $$ j= \sigma {T}^4$$ Now using the thermodynamic derivation of the energy radiated we can derive the above relation, which leads to $T^4$. But is there any intuitive reason for the $T^4$ term?

quantum field theory - Why is the beta function in RG usually defined as a "logarithmic" derivative?

What is the motivation behind defining the beta function as the logarithmic derivative of the coupling constant with respect to scale and not just the regular derivative?

general relativity - metric determinant and its partial and covariant derivative

question : $\nabla_a \nabla_b \sqrt{g} \phi =\partial_a \sqrt{g} \partial_b \phi$ is true ?

because $\nabla_a \sqrt{g}=0$ so we can write $\sqrt{g} \nabla_a \nabla_b \phi$ , but because metric determinant is not transfers like scalar, we can not write partial derivative instead of covariant derivative. what do think guys ?

riddle - I have two faces, but show you one

I have two faces, but show you one.
I once had guests, but now have none.
I'm rarely bloody and seldom blue.
I'm often promised and sometimes new.

What am I?

Answer

You are:

THE MOON

I have two faces, but show you one.

The Moon has two faces in the sense that from our perspective it has a "dark side" or "far side" (which we never see) and the side which faces us - the oddity is that we have this relationship with the Moon.

I once had guests, but now have none.

This refers to the astronauts which have visited the Moon - there are currently no humans on the Moon - nor have there been for quite a while.

I'm rarely bloody and seldom blue.

This refers to the red and blue moons. These are rare phenomena c.f. "once in a blue moon"

I'm often promised and sometimes new.

The "new" moon is about the lunar cycles. Finally, one often "promises the moon" i.e. makes crazy promises.

Ideas:

Why not something about honey :D

cosmology - Has the universe we live in started as a black hole that is imploding?

As discussed in this question, as far back as the 1960s it was suggested that the Schwarzschild metric can be smoothly joined to a de Sitter metric. For example the idea has been used by Lee Smolin in his ideas idea for black hole evolution.

If the suggestion is true it means that matter falling into a black hole would emerge into a de Sitter spacetime arguably comparable to our own.

So has the universe we live in started as a black hole of this type?

Is it possible to prove that the big bang was a black hole of this type?

Answer

Although there are suggestions that a black hole could lead into another expanding universe you should regard these as highly speculative. Although the metrics can be stitched together it requires a contribution from currently hypothetical quantum gravity effects. Until we have a working theory of quantum gravity there is no way to comment on how likely the scenario is.

To date the constructions link the Schwarzschild metric to a de Sitter universe, and our universe is not de Sitter (though in the future it will tend asymptotically towards de Sitter geometry). So on the face of it our universe does not look as though it is the far side of a black hole. Having said this, the effect of matter wasn't included in the original models as it makes the calculation too hard. It is conceivable that matter falling into the black hole would emerge to produce a universe that looks like ours. At the current state of the art it just isn't possible to comment definitively.

hilbert space - What exactly is meant by a quantum state in QM?

From what I have read, a quantum state of a quantum object contains all the properties of the quantum object. But I have read that the Pauli Exclusion principle states that two identical particle in a system cannot have the same quantum state simultaneously.

But since the quantum state is the combination of all the properties of the quantum object and that due to a quantum object's wave nature its position can be infinitely different, why does the PEP work? Is a quantum state really represented by a collection of variables or is it represented by one fundamental property? If is a collection of variables, then the simple change of position can make the quantum object unique, why can't that work? If its a single variable, then which one and why?

Answer

A quantum state is an element of a projectivized complex Hilbert space. A quantum object has (at any given moment) one state, and that state is a complete description of the object.

In practice, we often simplify matters by choosing to ignore certain observables in order to be able to work with a more manageable Hilbert space. So if you're interested in position or momentum and willing to ignore spin, you might take your Hilbert space to consist of square integrable functions from ${\mathbb R}^3$ to ${\mathbb C}^1$. If you're interested only in spin, you might get away with something finite dimensional. But those are clearly understood to be approximations. The basic setup of the theory is that if you choose the right Hilbert space, your object has (at any given moment) a state that is a single element of the projectivization of that Hilbert space, and that state fully describes everything about the object.

Of course every element of every Hilbert space can be written as a sum in infinitely many ways. So can, for example, the number 8. This does not make quantum states "infinitely different" from anything, any more than it makes the number 8 "infinitely different" (whatever that might mean).

acoustics - How does sound travel in space?

In relation to this question:

How can a black hole produce sound?

Which notes that the hole "produces" sound. The top answer states that:

What you think of as the hard vacuum of outer space could just as well be seen as a very, very, very diffuse, somewhat ionized gas. That gas can support sound waves as long as the wavelength is considerably longer than the mean free path of the atoms on the gas.

I get that there is "stuff" in space, what I don't get is how sound travels it. I learnt in high school that sound was a wave - as an example, you could fix a shoelace at one end and vibrate the other - voila, a wave forms.

And then sound kind of moves in the same way through, say, air, because of the slight molecular attraction between individual molecules is enough to create a similar waveform.

And then you get to space, and there aren't any molecules nearby to irritate each other, so there's no sound.

If that's the case, why does it matter how long the wavelength is, if there isn't a molecule nearby for the initial molecule to affect, how can sound travel?

I'd understand a little better if the particle a just hit particle b (like a pool ball) and b carried the sound - but that has little to do with waves and wavelengths.

Or is it that the "jet" that is thrown out travels along as a contiguous blob, with a sound wave embedded within it?

Answer

A vibrating shoelace is a poor analogy for a sound wave, because that would be a transverse wave whereas sound is a longitudinal wave.

What this means is your example of molecules hitting each other is perfect. A longitudinal wave is described by the variation of the density of the particles. The Wikipedia page has some nice animations to help visualize what that means. The wavelength in this context is the distance between two periodic fluctuations in the density of the material.

So in the case of the question you linked to, because the sound emitted has a massive wavelength (and consequently an extremely low frequency) that means the distance between the density fluctuations are large. So the particle literally moves and collides with another particle and so on, propagating the wave.

Is there conservation of information during quantum measurement?

Consider the following experiment. I take a spin-$\frac{1}{2}$ particle and make a $\sigma_x$ measurement (measure the spin in the $x$ direction), then make a $\sigma_y$ measurement, then another $\sigma_x$ one, then $\sigma_y$, and so on for $n$ measurements. The formalism of quantum mechanics tells us that the outcomes of these measurements will be random and independent. I now have a string of completely random bits, of length $n$. Briefly, my question is where does the information in this string come from?

The obvious answer is "quantum measurements are fundamentally indeterministic and it's simply a law of physics that you get a random string when you do that". The problem I have with this is that it can be shown that unitary evolution of quantum systems conserves von Neumann entropy, just as Hamiltonian evolution of a classical system conserves Shannon entropy. In the classical case this can be interpreted as "no process can create or destroy information on the microscopic level." It seems like the same should be true for the quantum case as well, but this seems hard to reconcile with the existence of "true" randomness in quantum measurement, which does seem to create information.

It's clear that there are some interpretations for which this isn't a problem. In particular, for a no-collapse interpretation the Universe just ends up in a superposition of $2^n$ states, each containing an observer looking at a different output string.

But I'm not a big fan of no-collapse interpretations, so I'm wondering how other quantum interpretations cope with this. In particular, in the "standard" interpretation (by which I mean the one that people adhere to when they say quantum mechanics doesn't need an interpretation), how is the indeterminacy of measurement reconciled with the conservation of von Neumann entropy? Is there an interpretation other than no-collapse that can solve it particularly well?

addendum

It seems worth summarising my current thinking on this, and having another go at making clear what I'm really asking.

I want to start by talking about the classical case, because only then can I make it clear where the analogy seems to break down. Let's consider a classical system that can take on one of $n$ discrete states (microstates). Since I don't initially know which state the system is in, I model the system with a probability distribution.

The system evolves over time. We model this by taking the vector $p$ of probabilities and multiplying it by a matrix T at each time step, i.e. $p_{t+1} = Tp_t$. The discrete analogue of Hamiltonian dynamics turns out to be the assumption that $T$ is a permutation matrix, i.e. it has exacly one 1 on each rown and column, and all its other entries are 0. (Note that permutation matrices are a subset of unitary matrices.) It turns out that, under this assumption, the Gibbs entropy (aka Shannon entropy) $H(p)$ does not change over time.

