homework and exercises - Problem understanding the symmetry factor in a feynman diagram

I am trying to understand a $1/2$ in the symmetry factor of the "cactus" diagram that appears in the bottom of page 92 In Peskin's book. This is the diagram in question (notice that we are in $\phi^4$ theory)

In the book it is claimed that the symmetry factor of the diagram is

$$3!\times{}4\dot{}3\times{}4\dot{}3\dot{}2\times{}4\dot{}3\times{}1/2$$

where it says that the $3!$ comes from interchanging the vertices, the first $4\dot{}3$ from the placement of contractions in the $z$ vertex, the following $4\dot{}3\dot{}2$ from the placement of contractions in the $w$ vertex, the last $4\dot{}3$ from the placement of contractions in the $u$ vertex and the final $1/2$ from the interchange of $w-u$ contractions.

It is this last $1/2$ that I don't understand. Can you be more explicit on where this comes from?

Answer

We choose one of the $4$ z-fields to contract with the single x-field. We then choose one of the remaining $3$ z-fields to contract with one of the $4$ w-fields. The remaining two z-fields just contract with themselves. Now choose one of the remaining $3$ w-fields to contract with the single y-field.

(Here is where we have to be careful). There are $2$ choices for the w-field contraction with one of the $4$ u-fields, and then $3$ choices for the other w-u contraction. In computing this last combination we have over counted by a factor of $2$.

To see this more clearly, consider one of the contractions,

$\phi_a(w)\phi_b(w) \quad\phi_a(u)\phi_b(u)\phi(u)\phi(u)$

The subscript denote which fields are contracted with which other fields (I'm not sure how to express contractions in Latex).

There are two ways to get this particular contraction: we could either choose the first w-field to be contracted with the first u-field, and THEN choose the second w-field to be contracted with the second u-field; OR we could choose the second w-field to be contracted with the second u-field, and THEN choose the first w-field to be contracted with the first u-field.

Clearly both of these are equivalent. However, in the combinatorics we have counted both of them, and so we must divide by a factor of $2$. So the total number of different contractions giving the same expression as $(4.45)$ is

$3! \, \times \, 4 \cdot 3 \, \times \, 4 \cdot 3 \cdot 2 \, \times \, 4 \cdot 3 \, \times \, 1/2$

Where the $3!$ comes from the interchange of vertices.

EDIT: If that isn't clear, think about the following scenario. There are two boxes, in the first there are two objects, $A$, and $B$, and in the second there are two more, $C$, and $D$. How many different ways are there to pair off the objects so that each object in the first box has a partner in the second box? Clearly the answer is two: $A,C$ and $B,D$; and $A,D$ and $B,C$.

One might think the answer is $2\cdot 2$, but we can see that this produces duplicates

\begin{array}{|r|r|} \hline First Pair & Remaining Pair \\ \hline A,C & B,D\\ \hline A,D & B,C\\ \hline B,C & A,D\\ \hline B,D & A,C\\ \hline \end{array}

So we must multiply by a factor of $1/2$ to fix the overcounting.

Hope that helps.

ideal gas - How does the bulk modulus of air change with rising pressure?

I can't seem to find the answer to what should be a trivial question:

I have a rigid air-tight container of fixed volume and I am pumping air inside. The pressure is increasing (very slowly) from ~100kPa to ~50MPa - is the bulk modulus of air constant throughout the process or does it increase/decrease with increasing pressure?

I am assuming that the bulk modulus of gas should increase with increasing pressure as there is more force acting inside the gas (more gas molecules interactions) and the fluid itself is increasing in density.

Can you please offer any advice or reference me to some link.

Answer

If the temperature of the gas is kept constant during the compression then the bulk modulus of an ideal gas is just equal to the pressure.

The definition of the bulk modulus is:

$$ K = -V\frac{dP}{dV} \tag{1} $$

For an ideal gas $PV = RT$, so $P = RT/V$. If the temperature is constant this gives:

$$ \frac{dP}{dV} = -\frac{RT}{V^2} \tag{2} $$

and substituting into (1) we get:

$$ K = V \frac{RT}{V^2} = \frac{RT}{V} $$

and $RT/V$ is just $P$ so we get:

$$ K = P $$

Note that if the compression is not isothermal, or the gas is not ideal, equation (2) will not apply and the bulk modulus will not be equal to the pressure.

atomic physics - How does quantum mechanics explain stability of electron orbitals?

According to classical physics, an electron orbiting the nucleus would emit electromagnetic radiation. Losing energy in that way, it would spiral into the nucleus and the atom would collapse. Quantum mechanics explains that the electron cannot be treated as a classical particle having a definite position and velocity. The best we can do is specify the probability of it manifesting itself at any point in space and then something about Heisenberg uncertainty principle. How does quantum mechanics explain why the orbiting electron doesn't emit EM radiation and the atom doesn't collapse?

electricity - Definition of electric charge and proper explanation

Is there a definition of electric charge and proper explanation of it?

It is said "Electric charge is the physical property of matter that causes it to experience a force when close to other electrically charged matter." How is it though that matter can get charged?

Defining charge as the property of feeling a force with other charged matter seems circular. What is charge? Is there a non-circular definition / explanation?

lie algebra - Spin matrix for various spacetime fields

Let $V^{\mu}$ be a vector field defined in a Minkowski spacetime and suppose it transforms under a Lorentz transformation $V'^{\mu} = \Lambda^{\mu}_{\,\,\,\nu}V^{\nu}$. We can write this like $V'^{\mu} = (e^{i\omega})^{\mu}_{\,\,\,\nu}V^{\nu}$ I think where $\omega$ denotes a rotation in some plane spanned by indices $\left\{\rho \sigma\right\}$, say. In 2D Euclidean space time, we can write the matrix representation of $\Lambda$ as $$\begin{pmatrix} \cos \omega & \sin \omega\\-\sin \omega&\cos \omega\end{pmatrix}$$ and in Minkowski space this changes to the 'hyperbolic' rotation. Linearising the above yields $$\begin{pmatrix}1&\omega\\-\omega&1\end{pmatrix} = \text{Id} + \begin{pmatrix} 0&\omega\\-\omega&0\end{pmatrix} = \text{Id} + \omega \begin{pmatrix} 0&1\\-1&0\end{pmatrix}$$

Now compare with the more general treatment: $V'^{\mu} = \Lambda^{\mu}_{\,\,\,\nu}V^{\nu} \approx (\delta^{\mu}_{\nu} + \omega^{\mu}_{\,\,\,\nu})V^{\nu}$, where $\omega^{\mu}_{\,\,\,\nu} \equiv (\omega^{\rho \sigma} S_{\rho \sigma})^{\mu}_{\,\,\,\nu}$ In 2D, the spin matrix $S$ when acting on vectors in 2D Euclidean space time is therefore the matrix multiplying $\omega$ above, which agrees with the single generator of the SO(2) group.

If we continue with the general analysis, we obtain $V'^{\mu} - V^{\mu} = \omega^{\mu}_{\,\,\,\nu}V^{\nu} = \eta^{\mu \rho}\omega_{\rho \nu} V^{\nu} = \omega^{\rho \sigma} \delta^{\mu}_{\rho} \eta_{\sigma \nu} V^{\nu}$ Now use the antisymmetry of $\omega_{\rho \sigma}$ gives $$2(V'^{\mu} - V^{\mu}) = \omega^{\rho \sigma}(\delta^{\mu}_{\rho} \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \eta_{\rho \nu})$$ from which we can identify $S$ to be $\delta^{\mu}_{\rho} \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \eta_{\rho \nu}$. I am wondering how this agrees with the matrix I obtained above.

Many thanks.

game - Face Up Poker with Alice and Bob

Alice and Bob are going to have a poker night. They have invented a variant where there are no secrets, and no randomness, called Face-Up Poker. Here's how it works.

A standard 52 card deck is spread face up on a table.

1. Alice picks up any 5 cards from the table.


2. Bob does the same.



3. Alice discards any number of cards, throwing them on the floor. She then picks up the same number of cards from the table.


4. Bob does the same (Bob can't pick up cards that Alice discarded).

They then compare hands, and the best hand wins, with Bob winning ties.

Which player can force a win? What is their winning strategy?

This puzzle assumes familiarity with the rankings of poker hands, but no other knowledge of the rules of poker. For a refresher these rankings, see this helpful page (the hands are listed worst to best from top to bottom).

Answer

There is a winning strategy for

Alice. For her starting hand, she will select all four 10's, along with another lower card (let's say the 2 of clubs for concreteness). This will make it impossible for Bob to end up with a royal flush. Bob needs to prevent Alice from making a royal flush when she discards, so he must select a starting hand containing a card of rank higher than 10 in each of the four suits. This means that for at least three of the four suits, Bob's starting hand cannot contain a card of rank lower than 10 in that suit.

When it is Alice's turn to discard, she chooses one of the suits in which Bob did not take a card of rank lower than 10. She keeps the 10 of that suit, discards her other four cards, and takes the 6, 7, 8, and 9 of the chosen suit (Bob did not select a card of rank lower than 10 in the chosen suit). Alice's final hand will be a straight flush to the 10. Since all of the other 10's are now on the floor, Bob cannot possibly make a straight flush whose high card is 10 or above, so Bob's hand cannot possibly tie or beat Alice's.

acoustics - Why is light produced when an underwater bubble is collapsed with a sound wave?

