big bang - Could Hyper-Massive Black Holes be due to Dark Matter in the Early Universe?

An interesting discussion started here: Is there a limit as to how fast a black hole can grow?

I am curious if Thompson Scattering and Eddington Luminosity have the same effect on Dark Matter (or alternatively, weakly-interacting massive particles) as they do with purely ionized hydrogen and other (known) particles that radiate (or can radiate due to decay, energy, friction, etc..).

Additionally, given dark matter's W.I.M.P.-like nature, would it not be possible that in the early universe where dark matter is believed to have materialized parallel to normal and anti-matter, and given its gravitational properties, could dark matter have contributed to the universe's earliest (and largest) black holes, and in turn explain black holes being larger than previously thought possible in a 13.7~ Billion Year history?

The properties of Dark Matter would render it virtually unaffected by the velocity and motion of the accretion disk. Since we are not yet sure if dark matter has a charge, it may also be unaffected by the black hole's electromagnetic field. It just pours into the event horizon virtually unhindered.

I do not expect an authoritative answer, as the only person who could provide such an answer would have earned a nobel prize already for dark matter's discovery (and thus be far too important to answer such a question), but a hypothetical answer based on what we do know about dark matter would be sufficient.

quantum mechanics - To calculate the correlation functions of an XX spin chain, Wick's theorem is used. But is it valid for a chain of any size?

The correlation functions found in Barouch and McCoy's paper (PRA 3, 2137 (1971)) for the XX spin chain use a method which uses Wick's theorem. For the zz correlation function, this gives

$\langle \sigma_l^z \sigma_{l+R}^z \rangle = \langle \sigma_l^z \rangle^2 - G_R^2$

where for $R=1$, $G_1 = -\langle \sigma_l^x \sigma_{l+1}^x+ \sigma_l^y \sigma_{l+1}^y \rangle/2$.

If I calculate $\langle \sigma_l^z \sigma_{l+1}^z \rangle$ both explicitly and using the equation above for 8 qubits, I get different answers.

So is Wick's theorem still valid for 8 qubits, which means I've just made a mistake? Or is it valid only in the thermodynamic limit?

Thanks

Edit:

Thanks for your replies everyone. @lcv However, I haven't used the analytical diagonalisation for this - I have simply used Mathematica to diagonalise the 8 qubit chain numerically after substituting arbitrary values for the coupling strength, magnetic field and temperature. Hence it can't be an error in the diagonalisation. It is the thermal average I have calculated, that is $\langle \sigma^z_l \rangle=tr(\rho \sigma^z_l )$ where $\rho=e^{−H/T}/tr(e^{−H/T})$ and T is temperature. But in doing this, I find that $\langle \sigma^z_l \sigma^z_{l+R} \rangle \neq \langle \sigma^z_l \rangle^2 - G_1^2$ where I've defined $G_1$ above.

Edit2 (@marek @lcv @Fitzsimons @Luboš) I'm going to try to clarify - The open XX Hamiltonian in a magnetic field is

\begin{equation} H=-\frac{J}{2}\sum_{l=1}^{N-1} (\sigma^x_l \sigma^x_{l+1} + \sigma^y_l \sigma^y_{l+1})- B \sum_{l=1}^N \sigma^z_l \end{equation}

In Mathematica, I have defined the Pauli spin matrices, then the Hamiltonian for 8 qubits. I then put in values for $J$, $B$ and $T$, and calculate the thermal density matrix,

\begin{equation} \rho = \frac{e^{-H/T}}{tr(e^{-H/T})} \end{equation}

So now I have numerical density matrix. I then calculate $\langle \sigma^z_l \sigma_{l+1}^z \rangle=tr(\rho \sigma^z_l \sigma_{l+1}^z )$ using the definitions of the Pauli spin matrices and $\rho$.

Next I calculate $\langle \sigma_l^z \sigma_{l+R}^z \rangle$ using the result from Wick's theorem which gives $\langle \sigma_l^z \rangle^2 - G_R^2$ where for $R=1$, $G_1 = -\langle \sigma_l^x \sigma_{l+1}^x+ \sigma_l^y \sigma_{l+1}^y \rangle/2$. I again use the Pauli spin matrices I defined and the same numerical $\rho$ to calculate them.

But I get a different (numerical) answer for each of these.

quantum field theory - What does Weinberg–Witten theorem want to express?

Weinberg-Witten theorem states that massless particles (either composite or elementary) with spin $j > 1/2$ cannot carry a Lorentz-covariant current, while massless particles with spin $j > 1$ cannot carry a Lorentz-covariant stress-energy. The theorem is usually interpreted to mean that the graviton ( $j = 2$ ) cannot be a composite particle in a relativistic quantum field theory.

Before I read its proof, I've not been able to understand this result. Because I can directly come up a counterexample, massless spin-2 field have a Lorentz covariant stress-energy tensor. For example the Lagrangian of massless spin-2 is massless Fierz-Pauli action:

$$S=\int d^4 x (-\frac{1}{2}\partial_a h_{bc}\partial^{a}h^{bc}+\partial_a h_{bc}\partial^b h^{ac}-\partial_a h^{ab}\partial_b h+\frac{1}{2}\partial_a h \partial^a h)$$

We can calculate its energy-stress tensor by $T_{ab}=\frac{-2}{\sqrt{-g}} \frac{\delta S}{\delta g^{ab}}$, so we get

$$T_{ab}=-\frac{1}{2}\partial_ah_{cd}\partial_bh^{cd}+\partial_a h_{cd}\partial^ch_b^d-\frac{1}{2}\partial_ah\partial^ch_{bc}-\frac{1}{2}\partial^ch\partial_ah_{bc}+\frac{1}{2}\partial_ah\partial_bh+\eta_{ab}\mathcal{L}$$ which is obviously a non-zero Lorentz covariant stress-energy tensor.

And for U(1) massless spin-1 field, we can also have the energy-stress tensor $$T^{ab}=F^{ac} F^{b}_{\ \ \ c}-\frac{1}{4}\eta^{ab}F^{cd}F_{cd}$$ so we can construct a Lorentz covariant current $J^a=\int d^3x T^{a 0}$ which is a Lorentz covariant current.

Therefore above two examples are seeming counterexamples of this theorem. I believe this theorem must be correct and I want to know why my above argument is wrong.

Answer

1. 
   

   The stress tensor for $h_{ab}$ is not Lorentz covariant, despite the fact that it looks like it is. This is because $h_{ab}$ itself is not a Lorentz tensor. Rather under Lorentz transformations $$ h_{ab} \to \Lambda_a{}^c \Lambda_b{}^d h_{cd} + \partial_a \zeta_b + \partial_b \zeta_a ~. $$ The extra term is present to make up for the fact that $h_{ab}$ is not a tensor of the Lorentz group. Plug this into the stress tensor and you will find that the stress tensor also transforms with a inhomogeneous piece thereby making it non-covariant.

   
   


2. 
   

   The photon is not charged under the $U(1)$ gauge symmetry. Thus, its $U(1)$ current is zero. The current you have defined is not the $U(1)$ current. Rather it is the current corresponding to translations. Weinberg-Witten theorem has nothing to say about this current.

   
   

electrostatics - What are the advantages of multipole expansion of potentials?