(It's also worth mentioning, as an aside, that instead of representing $p$ as a vector, I could choose to represent it as a diagonal matrix $P$, with $P_{ii}=p_i$. It then looks a lot like the density matrix formalism, with $P$ playing the role of $\rho$ and $T$ being equivalent to unitary evolution.)

Now let's say I make a measurement of the system. We'll assume that I don't disturb the system when I do this. For example, let's say the system has two states, and that initially I have no idea which of them the system is in, so $p=(\frac{1}{2},\frac{1}{2})$. After my measurement I know what state the system is in, so $p$ will become either $(1,0)$ or $(0,1)$ with equal probability. I have gained one bit of information about the system, and $H(p)$ has reduced by one bit. In the classical case these will always be equal, unless the system interacts with some other system whose state I don't precisely know (such as, for example, a heat bath).

Seen from this point of view, the change in von Neumann entropy when a quantum measurement is performed is not surprising. If entropy just represents a lack of information about a system then of course it should decrease when we get some information. In the comments below, where I refer to "subjective collapse" interpretations, I mean interpretations that try to interpret the "collapse" of a wavefunction as analogous to the "collapse" of the classical probability distribution as described above, and the von Neumann entropy as analogous to the Gibbs entropy. These are also called "$\psi$-epistemic" interpretations.

But there's a problem, which is this: in the experiment described at the beginning of this question, I'm getting one bit of information with every measurement, but the von Neumann entropy is remaining constant (at zero) instead of decreasing by one bit each time. In the classical case, "the total information I have gained about the system" + "uncertainty I have about the system" is constant, whereas in the quantum case it can increase. This is disturbing, and I suppose what I really want to know is whether there's any known interpretation in which this "extra" information is accounted for somehow (e.g. perhaps it could come from thermal degrees of freedom in the measuring apparatus), or in which it can be shown that something other than the von Neumann entropy plays a role analogous to the Gibbs entropy.

Do gravitational waves travel faster than light?

In Feb 12, 2016 edition of Times of India, an article read

[with the discovery of gravitational waves, we will be able to] Track Supernovas hours before they're visible to any telescope because the waves arrive Earth long before any light does, giving astronomers time to point telescopes like Hubble in that direction

See also page 13 of the paper.

Does this mean that gravitational waves reach us before light from a source? Can this be some printing mistake or am I interpreting it wrongly?

Edit: Can there be special cases (as explained in some answers) where gravitational waves seem to reach before light waves from a source (though not violating the speed limit)?

Answer

It's an incredibly misleading statement, so it's not you.

Gravitational waves propagate at the speed of light, so their detection by Earth-bound detectors is expected to correlate with the arrival of light from distant events assuming the source of light generation is identical (not spatially or temporally separated) to the source of the gravitational disturbance.

In the case of a supernova, it's actually a dynamic process instead of a flip of a switch, and so the change in the magnitude of light emission can indeed lag behind by several hours from the start of collapse of the star's core - the detection of gravitational waves could allow us to "buy back" that several hour window by detecting the gravitational waves produced by core collapse instead of having to wait for the light magnitude increase. There's no disconnect here, just sloppy reporting.

In many cases however, we infer gravitational events or influences have occurred or exist by witnessing a change in motion of light emitting (or reflecting) objects that are directly affected by the event/influence - think of a supermassive black hole at a galactic center that we can't observe directly, but infer its existence by the motion of stars in its vicinity. Or the orbital behavior of Neptune that suggested other massive objects yet to be found in our solar system.

Depending on the nature of the event, we may have to infer that a black hole merger, for example, has happened by observing the changes in motion of objects we can see with traditional telescopes. This introduces a time-lag in addition to the normal speed-of-light timelag we're bound by whenever we look up at the night sky:

Gravitational influence must travel at the speed of light from the site of the event to the light-emitting object that we can observe, and then the light from that object must travel to our telescopes, again at the speed of light. At the moment that the event happened, the light from the object we're observing with our telescopes had not yet felt the disturbance, so there's an additional lag in detection time that must be accounted for - we're not really observing the black hole in this example, we're observing a surrogate object.

The ability to detect gravitational waves may allow us to "buy back" this additional lag by now 'directly' observing the inciting events... bound by the speed of light, of course.

Conservation of quantum Noether current

The Noether current for a set of scalar fields $\varphi_a$ can classically be written as:

$$j^\mu(x)=\frac{\delta \mathcal L(x)}{\partial(\partial_{\mu}\varphi_a(x))}\delta \varphi_a(x)$$

The divergence of this current can then be written as: $$\partial_\mu j^\mu(x)=\delta \mathcal L(x)-\frac{\delta S}{\delta \varphi_a(x)}\delta \varphi_a(x)$$

If the classical field equations are satisfied the second term on the right hand side vanishes. However in quantum theory the classical field equations are not satisfied. Why is the current still conserved for a symmetry in this case?

logical deduction - The sadistic executioner (a.k.a the 100 prisoners)

100 prisoners are sent into a room to hear the warden say: "tomorrow I will line you up and put a hat on your head. The hat can be red, green, yellow, …" and he goes on enumerating colors. He then adds: "The executioner will then start from the back of the line and run up to the front, and to every prisoner he will ask what the color of his hat is. If the prisoner guesses the color correctly then he is freed, but if he gets it wrong he will be killed right away.

The following days, the prisoners are lined up, and the warden puts a colored hat on their head. Then, the executioner starts asking question. The unlucky prisoner at the back of the line guesses wrong and is killed. But by some magic, all the other prisoners guess the color of their hat correctly.

How did they do?

PS: I apologize for my English

Einstein's mass energy equivalence

Mass is a packed form of energy (as $E=mc^2$) suggests. Also we can convert mass energy into another forms such as in nuclear reactions. Does the mass energy remains packed until it is urged to get converted? Or is they are converting into one another such that an equilibrium has established?(just like dissolution and crystallization of sugar in water.)

Answer

$E=mc^2$ expresses an equivalence between mass and energy. It says nothing about how, when, or why one form of mass/energy is converted to another. For example it provides no reason for why, when two protons with enough KE collide, three protons and an antiproton may result. It just confirms that it's energetically possible.

Are stable orbits within the event horizon of a black hole possible?

This paper https://arxiv.org/abs/1103.6140 theorizes that orbits inside the event horizon of a rotating or charged black hole are not only possible but actually stable enough to potentially support life. How can this possibly be true?

Using the formula for the Schwarzchild radius of a black hole, it seems to me like any radius smaller than that is another event horizon. This would mean that any orbit inside a black hole can only be an inwards spiral, even if you negate electromagnetic or gravitational wave radiation. The paper seems to say this is not true if both the black hole and the planet were charged, but why is that? Even if they had an enormously strong charge, wouldn't the force of gravity still dominate?

Answer

Let's consider the charged non-rotating black hole since this is the simplest case. The geometry is described by the Reissner-Nordström metric, though we won't need to go into the gory details to get a basic idea of what is happening.

If you start with zero charge this is just the Schwarzschild metric. All the mass/energy is at the singularity so no matter how close you get to the centre of the black hole all the mass is still ahead of you pulling you inwards. The result is that once you pass through the horizon the $r$ coordinate becomes timelike and remains timelike as you approach and eventually crash into the singularity. No stable orbits are possible inside the horizon.

But once you charge the black hole the electric field is present both inside and outside the event horizon, and this field has an energy so it produces a gravitational force. When you're at some distance $r$ from the centre there is a part of the field behind you pulling you outwards. And the closer you get to the singularity the more of the electric field is behind you pulling you outwards.

The result is that a charged black hole has two horizons. As you pass through the outer horizon the radial coordinate becomes timelike and in this region you are doomed to fall inwards. However there is a second horizon marking the point at which the energy of the field outside balances out the mass at the singularity. As you pass through this horizon the radial coordinate becomes spacelike again and inside the second horizon you are not doomed to fall inwards. Indeed it's possible to find worldlines that travel in through the two horizons then turn round and travel back out again. For more on this see Entering a black hole, jumping into another universe---with questions.

The paper contends that inside the second horizon there are stable orbits where planets could orbit and life exist. I haven't been through the paper so I can't comment. It isn't immediately obvious that stable orbits exist inside the second horizon, after all a Schwarzschild black hole has no stable orbits for $r_s \lt r \lt 3r_s$ even though this is outside the horizon. So the fact the $r$ coordinate is spacelike doesn't guarantee stable orbits exist. However since the paper was peer reviewed I assume they have done the sums correctly.