Why is light produced when an underwater bubble is collapsed with a sound wave?

I have come across this fact on a page (similar to this) but can't understand "Why". I'm just curious about this interesting fact. And, I want to know what's the currently accepted model for this phenomenon and how good it holds. Any kind of help regarding this topic will be highly appreciated.

general relativity - Why does time stop in black holes?

Everyone says that time stops in the black hole. It's a "fact". However, I have never heard everyone explaining that.

Of course, I know that observer in weaker gravitational field sees that something in stronger gravitational field is experiencing slower time. However, slower and no at all is quite different.

I have no idea what equation is used to calculate dime dilatation, but it will use gamma and therefore division. And the only time division of non-zero constant yields zero is when you divide by infinity.

And although black holes are super heavy, super badass and super black, they posses finite energy and therefore finite gravitational acceleration (even at event horizon).

So shouldn't just the time be very slow, rather than just stop from our point of view?

Answer

Why does time stop in black holes?

Time according to whom?

The fact is that, in special and general relativity, there is no universal time. Indeed, time is a coordinate in relativity so one must be careful to specify the coordinate system when asking questions like this.

Now, every entity also has an associated proper time which is not a coordinate which means that it is coordinate independent (invariant). Think of your proper time as the time according to your 'wrist watch'.

In the context of the static black hole (Schwarzschild black hole) solution, there is a coordinate system (Schwarzschild coordinates) that we can associate with the observer at infinity. That is to say, the coordinate time corresponds to the proper time of a hypothetical entity arbitrarily far from the black hole.

In this coordinate system, we can roughly say that the coordinate time 'stops' at the event horizon (in fact, there is no finite value of this coordinate time to assign to events on the horizon).

However, there are coordinate systems with finite coordinate time at the horizon, e.g., Kruskal-Szekeres coordinates.

Moreover, for any entity falling freely towards the horizon, the proper time does not 'stop'. Indeed, the entity simply continues through the horizon towards the 'center' of the black hole and then ceases to exist at the singularity.

We interpret the fact that the Schwarzschild coordinate time does not extend to the horizon as follows: no observer outside the horizon can see an entity reach (or fall through) the horizon in finite time. This is simply understood as the fact that light emitted from (or inside) the horizon cannot propagate to any event outside the horizon.

Why? Because the spacetime curvature at the horizon is so great that there is no light-like world line the extends beyond the horizon. Indeed, the horizon is light-like. A photon emitted 'outward' at the horizon simply remains on the horizon.

Within the horizon, the spacetime curvature is such that there are no world lines that do not terminate on the singularity - the curvature is so great within the horizon that the future is in the direction of the singularity.

riddle - The Adventure of Jack #1

Jack's adventure starts at the top left of the table,

Hey, Vietnamese money should be removed!

So he adventures along to Rome, stopping for 50 minutes.

And suddenly he's back at the corner of the table!

So Jack carefully adds 1 minute to his timer,

And is transported to a new location!

What country is Jack in?

Hint:

From this statement, you will get a city, then find in which country that city is in.

Hint 2:

In the clue "So Jack carefully adds one minute to his timer," you are adding multiple letters to what you already have.

Hint 3:

The word ONE is what you are adding to Jack's timer.

I better not have to give you any more clues!

Answer

Answer:

The top left (as corrected) of the periodic table is HYDROGEN.

Hey, Vietnamese money should be removed!

remove HEY (hey!) and DONG (Vietnamese money) to get R

So he adventures along to Rome, stopping for 50 minutes.

The Roman 50 is L, so adding that gives RL

And suddenly he's back at the corner of the table!

As per @Jafe, back (reverse) to the periodic table, LR (lawrencium)
(though is it really a corner?)

So Jack carefully adds 1 minute to his timer

per hint 3, add ONE giving LRONE, or perhaps LORNE, which is a seaside village in Victoria Australia.

So Jack is in

Australia.

soft question - Why is the harmonic oscillator so important?

I've been wondering what makes the harmonic oscillator such an important model. What I came up with:

• 
  

  It is a (relatively) simple system, making it a perfect example for physics students to learn principles of classical and quantum mechanics.

  
  


• 
  

  The harmonic oscillator potential can be used as a model to approximate many physical phenomena quite well.

  
  

The first point is sort of meaningless though, I think the real reason is my second point. I'm looking for some materials to read about the different applications of the HO in different areas of physics.

Answer

To begin, note that there is more than one incarnation of "the" harmonic oscillator in physics, so before investigating its significance, it's probably beneficial to clarify what it is.

What is the harmonic oscillator?

There are at least two fundamental incarnations of "the" harmonic oscillator in physics: the classical harmonic oscillator and the quantum harmonic oscillator. Each of these is a mathematical thing that can be used to model part or all of certain physical systems in either an exact or approximate sense depending on the context.

The classical version is encapsulated in the following ordinary differential equation (ODE) for an unknown real-valued function $f$ of a real variable: \begin{align} f'' = -\omega^2 f \end{align} where primes here denote derivatives, and $\omega$ is a real number. The quantum version is encapsulated by the following commutation relation between an operator $a$ on a Hilbert space and its adjoint $a^\dagger$: \begin{align} [a, a^\dagger] = I. \end{align} It may not be obvious that these have anything to do with one another at this point, but they do, and instead of spoiling your fun, I invite you to investigate further if you are unfamiliar with the quantum harmonic oscillator. Often, as mentioned in the comments, $a$ and $a^\dagger$ are called ladder operators for reasons which we don't address here.

Every incarnation of harmonic oscillation that I can think of in physics boils down to understanding how one of these two mathematical things is relevant to a particular physical system, whether in an exact or approximate sense.

Why are these mathematical models important?

In short, the significance of both the classical and quantum harmonic oscillator comes from their ubiquity -- they are absolutely everywhere in physics. We could spend an enormous amount of time trying to understand why this is so, but I think it's more productive to just see the pervasiveness of these models with some examples. I'd like to remark that although it's certainly true that the harmonic oscillator is a simple an elegant model, I think that answering your question by saying that it's important because of this fact is kind of begging the question. Simplicity is not a sufficient condition for usefulness, but in this case, we're fortunate that the universe seems to really "like" this system.

Where do we find the classical harmonic oscillator?

(this is by no means an exhaustive list, and suggestions for additions are more than welcome!)

1. Mass on a Hooke's Law spring (the classic!). In this case, the classical harmonic oscillator equation describes the exact equation of motion of the system.


2. Many (but not all) classical situations in which a particle is moving near a local minimum of a potential (as rob writes in his answer). In these cases, the classical harmonic oscillator equation describes the approximate dynamics of the system provided its motion doesn't appreciably deviate from the local minimum of the potential.


3. Classical systems of coupled oscillators. In this case, if the couplings are linear (like when a bunch of masses are connected by Hooke's Law springs) one can use linear algebra magic (eigenvalues and eigenvectors) to determine normal modes of the system, each of which acts like a single classical harmonic oscillator. These normal modes can then be used to solve the general dynamics of the system. If the couplings are non-linear, then the harmonic oscillator becomes an approximation for small deviations from equilibrium.


4. Fourier analysis and PDEs. Recall that Fourier Series, which represent either periodic functions on the entire real line, or functions on a finite interval, and Fourier transforms are constructed using sines and cosines, and the set $\{\sin, \cos\}$ forms a basis for the solution space of the classical harmonic oscillator equation. In this sense, any time you are using Fourier analysis for signal processing or to solve a PDE, you are just using the classical harmonic oscillator on massively powerful steroids.


5. Classical electrodynamics. This actually falls under the last point since electromagnetic waves come from solving Maxwell's equations which in certain cases yields the wave equation which can be solved using Fourier analysis.

Where do we find the quantum harmonic oscillator?

1. Take any of the physical systems above, consider a quantum mechanical version of that system, and the resulting system will be governed by the quantum harmonic oscillator. For example, imagine a small system in which a particle is trapped in a quadratic potential. If the system is sufficiently small, then quantum effects will dominate, and the quantum harmonic oscillator will be needed to accurately describe its dynamics.


2. Lattice vibrations and phonons. (An example of what I assert in point 1 when applied to large systems of coupled oscillators.


3. Quantum fields. This is perhaps the most fundamental and important item on either of these two lists. It turns out that the most fundamental physical model we currently have, namely the Standard Model of particle physics, is ultimately based on quantizing classical fields (like electromagnetic fields) and realizing that particles basically just emerge from excitations of these fields, and these excitations are mathematically modeled as an infinite system of coupled, quantum harmonic oscillators.

renormalization - How is the apparent significance of (length) scales in physics explained?

From what I understand, especially from reading arguments on Physics.SE, different (length) scales of a system are extremely important. It's clear that if there are two scales $\delta,d,D,\Delta$ with say $$\delta < d \ll D < \Delta,$$ then effect, which happen on a length scale $d$ might be neglected if one is interested in effects of the length scale $D$.

However, I don't see which these points justify arguments involving the length scales per se. As far as I can see, the existence of a quantity itself doesn't imply that physical effects are of that size. If I have a natural length scale $l$, and problems of geometries involving a length $L$, then things could end up with depending on $$l\propto\frac{1}{16^{\pi^2}}L$$ or $$2^{\text{dim}}L\gg L.$$ For example, often people argue that if some characteristic scale $H$ is close to $\hbar$ then things will get problematic. I don't see why this would be a priori justified at all. Conversely, why does a model with some small quantifity automatically have to be suppressed at one point? And if I have a small and a big scale, why are the scales in between relevant, if their value is obviously just some real value times one other values.