When I see the equations of multipolar expansions they "look" to me harder than the original expressions. For example:

Multipole expansion - spherical form, in Wikipedia

I bet that this is not the case, but I want to know why. For example, when used as integrands, are these expressions advantageous for analytical or for numerical integration?

My main concern is about its usage in quantum mechanics (specifically about magnetic multipole expansions). I found many references about the math involved but I can't realize why it is useful. I've read the reference:

Ideas of quantum chemistry, Lucjan Piela (Elsevier, 2014). Appendix X; Google Books link

but it is not clear to me "why there is a loss of the accuracy" (as stated in the first page) by using the original expression and how the use of multipole expansion avoid that.

It seems to me that the advantage is in a interchange of variables but it is not clear to me.

Anticipated thanks to any help.

Answer

To me, the main advantages of multipolar expansions are typically conceptual ones, in that they help encapsulate and separate the effects of the shape of the charge distribution and the position of the test charge on the electrostatic interaction between the two.

Consider, for example, the interaction between a test charge at position $\mathbf r$ and a charge distribution $\rho(\mathbf r)$ that's localized inside a sphere of radius $R$ around the origin. The electrostatic potential for such a situation is given, in general, by the integral of the Coulomb kernel over the charge cloud, $$ V(\mathbf r)=\int_\Omega\frac{\rho(\mathbf r')\mathrm d\mathbf r'}{|\mathbf r-\mathbf r'|}. \tag 1 $$ This is a bit of a big, messy object, because it requires you to carry out a long, complicated three-dimensional numerical integral every time you want to know the value of $V(\mathbf r)$.

So how does a multipolar expansion help? I'm going to skip all the details of how we obtain it, but it's a pretty standard result that when $r>R$ you can express the potential in $(1)$ as $$ V(\mathbf r) =\int_\Omega\frac{\rho(\mathbf r')\mathrm d\mathbf r'}{|\mathbf r-\mathbf r'|} =\sum_{l=0}^\infty\sum_{m=-l}^l Q_{lm}\frac{Y_{lm}(\theta,\varphi)}{r^l} \tag 2 $$ where $$ Q_{lm}=\frac{4\pi}{2l+1}\int Y_{lm}^*(\theta',\varphi')r'^l\rho(\mathbf r')\mathrm d\mathbf r' \tag 3 $$ are the multipole moments of the system. At a first glance this looks about as ugly as before, and I've only traded the messy integral for a messy series, which I will still need to sum numerically whenever I want to calculate $V(\mathbf r)$.

The point here is that in multipolar series like $(2)$, it is typically only a few terms that contribute significantly, and this means that I only need to calculate a few of the multipole moments. This is important! It means I can precompute only a few of the full three-dimensional numerical integrals in $(3)$, and still have an excellent grasp of the potential.

The precomputation point is the important one. I have managed to isolate the few quantities of the charge distribution - the multipole moments $Q_{lm}$ - that control the weight of the different multipole components of the field, and after that all I need to do is calculate a few potentials (which are both numerically and conceptually simple) and superpose them to get the full interaction.

---

In addition to needing to calculate fewer integrals, these integrals tend to also be simpler and more benign numerically, and this ties in with the concern voiced by Piela in the book you link to. In particular, the integrals in $(1)$ often involve lots of terms that very nearly cancel out, which means that you need very high accuracy in both $a$ and $b$ to get only mediocre accuracy in $a-b$.

To ground this in an example, consider two charges $\pm q$ on the $z$ axis at $\pm \delta=1\:\mathrm{mm}$, and their interaction with a test particle on the same axis at $z=1\:\mathrm{km}$. If you want to calculate the electrostatic potential there, the equivalent of the integral in $(1)$ is to just add the two contributions directly: \begin{align} V &=\frac{q}{z+\delta}-\frac{q}{z-\delta} =\frac{q}{1000.001\:\mathrm m}-\frac{q}{999.999\:\mathrm m} \\&=\frac{q}{\mathrm{m}}\left(0.00099999900000999\cdots-0.00100000100000100\cdots\right) \\&\approx -2\times 10^{-9}\frac{q}{\mathrm{m}} \end{align} This means in particular that I need all nine decimals on each of the individual potentials - five significant figures - to get a single significant figure of accuracy in the result. If this is the only way to do things then it's mostly a frown-and-bear it sort of situation, but if I want to calculate the potential in a bunch of different positions then there will be a lot of frowning going on, particularly if instead of simple sums I need to be doing high-accuracy numerical integrals all the time, to get only a couple of significant figures in my potential.

In contrast, if I first calculate the dipole moment of the system and use the dipole field, I know that $$Q_{q,z}=q\delta-q(-\delta)=2q\delta=2q\times 1\:\mathrm{mm},$$ so I get the same single significant figure of accuracy for $V$ while using only a single significant figure of accuracy for $z$ and $\delta$. This kind of numerical tit-for-tat, where the level of desired accuracy in the result is well matched by the level of required accuracy in the calculation, is the hallmark of a desirable numerical scheme. As a bonus, I know that the next term up in the multipole series is on the order of $\delta/z\approx 10^{-6}$ with respect to the dipole term, so if you want a more accurate result then for the first five significant figures or so you know you should concentrate on the dipole term, and it's only after that that you need to include higher-order terms.

---

This leads me to a separate conceptual advantage of multipolar expansions, and it is the separation of scales that's present in $(2)$: each term up the ladder scales as a higher power of $1/r$ than its predecessor; moreover, the lower-$l$ terms tend to have a very 'fuzzy' angular structure, and it's only in higher $l$ that the angular detail comes in. This is the way to make precise the observation that, from far away, a charged system is pretty much indistinguishable from a single ball of charge, and that the details of the distribution of charge only become important as you close in.

This establishes a hierarchy of approximations that is very important in terms of how we conceptualize and apply electrodynamics in a broad range of contexts - from electrodynamics in continuous media to the interaction of atoms and molecules with radiation to the description of intermolecular forces to the focusing optics of electron microscopes and accelerators to a host of other settings. In all of these settings the separation of electromagnetic interactions into different multipolar orders brings conceptual clarity into the relative importance of different parts of the interaction, and that's its main value.

logical deduction - Self Referential Puzzle - medium 2

Does anyone have a solution for this puzzle?

I keep getting contradictions. If anyone knows the answer, I would be happy to see what logic failures I am making.

1. The first question whose answer is D is the question
A.  8
B.  7

C.  6
D.  5
E.  4

2. Identical answers have questions
A.  3 and 4
B.  4 and 5
C.  5 and 6
D.  6 and 7
E.  7 and 8


3. The number of questions with the answer E is
A.  1
B.  2
C.  3
D.  4
E.  5

4. The number of questions with the answer A is
A.  1

B.  2
C.  3
D.  4
E.  5

5. The number of questions with the answer A equals the number of questions with the answer
A.  A
B.  B
C.  C
D.  D

E.  none of the above

6. The last question whose answer is B is the question
A.  5
B.  6
C.  7
D.  8
E.  9

7. Alphabetically, the answer to this question and the answer to the following question are

A.  4 apart
B.  3 apart
C.  2 apart
D.  1 apart
E.  the same

8. The answer to this question is the same as the answer to the question
A.  1
B.  2
C.  3

D.  4
E.  5

9. The number of questions whose answers are consonants
A.  3
B.  4
C.  5
D.  6
E.  7


10. The answer to this question is
A.  A
B.  B
C.  C
D.  D
E.  E

optics - Optimal laser wavelength for heating air

Lets say I want to heat air with laser,what wavelenght should I chose,ultraviolet,infrared or something in visible spectrum? To clarify,I want the laser beam to lose power and get converted to heat in shortest amount of distance possible. I am looking for maximum absorbtion, to convert the laser into heat.