However we should note that it is extremely unlikely a black hole would ever accumulate enough charge to move the inner horizon any great distance from the singularity, so realistically this is never going to happen for a charged black hole. With a rotating black hole it is more feasible, though I'm unsure if the supermassive black holes at galaxy cores are rotating fast enough to make it possible.

mathematics - Determinant of Word Matrix

Challenge: Create a 3x3 matrix of letters (NO REPEATS) such that the determinant of the matrix forms words from the 'multiplications'. For example, in a simple 2x2 case, take the matrix

$\begin{bmatrix}O & I\\N & R\end{bmatrix}$

the determinant of which is OR-IN. This forms 2 2-letter words. In the case of a 3x3, you will end up with 6 3-letter words if done correctly.

Don't know how to find the determinant of a matrix? Fear not, here's the guide (or read a more detailed explanation here):

$|A| = \begin{vmatrix} a & b & c\\d & e & f\\g & h & i \end{vmatrix} = a\begin{vmatrix} e & f\\h & i \end{vmatrix} - b\begin{vmatrix} d & f\\g & i \end{vmatrix} + c\begin{vmatrix} d & e\\g & h \end{vmatrix} = aei+bfg+cdh-ceg-bdi-afh . $

If that looks too complicated, here's another method that some find easier.

Don't worry about the additions and subtractions, I'm only concerned with the words formed. That said, if your words could be added and subtracted as base 26 numbers according to the determinant and result in a valid word, I would be incredibly impressed (a 100 reputation impressed, even). (This is mathematically impossible, determinant is always 0; sorry to get your hopes up). As usual, words are considered valid iff they are English words found on dictionary.com (read: this does not include (pre/suf)fixes, abbreviations, acronyms, etc.)

Answer

Try

$\begin{bmatrix} B & S & L\\I & A & U\\M & D & T \end{bmatrix}$

This produces the following words:

BAT
SUM
LID
LAM
BUD

SIT

general relativity - Cosmology - how to do the $V/V_{max}$ test?

In Cosmology, we have the co-moving distance (assuming $\Omega_k=0$), $$D_C=\frac c{H_0}\int_0^z\frac{dz'}{\sqrt{\Omega_m(1+z)^3+\Omega_\Lambda}}$$ and we also have the total co-moving volume formula $$V=\frac{4\pi}3 D_C^3$$Then we can use what is called the $\langle V/V_{max}\rangle$ test to test if a sample of objects has uniform co-moving density & luminosity that is constant in time. The number of objects per unit co-moving volume with luminosity in the range $(L,L+dL)$ is given by $\Phi(L)$. Then the total number of objects in the sample is $$\int_0^\infty\Phi(L)\int_0^{V_{max}(L)}\mathrm dV\,\mathrm dL$$ If we have a uniform distribution of objects, then the value of this expectation should be $1/2$, as described by this similar question asked on Student Room, which did not get any responses.

This test is supposedly well known, but I can't find any questions about it here, nor can I find a simple article about the actual test, or this result. There are many articles online instead showing generalisations to this test which seem very abstract to me.

Question: From these definitions, I don't know how to get a value for $\langle V/V_{max}\rangle$, nor do I know what the explicit formula is. What is the explicit formula for $\langle V/V_{max}\rangle$? Is it that integral? Clearly, $V$ depends only on the value of $z$, but I don't know what a uniform distribution of objects implies about the distribution of $z$. Can someone help me understand this a bit better?

Answer

I am not so sure about the cosmological application, but the principle is straightforward.

If you have an estimated distance $D$ to an object, then that defines a volume of $$ V =\frac{4\pi}{3} D^3$$

If your survey is capable of detecting such objects to a distance $D_{\rm max}$, then this defines a volume $V_{\rm max}$.

So for each object you can calculate $V/V_{\rm max}$. If the source population is uniform in space (and hence in time for cosmological sources), then the average $\langle V/V_{\rm max}\rangle = 0.5$. In fact you can go further and say that $V/V_{\rm max}$ ought to be uniformly distributed between 0 and 1.

This can be done as a function of source type, or luminosity or whatever.

calculation puzzle - Does this need to be fixed?

I found this device the other day - it seems like a calculator, but the results it's giving me don't make any sense. Here, if I try this...

$6 + 3 = 38$

See! It doesn't make any sense at all. Should be nine. And this should be seven, right?

$2 - -5 = 37$

Nope. ... Both thirty something huh. It isn't always. I'll try again.

$4 \times -3 = 17$

And once more...

$4 \times -9 = 5$

I think this thing is broken. Any ideas?

If I entered $-2 \times 5$, what would the output be? Why?

Hint 1:

I tried plugging is some more calculations. Also, the device reset or something before I started this, whatever that means. Anyways.
$24 + 3 = 6$

$-3 \times 1 = 9$
$-4 \div 2 = 2$
[No reset occurs between the main sequence and $-2 \times 5$ being entered in the puzzle itself]

Hint 2:

Since the device is a computer, I added the computer-science tag.
[You don't need to know any computer stuff to work this out - but this might help guide you in the right direction.]

Answer

You are using an unusual real-world device called

an RPN calculator.

Calculations:

The stack starts off containing 32 due to whatever you did before the puzzle, and the top-of-stack value is displayed on-screen (oddly the screen seems to show you the bottom of the stack not the top; I'm not sure about that bit). I have omitted some of the intermediate states. Stack contents are shown as [bottom ... top].
6 + [38]
3 2 - [38 1]
- [37]

5 4 * [37 20]
- [17]
3 4 * [17 12]
- [5]
9 [5 9]
The = key seems to allow you to enter two consecutive numbers without them getting concatenated into one, so "3 2" would give you ones values on the stack: [32], but "3 = 2" puts two values on: [3 2]. Or maybe "=" causes the display of the bottom-of-stack rather than the top-of-stack that I'd expect from an RPN calculator.

And for the question:

- [-4]

2 * [-8]
5 [-8 5]
so "-8" would be displayed on-screen unless the "=" key is what causes the bottom-of-stack to be displayed, in which case the screen would show "5", because this final calculation doesn't have an "=".

rhyme - A poem of dynamic words

In many great novels, to the reader I appear

Adjust me, and I describe things that are near

Once more, and my meaning changes only slightly
Again, and you excercised your free will, rightly

Yet another, and I'm something we all ought to do
Again, I'm a boundary with a beautiful view

Again, I'm a place to which many must go
And again, I'm a thing wise men shouldn't throw

Now behead me, I'm a thing you must watch while you speak
Your next action depends on the word that you seek:

Behead me once more, and my use is widespread

Or instead take my foot, of hard things I am said

Answer

In many great novels, to the reader I appear

Great novels have a very compelling theme.

Adjust me, and I describe things that are near

Would you perhaps be referencing these, over here? (m -> s)

Once more, and my meaning changes only slightly

Or perhaps you mean those? (e -> o)

Again, and you exercised your free will, rightly

If you exercised your free will, you chose something. (t -> c)

Yet another, and I'm something we all ought to do

I thought this was share, but it turns out we should all do a chore once in awhile. (s -> r)

Again, I'm a boundary with a beautiful view

Especially beautiful at sunset or sunrise, this references the shore. (c -> s)

Again, I'm a place to which many must go

If you need to buy something, you have to go to the store. (h -> t)

And again, I'm a thing wise men shouldn't throw

If you're wise, you will throw no stone. (r -> n)

Now behead me, I'm a thing you must watch while you speak

I know I've heard this from my mother: you better watch your tone! (goodbye, s)

Behead me once more, and my use is widespread

The number one is a pretty widespread number, I'd say. (goodbye, t)

Or instead take my foot, of hard things I am said

That thing must weigh a ton! (goodbye, e)

riddle - An Odd New Technology

Since it's Friday, and I'm really looking forward to the weekend, I've created yet another riddle for everyone to try and solve.

The Puzzle

I was in my chemistry class the other day, and I noticed something rather odd about the periodic table.

I couldn't place my finger on it, yet I could see it clear as day. The professor stated that it was a new technology that he had created and it disproved a lot of the fundamental teachings for math and science. He stated that with the fundamentals of his technology, a basic math equation such as $4 + 6 = 10$ was now wrong. I asked how this could be, and he simply responded:

The very fabric of our knowledge is built on a lie that was established thousands of years ago.

How, seriously, how can this be correct? The lie that he speaks of has been proven over the course of these thousands of years.