So how is the apparent predominance in reasoning with length scales to be explained? I'm specifically thinking of field theories here, but not only.

Answer

The reason is that computations producing large dimensionless numbers are generally complex, in a computational sense, meaning that you have to have a plan, or a specific intent, to produce these large numbers. Physical systems are generally dumb, so that they rely on very simple computations at their core, and these computations do not produce large dimensionless numbers easily.

This is only true for exponential computations, you can generate dimensionless large numbers which are polynomially large relatively naturally. For example, when considering atoms, you might believe that the scale of the wavelength of visible light, which about 10,000 times bigger than the radius of the atom, is too big to have appreciable scattering off an atom. But this is not so if the atom has an absorption peak at a certain wavelength. In this case, the resonance for frequencies close to the maximum peak frequency can amplify the scattering by a factor which goes as $f_0/(f-f_0)$, where $f_0$ is the resonant frequency and f is the frequency of the light. This is a simple polynomial amplification factor, but it can easily give you a factor of 10,000 if f is tuned to be within 1 part in 10,000 of the resonant frequency.

In this case, which is typical, there is a small dimensionless parameter, the detuning parameter of the resonance, which serves to amplify the scattering cross section so that it can be many orders of magnitude bigger than the atomic radius. But absent this fine tuning, the general argument that light is bigger than an atom is correct, so that the light scatters off the atom only a small amount.

To get an exponentially enormous dimensionless number, like $10^{10^2}$, you need a machine that computes exponentials, and it is difficult to think up a physical system which computes exponentials. You are constrained by conservation of energy, so you generally can't have replication like bacterial growth.

general relativity - Is Mach's Principle Wrong?

This question was prompted by another question about a paper by Woodward (not mine). IMO Mach's principle is very problematic (?wrong) thinking. Mach was obviously influenced by Leibniz. Empty space solutions in GR would result in a Minkowski metric and would suggest no inertia. Mach's principle seems incompatible with GR. Gravitational waves could also be a problem. I had thought that papers like one by Wolfgang Rindler had more or less marginalised the Mach Principle, but I see lots of Internet discussion of it. Is it correct? Wrong? Is there evidence? (frame dragging experiments)?

Let's use this definition from ScienceWorld.Wolfram.com:

In his book The Science of Mechanics (1893), Ernst Mach put forth the idea that it did not make sense to speak of the acceleration of a mass relative to absolute space. Rather, one would do better to speak of acceleration relative to the distant stars. What this implies is that the inertia of a body here is influenced by matter far distant.

What is "mass" in particle physics?

It's clear, from reading pop-science articles about the Higgs boson, that particle physicists have something very specific in mind when they say "mass". In classical physics the mass of a particle is just a given value, but in the context of particle physics you hear things like "computing the mass" or "such and such interaction gives some mass".

What does "mass" mean to a particle physicist? How does it relate-to/explain the classical notion of mass, and how does it differ?

I'm not sure what level of explanation I'm looking for. I don't know particle physics, obviously, so any grad-level equations are likely to go way over my head. But explanations like "bouncing back and forth all the time" from this minute physics video are too simplified. I'm left wondering, for example, how coupling with the Higgs field makes electrons bounce back and forth? And why wouldn't it just scatter them? And how is this process Lorentz invariant? And why does the bouncing resist/react-to forces?

electromagnetism - Does electricity flow on the surface of a wire or in the interior?

I was having a conversation with my father and father-in-law, both of whom are in electric related work, and we came to a point where none of us knew how to proceed. I was under the impression that electricity travels on the surface while they thought it traveled through the interior. I said that traveling over the surface would make the fact that they regularly use stranded wire instead of a single large wire to transport electricity make sense.

If anyone could please explain this for some non-physics but electricly incline people, I would be very appreciated.

Answer

It depends on the frequency. DC electricity travels through the bulk cross section of the wire.

A changing electrical current (AC) experiences the skin-effect where the electricity flows more easily in the surface layers. The higher the frequency the thinner the surface layer that is usable in a wire. At normal household AC (50/60hz) the skin depth is about 8-10mm but at microwave frequencies the depth of the metal that the current flows in is about the same as a wavelength of visible light

edit: Interesting point from Navin - the individual strands have to be insulated from each other for the skin effect to apply to each individually. That is the reason for the widely separated pairs of wires in this question What are all the lines on a double circuit tower?

particle physics - Breit-Wigner vs propagator?

Consider the two Feynman diagrams below:

I know for the second that we should use the Breit-wigner curve, but what about the first? If we go about using the propagator: \[\frac{1}{\tilde q^2-m_z^2c^2}\text{, (+ - - -) signature}\] then we a singularity when $\tilde q^2=m_z^2c^2$ which is subverted when the Breit-Wigner curve is used. Thus my question is as follows: When is it appropriate to use the propagator as given above and when to use the Breit-Wigner curve.

quantum field theory - Gauge fixing and degrees of freedom

Today, my friend (@Will) posed a very intriguing question -

Consider a complex scalar field theory with a $U(1)$ gauge field $(A_\mu, \phi, \phi^*)$. The idea of gauge freedom is that two solutions related by a gauge transformation are identified (unlike a global transformation where the solutions are different but give rise to the exact same observables), i.e. $$(A_\mu(x), \phi(x), \phi^*(x)) ~\sim~ (A_\mu(x) + \partial_\mu \alpha(x), e^{i \alpha(x)}\phi, e^{-i \alpha(x)}\phi^*(x)).$$ The process of "gauge fixing" is to pick one out of the many equivalent solutions related via gauge transformation. The usual procedure of gauge fixing is to impose a condition on $A_\mu$ so that one picks out one of the solutions. His question was the following:

Instead of imposing a gauge condition on $A_\mu$, why do we not impose a gauge condition on $\phi$? Wouldn't this also pick out one the many equivalent solutions? Shouldn't this also give us the same observables? If so, why do we not do this in practice?

After a bit of discussion, we came to the following conclusion:

The idea of gauge symmetry comes from the requirement that a quantum theory involving fields $(A_\mu, \phi, \phi^*)$ have a particle interpretation in terms of a massless spin-1 particles and 2 spin-0 particles. However, prior to gauge fixing, the on-shell degrees of freedom include those of a massless spin-1 particle and 3 spin-0 fields ($A_\mu \equiv 1 \otimes 0,~\phi,\phi^* \equiv 0$). We would now like to impose a gauge condition to get rid of one scalar degree of freedom. There are two ways to do this -

1. 
   

   Impose gauge condition on $A_\mu$ so that $A_\mu \equiv 1$. Now, $A_\mu$ corresponds to a massless spin-1 particle and the complex scalar corresponds to two spin-0 particles. This is what is usually done.

   
   


2. 
   

   Impose a gauge condition on $\phi$. For instance, one can require that $\phi = \phi^*$. We now have a real field corresponding to a spin-0 particle. However, $A_\mu$ still contains the degrees of freedom of both a massless spin-1 and a spin-0 particle.

   
   

I claimed that the second gauge fixing procedure is completely EQUIVALENT to the first one. However, the operator that now creates a massless spin-1 particle is some nasty, possibly non-Lorentz Invariant combination of $A_0, A_1, A_2$ and $A_3$. A similar statement holds for the spin-0 d.o.f. in $A_\mu$. Thus, the operators on the Hilbert space corresponding to the particles of interest are not nice. It is therefore, not pleasant to work with such a gauge fixing procedure.

In summary, both gauge fixing procedures work. The first one is "nice". The second is not.

Is this conclusion correct?

NOTE: By the statement $A_\mu \equiv 1$, I mean that $A_\mu$ contains only a massless spin-1 d.o.f.

Answer

If $\phi$ is non-zero, fixing the phase of $\phi$ is a perfectly valid gauge condition. It's used frequently in Standard Model calcuations involving the Higgs field, where it goes by the name unitarity gauge. This is a nice gauge in some ways, because it makes manifest the fact that there's a massive vector field in the system.

Edit: Some caution is required with unitary gauge. It's a complete gauge when you can reasonably treat $\phi$ as non zero, because it uses every degree of freedom in the gauge transformation. This means for example that it's ok to use in perturbative calculations around a Higgs condensate. But when $\phi$ can vanish, the phase function isn't uniquely defined, which means the gauge transformation is not invertible. This gauge isn't quite a gauge.

electromagnetism - How do we apply Ampère's law for non-planar loops?

How do we apply Ampère's (magnetism) law for non-planar loops?

Its most general form(or you can say the only one I know) is $$ \oint_C \mathbf B\cdot\mathrm d\mathbf l = \mu_0 \iint_S\mathbf J\cdot \mathrm d\mathbf S $$ But what would current enclosed mean in case of non planar loops. I mean infinite amount of curves can contain such loop. As a result while the right-hand side (line integral of B field) would same in each but the integral of current density would be different for each curve (surface or manifold).

mathematical physics - What are Grassmann (even/odd) numbers used in superalgebras?

Are Grassmann numbers a concept of graded Lie algebras or is something specific to superalgebras? What are they (i.e: how are they defined, important properties, etc.)? Is there a reasonable introduction to them?