When I think about it,two different wavelenght may produce equal heating of air,but the thing is one that have short range will produce heating that is more concentrated in space while other will heat air over longer distance so the energy will be spread over greater amount of air,I want that short range concetrated type heating so minimum quantity is heated to high as possible peak temperature.

story - Dance a Jigsaw - Clue Fifteen

_{<<---First clue
<---Previous clue}

---

Now that you had your clue, you have a feeling that it might be time to get out. You look around for a door, and... don't see one. However, you do see something that you hadn't noticed before - a rectangle with pieces lying next to it. You go to take a closer look.

On closer inspection, it is revealed to be a jigsaw puzzle. However, it doesn't have the regular edges - all of them have smooth edges. It looks like this will take a while...

---

_{Next clue--->}

Answer

It looks like this ->

Image Source

And here's the label of the Jigsaw Drink

Text -> "Old Dutch Apple Cider Vinegar"

So the Clue is

Vinegar

waves - What is the difference between phase difference and path difference?

The path difference is the difference between the distances travelled by two waves meeting at a point. Given the path difference, how does one calculate the phase difference?

movies - Shakespeare. William Shakespeare

Can you determine these popular quotes as retold by Shakespeare? Before anyone points out, yes I'm aware that Bill used predominantly iambic pentameter, but I've never been a huge fan of that, so I'm going with iambic tetrameter instead.

1.

How oft the potion, orange in hue
delights my sense among the dew.

2.

Indeed this land can scarce contain
Our presence should we both remain.

3.

From first I heard thine mouth's sweet spell
A victim helpless I befell.

4.

This vessel shall contain our dirge
Should storm or monstrous beast resurge.

5.

His stature lacking to impress
Beseech you greet him none the less.

Answer

Here's a shot. All but the 2nd are on this nifty Wikilist:

How oft the potion, orange in hue
delights my sense among the dew.

"I love the smell of napalm in the morning." -- Apocalypse Now

Indeed this land can scarce contain
Our presence should we both remain.

"This town ain't big enough for the both of us", -- The Western Code My thanks to dperry!

From first I heard thine mouth's sweet spell
A victim helpless I befell.

"You had me at 'hello.'" -- Jerry Maguire

This vessel shall contain our dirge
Should storm or monstrous beast resurge.

"You're gonna need a bigger boat." -- Jaws

His stature lacking to impress
Beseech you greet him none the less.

"Say 'hello' to my little friend!" -- Scarface

And for flavor, the title references...

"Bond. James Bond" -- first said in Dr. No

riddle - Powers of two my children be

Powers of two my children be,
Arriving in birth years separately.

Their sum is now the reverse of me
And the years between me and a power of three.

My digit difference is the number you see
And my digits are powers of binary.

And now the question I ask of thee:
Their number and age, and the age that I be.

A little riddle I came up with for my sons birthday.

Answer

Powers of two my children be,

Arriving in birth years separately.

Unspecified number of children but no twins and their age is 2^n.

Their sum is now the reverse of me

Their sum can be any number since any number can be expressed as a sum of powers of two. Your age can be written 10d + u and the sum of their age 10u + d.

And the years between me and a power of three.

10d + u - 3^x = 10u + d or 3^x - (10d + u) = 10u + d
Therefore 3^x = 9(u - d) or 3^x = 11(u + d)
The latter has no solution but the former has many solutions: (u - d) just needs to be a power of three

My digit difference is the number you see

The difference is the number of children: you have either three or nine children.

And my digits are powers of binary.

There are four possible digits: 1, 2, 4 and 8
If their difference is a power of three (= 3), they can only be 1 and 4.

And now the question I ask of thee:
Their number and age, and the age that I be.

You are 41 and the sum of their age is 14 (1110 in binary). Therefore you have three children and they are 8, 4 and 2 years old.

geometry - Triangle in a circle

Suppose three points are chosen at random in a circle. A triangle is made with these three points as vertices.

What's the probability that the triangle contains the origin of the circle?

(Although I have an answer, I thought I would throw it out there and see what others make of it!)

Answer

First, observe

That the triangle does not contain the center if and only if all three points are contained in a half-circle. So, only the polar angles of the points matter, not their distance to the center.

Next, let's

Label the points $A$, $B$, $C$. If we starting at point $A$ and go clockwise for half a circle, there's a $1/4$ chance that both point $B$ and point $C$ are in that half-circle, putting all the points on a half-circle. Likewise for starting at point $B$ and starting at point $C$.

These events are

Disjoint (well, overlap with probability 0), and comprise all the way the three points can be on a half-circle, since there's always exactly one for the other two are within half a circle clockwise. So, the probability of them being on a half-circle is $$3/4 = 1/4 + 1/4 + 1/4 $$and so the probability the triangle contains the center is the complement $$1 - 3/4 = 1/4$$ By the same argument, if we generalize to choosing $n$ points and connecting them in a polygon in cyclic order of polar angle, the probability that it contains the center is $$ 1 - \frac{n}{2^{n-1}}$$.

notation - Higgs mechanism in QED

I'm trying to understand the Higgs mechanics. For that matter, I'm exploring the possibility of giving mass to the photon in a gauge-invariant way. So, if we introduce a complex scalar field:

$$ \phi=\frac{1}{\sqrt{2}}(\phi_{1}+i\phi_{2}) $$

with the following Lagrangian density (from now on, just Lagrangian)

$$ \mathcal{L}=(\partial_{\mu} \phi)^{\star}(\partial^{\mu} \phi)-\mu^2(\phi^{\star}\phi)+\lambda(\phi^{\star}\phi)^2$$

and $\mu^{2}<0$.

We note that the potential for the scalar particle has an infinity of vacuums all of them in a circle of radius $v$ around (0,0). We introduce two auxiliary fields $\eta,\xi$ to express the perturbations around the vacuum

$$ \phi_0=\frac{1}{\sqrt{2}}[(v+\eta)+i \xi ]$$

Introducing the covariant derivative and the photon field, I have to compute the following thing

$$(D^{\mu} \phi)^{\dagger}(D_{\mu} \phi) $$

The derivatives included in $(D^{\mu}\phi)^{\dagger}$ are supposed to act upon the $(D_{\mu} \phi)$?

Answer

The answer is no. Just as in the case without a gauge field, it is just a product of two derivatives of the field $\phi$. You might be interested in the chapter "Scalar Electrodynamics" in Srednicki's book.

thermodynamics - Irreversible expansion and time reversal symmetry

Suppose there are N non-interacting classical particles in a box, so their state can be described by the $\{\mathbf{x}_i(t), \mathbf{p}_i(t) \}$. If the particles are initially at the left of the box, they can eventually occupy the whole box according to the Newtons law. In this case, we call the gas expand and this process is irreversible.

Nevertheless, Newtons laws also have time-reversal symmetry, so we should be able to construct an initial condition such that particles occupy the whole box (i.e. not all particles in the left chamber) will all move to the left chamber. Here are the simple questions:

1. 
   