Well, I decided to dive further into his periodic table, and couldn't find any of the essential building blocks such as hydrogen or helium. No matter how I did the math, I just couldn't do it. However, I was able to prove the following formulas true:

$4 + 20 = 61$

$13c + 6 = 36$

$15 + 4 = 12$

I was told there is another accurate formula that supports the professor's claims in his thesis, and that to find it I would have to solve a nested riddle which has also proven quite difficult.

The proof that you seek, is hidden in code; then scrambled again, the message unknown.

The key that you'll need, is hidden as well; seventy-seven minus all will tell.

Built on three, plus four you will see; the message that's hidden, not given, not free.

He gave me a piece of paper with a message scribbled on it; and told me to visit this website. He said that I shouldn't change the available mode options, but that the riddle above will tell me what to do. Here is the text:

K561YMuzen679PIISrSAYz2EQK+3uEPK7hrW7jKhlJH2FUCCWS63EZPolPvcygAcWIIKKiEMv11v+Cmy00TnOn/R/iwOvAOYq8wrk4EuCFLbXEZke3cPj7g8qzG9Fwq+B0g5T4ZgtBkPJfQ6411/V/JWq+bbLpBzTNo65EPG7LXyPzqnVuq4QVS51MOy8c7Jzf9I8Zy4YzfyizmGfRXdvHgDbW7AdIkNZPha/UlHup91KQxiWcWCl6/ayjzfvk9E0bNny8A5dysanBtDNbkkJ2+WMCxgdhSHlHgMP+ErTzpfr6IZo0DNj6VzdjM6JOE=

He also said that he found another formula quite useful on his journey to creating this new technology, and said that it might prove to be useful to me:

$PL - P$

I'm not really sure what to make of it; hopefully you guys can help me out here. What is the professor's thesis, and what is the final formula that supports it? Also, what’s up with the weird expression he said would help?

Notes and Hints

---

I updated the message to be more fitting to a proper answer. For those who haven’t managed to decrypt yet, the unencrypted portions of the puzzle still apply. Just the encrypted portions have changed. I didn’t like the original answer as I thought it didn’t do the puzzle justice. The old version is still available in the edit history if you so choose.

Encryption Hint: The encryption method is difficult to figure out based on the minimal information given; however, once you find it you'll have one of those "ohhhhhhh okay" moments.

The message encrypted, electrical ties; would you like McChicken, or a burger with fries?

Key Hint: The key, similar to the encryption method is also difficult to find due to the nature of the riddle. I can tell you it is all lower case, and is more than one word!

The key you all need, is trapped within space; where it can seem hard, to keep up your pace.

Final Hint: Since there is already an answer that has made it to the final stages, I've decided to give one last hint to really help everyone.

The paragraphs start, after the art; include everything but math.

Clarifications

---

• @user477343's comment in Rot13 is correct, also brought up this post.
  
  
  
  • The linked puzzle has no relation to this one.
  
  
  
  



• @Hugh wondered if the duplication of elements was significant.
  
  
  
  • In response, the only significance is that one provides a correct answer, where the other does not. I used duplicated elements to throw people off.
  
  
  
  


• @SteveV pointed out the duplication of column headers 3 and 10 being IIIA.
  
  
  
  • This is purely coincidence and has no significance. Also, great reference to 32, 18, and 17. I thought about using those in my puzzle but decided against it.
  
  
  
  


-

If you have any questions along your journey, feel free to ask; and as always, if you down-vote, please explain in a comment how I can improve my post.

Answer

I think that

The Professor's theory is a lie

I got there by decrypting the code with these settings

Encryption Method: Arcfour (Arc from electricity, four from 4)

Encryption Key: zero gravity

This reveals a new challenge

Wkh sxccoh frqwdlqv wkh dqvzhuv, orrn ehbrqg wkh fkduw; wklv whfkqrorjb lv qrw frqwdlqhg, zlwklq wkh zrun ri duw. Vhyhudo qxpehuv jlyhq, wkh zrugv fdq vxuhob whoo; frxqw wkh vwduv, wklv surihvvru lvq’w zhoo.
1a16
5a12
5a16
4a5
6a4&9

T11

Which can be solved by

A caesar cypher. PL (playdoh 42) - P (plaque 19) = 23, which is the cypher key

This reveals the final puzzle:

the puzzle contains the answers, look beyond the chart; this technology is not contained, within the work of art. several numbers given, the words can surely tell; count the stars, this professor isn’t well.
1x16

5x12
5x16
4x5
6x4&9
q11

You can use these clues to

Get the 16th word of the first paragraph, 12th word of the 5th paragraph, etc.

Which reveals

The professor's thesis is a lie

Unless I messed up that last part, which is very likely

smallest number obtainable from 2020

If only the four basic operations, concatenation and parenthesis are allowed, the largest number which can be obtained from $2$ $0$ $2$ $0$ is... $2020$ :-) (If exponentials were allowed, $20^{20}$ would be much higher, of course). But what is the smallest number which can be obtained?

Clarifications (note that many answers were written before these were added):

• The numbers 2, 0, 2, 0 must be used in that order.



• None may be omitted.


• The "four basic operations" do not include unary + (which would be a no-op in any case) or unary - (which would e.g. allow -2020 as an answer).


• "Smallest" means "most negative", not "closest to zero".


• Concatenation may only be applied to literal digits.


• Exponentiation is not allowed, even though it is written without any explicit operators.


• Inserting decimal points is not allowed.

Answer

If only addition ($+$), subtraction ($−$), multiplication ($\times$) and division ($/$), without unary minus, then

$2 \times ( 0 - 20) = -40$

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

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

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

Answer

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

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

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

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

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

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

thermodynamics - What is meant by boiling off electrons in a heater coil?

In my electricity and magnetism course, we used a Thompson tube to produce an electron beam. There is a heating element at the back of the tube and the lab manual claims that "electrons are boiled off." Can someone explain what is meant by this? Boiled off seems an odd term to use but maybe if I understand it meaning, it won't seem so odd.

Answer

The heated filament produces electrons by thermionic emission.

It takes energy to remove an electron from a metal surface, and this energy is called the work function. In a metal the electrons are continuously scattering off lattice vibrations, i.e. the vibration of the metal atoms due to thermal energy. Typically the energy the electron gains from the scattering is of order $kT$, where $k$ is Boltzmann's constant and $T$ is the temperature. At room temperature $kT$ is about 0.025eV while metal work functions are generally around 2 - 5eV, so it would appear that electrons aren't scattered with enough energy to escape the metal.

However the scattering is a random process, and occasionally an electron will be scattered with an energy greater than the work function. If this electron is near the metal surface it can escape. At room temperature this is very unlikely, but as you heat the metal the energy of the lattice vibrations increases and consequently the electrons are scattered with greater energy. At some point the scattering energy gets big enough that a significant number of electrons are emitted, and this is thermionic emission.

It's important to remember that the probability that an electron is scattered with enough energy to escape the metal remains small even when the metal is red hot. Only a very small percentage of the electrons in the metal escape, but this small escape rate is enough to generate a measurable current.

Anyhow, you can see why electrons are described as being boiled off, because it's the thermal vibrations of the metal that eject the electrons.

What would happen if two protons collide?

I am an High School student just curious about Physics. Im am planning to study it at University, but for now my knowledge is clearly very poor and non technical.

I was thinking...what would happen if two protons collide? I am not sure that my question has sense, so my apologies for that.

Thank you in advance.

Answer

The answer depends on how exactly the protons collide. In many circumstances the results of the collision can be explained in terms of electromagnetic forces and classical scattering theory. As long as the protons remain far enough apart during the collision, then the protons can be treated like two positively charged classical point particles. Even in a "head-on" collision, if the collision is low-energy, the protons may very well remain far enough apart to be treated classically since (classically) the closest they can get to one another is approximately $d\sim \frac{ke^2}{E}$, where e is the unit electric charge, E is the initial kinetic energy of the colliding particles, and k is the electric force constant (i.e., the magnitude of the force between two charged particles a distance $R$ apart is $\frac{kQ_1Q_2}{R^2}$).

On the other hand, if the collision is energetic enough that the protons can approach each other more closely that about $10^{-15}$ meters, then the so-called "strong force" comes into play, and a lot of interesting things can start to happen. This regime is the subject of advanced quantum mechanics and particle physics. You can find a fairly accessible (relatively speaking) description of this topic in Griffith's book "Introduction to Elementary Particle".

Cheers.

terminology - The meaning of 'postulate' in physics?

What does postulate mean in physics? What is its role in physical theories?

Is it possible to break physical postulates?

newtonian mechanics - How can tangential acceleration from a radial force be explained?