I think that what makes me wonder a little is, since there does not seem to be a sensible constructivist approach to these entities (other than accepting them as the entities that satisfy the required properties) is there nothing that stops someone from going into 'constructing' meta-superalgebras by defining 'numbers' $\kappa_{i}$, such that, e.g.,

$$\kappa_{i} \kappa_{j} = \theta_{k} \quad (\leftarrow \text{Grassmann odd}),$$ $$\kappa_{i} \kappa_{j}\kappa_{m} \kappa_{n} = \theta_{p}\quad (\leftarrow \text{Grassmann even}).$$

So I define such numbers as 'square-roots' of grassmann $a$-numbers. It seems nothing stops this process ad infinitum. Maybe there is some property I'm missing that will allow the algebra to be closed but I don't know what that could be.

Btw, I think this is a great reference Phys.SE question regarding this topic: "Velvet way" to Grassmann numbers.

Answer

Here I will just make a couple of general remarks.

1) Graded algebras usually refer to $\mathbb{Z}$- or $\mathbb{N}$-graded algebras, while superalgebras are $\mathbb{Z}_2$-graded algebras.

2) Grassmann numbers are oddly graded supernumbers.

Please click on the links for further information, important properties and references.

References:

1) Bryce De Witt, "Supermanifolds", Cambridge Univ. Press, 1992.

2) Deligne, Pierre and John W. Morgan, "Notes on Supersymmetry (following Joseph Bernstein)". Quantum Fields and Strings: A Course for Mathematicians (1999). American Mathematical Society. pp. 41–97. ISBN 0-8218-2012-5.

---

Concerning v3 of the question. The $\kappa_i$'s correspond to a $\mathbb{Z}_4$ grading, and there are indeed research works in that direction. However, many properties of numbers and supernumbers do not generalize easily to $\mathbb{Z}_n$-grading with $n>$2. For instance, I think that already Berezin showed that it is not possible to define a useful notion of (Berezin) determinant of matrices with $\mathbb{Z}_n$-graded entries if $n>$2.

no computers - De-zephyr The Solution!

Here's another original chess problem of mine for you all to solve today!

In the following position, White wants to stalemate Black even though they can win, for unknown, unclear, unexplained, and illogical reasons.

In how many moves can White forcibly stalemate Black? You may only use your 100%, organic, al naturale brain power!

Answer

The title means

that we are going to use the windmill technique (Zephyr is the name of a wind).

Here is a solution: apronus link.

PGN:

[FEN "5Nqk/pbppp1r1/4NQR1/6pK/6pn/2b3p1/1B4p1/r5n1 w - -"] 1. Rh6+ Qh7 2. Rxh7+ Kg8 3. Rxg7+ Kh8 4. Rxe7+ Kg8 5. Rg7+ Kh8 6. Rxd7+ Kg8 7. Rg7+ Kh8 8. Rxc7+ Kg8 9. Rg7+ Kh8 10. Rxb7+ Kg8 11. Rg7+ Kh8 12. Rxa7+ Kg8 13. Rg7+ Kh8 14. Ng6+ Nxg6 15. Rxg6+ Kh7 16. Rg7+ Kh8 17. Rxg5+ Kh7 18. Rg7+ Kh8 19. Rxg4+ Kh7 20. Rg7+ Kh8 21. Rxg3+ Kh7 22. Rg7+ Kh8 23. Rxg2+ Kh7 24. Rg7+ Kh8 25. Rxg1+ Kh7 26. Qg7+ Bxg7 27. Rxg7+ Kh8 28. Bxa1

Note that

the bishop could capture any time during the second phase, but this would result in an even faster stalemate. See this for instance. If we assume that the opponent tries to make it as long as they can, then the above is the solution, in 27 moves and a half.

visible light - Why is the sky never green? It can be blue or orange, and green is in between!

I, like everybody I suppose, have read the explanations why the colour of the sky is blue:

... the two most common types of matter present in the atmosphere are gaseous nitrogen and oxygen. These particles are most effective in scattering the higher frequency and shorter wavelength portions of the visible light spectrum. This scattering process involves the absorption of a light wave by an atom followed by reemission of a light wave in a variety of directions. The amount of multidirectional scattering that occurs is dependent upon the frequency of the light. ... So as white light.. from the sun passes through our atmosphere, the high frequencies become scattered by atmospheric particles while the lower frequencies are most likely to pass through the atmosphere without a significant alteration in their direction. This scattering of the higher frequencies of light illuminates the skies with light on the BIV end of the visible spectrum. Compared to blue light, violet light is most easily scattered by atmospheric particles. However, our eyes are more sensitive to light with blue frequencies. Thus, we view the skies as being blue in color.

and why sunsets are red:

... the light that is not scattered is able to pass through our atmosphere and reach our eyes in a rather non-interrupted path. The lower frequencies of sunlight (ROY) tend to reach our eyes as we sight directly at the sun during midday. While sunlight consists of the entire range of frequencies of visible light, not all frequencies are equally intense. In fact, sunlight tends to be most rich with yellow light frequencies. For these reasons, the sun appears yellow during midday due to the direct passage of dominant amounts of yellow frequencies through our atmosphere and to our eyes.

The appearance of the sun changes with the time of day. While it may be yellow during midday, it is often found to gradually turn color as it approaches sunset. This can be explained by light scattering. As the sun approaches the horizon line, sunlight must traverse a greater distance through our atmosphere; this is demonstrated in the diagram below. As the path that sunlight takes through our atmosphere increases in length, ROYGBIV encounters more and more atmospheric particles. This results in the scattering of greater and greater amounts of yellow light. During sunset hours, the light passing through our atmosphere to our eyes tends to be most concentrated with red and orange frequencies of light. For this reason, the sunsets have a reddish-orange hue. The effect of a red sunset becomes more pronounced if the atmosphere contains more and more particles.

Can you explain why the colour of the sky passes from blue to orange/red skipping altogether the whole range of green frequencies?

I have only heard of the legendary 'green, emerald line/ flash'

that appears in particular circumstances

Green flashes are enhanced by mirage, which increase refraction... is more likely to be seen in stable, clear air,... One might expect to see a blue flash, since blue light is refracted most of all, and ... is therefore the very last to disappear below the horizon, but the blue is preferentially scattered out of the line of sight, and the remaining light ends up appearing green

but I have never seen it, nor do I know anybody who ever did.

Answer

The sky does not skip over the green range of frequencies. The sky is green. Remove the scattered light from the Sun and the Moon and even the starlight, if you so wish, and you'll be left with something called airglow (check out the link, it's awesome, great pics, and nice explanation).

Because the link does such a good job explaining airglow, I'll skip the nitty gritty.

So you might be thinking, "Jim, you half-insane ceiling fan, everybody knows that the night sky is black!" Well, you're only half right. The night sky isn't black. The link above explains the science of it, but if that's not good enough, try to remember back to a time when you might have been out in the countryside. No bright city lights, just the night sky and trees. Now when you look at the horizon, can you see the trees? Yes, they're black silhouettes against the night sky. But how could you see black against black? The night sky isn't black. It's green thanks to airglow (or, if you're near a city, orange thanks to light pollution).

Stop, it's picture time. Here's an above the atmosphere view of the night sky from Wikipedia:

And one from the link I posted, just in case you didn't check it out:

See, don't be worried about green. The sky gets around to being green all the time.

special relativity - Why can't we make measurements in a photon's rest frame when loop diagrams make measurements possible?

It is one of the axioms of special relativity that the photon has no rest frame; light travels at speed c when measured in any inertial frame of reference. As a corollary, it is often said that if one were in a photon's rest frame, infinite time-dilation and length-contraction would make the universe appear (from a photon's perspective) to be unchanging, with zero length in its direction of motion. In other words a measurement would not be possible in the reference frame of a photon, because there would be no time or space in which one could conduct it.

My question is: how does this accord with the fact that parallel-moving photons can interact with each other via loop diagrams[*]? Photons can therefore be used to perform measurements in the photon's rest frame. I suppose an answer may be "the act of measurement changes the photons momentum vector and therefore its co-moving frame is non-inertial" however this can be said of any measurement.

[*] Or can they? I can't find an explicit amplitude to check this (for, say, the photon-photon "box" diagram), however as far as I can tell from the Euler-Heisenberg Lagrangian nonlinear effects should be present for parallel photons.

solid state physics - How do we get a 2DEG in a remote doped heterostructure?

I have a question regarding the way in which one often constructs a two-dimensional electron gas in heterostructures. I have a specific example in mind, although I believe this is quite a common way to build the system if I can trust the literature. As shown in the diagram below (borrowed from Semiconductor Nanostructures by Thomas Ihn) we have a type 1 heterostructure with GaAs and AlGaAs. On top of that there is a $\delta$ doped donor layer (so there is one layer with heavy doping), above which there is another GaAs layer, an AlAs layer and another GaAs layer.

Now, I have a few questions about this. First of all I'm not entirely sure why we don't just make the 2DEG by sandwiching the GaAs layer between two AlGaAs layers (ABA) so that you have a quantum well in your conduction band, which then gives you your 2DEG, like in the diagram below. In my mind this is the most simple way of going about such a problem, but perhaps it is just too simplistic. You're not taking the surface into account and such.