   How to select the initial condition $\{\mathbf{x}_i(0), \mathbf{p}_i(0) \}$ if you can solve the set of equation of motion.

   
   


2. 
   

   Why the gas expansion is irreversible even though you can select the initial condition above.

   
   

Answer

1/ OK, let's start from an initial condition where all the particles are made to fit a tiny little corner of the room and their initial velocities are chosen randomly, according to a Maxwell-Boltzmann distribution for instance. As we let the system evolve, the gas will expand, that is true because it corresponds to the behaviour the Maxwell-Boltzmann predicts, that of an ideal gas near equilibrium and therefore the gas will try to fill the room.

Now, suppose at some moment we stop the evolution of the system, for instance after the gas has filled the entire room. Now, if we reverse all the velocities of the particles but do not change their positions, it is clear that by construction, we have found a solution that will evolve backwards to the corner where the gas was originally. So, it is indeed possible to find an initial configuration and in fact several initial configurations that satisfy your requirement. (But it takes a Maxwell Demon to realize them in practice.)

2/ Now, why doesn't that invalidate the macroscopic irreversibility of the system? Well, we should look at how many microstates corresponding to the macrostate "box entirely filled" do go back to the corner. Say the box has volume $V$. Now, assuming we have $N$ particles, the total number of microscopic configurations compatible with the box filled will be proportional to $V^N$. (I am omitting the velocities in this analysis, they will only make the analysis less simple to follow.)

How many of these configurations go back to the corner? Well, to compute that, remember that we can easily construct the solutions which go back to the corner by taking the configurations that start from the corner and expand to fill the entire room and then reverse the velocities. So, let's say we call the volume of the corner $W$. So the amount of trajectories that go back to the corner will be proportional to $W^N$. So, the fraction of trajectories that go back to the corner among the trajectories that fill the entire volume will be $W^N/V^N$. Since $W/V<1$ and $N$ is a large number, of the order of the Avogadro number, you can see why you never actually observe these trajectories in real life.

But, if you make your system small enough, less than 10 particles, and your corner is a half of the room, then the probability is about $1/2^{10} \approx 0.001$, I'd say it is worth the wait. ;p

OK, I wanted to add something to this explanation. The apparant paradox that is proposed in the OP is originally attributed to Loschmidt. The paradox Marek is talking about which is related to the ergodicity of the system is due to Zermelo, they are different paradoxes and require different answers.

electromagnetism - Definition of magnetic field

Can we define magnetic field at a point as:

Force on a point magnetic north pole at that point divided by its pole strength.

Anything wrong in this definition?

(The concept of point magnetic pole is an idealization as point magnetic poles don’t exist in nature and also there are no magnetic monopoles observed in nature.)

Question edited:

So does the above definition mean $\vec{B}$ or $\vec{H}$?

I think my definition does not represent $\vec{H}$ because we have not divided $\mu_0$ anywhere. So it represents $\vec{B}$ in free space. Am I correct?

Answer

Your definition is exactly that of magnetic field strength, H, on the old cgs system. The oersted was its unit, a magnetic field strength of 1 dyne per unit pole. [A unit pole was such that if two unit poles were placed 1 cm apart in vacuo, there would be a force of 1 dyne between them.]

The study of electromagnetism was built on this foundation, and produced the edifice of electromagnetic theory that we have today – even though we may now choose to define things differently. So one answer to your question would be that there's nothing wrong with the 'per unit pole' definition. It has delivered the goods.

Another view is that it's odd to use the non-existent monopole as the basis for electromagnetism. In fact the subject used to be taught using long ball-ended (dumb-bell shaped) magnets. The ball at the 'North' (or South) end produced a magnetic field in the surrounding air radially outwards (or inwards) varying with an inverse square law. So the balls seemed to be behaving as monopoles. But there's a very important caveat… The net magnetic flux from the North pole ball (unlike from a North monopole) is zero! This is because as much flux approaches the ball through the 'bar' of the magnet as leaves through the air, and as much flux leaves the South pole ball through the bar of the magnet as enters through the air. One of the consequences is that we have to be especially careful studying magnetic materials using this approach. Arguably an approach based on Ampère's current loops or the corresponding quantum concept is less contrived.

situation - Fire, guns or lions?

A murderer is condemned to death.
He has to choose between three rooms.
The first is full of raging fires.
The second is full of assassins with loaded guns.
The third is full of lions that haven't eaten in 3 years.

Which room is safest for him?

(Didn't make it up myself, but thought you might like it)

Answer

Obviously its

The lions who hasn't eaten in three years as it would be dead by now

quantum mechanics - Hilbert space vs. Projective Hilbert space

Hilbert space and rays:

In a very general sense, we say that quantum states of a quantum mechanical system correspond to rays in the Hilbert space $\mathcal{H}$, such that for any $c∈ℂ$ the state $\psi$ and $c\psi$ map to the same ray and hence are taken as equivalent states.

1. 
   

   How should one interpret the above in order to understand why $\psi$ and $c\psi$ are the same states? Clearly for $c= 0$ it doesn't hold, and for $c=1$ it is trivial, but why should this equivalence hold for any other $c$?

   
   
   


2. 
   

   Knowing Hilbert space is a complex vector space with inner product, is ray just another way of saying vectors?

   
   


3. 
   

   In the case that $c$ just corresponds to a phase factor of type $e^{i\phi}$ with $\phi \in \mathbb{R},$ then obviously $|\psi|=|e^{i\phi}\psi|,$ i.e. the norms didn't change, then what is the influence of $e^{i\phi}$ at all? In other words what does the added phase of $\phi$ to the default phase of $\psi$ change in terms of the state of the system?

   
   

Projective Hilbert space:

Furthermore, through a process of projectivization of the Hilbert space $\mathcal{H},$ it is possible to obtain a finite dimensional projective Hilbert space $P(\mathcal{H}).$ In the Projective Hilbert space, every point corresponds to a distinct state and one cannot talk in terms of rays anymore.

4. 
   

   What does such projectivization entail in a conceptual sense? I guess in other words, how are rays projected to single points in the process? and what implies the distinctness? Is such process in any way analogous to the Gram–Schmidt process used to orthonormalise a set of vectors in linear algebra?

   
   


5. 
   

   When one limits the Hilbert space to that of a certain observable of the system at hand, e.g. momentum or spin space (in order to measure the momentum and spin of a system respectively), does that mean we're talking about projective spaces already? (e.g. is the spin space spanned by up $\left|\uparrow\rangle\right.$ and down $\left|\downarrow\rangle\right.$ spins states of a system referred to as projective spin Hilbert space?)

   
   

The aim is to develop a better and clearer understanding of such fundamental concepts in quantum mechanics.

Answer

Why are states rays?

^{(Answer to OP's 1. and 2.)}

One of the fundamental tenets of quantum mechanics is that states of a physical system correspond (not necessarily uniquely - this is what projective spaces in QM are all about!) to vectors in a Hilbert space $\mathcal{H}$, and that the Born rule gives the probability for a system in state $\lvert \psi \rangle$ to be in state $\lvert \phi \rangle$ by

$$ P(\psi,\phi) = \frac{\lvert\langle \psi \vert \phi \rangle \rvert^2}{\lvert \langle \psi \vert \psi \rangle \langle \phi \vert \phi \rangle \rvert}$$

^{(Note that the habit of talking about normalised state vectors is because then the denominator of the Born rule is simply unity, and the formula is simpler to evaluate. This is all there is to normalisation.)}

Now, for any $c \in \mathbb{C} - \{0\}$, $P(c\psi,\phi) = P(\psi,c\phi) = P(\psi,\phi)$, as may be easily checked. Therefore, especially $P(\psi,\psi) = P(\psi,c\psi) = 1$ holds, and hence $c\lvert \psi \rangle$ is the same states as $\lvert \psi \rangle$, since that is what having probability 1 to be in a state means.