A mass is attached to a rope, and put into a circular motion. If I pull the string from the center, the tangential speed of the mass will increase (by conservation of angular momentum).

I am applying a force only in the radial direction, so how can the tangential velocity increase if there is no tangential force?

wordplay - How long can two words be if they make up a larger word (challenge)

I came up with this little game. Basically we have to write a word using small caps and big caps so that the small caps write a word and the big caps write a word (in their original order), for example:

STRINginG, the words are STRING and gin. ( or STringING with STING and ring)

MoneY, the words are MY and one.

THEoreM, the words are THEM and ore.

The score of each word is the length of the smallest word.

So STRINginG has score $3$, MoneY has score $2$ and THEoreM has score $3$.

However I do not want to accept the cases in which all the big caps are together. So things like BUTTERfly gentleMAN or CARrot are not allowed.

I would appreciate a lot if you could give me examples of words with high scores (the more the merrier), something especially awesome would be if the words could have related meanings. Thank you in advance.

cosmology - ElectroWeak symmetry breaking and the early universe

I'm trying to piece together a coherent understanding of the EW symmetry breaking and its effect on early cosmology, from a number of somewhat incomplete sources. Any corrections to the following presumed steps or conclusion would be greatly appreciated. i) In the early universe, before freeze-out, the Higgs field energy was way up on the sides of its "sombrero" potential curve (very high energy level.) ii) Therefore no vacuum expectation value for the Higgs field iii) Thus no EW symmetry breaking iv) All the SU(2)XU(1) symmetric intermediate vector bosons (B and original W) were massless, weak mixing angle = 0 v) Thus the EW force was long range during this time (Until the Higgs field energy falls far enough to acquire the VEV and break the symmetry.) vi) Fermions massless

That seems to me to require a huge recomputation of most SM results and "constants" that would be very different from what we see in experiment today. Little or none of today's data and theory could be applied to that period. Do calculations in cosmology for this period account for this? Is there something missing or wrong in this line of reasoning?

special relativity - Where are all the slow neutrinos?

The conventional way physicists describe neutrinos is that they have a very small amount of mass which entails they are traveling close to the speed of light. Here's a Wikipedia quote which is also reflected in many textbooks:

It was assumed for a long time in the framework of the standard model of particle physics, that neutrinos are massless. Thus they should travel at exactly the speed of light according to special relativity. However, since the discovery of neutrino oscillations it is assumed that they possess some small amount of mass.1 Thus they should travel slightly slower than the speed of light... -- Wikipedia (Measurements of Neutrino Speed)

Taken at face value, this language is very misleading. If a particle has mass (no matter how small), its speed is completely relative, and to say that neutrinos travel close to the speed of light, without qualification, is just as incorrect as saying electrons or billiard balls travel close to the speed of light.

So what is the reason everyone repeats this description? Is it because all the neutrinos we detect in practice travel close to the speed of light? If so, then I have this question:

Neutrinos come at us from all directions and from all sorts of sources (stars, nuclear reactors, particle accelerators, etc.), and since they have mass, just like electrons, I would have thought we should see them traveling at all sorts of speeds. (Surely some cosmic neutrino sources are traveling away from the earth at very high speeds, for example. Or what about neutrinos emitted from particles in accelerators?)

So like I said at the start: Where are all the slow neutrinos? And why do we perpetuate the misleading phrase: 'close to the speed of light' (i.e. without contextual qualification)?

Answer

Strictly speaking, it is indeed incorrect that neutrinos travel at "close to the speed of light". As you said, since they have mass they can be treated just like any other massive object, like billiard balls. And as such they are only traveling at nearly the speed of light relative to something. Relative to another co-moving neutrino it would be at rest.

However, the statement is still true for almost all practical purposes. And it doesn't even matter in which reference frame you look at a neutrino. The reason is that a non-relativistic neutrino doesn't interact with anything. Or in other words: all the neutrinos you can detect necessarily have to have relativistic speeds.

Let me elaborate. Since neutrinos only interact weakly they are already extremely hard to detect, even if they have high energies (> GeV). If you go to ever lower energies the interaction cross-section also decreases more and more. But there is another important point. Most neutrino interaction processes have an energy threshold to occur. For example, the inverse beta decay

$$ \bar\nu_e + p^+ \rightarrow n + e^+$$

in which an antineutrino converts a proton into a neutron and a positron, and which is often used as a detection process for neutrinos, has a threshold of 1.8 MeV antineutrino energy. The neutron and the positron are more massive than the antineutrino and the proton, so the antinneutrino must have enough energy to produce the excess mass of the final state (1.8 MeV). Below that energy the (anti)neutrino cannot undergo this reaction any more.

A reaction with a particularly low threshold is the elastic scattering off an electron in an atom. This only requires a threshold energy of the order of eV (which is needed to put the electron into a higher atomic energy level). But a neutrino with eV energies would still be relativistic!

Assuming that a neutrino has a mass of around 0.1 eV, this would still mean a gamma factor of $\gamma\approx 10$. For a neutrino to be non-relativistic it would have to have a kinetic energy in the milli-eV range and below. This is the expected energy range of Cosmic Background Neutrinos, relics from the earliest times of the universe. They are so to say the neutrino version of the Cosmic Microwave Background. So not only do non-relativistic neutrinos exist (according to mainstream cosmological models), they are also all around us. In fact, their density at Earth is $\approx$50 times larger than neutrinos from the Sun!

There is a big debate if they can ever be detected experimentally. There are a few suggestions (and even one prototype experiment), but there are differing opinions about the practical feasibility of such attempts. The only process left for neutrinos at such small energies is neutrino-induced decay of unstable nuclei. If you have an already radioactive isotope, it's like the neutrino would give it a little "push over the edge". The $\beta$-electron released in the induced decay would then receive a slightly larger energy than the Q-value of the spontaneous decay and the experimental signature would be a tiny peak to the right of the normal $\beta$-spectrum. This will still be an extremely rare process and the big problem is to build an apparatus with a good enough energy resolution so that the peak can be distinguished from the spectrum of normal spontaneous nuclear decay (amidst all the background). The Katrin experiment is trying to measure the endpoint of $\beta$-spectrum of Tritium in order to determine the neutrino mass. But under very favorable circumstances they even have some chance to detect such a signature of cosmic background neutrinos.

TL;DR: In fact there are non-relativistic neutrinos all over the place, but they they interact so tremendously little that they seem to not exist at all.

quantum mechanics - Relative weights in rotational bands of symmetric diatomic molecules

In an old paper, Ehrenfest 1931, the introduction starts off as follows:

The band spectra of symmetric diatomic molecules show certain striking differences from those of asymmetric molecules. For when the two nuclei of the molecule are identical, the intensity of the individual lines of a band, instead of varying smoothly from line to line, alternates more or less markedly.

There is something called the relative weight, defined by the ratio of the number of states that are symmetric under interchange of the two nuclei and the number that are asymmetric. In modern language, let the degeneracy of the nuclear ground state be $g$. Then the relative weight is $(g+1)/(g-1)$ if the nuclei are bosons, $(g-1)/(g+1)$ if fermions. If I'm understanding correctly, then if the rotational state of the molecule is $J$, then depending on whether $J$ is odd or even, the two nuclei can pick up a relative phase $(-1)^J$, so the degeneracy of the odd-$J$ and even-$J$ states alternates according to the relative weights.

Assuming the above is right, then there's only one thing I really don't understand. How does this show up experimentally as an alternating pattern of intensities? What kind of transitions are we talking about? M1? E2? Are these absorption spectra? Emission spectra? Suppose you populate a molecular state with some spin. Then we're just hopping down a ladder, and it seems to me that by simple conservation of probability there can't be any alternation of intensity. Or by intensity do they actually mean transition rate rather than what an experimentalist would call intensity? (If so, then I'm curious how one would determine these transition rates. From natural line widths? From competition between radiation and collisions in a sample of gas?)

P. Ehrenfest and J. R. Oppenheimer, "Note on the Statistics of Nuclei," Phys. Rev. 37 (1931) 333, http://link.aps.org/doi/10.1103/PhysRev.37.333 , DOI: 10.1103/PhysRev.37.333

solid state physics - How can the Hall effect ever show positive charge carriers?

The Hall effect can be used to determine the sign of the charge carriers, as a positive particle drifting along the wire and a negative particle drifting the other direction get deflected the same (as $F = q \vec{v}\times\vec{B} = (-q) (-\vec{v})\times\vec{B}$). But I don't understand how positive charge carriers are ever possible.