But okay, let us continue. On top of the AlGaAs layer we have the $\delta$ donor layer. Why do we use this? I know that the idea of sheet doping is that you create an electrostatic potential and thus that (in the absence of other factors) you have donor electrons that are bound to the plane. But now in this system, because the layer is on top of the AlGaAs which has a larger band gap than GaAs, it is energetically favorable for the donor electrons to move towards the GaAs. On the other hand the positively charged donors also pull on these electrons. Is this then how we get our 2DEG? The donor electrons somehow get into a bound state at the interface?

I am not sure if this is the case, but perhaps it is. My final question would then be, if this is indeed so, why do we have all these additional layers on top? The GaAs, the AlAs, more AlGaAs, and more GaAs. I don't get the purpose of this. Perhaps it is related to surface states and Fermi level pinning or something?

Answer

Regarding your question on sandwiching GaAs between two AlGaAs barriers:

If you do this for a narrow quantum well (like you sketched above), the electron wavefunction protrudes into the barrier quite a bit. As the barrier material is a ternary alloy, the electrons are exposed to alloy scattering. This is simply due to the fact that Ga and Al atoms are assumed to be randomly distributed. A second important contribution is interface roughness scattering, caused by an imperfect interface. Even in highest quality structures, you will still have some roughness at the order of one monolayer (RMS ~ 0.3 nm). So these are the arguments against a more complex heterostructure.

Now, in fact state-of-the-art high-mobility 2DEGs are realized with quantum wells, thus barriers on both sides. This is due to the necessity of getting as many electrons as possible into the channel. It sounds contradictory, but within certain limits, more electrons in the channel lead to higher mobility. Due to technical limits, you can only put a certain amount of doping in one sheet. Therefore, people used two so called delta-dopings on either side of the GaAs channel. Why does this work then? They simply use a rather wide quantum well (at the order of 50 nm), which only yields small electronic confinement. The electronic wavefunction is then mostly localized in the GaAs, which is why this still can be a high mobility structure.

In any case, you need to take the surface into account. GaAs doped ~ 1e16/cm³ leads to a depleted region, which is roughly 1 µm thick from the surface. Now if you talk about high mobility structures, your background doping is far below 1e16, therefore your depleted layer grows in thickness. If you want realistic structures, where you can fabricate contacts and gate electrodes, you can not bury the 2DEG infinitely deep. Therefore, you can't neglect surface effects. Usually, you compensate surface states by an additional doping sheet closer to the surface.

The band bending issue was discussed already in the answer by @ignacio. Basically, you have to solve the Schrödinger-Poisson equation in a self-consistent way to obtain the correct band profile. Basically, you solve Schrödinger's equation, which gives you bound states. Based on these, you distribute your electrons, which then allows you to solve Poissons equation, which gives you band bending and therefore a new conduction band profile on which you can solve Schrödinger's equation again. You do this in an iterative way, until your solution converges to the hopefully correct one. This is somewhat tedious on paper, but there are free programs available. E.g. the famous solver by Greg Snider. But it's also not too difficult to write such a simulation by yourself. Actually its a good occasion to practice and see, if you understood the problem.

The structure, you sketched here, is a bit more complex. Obviously, they use an AlAs layer to prevent electrons from leaking to the surface and to push them towards the 2D channel.

calculation puzzle - Fuzzy the Fuzzball

Fuzzy the fuzzball was very sad.
He had lost his fuzz.
Everyone knew if you lost your Fuzz you could no longer be a fuzzball.

Fuzzy went to this friend Frizzy and shared his dilemma.
"Don't worry, Fuzzy," Frizzy said. "You can buy new fuzz in the FuzzyFuzz store."

Fuzzy went to the FuzzyFuzz store and asked to buy Fuzz.
"How much Fuzz do you need?" the clerk asked.
"Enough to make me fuzzy," Fuzzy said. "I'm a perfect sphere, 2.3456" in diameter."

"We sell three types of Fuzz," said the clerk. "Spartan fuzz is \$2.34 per cm². Standard fuzz is \$3.45 per cm². Super Fuzz is \$4.56 per cm²."

"Ok," said Fuzzy. "I have \$543.21. I will spend up to 67% of my money to buy the best Fuzz I can afford. I sure hope I can afford to become Super Fuzzy!"

How fuzzy is Fuzzy the fuzzball after his purchase?

Answer

So Fuzzy's total surface area is

$17.28$

Then

we multiply that by $2.54^2$ to get cm.
$111.483648$

We now find 67% of 543.21

363.9507

Now we divide the money into parts. How many $cm^2$ total for each?

Super: 79.91375 cm$^2$ for \$363.9507
Regular: 105.49
Spartan: 155.53

So,

since only one of them is obviously greater, we can say he can be completely covered in Spartan Fuzz.

black holes - Hawking Radiation and virtual particles

Why is energy/ mass taken away from a black hole? Doesn't the energy coming from the virtual particle come from the vacuum energy? ...if this is so, why does the black hole have to pay the energy debt?

If the black hole creates the particles by its energy, what energy was used from the black hole to create them?

electromagnetism - Definition of Ampere

On Wikipedia it says:

This force is used in the formal definition of the ampere, which states that it is "the constant current that will produce an attractive force of $2 × 10^{-7}$ newton per metre of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one metre apart in a vacuum."

In reference to the definition of an Ampere, why was $2 × 10^{–7}$ chosen?

Answer

The definition of Ampere is obtained by the below equation of force between two infinitely long parallel current carrying conductors.

Where $F$ is force, $\triangle{L}$ is small length element, $\mu_0$ is absolute permeability of vaccum or free space, $I_1, I_2$ are current flowing through two conductors.

By calculation we can obtain that $\frac{\mu_0}{4\pi}=10^{-7} T A^{-1} m$

When $\triangle{L}=1m, I_1=I_2=1A, r=1m$. By substituting the values in the above equation given in the figure, we obtain $\frac{F}{\triangle{L}}$ to be equal to $2X10 ^{-7}N m{-1}$.

Thus we have the definition of one ampere as: One ampere is that current, which when flowing through each of the two parallel conductors of infinite length and placed in free space of one metre from each other, produces between them a force of $2X10^{-7}$ newton per metre of their lengths.

word - Four degrees of separation (I make billions #8)

(I make billions #8)

In a typical two part connect-wall puzzle, you are given 16 words which have 2 degrees of separation from the final answer. The first degree of separation is that each clue word is directly related to a group word. The second degree of separation is that each group word is directly related to the final word.

In this puzzle you are given 64 clue words that have 4 degrees of separation from the final word. The puzzle has two stages. In stage 1, you will solve 16 mini-puzzles involving two degrees of separation to obtain 16 new words. In stage 2, the 16 new words are used as a two part connect-wall puzzle to find the final word.

Stage 1

The objective of this stage is to find 16 missing words. Each missing word can be found through a mini-puzzle, where 4 clue words are given. The difficulty is that each clue word is two degrees separated from the missing word, meaning each clue word doesn't relate to the missing word directly, but only indirectly through a connecting word. For example, if the missing word were guard, possible clue words could be busy as in busy body guard or bone as in guard dog bone.

The rules of word connections are:

1. Each word connection must be one of the following (from most common to least):
   
   
   
   • A two word phrase/expression such as guard dog honor guard
   
   
   • Two parts of a compound word such as bodyguard rainfall
   
   
   
   • The title of a movie/tv show/song/etc or a location. All words will be common words (or names that are common words), and may be capitalized as necessary to form the connection. For example, hunger games green bay toy story trump tower.
   
   
   
   


2. The clue word and missing word must sit on opposite sides of the connecting word. Therefore, busy body guard is allowed because busybody and bodyguard are valid connections. But bag body guard is not allowed because although body bag is a connection, it doesn't have the right word order.

Each mini-puzzle looks like this and gives further clues to help solve the missing word:

 1. camp    (4)  drama    (5)  _____3_____ (4) animal   (4) climbing

• The long blank line _____3_____ represents the missing word you are trying to find. The 3 means that this word has 3 letters in it.



• Clues to the left of the missing word look like camp (4), which means that camp is connected on the left through a four letter connecting word. So you would be looking for camp xxxx yyy where xxxx is a four letter connecting word and yyy is the three letter missing word. One example solution could be camp site map.


• Similarly, clues to the right of the missing word look like (4) animal, which means that animal is connected on the right through a four letter connecting word, e.g. yyy xxxx animal. One example solution could be mud pack animal.

When you solve a mini-puzzle, you should write your solution like this (the solution to the above):

1. camp fire drama queen ant farm animal hill climbing

Without further ado, here are the 16 mini-puzzles of stage 1:

 1. bath    (4)  _____4_____  (5) flour    (4) pressure (4) start
 2. brick   (4)  _____6_____  (7) jet      (7) lot      (6) stick

 3. french  (7)  _____5_____  (4) room     (4) fish     (4) square
 4. self    (4)  _____5_____  (4) dance    (6) rights   (5) shirt
 5. public  (7)  _____5_____  (6) agent    (4) coat     (4) shape
 6. atomic  (5)  twisted (4)  _____4_____  (3) cave     (5) shot
 7. bomb    (5)  trouble (5)  _____8_____  (6) collar   (5) thin
 8. cargo   (3)  foot    (4)  _____4_____  (4) blue     (3) rat
 9. check   (4)  swim    (4)  _____4_____  (4) bearing  (6) circuit
10. common  (6)  medical (5)  _____5_____  (4) keeper   (5) printer
11. daily   (6)  diesel  (5)  _____5_____  (7) music    (4) style
12. dark    (5)  smoke   (6)  _____4_____  (4) fold     (4) palm

13. day     (3)  film    (5)  _____6_____  (6) block    (7) lesson
14. dish    (4)  heart   (4)  _____3_____  (5) meat     (6) roll
15. drug    (5)  stone   (4)  _____5_____  (4) bell     (4) jockey
16. garlic  (6)  phone   (4)  _____5_____  (5) club     (5) runner

Stage 2

Once you find the 16 missing words from stage 1, these words will form a two part connect-wall puzzle. The 16 words are to be grouped into four groups of four words each. Each word in a group is directly connected to a group word using the same type of word connections as in stage 1 (i.e. two word phrase, compound word, title/location). The four group words are connected to the final word in the same way. When you solve the connect wall, you can answer the question:

I make billions. Who / what am I?