A ray is now the set of all vectors describing the same state by this logic - it is just the one-dimensional subspace spanned by any of them: For $\lvert \psi \rangle$, the associated ray is the set

$$ R_\psi := \{\lvert \phi \rangle \in \mathcal{H} \vert \exists c \in\mathbb{C}: \lvert \phi \rangle = c\lvert \psi \rangle \}$$

Any member of this set will yield the same results when we use it in the Born rule, hence they are physically indistiguishable.

Why are phases still relevant?

^{(Answer to OP's 3.)}

For a single state, a phase $\mathrm{e}^{\mathrm{i}\alpha},\alpha \in \mathbb{R}$ has therefore no effect on the system, it stays the same. Observe, though, that "phases" are essentially the dynamics of the system, since the Schrödinger equation tells you that every energy eigenstate $\lvert E_i \rangle$ evolves with the phase $\mathrm{e}^{\mathrm{i}E_i t}$.

Obviously, this means energy eigenstates don't change, which is why they are called stationary states. The picture changes when we have sums of such states, though: $\lvert E_1 \rangle + \lvert E_2 \rangle$ will, if $E_1 \neq E_2$, evolve differently from an overall multiplication with a complex phase (or even number), and hence leave its ray in the course of the dynamics! It is worthwhile to convince yourself that the evolution does not depend on the representant of the ray we chose: For any non-zero complex $c$, $c \cdot (\lvert E_1 \rangle + \lvert E_2 \rangle)$ will visit exactly the same rays at exactly the same times as any other multiple, again showing that rays are the proper notion of state.

The projective space is the space of rays

^{(Answer to OP's 4. and 5. as well as some further remarks)}

After noting, again and again, that the physically relevant entities are the rays, and not the vectors themselves, one is naturally led to the idea of considering the space of rays. Fortunately, it is easy to construct: "Belonging to a ray" is an equivalence relation on the Hilbert space, and hence can be divided out in the sense that we simply say two vectors are the same object in the space of rays if they lie in the same ray - the rays are the equivalence classes. Formally, we set up the relation

$$ \psi \sim \phi \Leftrightarrow \psi \in R_\phi$$

and define the space of rays or projective Hilbert space to be

$$ \mathcal{P}(\mathcal{H}) := (\mathcal{H} - \{0\}) / \sim$$

This has nothing to do with the Gram-Schmidt way of finding a new basis for a vector space! This isn't even a vector space anymore! (Note that, in particular, it has no zero) The nice thing is, though, that we can now be sure that every element of this space represents a distinct state, since every element is actually a different ray.¹

(Side note (see also orbifold's answer): A direct, and important, consequence is that we need to revisit our notion of what kinds of representations we seek for symmetry groups - initially, on the Hilbert space, we would have sought unitary representations, since we want to preserve the vector structure of the space as well as the inner product structure (since the Born rule relies on it). Now, we know it is enough to seek projective representations, which are, for many Lie groups, in bijection to the linear representations of their universal cover, which is how, quantumly, $\mathrm{SU}(2)$ as the "spin group" arises from the classical rotation group $\mathrm{SO}(3)$.)

OP's fifth question

When one limits the Hilbert space to that of a certain observable of the system at hand, e.g. momentum or spin space (in order to measure the momentum and spin of a system respectively), does that mean we're talking about projective spaces already? (e.g. is the spin space spanned by up |↑⟩ and down |↓⟩ spins states of a system referred to as projective spin Hilbert space?)

is not very well posed, but strikes at the heart of what the projectivization does for us: When we talk of "momentum space" $\mathcal{H}_p$ and "spin space" $\mathcal{H}_s$, it is implicitly understood that the "total space" is the tensor product $\mathcal{H}_p \otimes \mathcal{H}_s$. That the total/combined space is the tensor product and not the ordinary product follows from the fact that the categorial notion of a product (let's call it $\times_\text{cat}$) for projective spaces is

$$ \mathcal{P}(\mathcal{H}_1) \times_\text{cat} \mathcal{P}(\mathcal{H}_2) = \mathcal{P}(\mathcal{H}_1\otimes\mathcal{H}_2)$$

For motivations why this is a sensible notion of product to consider, see some other questions/answers (e.g. this answer of mine or this question and its answers).

Let us stress again that the projective space is not a vector space, and hence not "spanned" by anything, as the fifth question seems to think.

---

^{1The inquiring reader may protest, and rightly so: If our description of the system on the Hilbert space has an additional gauge symmetry, it will occur that there are distinct rays representing the same physical state, but this shall not concern us here.}

What does the solution to a wave equation represent?

What does the solution of a wave equation represent? When one finds a solution to a wave equation, how can one interpret this solution? Does the solution show the behavior of a wave?

Answer

For the usual wave equation,

$$\frac{\partial^2 u}{\partial t^2} - v^2 \nabla^2 u = 0$$

$u = u(x,t)$ is taken to be some property of the wave, at position $x$ and time $t$. For example, for the electromagnetic wave equation, $E(x,t)$ would be the force per unit charge at $(x,t)$.

If we had a rope, and a wave $u(x,t)$ propagated along the rope, then we could define $u(x,t)$ as the displacement of the infinitesimal piece of rope along it at position $x$, at the instant $t$.

general relativity - What is known about the topological structure of spacetime?

General relativity says that spacetime is a Lorentzian 4-manifold $M$ whose metric satisfies Einstein's field equations. I have two questions:

1. 
   

   What topological restrictions do Einstein's equations put on the manifold? For instance, the existence of a Lorentz metric implies some topological things, like the Euler characteristic vanishing.

   
   


2. 
   

   Are there any experiments being done or even any hypothetical experiments that can give information on the topology? E.g. is there a group of graduate students out there trying to contract loops to discover the fundamental group of the universe?

   
   

newtonian mechanics - Why does Friction act Parallel to the Surface (Microscopic Level)?

If we assume an object (blue layer) sliding towards the right across a surface (black layer). According to Newton's 3rd Law, the frictional force would act to the left (resisting the object's relative motion).

To prevent confusion (as this could lead to one). Imagine a simple inclined plane and a box sliding down due to its own Weight, we know that the force acting on the Plane is the force component that is perpendicular to the Weight, and in accordance to Newton's 3rd law, the box would experience the same amount of force but opposite in direction (which in this case is known as the Normal Force.)

Now moving on to my main problem. The white arrows I drew are perpendicular to the surface of the tiny ridges/asperities (black layer), which also indicates these are the components of the 'pushing' force (red arrow) that would actually act on the ridges, and the pink arrows (equal in magnitude but opposite in direction) are the resistive/reaction force that the object (blue layer) would experience.