As I've understood, a positive hole is nothing more than the absence of an electron. As all the electrons in the valence band are still negatively charged, why would this hole behave in a magnetic field as if it were positive?

Also, a hole is created if an electron is excited into the conductance band. If there is always the same numbers of holes as electrons, how can any Hall effect ever occur?

Thank you very much

Answer

There are two essential facts that make a hole a hole: Fact (1) The valence band is almost full of electrons (unlike the conduction band which is almost empty); Fact (2) The dispersion relation near the valence band maximum curves in the opposite direction to a normal electron or a conduction-band electron. Fact (2) is often omitted in simplistic explanations, but it's crucial, so I'll elaborate.

STEP 1: Dispersion relation determines how electrons respond to forces (via the concept of effective mass)

EXPLANATION: A dispersion relation is the relationship between wavevector (k-vector) and energy in a band, part of the band structure. Remember, in quantum mechanics, the electrons are waves, and energy is the wave frequency. A localized electron is a wavepacket, and the motion of an electron is given by the formula for the group velocity of a wave. An electric field affects an electron by gradually shifting all the wavevectors in the wavepacket, and the electron moves because its wave group velocity changes. Again, the way an electron responds to forces is entirely determined by its dispersion relation. A free electron has the dispersion relation $E=\frac{\hbar^2k^2}{2m}$, where m is the (real) electron mass. In the conduction band, the dispersion relation is $E=\frac{\hbar^2k^2}{2m^*}$ ($m^*$ is the "effective mass"), so the electron responds to forces as if it had the mass $m^*$.

STEP 2: Electrons near the top of the valence band behave like they have negative mass.

EXPLANATION: The dispersion relation near the top of the valence band is $E=\frac{\hbar^2k^2}{2m^*}$ with negative effective mass. So electrons near the top of the valence band behave like they have negative mass. When a force pulls the electrons to the right, these electrons actually move left!! I want to emphasize again that this is solely due to Fact (2) above, not Fact (1). If you could somehow empty out the valence band and just put one electron near the valence band maximum (an unstable situation of course), this electron would really move the "wrong way" in response to forces.

STEP 3: What is a hole, and why does it carry positive charge?

EXPLANATION: Here we're finally invoking Fact (1). A hole is a state without an electron in an otherwise-almost-full valence band. Since a full valence band doesn't do anything (can't carry current), we can calculate currents by starting with a full valence band and subtracting the motion of the electrons that would be in the hole state if it wasn't a hole. Subtracting the current from a negative charge moving is the same as adding the current from a positive charge moving on the same path.

STEP 4: A hole near the top of the valence band move the same way as an electron near the top of the valence band would move.

EXPLANATION: This is blindingly obvious from the definition of a hole. But many people deny it anyway, with the "parking lot example". In a parking lot, it is true, when a car moves right, an empty space moves left. But electrons are not in a parking lot. A better analogy is a bubble underwater in a river: The bubble moves the same direction as the water, not opposite.

STEP 5: Put it all together. From Steps 2 and 4, a hole responds to electromagnetic forces in the exact opposite direction that a normal electron would. But wait, that's the same response as it would have if it were a normal particle with positive charge. Also, from Step 3, a hole in fact carries a positive charge. So to sum up, holes (A) carry a positive charge, and (B) respond to electric and magnetic fields as if they have a positive charge. That explains why we can completely treat them as real mobile positive charges in their response to both electric and magnetic fields. So it's no surprise that the Hall effect can show the signs of mobile positive charges.

quantum field theory - Why Zeta regularization is not valid for multiple-loops?

Why zeta regularization only valid at one-loop?

I mean there are zeta regularizations for multiple zeta sums. Also we could use the zeta regularization iteratively on each variable to obtain finite corrections to multiple loop diagrams.

quantum mechanics - What's wrong with this derivation that $ihbar = 0$?

Let $\hat{x} = x$ and $\hat{p} = -i \hbar \frac {\partial} {\partial x}$ be the position and momentum operators, respectively, and $|\psi_p\rangle$ be the eigenfunction of $\hat{p}$ and therefore $$\hat{p} |\psi_p\rangle = p |\psi_p\rangle,$$ where $p$ is the eigenvalue of $\hat{p}$. Then, we have $$ [\hat{x},\hat{p}] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar.$$ From the above equation, denoting by $\langle\cdot\rangle$ an expectation value, we get, on the one hand $$\langle i\hbar\rangle = \langle\psi_p| i \hbar | \psi_p\rangle = i \hbar \langle \psi_p | \psi_p \rangle = i \hbar$$ and, on the other $$\langle [\hat{x},\hat{p}] \rangle = \langle\psi_p| (\hat{x}\hat{p} - \hat{p}\hat{x}) |\psi_p\rangle = \langle\psi_p|\hat{x} |\psi_p\rangle p - p\langle\psi_p|\hat{x} |\psi_p\rangle = 0$$ This suggests that $i \hbar = 0$. What went wrong?

terminology - Do gauge bosons really transform according to the adjoint representation of the gauge group?

Its commonly said that gauge bosons transform according to the adjoint representation of the corresponding gauge group. For example, for $SU(2)$ the gauge bosons live in the adjoint $3$ dimensional representation and the gluons in the $8$ dimensional adjoint of $SU(3)$.

Nevertheless, they transform according to

$$ A_μ→A′_μ=UA_μU^\dagger−ig(∂_μU)U^\dagger ,$$

which is not the transformation law for some object in the adjoint representation. For example the $W$ bosons transform according to

$$ (W_μ)_i=(W_μ)_i+∂_μa_i(x)+\epsilon_{ijk}a_j(x)(W_μ)_k.$$

Answer

A gauge field transforms in the adjoint of the gauge group, but not in the adjoint (or any other) representation of the group of gauge transformations.

In detail:

Let $G$ be the gauge group, and $\mathcal{G} = \{g : \mathcal{M} \to G \vert g \text{ smooth}\}$ the group of all gauge transformations.

A gauge field $A$ is a connection form on a $G$-principal bundle over the spacetime $\mathcal{M}$, which transforms as $$ A \mapsto g^{-1}Ag + g^{-1}\mathrm{d}g$$ for any smooth $g : \mathcal{M} \to G$. If $g$ is constant, i.e. not only an element of $\mathcal{G}$, but of $G$ itself, this obviously reduces to the adjoint action, so $A$ does transform in the adjoint of $G$, but not in the adjoint of $\mathcal{G}$. With respect to $\mathcal{G}$, it does not transform in any proper linear (or projective) representation in the usual sense, but like an element of a Jet bundle.

cosmology - Why global warming

According to big bang , the universe is expanding . By this we can say that distance between sun and earth is increasing so it should happen that earth temp. Decrease but in real life what we see is global warming , why?

Answer

The Universe has indeed been in the state of global cooling since its very birth. It started very, very hot. Right now, the cosmic microwave radiation that penetrates the space (it is a thermal radiation of a sort, like the red light you get from a heated piece of metal in your oven) has the temperature of 2.7 kelvins (-270 degrees Celsius or so).

Every ten billion years or so (the age of the Universe is 13.8 billion years), this temperature will drop to one-half; the drop is exponential (one-half, then one-quarter, one-eighth, and so on) because the expansion of the Universe is approximately exponential in time due to the positive cosmological constant that accelerates the expansion.

One must correct you: the expansion of the Universe doesn't mean that the Earth-Sun distance is increasing. The Solar System is a gravitationally bound one, by local forces, and those are not affected by the expansion of the Universe. The Earth-Sun distance may be extremely slowly changing due to other effects but the expansion of the Universe isn't one of them. So the expansion of the Universe really means that you may "store" an increasing number of solar systems into the Universe!

At any rate, the temperature 2.7 kelvins of the outer space is very cold and doesn't impact the Earth's temperatures much. The Earth's temperature is much higher than 2.7 kelvins – about 300 kelvins in average (15 Celsius degrees or so in average) – and it is this pleasantly warm because of the equilibrium between the incoming energy and heat from the Sun; and the outgoing thermal radiation emitted by the Earth.

It may be calculated that the equilibrium temperature at which the thermal energy emitted by the Earth is precisely equal to the incoming energy from the Sun is about 300 kelvins. The temperature depends on the place (especially the latitude: warm equator vs cold polar regions), daytime (day vs night), season (summer vs winter), and other factors.