Note: A complete solution should include 64 connecting words and 16 missing words (in the format described above), 4 group words, and the 1 final word/answer. Also, I found that the mini-puzzles can be quite hard, so collaboration is allowed and encouraged.

_{Previous puzzles in this series: #1 #2 #3 #4 #5 #6 #7}

Answer

The 16 resolved connecting groups are as follows:

1. bath SALT FLAT BREAD flour TYRE pressure HEAD start
2. brick WALL STREET FIGHTER jet PARKING lot HOCKEY stick
3. french QUARTER FINAL EXAM room BLOW fish FOUR square
4. self PITY PARTY LINE dance ANIMAL rights DRESS shirt
5. public DOMAIN SPACE TRAVEL agent DUST coat SHIP shape
6. atomic CLOCK twisted PAIR WISE MAN cave CRACK shot
7. bomb SCARE trouble SHOOT STRAIGHT JACKET collar RAZOR thin

8. cargo BAY foot REST AREA CODE blue RUG rat (courtesy of Mohirl and tmpearce)
9. check BOOK swim SUIT CASE LOAD bearing CLOSED circuit
10. common GROUND medical WASTE WATER GATE keeper COLOR printer
11. daily DOUBLE diesel MOTOR CROSS COUNTRY music HAIR style
12. dark HORSE smoke SCREEN PLAY BILL fold DATE palm
13. day JOB film STRIP SEARCH ENGINE block HISTORY lesson
14. dish SOAP heart BEAT BOX LUNCH meat SPRING roll
15. drug STORE stone COLD FRONT DOOR bell DESK jockey
16. garlic BUTTER phone JACK KNIFE FIGHT club BLADE runner

This provides us with a word wall which looks like this:

----------------------------------------------  
|  FLAT    |  STREET  |  FINAL     |  PARTY  |  
----------------------------------------------  
|  SPACE   |  WISE    |  STRAIGHT  |  AREA   |  
----------------------------------------------  
|  CASE    |  WATER   |  CROSS     |  PLAY   |  
----------------------------------------------  
|  SEARCH  |  BOX     |  FRONT     |  KNIFE  |  
----------------------------------------------

This in itself can be resolved into the following four connecting groups:

WORD = Play (wordplay), Search (wordsearch), Cross (crossword), Final (the final word)
OFFICE = Space (Office Space), Box, (box office), Party (office party), Front (front office)
EDGE = Knife (knife-edge), Wise (edgewise), Straight (straight edge), Case (edge case)
SURFACE = Street (street surface), Water (surface water), Flat (flat surface), Area (surface area)

All of which connect with:

MICROSOFT - as the names of products made by the multinational tech company which is worth not just billions, but trillions even!

logical deduction - Nine Bottles of Wine, One Poisoned, Two Mice, Two Tests

Came across this puzzle and couldn't think of a solution:

You are given nine bottles of wine out of which one is poisoned. Given two mice (that will die when taken a sip of poisoned wine), how can you find out which bottle is the poisoned one by only performing two tests?

Any ideas?

Edit: I misunderstood the exact meaning of test. I thought that each “drink/sip” was a test. Thanks for all the answers.

Answer

Number the bottles 0 through 8; these can be represented as $2$ digit ternary numbers (example, $7=21_3$). Call the mice Alice and Bob.

• For the first experiment, have Alice drink from every bottle whose first ternary digit is $0$, and have Bob drink from those whose second ternary digit is $0$.


• For the second experiment, replace every $0$ in the previous sentence with $1$.

After the two experiments, the experiment during which Alice died (if any) tells you the first ternary digit of the poisoned bottle: if she died on the first experiment, it is $0$, on the second, it is $1$, if never, it is $2$. Similarly, Bob's lifespan tells you the second digit. Thus, afterwards you will know the bottle.

Throwing a micro black hole into the sun: does it collapse into a black hole or does it result in a supernova?

What do we know about accretion rates of micro black holes? Suppose a relative small black hole (mass about $10^9$ kilograms) would be thrown into the sun. Eventually this black hole will swallow all matter into the star, but how much time will pass before this happens?

Are there any circumstances where the black hole would trigger a gravitational collapse in the core, and result in a supernova?

There seems to be some margin for the accretion heating to counter or exceed the heating from fusion, so it could throw the star over the temperature threshold for carbon-12 fusion and above. The black hole is converting nearly 80% - 90% of the rest-mass of the accretion matter to heat, while fusion is barely getting about 0.5% - 1%.

Bonus question: Could this be used to estimate a bound on primordial micro black holes with the fraction of low-mass stars going supernova?

Answer

The micro black hole would be unable to accrete very quickly at all due to intense radiation pressure.

The intense Hawking radiation would have an luminosity of $3.6 \times 10^{14}$ W, and a roughly isotropic flux at the event horizon of $\sim 10^{48}$ W m$^{-2}$.

The Eddington limit for such an object is only $6 \times 10^{9}$ W. In other words, at this luminosity (or above), the accretion stalls as matter is driven away by radiation pressure. There is no way that any matter from the Sun would get anywhere near the event horizon. If the black hole was rotating close to the maximum possible then the Hawking radiation would be suppressed and accretion at the Eddington rate would be allowed. But this would then drop the black hole below its maximum spin rate, leading to swiftly increasing Hawking radiation again.

As the black hole evaporates, the luminosity increases, so the accretion problem could only become more severe. The black hole will entirely evaporate in about 2000 years. Its final seconds would minutely increase the amount of power generated inside the Sun, but assuming that the ultra-high energy gamma rays thermalised, this would be undetectable.

EDIT: The Eddington limit may not be the appropriate number to consider, since we might think that the external pressure of gas inside the Sun might be capable of squeezing material into the black hole. The usual Eddington limit is calculated assuming that the gas pressure is small compared with the radiation pressure. And indeed that is probably the case here. The gas pressure inside the Sun is $2.6 \times 10^{16}$ Pa. The outward radiation pressure near the event horizon would be $\sim 10^{40}$ Pa. The problem is that the length scales are so small here that it is unclear to me that these classical arguments will work at all. However, even if we were to go for a more macroscopic 1 micron from the black hole, the radiation pressure still significantly exceeds the external gas pressure.

Short answer: we wouldn't even notice - nothing would happen.

Bonus Question: The answer to this is it doesn't have a bearing on the supernova rate, because the mechanism wouldn't cause supernovae. Even if the black hole were more massive and could grow, the growth rate would be slow and no explosive nucleosynthesis would occur because the gas would not be dense enough to be degenerate.

Things change in a degenerate white dwarf, where the enhanced temperatures around an accreting mini-black hole could set off runaway thermonuclear fusion of carbon, since the pressure in a degenerate gas is largely independent of temperature. This possibility has been explored by Graham et al (2015) (thanks Timmy), who indeed conclude that type Ia supernova rates could constrain the density of micro black holes in the range $10^{16}$ to $10^{21}$ kg.

Difference between spinor and vector field

How do we distinguish spinors and vector fields? I want to know it in terms of physics with mathematical argument.

Answer

The answer to your question depends on the context, but the basic unifying theme distinguishing different kinds of fields (like vector fields, scalar fields, etc.) is how these fields transform when they are acted on by Lie Groups (and or Lie Algebras) which falls under the mathematical subject of representation theory of Lie groups and Lie algebras. Here are some examples to illustrate what I mean:

Consider a real field $V^\mu(x)$ defined on 4-dimensional Minkowski space $\mathbb R^{3,1}$. We say that this field is a Lorentz 4-vector field provided when it is acted on by a Lorentz transformation $\Lambda = (\Lambda^{\mu}_{\phantom\mu\nu})$ (an element of the Lorentz Group $\mathrm{SO}(3,1)$), it transforms as follows: $$ V^\mu(x) \to V'^\mu(x) = \Lambda^{\mu}_{\phantom\mu\nu} V^\nu(\Lambda^{-1}x) $$

Now Consider instead a field $\Psi^a(x)$ defined on 4-dimensional Minkowski space $\mathbb R^{3,1}$. We say that such a field is a Dirac Spinor provided when it is acted on by Lorentz transformations, it transforms as follows: $$ \Psi^a(x) \to \Psi'^a(x) = R_{\mathrm{dir}}(\Lambda)^a_{\phantom a b}\Psi^b(\Lambda^{-1}x) $$ where $R_{\mathrm{dir}}$ is called the Dirac spinor representation of the Lorentz Group.