Looking back at the direction of the arrows, I notice that they are not opposite to the direction of the object's movement. At first I thought this a contradiction to what is usually told that friction acts opposite to the sliding motion, but using the inclined plane analogy, assuming the weight to be the 'pushing force' and the frictional/resistive force I drew & explained as the 'inclined plane's normal force, which aren't also opposite to each other, this makes sense and therefore Newton's law isn't violated.

Now the question is, Why do we assume the frictional force to be acting exactly opposite of the object's relative motion?

*I might be too 'microscopic' but what I think might be the reason for this is probably, when all these different-directional frictional forces for every ridges/asperinties are 'averaged', it's direction would be very close to being opposite of the object's motion. Thus for simplicity sake we all assume it to be just opposite. Can anyone verify whether this is true? And if not, please kindly correct me. Any clarification would be very helpful.

visual - Counter Flipping

A counter flip game is played in the following way:

• In each move, you can flip an entire row or column


• Flipping a row or column inverts all the counters on that row or column from $X$ to $O$ or vice versa

For example,

\begin{gather*} XOX\\OXO\\XOX \end{gather*} can be converted to \begin{gather*} OOO\\OOO\\OOO \end{gather*} by three moves:

1. Flip first row: \begin{gather*} OXO\\OXO\\XOX \end{gather*}


2. Flip third row: \begin{gather*} OXO\\OXO\\OXO \end{gather*}



3. Flip middle column: \begin{gather*} OOO\\OOO\\OOO \end{gather*}

If you start with \begin{gather*} OOO\\XOX\\OOO, \end{gather*} prove whether or not it is possible to finish with: \begin{gather*} XOO\\OXX\\OOX \end{gather*}

Answer

Consider only the first two cells of the top row and the first two of the bottom row. Every move that changes them flips exactly two. They start off all O, so after every move 0, 2, or 4 of them will be an O. The goal has three O's so is not possible.

Edit: Now that I have a bit more time, and my answer is already the accepted one, I'll add a short summary of the theory of this type of game which most of the other answers have touched on.

• This is a variant of the Lights Out game.



• There are 9 lights, which can be on (O) or off (X).


• There are 6 moves available - flipping any column or any row.


• Applying any move twice is the same as doing nothing.


• You can switch the order the moves, and they still have the same effect.

These last two facts mean that you will have to apply any move at most once, since if you did it more often you can rearrange the moves so that the repeated moves happen successively, after which pairs of them cancel leaving at most one of each.

In this particular puzzle, there are $2^9$ potential states that the 9 lights can have.

The 6 moves are not independent: Flipping all three rows is the same as flipping all three columns. Therefore, we need never use one of the six moves. For example, instead of flipping the first row, you could instead flip all the columns and the other two rows.

That leaves only 5 types of move to play with. Therefore there can only ever be $2^5$ states that we can reach with these moves (as each is applied at most once in any order).

With these 5 available moves we can set the first row and column to any state we like. Use the three column moves to set the states of the first row, and then moves on the bottom two rows to set the rest of the first column. The state of the other four lights cannot be changed independently of the top row and left column.

Consider any of the following sets of four lights:

  x x .    x . x    x x .    x . x
  x x .    x . x    . . .    . . .
  . . .    . . .    x x .    x . x

Every move toggles exactly two of the lights (or none at all). Therefore, the number of lights that are on (of the four you are looking at) can only change by an even number by any move, or by any sequence of moves. If your starting position has an even number of the four lights on, then so will the end position. If it is odd at the start, then it's odd at the end. From this you can deduce what the state of the bottom-right 2x2 square of lights will have to be after you have set the top row and left column to the states you want.

There is a large amount of mathematical literature on the subject of Lights Out games. It can be solved and analysed using Linear Algebra. You can find an explanation of those techniques on my web page here.

newtonian mechanics - Real world examples for projectile thrown upwards or downwards

I am preparing a physics course for high school about projectile motions. If a projectile moves with initial velocity $v_0$ in the gravitational field of the earth, the equation

$$ s(t) = \frac{1}{2} g t^2 + v_0 t $$

holds, where $s(t)$ is the travelled distance at time $t$. Now I am looking for real world applications of this equation (or the corresponding equation for the velocity).

More specifically I don't want any problems where the equation is somehow artificially embedded into a real world situation, for example

"you fire the projectile of a signal pistol with initial velocity $v_0=...$, at which height is the projectile at time $t=...$, what is the maximal height..." projectile signal pistol In that example it is not clear why you know the initial velocity or why you want to calculate the maximal height (indeed, I think in most cases the manufacturer writes the maximal height into the manual, but then why should you be interested in calculating the initial velocity?)

So the point is, in the problems/examples I am looking for, it should be clear, why one has the input data and why one wants to calculate other things using the equation of motion.

Please stick to one problem/example per answer. Further references for the context of the example would be nice.

Edit: I should make clear, that I want examples where $v_0 \neq 0$. I am only intersted in the upward - downward -case.

One example I was thinking of, was that of a vulcano that ejects stones. However I don't know much about vulcanos, so I don't know which initial data are known, and what people want so calculate...

homework and exercises - The relation between Hamiltonian and Energy

I know Hamiltonian can be energy and be a constant of motion if and only if:

1. Lagrangian be time-independent,


2. potential be independent of velocity,


3. coordinate be time independent.

Otherwise $$H\neq E\neq {\rm const},$$ or $$H=E\neq {\rm const},$$ or $$H\neq E={\rm const}.$$

I am looking for examples of these three situation.

general relativity - Why is GR renormalizable to one loop?

I have read in a few places that GR is renormalizable at one loop. (hep-th/9809169 for example, second sentence, although they don't seem to develop this point at all). Is this do to some hidden symmetry in the theory? Naively we need new counter terms at all orders, even one loop, in perturbation theory, right?

quantum mechanics - Franck-Hertz experiment what transition takes place?

For the Franck-Hertz experiment there is a voltage drop at $4.9\rm\,V$. What transition does this represent in the mercury? Looking at the energy levels it seems to be from the ground to the 2nd excited state. But would this state not be occupied by other electrons? Any references would also be great.

Answer

According to a list of levels from NIST, the ground state for mercury has quantum numbers $^1S_0$ (with the electron in the 6th shell). I usually have to look up how to read those symbols: the $^1$ tells you it's a spin singlet, the $S$ tells you that the orbital angular momentum $L=0$, and the $_0$ tells you the value of the total angular momentum quantum number $J$. The first excited states are a $^3P$ triplet (spin triplet, $L=1$) with \begin{align} \begin{array}{cc} J & \text{Wavenumber }\mathrm{(cm^{-1})} & \text{Energy (eV)} \\ \hline 0 & 37\,600 & 4.66 \\ 1 & 39\,400 & 4.88 \\ 2 & 44\,000 & 5.45 \end{array} \end{align} The selection rules for atomic transitions forbid transitions where $J$ goes from zero to zero. In photon-mediated transitions this is pretty easy to understand: the photon must carry away one unit of angular momentum. (You can have transitions with $\Delta J=0$ for nonzero $J$; in that case you can imagine that the orientation $m_J$ of the atom's spin must change.) Apparently the same rules apply to the collisional transitions seen in the Franck-Hertz experiment — otherwise you could excited the $^3P_0$ first excited state and the energy involved would be $4.7\rm\,V$. Very interesting observation!

In your question you ask,

would this state not be occupied by other electrons?