Some of the factors are "systemic" and may change for extended periods of time. The reflectivity of the Earth's surface (albedo) affects the equilibrium temperature (a darker Earth would be warmer, like a dark asphalt in the summer), and so does the amount of "greenhouse gases" (gases that prevent the outgoing thermal radiation of the Earth to escape from its atmosphere). The greenhouse gases act like a "sweater" or a "blanket" that prevents you from freezing in a cold weather by allowing you to "keep" the body heat you produce.

The term "global warming" is a catchword for a nearly negligible expected (and perhaps partially observed, but this is controversial) rise of the temperature caused by the increasing concentration of the minor greenhouse gases in the atmosphere, especially carbon dioxide (CO2). The concentration is rising because people burn coal or oil, so carbon plus oxygen in the air creates carbon dioxide which is still gas that spends a century or so in the atmosphere.

The main greenhouse gas, water vapor (H2O), has concentrations that are almost exactly dictated by the temperature and other "external drivers". The water vapor as a greenhouse gas is responsible for about 30 Celsius degrees of extra warming on the Earth's surface because it prevents some of the outgoing heat from escaping. The Earth would be really cold and inhospitable without water vapor in the atmosphere.

Carbon dioxide (the main plant food - trees are not built out of the "soil" but largely out of the carbon dioxide they catch from the air, so carbon dioxide is, along with oxygen and water, the most important gas/liquid for life) is only causing 10 times less than that, about 3 degrees, and the dependence of the total warming caused by this gas is only increasing logarithmically. The more CO2 we have, the less important the newly added CO2 becomes. Since 1750, the concentration of CO2 in the atmosphere has increased by 40 percent and this has caused some increase of the average surface temperature on the Earth because the extra CO2 has prevented some heat from escaping from the Earth, too.

The total warming caused by the newly added CO2 since the beginning of the industrial revolution could have been something between 0.2 deg C and 0.8 deg C – the latter is the whole observed or estimated temperature change. The exact value is not known (it is about 1/2 of the so-called climate sensitivity) and the value, with potential political implications, is a highly controversial number.

Regardless of the exact value, this change in 250 years is a small fraction of a percent of the absolute temperature 300 kelvins on Earth and these sub-degree temperature changes cannot be felt by human beings even if they occur immediately. They surely cannot be felt if this change is divided to 100 or 250 years. One may only demonstrate that something about the temperature has been slightly changing if he carefully observes pretty much the whole globe for whole days, whole years, and for 50 years, with very accurate thermometers and evaluate the data with extremely careful and accurate algorithms. If any condition fails to be met, the changes of the temperatures up and down become equally likely.

So global warming is an academic contribution to the time derivative (change in time) of the global mean (average) temperature. Despite what you may hear in the media, it is negligible for practical purposes, there are many other sources with both signs the contribute to the temperature changes (solar activity and sunspots, ocean cycles, El Nino, volcano eruptions, changing composition of plants, and so on), but despite the negligibility of the "global warming", the temperature changes caused by the changing greenhouse gases and other factors are still many orders of magnitude (millions of times) greater than the changes of the Earth's temperature induced by the evolution of the whole cosmos, e.g. by its expansion. The Cosmos is large, heavy, and has many interesting properties, but for practical purposes, it doesn't really matter on the Earth. Only the Sun, the Earth's atmosphere, and oceans (and biosphere etc.) matter for the Earth's surface temperatures.

homework and exercises - How to find the electrostatic potential of a hydrogen-like charge density?

I've been trying to find the scalar potential that would correspond to the charge density of a ground state hydrogen atom. The result is known, and the inverse of my problem can be found e.g. in Jackson's electrodynamics problem 1.5 or many questions here on this site.

The problem asks you to find the charge density that corresponds to the following potential: $$ \Phi(r) = \frac{q \exp{(-\alpha r)}}{4 \pi \epsilon _0 r}\left(1+\frac{\alpha r}{2}\right) .$$

As far as I can tell, according to Poisson's equation you basically just need to get the Laplacian of this potential. This is not hard to do, and the result is what you would expect:

$$ \rho(r) = \frac{-q\alpha^3}{8\pi}\exp{(-\alpha r)}. $$

• My question is the inverse: given $\rho(r)$, how would you find $\Phi(r)$?

The most obvious approach that I had is to use the integral you get from Coulomb's law:

$$ \Phi(r) = \int \frac{\rho(r')}{|r-r'|}d^3r' $$

However, I have not been able to solve the integral by hand and Mathematica can not tell me the result either. My guess is that this potential never goes to zero, so the direct integration is not feasible? If so, then how else would you go about solving this problem?

(My next task would be to solve the same thing for a Gaussian density, for which I can again find the result on Wikipedia. Is that problem easier or harder than this one?)

Answer

We don't provide complete answers to homework-like questions, even for a bounty.

You've got the wrong charge density because, when you took the Laplacian of the potential, you didn't take into account the fact that

$$\nabla^2\frac{1}{r}=-4\pi\delta^3(\vec r).$$

To understand this, think of the potential of a point charge.

If you simply use the spherical-coordinates expression for the Laplacian,

$$\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+...$$

you would calculate that

$$\nabla^2\frac{1}{r}=0$$

and this is incorrect. Basically, the spherical-coordinates expression for the Laplacian isn't valid at $r=0$.

The correct charge density is

$$\rho(\vec r)=q\delta^3(\vec r)-\frac{q\alpha^3}{8\pi}e^{-\alpha r},$$

where the Dirac delta function represents the positive charge density of the proton and the second term the negative charge density of the electron cloud.

Note that if you integrate this over all space, you get zero; a hydrogen atom has no net charge.

The integral you want to do is

$$\Phi(\vec r)=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec r')}{|\vec{r}-\vec{r}'|}d^3\vec r'.$$

Note the vector signs which you left out. They're important; $|\vec{r}-\vec{r}'|$ and $|r-r'|$ are two different things. Note also the $1/4\pi\epsilon_0$ that you omitted.

Using spherical polar coordinates for $\vec r'$, with the polar axis through $\vec r$, this is

$$\Phi(\vec r)=\frac{q}{4\pi\epsilon_0}\left\{\frac{1}{r}-\frac{\alpha^3}{8\pi}\int_0^\infty r'^2 dr' \int_0^\pi \sin{\theta'}d\theta' \int_0^{2\pi} d\phi' \frac{e^{-\alpha r'}}{(r^2+r'^2-2r r'\cos\theta')^{1/2}}\right\}.$$

The $\phi'$ integration is trivial.

The $\theta'$ integration can be performed by letting $u=\cos\theta'$. The result will involve the absolute value $|r-r'|$.

The $r'$ integration can be performed by splitting the integral into two parts,

$$\int_0^r dr'...+\int_r^\infty dr'...$$

so that you can take $|r-r'|$ to be either $r-r'$ or $r'-r$.

With all these hints, you can fill in the details.

Since the charge density is spherically symmetric, another approach would be to use Gauss’ Law to compute the field, and then integrate a radial path integral in from infinity to compute the potential.

wordplay - Getting All Riled Up

Recently, there's been an awful lot of Riley riddles across PSE. Why are they so prolific? I think there's three main reasons: they're quick to write, quick to solve, and apart from its minor gimmick they're very accessible. This crossword's also like a Riley riddle!

Firstly, one-third of the clues are quick to write: turns out writing clues is quicker when you leave out one letter from each of them. The dropped letters will spell out my thoughts on some of the Riley riddles in PSE.

Another one-third of the clues give answers that are quick to solve: when entering them into the grid, you can skip writing one entire section of them! In the spirit of Riley riddles, each of these entries is missing either its prefix, its suffix, or its infix, leaving behind a word (of varying degress of commonness). Looking at the missing parts should give you an idea of my attitude towards Riley riddles.

Apart from the minor gimmicks, the last one-third of the clues are more accessible: they're read and entered as directed.