There are also other kinds of spinors, all of whom transform according to different representations of different Lie groups.

riddle - Four Directions #6 - Where and what am I?

Inspired by metazen's riddles, I have decided to create my own

To my west are a bunch of crooks

To my north is a special fruit

To my east is somewhere everyone wants to be

To my south is a giant clock

I am in a collection room

Where am I?

Hint 1:

Look at the tags

pattern - Procedurally generated riddles

Last morning, my AI friend wrote me a letter.

hello thomas. how are you? i am good. you said you like riddles so I made some riddles for you. please enjoy.

After some time, though I enjoyed the riddles, I still wasn't sure about my answers. So I asked my AI friend to send me a list of answers, so I could check with my own. He instantly answered with two letters.

1. A rooster


2. A person that can't see in the dark


3. A werewolf


4. A mutant who is also a good surgeon


5. A mutant


6. A circus bear that escaped back into the wild

...

13. A slightly better fighting machine


14. A good fighting machine


15. A fighting machine that gets upgrades for good performance


16. A veteran cat that is still at war


17. A veteran cat


18. A veteran cat with a bad memory

...

25. A reverse-werewolf


26. A cat that stumbled in the dark


27. A cat

As you can see, this letter didn't have all the answers. But I soon got another.

sorry, i happened to have lost the answers to other puzzles. but i happened to save them in my cache. hope you enjoy.

? A bad fighting machine

? A bear pretending to be a human

? A cat at a hell of a war

? A cat at war

? A cat with a heat-sensitive paw

? A person who is too good at pretending to be a bear

? A self-repairing fighting machine

? An easily frightened inventor of a cloning machine

? An ever-evolving fighting machine

? An inventor of a cloning machine

? Puss in the boots

Having made my count, I answered him

It seems like one answer is missing here?

To which he answered in his usual manner:

yes

---

The problem

Help me put the last 11 answers into proper order and answer the unanswered riddle!

Hint:

Please google "fighting machine". I mean, it's a thing.

Hint 2:

Inspect the breaking points: riddle 1, riddle 14 (middle), riddle 27. It's as simlpe as that. What's the difference between: a rooster, a good fighting machine, and a cat (in that order)?

Hint 3:

If you're into riddles, I assure you you've already heard this one!

Answer

Starting over after hint #3. This is based on the old riddle...

What has four legs in the morning, two legs in the day, and three legs in the evening?

The catch is that since this is an AI generating riddles...

It uses all possible number combinations.

Full list:

1. (2, 2, 2) A rooster
2. (2, 2, 3) A person that can't see in the dark
3. (2, 2, 4) A werewolf
4. (2, 3, 2) A mutant who is also a good surgeon

5. (2, 3, 3) A mutant
6. (2, 3, 4) A circus bear that escaped back into the wild
7. (2, 4, 2) A person who is too good at pretending to be a bear
8. (2, 4, 3) An easily frightened inventor of a cloning machine
9. (2, 4, 4) An inventor of a cloning machine
10. (3, 2, 2) A bad fighting machine
11. (3, 2, 3) A self-repairing fighting machine
12. (3, 2, 4) An ever-evolving fighting machine
13. (3, 3, 2) A slightly better fighting machine
14. (3, 3, 3) A good fighting machine

15. (3, 3, 4) A fighting machine that gets upgrades for good performance
16. (3, 4, 2) A veteran cat that is still at war
17. (3, 4, 3) A veteran cat
18. (3, 4, 4) A veteran cat with a bad memory
19. (4, 2, 2) A bear pretending to be a human
20. (4, 2, 3) ???
21. (4, 2, 4) Puss in the boots
22. (4, 3, 2) A cat at a hell of a war
23. (4, 3, 3) A cat at war
24. (4, 3, 4) A cat with a heat-sensitive paw

25. (4, 4, 2) A reverse-werewolf
26. (4, 4, 3) A cat that stumbled in the dark
27. (4, 4, 4) A cat

The missing one is the original riddle, whose answer is...

A man (four legs in infancy, two legs in adulthood, two legs plus a cane in old age).

visual - Which way is the bus going?

If I'm not mistaken, this is a children's puzzle, but it got me thinking for a while.

The bus is driving on the autobahn. Relative to you, which way is the bus going? Left or Right?

Answer

I saw this problem on a web page once.

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The initial thought that most people have is that you can't tell, because the bus is totally symmetrical. However, this misses one important detail that apparently children will imply in their minds more often than adults.

The canonical answer is that this picture of a bus is missing its door, and therefore the bus must be going left, because the door is on the right side of the bus and there's no door visible on the side that's facing you in the picture.

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Of course, as Gilles pointed out, depending on which country you're in, the bus could be driving on either side of the road. If you're in Britain or some of the Commonwealth countries, then the bus must be going right, because the door is on the left side of the bus.

And depending on where from the bus you're exiting, the door could be on the top or bottom as well. I've never seen a bus that had its wheels so far below the chassis like on this one, and maybe there's a stairwell that opens from the bottom to let people in and out.

Alternatively, it could be a maglev train given that the car appears to be floating above two round things. A maglev train has doors on both sides. In both of the above cases, you can't tell which way the vehicle is going without further information.

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The question was updated to indicate that the bus was driving on the Autobahn. However, a bus on the Autobahn sounds really unsafe, and so we might actually be viewing a tipped-over bus from a bird's eye view (it would certainly explain the detached wheels). In such a case, you still couldn't tell which way the bus was going unless you knew which side of the highway it was driving on.

quantum electrodynamics - How is the EM force exchanged over long distances?

The Situation

Imagine we place two charged objects a very far distance apart, essentially making them point charges. How does the EM force interact between the two point charges if virtual photons cannot exist for long periods?

What I know

According to particle theory, the EM force is "activated" by the exchange of virtual photons. According to QM and the Uncertainty Principle: ΔE*Δt > h (approx). Virtual photons can only come about if they exist for a very, very short amount of time because the error in the energy becomes very large, and thus the conservation of Mass and Energy is not violated. If we place the two point charges very far away, the time component becomes very large, thus making the energy component have a small deviation, narrowing it down to a specific energy.

Thought Experiment

Place two point charges 1 light year apart, and give the two points equal charges such that the total electrostatic force is 1 newton between them. In order for the force to be exchanged, the "virtual particles" must now live for 1 year (Δt = 3.15569e+7 seconds), giving each photon a very specific energy and thus a very specific wavelength. If we look at the space in between these two objects, will we see light in that void?

Tying it all together

Sorry for the block of text, but this is how I thought of this problem.

So in conclusion:

1. 
   

   Can the EM force be exchanged by real photons and NOT virtual photons?

   
   


2. 
   

   Can virtual photons change into real photons?

   
   
   


3. 
   

   Over long distances, will we see radiation from the photons coming from the electromagnetic force?

   
   

Answer

Virtual particles are not real. They come, as I've said in many answers on this site, from a naive interpretation of Feynman diagrams which should not be taken as an actual, exact description of how the physics works.

The actual description of an interaction in the quantum field theory is more complicated than "photons are exchanged". In particular, "virtual particles" are perturbative "objects", since they are associated to lines in a Feynman diagram, which are mnemonic devices to remember which integrals are to be taken in what fashion. They are not how a force is defined in the quantum field theory, and not its fundamental, exact mode of transferral (such a thing does not exist because the quantum notion of force is not easily connected with the classical notion of force).

Nevertheless, on long ranges and at low energies, we may recover the classical electromagnetism, and, for instance, Coulomb's law from the quantum field theory (and in particular from the diagram that means "a virtual photon is exchanged") as I briefly outline in this answer. Hence, it is actually perfectly feasible to speak of an "exchange of virtual photons" also at long distances, if you also speak of it at short distances (I would advise against using this terminology at all). But these still are virtual photons, not actual photons, so you won't see any radiation, because nothing detectable, as far as we know, is actually "exchanged". Real photons do not mediate the electromagnetic force, they are light.

energy - Can electromagnetic radiation (i.e. photons) produce gravity?

I don't want to play with physical laws in a frivolous way.

Assuming that the nature of matter and energy is the same, can a high density of highly energetic photons produce a gravity force?

We do know that radiation is affected by space-time distortions, or in another way "feels gravity". Why do photons can (or cannot) produce a gravity field?

Answer

First of all, photons are not a pure form of energy. They are particles without rest mass, which means that they travel at the speed of light. Energy is a property of physical systems, the statement that something is energy makes no sense.

To answer your question: yes, electromagnetic radiation/photons contribute/s to the curvature of spacetime and therefore to gravity. The Einstein equations of general relativity related the curvature of spacetime to the stress-energy tensor, which contains not only mass but also energy. This also includes the energy of a photon.

quantum electrodynamics - Calculating the ionization rate

here is a kind of silly question: in quantumn physics, we assume that the charge is reserved. So if we use a laser, which is strong enough to make ionization happen, to shot at the system, how can we calculate the decreasing rate of the electrons? Or the ionization rate can be calculated by other ways?

I will very appreciate for helping me understand this question!

solar system - Why doesn't dark matter affect planetary motion?

If the universe is made up of ~95% dark matter, and it interacts only gravitationally then why didn't Newton and Kepler discover it before ? Why does it show itself only in the radial velocity profile of stars in galaxies and not in that of planets around the Sun ?

special relativity - A sees B's clock running slow and B sees A's clock running slow?