Here the answer is no. In that same list of levels you can see that the ground state configuration is $5d^{10}({}^1S)6s^2$, while the first excited states have $5d^{10}({}^1S)6s6p$ (on top of a xenon-like core of 54 inert electrons).

riddle - Three rings to rule them all

Three rings to rule them all,
Nine men fighting against nine.
Closing the lines, capture is done,
I win if seven are mine.

Three men are flying, but more are not.
But on a plane they are all.
Now can you tell from this what I am?
For answers to this I now call.

Answer

I can't solve it completely, but I think it is

Nine men's morris (lots of references to the rules of the game: nine pieces on each side, three on the same line means capture, capture seven pieces and you win, when only three pieces are left, "flying" is allowed).

particle physics - The delay between neutrinos and gammas in a supernova, and the absolute mass scale of neutrinos

In a supernova explosion (of some type), there is a huge amount of neutrinos and gamma rays produced by a runaway nuclear reaction at the stellar core. In a recent comment, dmckee noted that the neutrinos actually make it out of the supernova earlier than the gamma rays because the latter do scatter in trying to get out, and that the precise details of this delay, if measured, could shed light on just how small the (nonzero) neutrino mass is.

Is there some simple way of explaining the different models at stake, or what the essential physics is behind this? Is there some accessible reference about this?

Answer

The time delay between neutrinos and photons does not tell us directly about the absolute mass of the neutrinos. The time delay between a neutrino of mass $m$ and a massless neutrino does: $$ \Delta t = \frac{d}{v} - \frac{d}{c} \approx 0.5 \left(\frac{mc^2}{E}\right)^2 d, $$ where $\Delta t$ is the time delay in seconds, $v \approx c[1-\frac{1}{2}\left(\frac{mc^2}{E}\right)^2]$ from relativistic calculation, $d$ the distance to the supernova in 10 kpc (a "typical" value for Galactic supernova), neutrino rest mass $m$ in eV and neutrino energy in MeV. So it is a propagation-speed problem.

As for the reason of the delay between the neutrinos and the photons, several pieces of information should be helpful:

1. It is hot in the core of the supernova -- so hot that the photons scatter a lot with free electrons before escaping from the core (see comment for a more detailed version); (electromagnetic interaction)


2. Yet it is not hot enough to influence two of the three neutrinos, $\nu_{\mu}$ and $\nu_{\tau}$, almost at all, and only slightly for another ($\nu_{e}$); they just fly out of the core. (weak interaction)


3. So this is a different-interaction (or, cross-section) problem.

Take SN1987A for example, the neutrinos arrived 2-3 hours before the photons. On the other hand, $\Delta t$ calculated above, if detected, would be $\leq 1$ sec. With these two values and a bit more thinking, you might conclude that this "time of flight" method could not constrain very well the neutrinos with smaller-than-eV masses; and you would be right. There are better ways to do that. Here is a reference: http://arxiv.org/pdf/astro-ph/0701677v2.pdf.

lagrangian formalism - Confusing with the equation $(2.4)$ and $(2.5)$ of Landau and Lifshitz, Mechanics, Chapter 1, The principle of Least Action

I'm a 12th Grader and I'm interested in Lagrangian Mechanics and having a bit of knowledge about the Newtonian Mechanics. So, I found a book of Landau and Lifshitz's Mechanics and started reading from the very first chapter, but I've encountered some serious doubts here!

I'm just writing the symbol/variable meaning and conventions here first!

For instance, let's imagine a particle. $q$ is it's radius vector magnitude (it's scalar), $\dot q$ is the derivative of position vector or velocity (scalar), $t$ as time duration, $S$ as action.

So, $$S = \int \limits_{t_1}^{t_2} L(q, \dot q, t) dt.\tag{2.1}$$

Let variation of function be $\delta q(t)$, so now

$$\delta q(t_1) = \delta q (t_2) =0 \tag{2.3}$$

$$\Rightarrow \qquad\delta S = \int \limits_{t_1}^{t_2} L(q+ \delta q, \dot q + \delta \dot q, t)dt - \int \limits_{t_1}^{t_2} L (q, \dot q, t)dt = 0.\tag{2.3b}$$

So, after the next few lines, it changes to:

$$\delta S = \delta \int \limits_{t_1}^{t_2} L(q, \dot q, t)dt =0.\tag{2.4}$$

This is doubtful as $\delta$ isn't a number which can be multiplied both sides of equal sign both ways,

$$\int \left ( \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot q} \delta \dot q \right)dt = 0.\tag{2.4b}$$

From which multiverse the above thing in concluded even I can't understand after hours of thinking, please help me with these concepts.

---

Ref: https://archive.org/details/Mechanics_541/page/n11

general relativity - Behavior of black holes in higher- and lower-dimensional space-times

The behavior of black holes in 3+1 dimensional space-time as our own is rather well known: formation, event-horizon size, mass, spin, radiation etc.

However, my question is what would black holes behave like in universes with different number of space-time dimensions, such as 2+1 dimensional or 4+1 dimensional space-time?

Would they be able to form? Would the event horizon radius be smaller or larger? Would their densities be higher or lower? Would they spin, and how quickly? Would they emit Hawking radiation? How would the space-time behave close to it compared to in 3+1 dimensional space-time?

Answer

First of all, in lower dimensions (2+1 and 1+1) the gravity is much simpler. This is because in 3d curvature tensor is completely defined by Ricci tensor (and metric at a given point) while in 2d curvature tensor is completely defined by scalar curvature. This means that there are no purely gravitational dynamical degrees of freedom, in particular no gravitational waves.

General note: horizon (which is the defining feature of black hole) representing our inability to obtain information about events under it would always imply the entropy of corresponding solution. So, in all of black hole models there is some black hole thermodynamics. For Hawking radiation one needs to include quantum effects into consideration and also radiative degrees of freedom (if there are no gravitons or photons or any other '-ons' than nothing would radiate).

Let us start with case of 3d (that is 2+1). The Einstein equations in 2+1 spacetime without any matter fields would simply imply that spacetime is flat, that is 'constructed' from pieces of Minkowski spacetime. It may have nontrivial topology, so 2+1 gravity is a topological theory, but no black hole solutions exist. This model (in mathematical sense) is exactly solvable.

To introduce non-trivial 2+1 solutions we can add matter or cosmological constant (which could be considered the simplest form of matter). It turns out that the spacetimes with negative cosmological constant (which would locally be composed of pieces of anti-de-Sitter spacetimes) do admit the black hole solution: BTZ black hole (name after authors of original paper). This solution shares many of the characteristics of the Kerr black hole: it has mass and angular momentum, it has an event horizon, an inner horizon, and an ergosphere; it occurs as an endpoint of gravitational collapse (for that, of course, we need to include matter beyond cosmological constant in the consideration); and it has a nonvanishing Hawking temperature and interesting thermodynamic properties (see, for instance, paper by S. Carlip). The Hawking temperature of BTZ black hole $T\sim M^{1/2}$ which, in contrast to the (3+1)-dimensional case, goes to zero as $M$ decreases. Additionally, the simplicity of the model allows quantum treatment of it including statistical computation of the entropy (see references in paper by E.Witten).

There are many other variations of solutions in 2+1 gravity theories (for instance by including dilaton and EM fields, scalar fields etc.) but all of them require negative cosmological constant. This is because dominant energy condition forbids the existence of a black holes in 2+1 dimensions (see here).