Across

1 Returning partial cred is no certification? Think again! (7)

5 Period I must perhaps be stretched across the rack (7)
9 Girl's part of "I'm A Believer" (5)
10 Puppet beheaded with stiff bristle (3)
11 Sort repeated phrases by condition in empty rooms (5)
12 Waiting year to enter drug circle (8)
13 Chance of company backing dog (5)
15 Did carer really entrance master elder's heart? (5)
16 Satellite laboratory in terrible redundancy (6)
17 Clear head with books (2)
19, 20 Ling in a line dance or in lace? (8)

20 See 19
22 Home projects about mountains put back into shows (5)
24 Fits gun, almost, into Eve's caress (5)
25 Patchy plant glove with first sign of tearing (8)
28 Perhaps newspaper journalist in untraceable state (5)
29 Part of a yucky fish (3)
30 Bad design of headless little man nonetheless initially produced (5)
31 Sever head of donkey to freak out racist (7)
32 Pleas of female model left in state (7)

Down

1 Turn back time with Shelter to follow heart of Perry's Hot N Cold, perhaps? (5)
2 Farm out of corn, but acts bustling? (8)
3 Deploy, indeed allied (7)
4 Break barmen's rules without end—it's not complex (4, 6)
5 Argue Mark's clothing is sloppy in front (4)
6 Halo effect of light first seen in air (7)
7 Clean record encompassing fine play by Tottenham forward (6)
8 Incorrectly medicates, perhaps, mange acquired by female (9)
14 Perhaps some making light of faulty status a gate returned into? (10)
15 Took back out miracle talking animal, perhaps? (9)

18 Noticed crowds of those that make planks? (8)
21 Fickle worker following idol holding new way (7)
22 Anticipate rep messed with van briefly (3-4)
23 Durant flees to wasteland (6)
26 Sour and angsty, stupidly missing point (5)
27 The French ellipse, for one, is elliptical (4)

---

_{This is my first big variety cryptic! Let me know how I did!}

Answer

I believe this is the grid:

The "quick to write" clues are

5a (track), 11a (short), 15a (career), 19a (lying), 24a (evens), 31a (severe), 32a (please), 3d (dallied), 4d (ends), 8d (manage), 14d (flight), 22a (vain). This appears to spell THEYNEEDSAFI so it seems like something has gone wrong at the end. [EDITED to add:] Aha! ManyPinkHats suggests in TSL chat that 4d is not meant to be missing a letter (though I do think the clue works much better with "ends" than with "end") and that 23d should have "flexes", so we get THEY NEED A FIX. I bet this is right.

The "quick to solve" clues are

1a (CONSIDER), 16a (DISMISSAL), 17a (PATENT), 22a (DISPLAYS), 25a (INTERMITTENT), 30a (MALINTENT), 1d (REMITTENT), 2d (SUBCONTRACT), 5d (DISSENT), 7d (DISINFECT), 21d (INCONSTANT), 27d (LACONIC). It would appear that Level 51 is DISCONTENT about all the Rileys.

Level 51 says "Let me know how I did!", so ...

This was a lot of fun. There are a few clues I don't wholly understand and a few elements I'm not in love with, but on the whole I thought it was very good indeed. Here's a list of quibbles; I must reiterate that I thought this was a really good puzzle and plenty of newspaper-published crosswords would get similar lists from me. 11a: not wild about "condition" cluing "iff"; seems a bit loose (and also maybe a bit niche-y, though this place is crawling with mathematicians so maybe that's OK). 12a: presumably "drug" clues TAR meaning black tar heroin? I'm not sure it's correct to call that just "tar". Then again, pretty much every short noun has been used to mean some drug or other, so it's probably fine :-). 24a: this is the one and only place where there needs to be an actual change in the clue to accommodate the dropped letter (evens -> Eve's, with a capital letter and an apostrophe). That's not wrong but it feels a bit odd: I'd rather either embrace the idea that clues might get substantially changed about and make sure that several do, or else find a way to make none of them change at all. 29a: a bit of a shame that "part of" turns out to mean "initial part of". 4d: I think this does really need "ends" rather than "end", though of course that spoils the surface. 5d: I confess I don't understand the wordplay for this. (Maybe "Mark's clothing" is NT -- the gospel of Mark! -- but I don't see how "sloppy" would clue DISSE or CONSE.) 15d: I don't really like "out" as a prefix rather than suffix anagrind. 21d: I suspect a lot of Orthodox Christians would get cross about "idol" cluing ICON, though I suppose it works OK if they're both taken metaphorically.

strategy - Spiders and Ant on a Cube

There are three spiders and an ant, crawling on the edges of a wire frame cube. The spiders can crawl exactly one third as fast as the ants. Show that the spiders can catch the ant.

Think of bugs as point masses, and a spider catches an ant when they are in the exact same location. Any bug can instantaneously change direction. Everyone can see each other at all times.

Answer

A more simplified/intuitive way to think about it is if you have the following:

No matter where the ant and spider start, the spiders go into this formation:

Where the red dots are the spiders, and the green dot is the ant. Notice how the two red spiders on the edges of the cube will only move up and down. This prevents the ant from getting across to the other side without being eaten!

The spiders will have to move at most 1/2 a cube edge length to get to the corner. This means the ant can move a total of 1.5 cube edge lengths before the spider blocks its path. The ant being smart can try to prevent this setup by not going on the edge center (as shown in the picture), however the spiders only need to maintain/setup the ratio with the ant. So the ant may pass the half of the cube a few time, but then the spiders will have be able to 'sync to the ratio' with the ant.

How do the spiders get in sync you ask? Well, every time the ant walks one direction, the spider is crawling its edge to try and reach this magic ratio (which means he is behind where he should be). But as soon as the ant reverses direction, the spider can stand still until he is no longer 'behind' where he is trying to get to and is in sync with the ratio.

Now that the ant is stuck on the one half of the cube, the third spider can walk along towards the ant and trap him towards another spider!

mathematics - Rotating teams through stations

How can 14 teams rotate through 7 stations (2 teams at a station) without overlaps?

NOTE:
Every team participates in every station exactly once

general relativity - In GR, why should the spacetime manifold be differentiable?

In general relativity (GR), spacetime is viewed as a differentiable manifold of dimension $D$ with a metric of Lorentzian signature $(-,+,+,...,+)$.

My question is why differentiable?

story - Let's help Sherlock, shall we?

As usual, Sherlock Holmes was getting bored by the easy cases coming to him. He decided to learn some new languages to kill some time. Being the genius he is, he learnt five languages on that mundane Sunday. And as he was practising his newly acquired linguistic skills, he heard a knock on the door. 221B came alive...

It was a woman, in her mid-fifties, dressed well enough to be in the upper middle class; hair was tied -- had taken her time to decide to come to meet him -- but looked in distress as if she had seen a murder. Deductions aside, Sherlock asked her to sit and tell what had happened. She started telling her story, controlling her tears.

"My husband was in military."

"Sorry to interrupt you, but 'was'?" said Holmes.

"Yes, he died last night," she replied, barely controlling her tears.

"Oh, I am sorry. Please continue..." Holmes said whilst sitting on a chair next to her.

"He was a very good man, a great husband. We had a happy marriage, been married for almost thirty years. He had come home after retiring about a week ago. He was happy to have served the nation with all his intelligence and strength. After taking a rest the whole day, we had a good dinner last night at the best restaurant in the city. He started watching a movie, I can't remember the name, but it was about Indo-Pak war. He was all nostalgic about his glorious epoch in army, also boasted about the good direction and appreciated the dialogues. I can't understand Hindi, but I agreed with whatever he said. After the movie ended, we went to sleep. He was having a dream about his life in Africa. After that, I too dozed off. The first thing I did after waking up in the morning was kiss him on his forehead. To my surprise, his skin was cold. I tried to wake him up, but he didn't respond. I tried to feel his heartbeat, he had none. He probably had a heart attack. I immediately called a doctor, who came after an hour. After the checkup, the feel of his face told a horror story. My husband was dead. The doctor said it was indeed a cardiac arrest. I am not able to digest that the person who was so happy in the day, who had no medical history of any disease whatsoever, died of a heart attack. I have no hope left. I think someone murdered him. He had enemies who have access to military level tech. They must have killed him for something. Please help me."

After hearing all this, Holmes, who was already suffering from linguistic information overflow couldn't speak. All his mastery over language came to a standstill. Coming back to his senses, he asked whether they had any other family members. She replied that all they had was a cook, who was like their son. He was with them for the last twenty years, helped in household work and cooked great food. It was not possible that someone had bribed him to murder her husband.

All this was too much information, even for someone like Sherlock Holmes to process.

He knows that there is an awesome puzzle solving community on StackExchange that can help him to solve this case which is a bit awkward for his tired brain to handle.

So, can you help him solve the case?

Answer

She poisoned him at the restaurant.

Reasons:

1. They didn't eat at home that night.
2. She couldn't have known what he was dreaming.
3. She assumed a heart attack before calling the doctor.
4. Dissonant mention of assuming a heart attack, then wondering why.
5. Utilizing the cook and stating "I could not digest". A slip leading to the idea that there was something she should not ingest to begin with.

EDIT: Glossed over a detail that would have been #6