This paradox is very common it seems, in which A sees B's clock running slow and B sees A's clock running slow. Here is the question a little more concretely.

Let's say B flies by A's spaceship. If B's clock advances $t$, then in that time A's clock advances $\gamma t$ through time dilation (This is easily shown if there is a light clock next to B). Now from A's perspective (assume there is a light clock next to A), during that same time interval, he sees B's clock run slower and thus this time interval must be $\gamma^2 t$. Equating the time intervals gives $t = \gamma^2 t$ and $\gamma = 1$, which is clearly wrong. How is this reconciled?

Answer

It's reconciled by the relativity of simultaneity. Each observer is assumed to measure the time of events using local readings on a network of clocks at rest relative to that observer, which are defined as "synchronized" in that observer's own rest frame using the Einstein synchronization convention. But because this convention is based on each observer assuming that light moves at a constant speed relative to themselves, different observers will disagree about simultaneity, each thinking the clocks of the other are out-of-sync.

So, if one of your clocks is moving at 0.6c relative to me, I might first measure your clock reading T=0 seconds as it passed a clock of mine also reading t=0 s, then later I might measure your clock reading T=20 s as it passed a clock of mine reading t=25 s, at a distance of 15 light-seconds to the right of my other clock according to my ruler. From this I conclude that your clock only ticked 20 seconds in 25 seconds of my time, meaning your clock was slowed down by a factor of 0.8--but this depends on the assumption that my two clocks were synchronized, which isn't true in your own frame. If two clocks have a distance of L between them in their rest frame and are synchronized in that frame, then from the perspective of another frame which sees them moving at speed v, at any given moment the clock in the rear will have a time that's ahead of the clock in the lead by an amount $vL/c^2$, so in this example where the two clocks are 15 light-seconds apart in my rest frame, and you see them moving at 0.6c, the rear clock will have a time that's ahead of the lead clock's time by (0.6 light-seconds/second)(15 light-seconds)/(1 light-second/second)^2 = 9 seconds. So in your frame, at the same moment your clock was reading T=0 s and passing my left clock reading t=0 s, my right clock (which is in the 'rear' from your point of view, since both are traveling left in your frame) read t=9 seconds. Then 20 seconds later in your frame, your clock passes my right clock, and in your frame it's my clocks that are running slow by a factor of 0.8, so after 20 seconds my clocks have only ticked forward by 16 seconds. But since the right clock started at t=9 seconds, then after ticking an additional 16 seconds it will have reached t=25 seconds. So this is how you can consistently explain the fact that when your clock passed my right clock, your clock read T=20 s and mine read t=25 s, even though my clocks were running slower than yours in your own frame.

Here are some diagrams showing in a more visual way how observers using their own rulers and clocks can each measure the other's rulers as length-contracted and the other's clocks as slowed-down and out-of-sync, in a completely symmetrical way. In this example, we have two rulers with clocks mounted on them moving alongside each other, and in order to make the math work out neatly, the relative velocity of the two rulers is (square root of 3)/2 * light speed, or about 259.628 meters per microsecond. This means that each ruler will observe the other one’s clocks tick exactly half as fast as their own, and will see the other ruler's distance-markings to be squashed by a factor of two. Also, I have drawn the markings on the rulers at intervals of 173.085 meters apart—the reason for this is again just to make things work out neatly, it will mean that observers on each ruler will see the other ruler moving at 1.5 markings/microsecond relative to themselves, and that an observer on one ruler will see clocks on the other ruler that are this distance apart (as measured by his own ruler) to be out-of-sync by exactly 1 microsecond, some more nice round numbers.

Given all this, here is how the situation would look at 0 microseconds, 1 microsecond, and 2 microseconds, in the frame of ruler A:

And here’s how the situation would look at 0 microseconds, 1 microsecond, and 2 microseconds, in the frame of ruler B:

Some things to notice in these diagrams:

1. 
   

   in each ruler's frame, it is at rest while the other ruler is moving sideways at 259.6 meters/microsecond (ruler A sees ruler B moving to the right, while ruler B sees ruler A moving to the left).

   
   


2. 
   

   In each ruler's frame, its own clocks are all synchronized, but the other ruler's clocks are all out-of-sync.

   
   


3. 
   

   In each ruler's frame, each individual clock on the other ruler ticks at half the normal rate. For example, in the diagram of ruler A’s frame, look at the clock with the green hand on the -519.3 meter mark on ruler B--this clock first reads 1.5 microseconds, then 2 microseconds, then 2.5 microseconds. Likewise, in the diagram of ruler B’s frame, look at the clock with the green hand on the 519.3 meter mark on ruler A—this clock also goes from 1.5 microseconds to 2 microseconds to 2.5 microseconds.

   
   


4. 
   
   

   Despite these differences, they always agree on which events on their own ruler coincide in time and location with which events on the other. If you have a particular clock at a particular location on one ruler showing a particular time, then if you look at the clock right next to it on the other ruler at that moment, you will get the same answer to what that other clock reads and what marking it’s on regardless of which frame you’re using. Here’s one example:

   
   

You can look at the diagrams to see other examples, or extrapolate them further in a given direction then I actually drew them, and to later times, to find even more. In every case, two events which coincide in one reference frame also coincide in the other.

rotational dynamics - Instantaneous axis of rotation of a rigid body

For the description of rigid body motion, any point $O$ of the rigid body could be taken as reference, since the velocity of a generic point $P$ can be written in function of the angular velocity $\Omega$ and of the velocity of $O$, independently of the choice of $O$.

$$\dot{P} = \dot{ O} + \Omega∧(P −O)\tag{1}$$

That means that the motion of $P$ is seen as the composition of the traslation of $O$, plus a rotational motion about an axis passing through $O$: let's call this axis $\gamma$.

In which cases it is correct to say that $\gamma$ is an instantaneous axis of rotation of the rigid body?

In order to do that must the point $O$ (on the axis $\gamma$) have zero velocity (i.e. $\dot{O}=0$)? Or can I define $\gamma$ as an instantaneous axis of rotation in any case when I write down $(1)$?

Answer

A body can have parallel velocity on the instantaneous axis of rotation. This parallel velocity is sometimes designated with a scalar pitch value $h$, such that $\vec{v}_O = h \vec{\omega}$

Consider a point O on the IAR and a point P outside of it. You have

$$ \vec{v}_P = \vec{v}_O + (\vec{r}_P-\vec{r}_O) \times \vec{\omega} $$

Now I can prove that given the motion $\vec{v}_P$ of an arbitrary point P and the rotational velocity $\vec{\omega}$. The motion can always be decomposed into an axis of rotation point $\vec{r}_O$ and a parallel velocity $\vec{v}_O$.

Lemma Given $\vec{v}_P$ and $\vec{\omega}$ there exists a unique (relative) location $\vec{r}$ such that $$ \vec{v}_O = \vec{v}_P + \vec{\omega}\times \vec{r} = h \vec{\omega} $$ so the velocity $\vec{v}_O = h \vec{\omega} $ is parallel to $\vec{\omega}$ only. This location is on the instantaneous axis of rotation closest to P such that $\vec{r}_O = \vec{r}_P + \vec{r}$.

Proof

Take $\vec{r}= \frac{\vec{\omega} \times \vec{v}_P}{\| \vec{\omega}\|^2}$ in the transformation and expand out the terms

$$ \vec{v}_O = \vec{v}_P + \vec{\omega}\times \left( \frac{\vec{\omega} \times \vec{v}_P}{\| \vec{\omega}\|^2} \right)$$

Now use the vector triple product identity $a \times (b \times c) = b (a\cdot c) - c (a \cdot b)$

$$\require{cancel} \vec{v}_O =\vec{v}_P +\frac{\vec{\omega}(\vec{\omega}\cdot \vec{v}_P) - \vec{v}_P (\vec{\omega}\cdot \vec{\omega})}{\| \vec{\omega}\|^2} = \cancel{\vec{v}_P} + \frac{\vec{\omega}(\vec{\omega}\cdot \vec{v}_P)}{\| \vec{\omega}\|^2}-\cancel{\vec{v}_P} $$

Now define the scalar pitch as $$h = \frac{\vec{\omega} \cdot \vec{v}_P}{\| \vec{\omega} \|^2}$$ and the above becomes $$\vec{v}_O = h \vec{\omega}$$

So the velocity on O is parallel to the rotation only.

Example A body rotates by $\vec{\omega} = (0,0,10)$ and a point P located at $\vec{r}_P=(0.8,0.2,0)$ has velocity $\vec{v}_P = (2,-3,1)$. Find the IAR point O and pitch value.$h$.

$$ \vec{r}_O = \vec{r}_P + \frac{\vec{v}_P \times \vec{\omega}}{\| \vec{\omega} \|^2} = (0.8,0.2,0) + \frac{(-30,20,0)}{10^2} = (0.5,0,0)$$

$$h = \frac{\vec{\omega} \cdot \vec{v}_P}{\|\vec{\omega}\|^2} = \frac{10}{10^2} = 0.1$$

Confirmation

$$\vec{v}_P = h \vec{\omega} + (\vec{r}_P - \vec{r}_O) \times \vec{\omega} = (0,0,1) + (0.3,0.2,0) \times (0,0,10) = (2,3,-1) \; \checkmark $$

_{NOTES: See also related answer to question Why is angular velocity of any point about any other point of a rigid body always the same?}