Now to 1+1 dimensions. Locally all GR models in 1+1D are flat. So to include nontrivial spacetime geometry we need to modify gravity. This can be done by including dilaton field. The resulting models often admit nontrivial geometries with black holes (see paper by Brown, Hennaux, Teitelboim, wiki page on CGHS model, paper by Witten on BH in gauged WZW model, and this review). These black hole solutions also admit nontrivial thermodynamics and Hawking radiation. In particular the Hawking temperature is proportional to mass, so as the black hole evaporates it becomes colder (unlike 4D case where $T \sim M^{-1}$).

Now to higher dimensional gravity. Gravity itself is much richer than in lower dimensional cases, so analogues of all 4D black holes also exist in higher dimensions, as well as some new black hole-like solutions such as black strings and black p-branes. There are also multi-black hole configurations where multiple black holes are placed along the ring or line such that the total force on each of them is zero, resulting in equilibrium configuration. Since many uniqueness theorems for black holes only work for 3+1 dimensions there are even solutions with nontrivial horizon topologies such as black rings.

I suggest to look at the Living Review recommended by Ben Crowell or to this lectures by N. Obers.

The simplest black hole would be Schwarzschild–Tangherlini solution (analogue of Schwarzschild black hole) which is vacuum solution to Einstein field equations:

Here $\mu = R_s^{d-3} = \frac{16 \pi G M}{(d-2)\Omega_{d-2}}$ is mass parameter. This gives us the relationship between mass and Schwarzschild radius: $R_s \sim M^{1/(d-3)}$. The entropy is given by Bekenstein-Hawking formula:

$$ S = \frac {\cal A}{4G}=\frac 14 \left(\frac{\Omega_{d-2} R_s^{d-2}}{ G} \right). $$ Temperature could be found from the first law $ dS = d M / T $: $$T = \frac{d-3}{4 \pi R_s}.$$

Rotating solution (generalization of Kerr metric) would be Myers-Perry metrics. Note, that rotations in higher dimensions are more complex, so the angular momentum is represented by several parameters.

Also note, that many solutions with horizons elongated in one direction (such as black strings or black rings) turn out to be unstable via the Gregory-Laflamme instability, where the smooth 'tubular' horizon evolve growing perturbations of certain wavelengths. So possibly black strings and black rings would tend to decay into droplets-like black hole along them (the exact mechanics is yet unknown). But of course, the second law of thermodynamics would be observed, meaning that total area of the horizons would increase.

enigmatic puzzle - Prevent a Terror Attack

Agents of P.U.Z.Z.L.E. (Part-time United Zippy Zealots Loving Enigmas), we need your help.

Terrorists are planning an attack on an underground railway station. One of our undercover agents placed an SD card in one of our dead letter boxes, which is supposed to reveal the target of the attack.

The SD card contained these three images:

Your task as part of our analytics department is to find out, which undergound railway station was chosen as the target for an attack.

Hints:

Remember the topic of the challenge, the "visual" part doesn't end with the three images.

Credits:

The second image is from Wikimedia Commons.

Answer

The attack is going to be at the

Spittelmarkt station.

There are several hints pointing to one thing:

Name of the folder is an anagram of ZODIAC.
The image names all begin with "CNC", "Cancer".
22/06, the modified date of all images, is the day the Cancer sign begins.
In the background of all images, there is a faint but distinguishable Cancer symbol (see Khale_Kitha's answer).

So, what if we...

Overlay the Cancer constellation on top of the Berlin map?

We already have 4 points, the box and the three landmarks, and the constellation has 5 stars. If we put 4 stars on top of the points we already have, the fifth star should point to where the attack will be:

pattern - Find the next 3 items in the sequence

A sequence goes as follows:

A

AA
BA
ABAA
AAABBA
CABBAA
ACAABBBA
AAACBACBAA
CAACABAAACABBA
...

It's an altered version of a sequence you all know! The answer should be in the above format.

Additionally, will D ever appear?

Answer

If A = 1, B = 2, C = 3, etc...

ACBAACAAABCAACAABBAA
AAACABBAACCAABACBAACBABBBA
CAACAABBBABCBAABAAACABBAACABAACBAA

It's the same kind of sequence where the next item describes the item above it in [quantity of number][number] format.

EDIT: The requested number format.

13211311123113112211

11131221133112132113212221
3113112221232112111312211312113211

homework and exercises - What is the meniscus shape?

What is the shape of a meniscus?

I suppose that the problem is very complex, but is the solution known at least for a liquid that wets the wall in a big vessel? (exponential, maybe?)

english - Word sets with no repeating letters

I know a set of six 4-letter words without repeating a letter (i.e. using 24 different letters).

Here is an example:

gasp, verb, jinx, flow, duck, myth

Are there sets of words with no repeated letters and having

1. four 6-letter words?


2. five 5-letter words?


3. three 8-letter words?

Bonus: Find a set of words (non-repeating letters) of sizes 1,2,3,4,5,6.

Answer

Using words from the SOWPODS word list, it's possible to solve Q1 and Q2 but sadly not Q3. I checked this by creating a graph of k-letter isograms, connecting words with no shared letters, and searching for cliques.

Five 5-letter words

Lots of answers, all using obscure words: e.g.

brick, jumpy, vozhd, glent, waqfs
vibex, fjord, nymph, waltz, gucks
bling, jumpy, vozhd, treck, waqfs

Four 6-letter words

Lots of answers, some using slightly less obscure words: e.g.

jawbox, kvetch, flumps, drying
jumbly, dwarfs, kvetch, poxing
jawbox, fledgy, skrump, chintz

Three 7-letter words

Lots and lots of answers: e.g.

stumped, flyback, whoring
mucking, batfowl, zephyrs
jordans, phlegmy, fuckwit
overply, dumbing, thwacks

Two 10-letter words

blacksmith, gunpowdery

Electron-Photon entanglement

Are the electron and photon emerging from a Compton scatter entangled? If so, what experiment can be done on one that will give information about the other?

astronomy - How do astronomers measure the distance to a star or other celestial object?

How do scientists measure the distance between objects in space? For example, Alpha Centauri is 4.3 light years away.

Answer

There are a variety of methods used to measure distance, each one building on the one before and forming a cosmic distance ladder.

The first, which is actually only usable inside the solar system, is basic Radar and LIDAR. LIDAR is really only used to measure distance to the moon. This is done by flashing a bright laser through a big telescope (such as the 3.5 m on Apache Point in New Mexico (USA), see the Apollo Project) and then measuring the faint return pulse with that telescope from the various corner reflectors placed there by the Apollo moon missions.

This allows us to measure the distance to the Moon very accurately (down to centimeters I believe). Radar has been used at least out to Saturn by using the 305 m Arecibo radio dish as both a transmitter and receiver to bounce radio waves off of Saturn's moons. Round trip radio time is on the order of almost 3 hours.

If you want to get distances to things beyond our solar system, the first rung on the distance ladder is, as Wedge described in his answer, triangulation, or as it is called in astronomy, parallax. To measure distance in this manner, you take two images of a star field, one on each side of the Earth's orbit so you effectively have a baseline of 300 million kilometers. The closer stars will shift relative to the more distant background stars and by measuring the size of the shift, you can determine the distance to the stars. This method only works for the closest stars for which you can measure the shift. However, given today's technology, that is actually quite a few stars. The current best parallax catalog is the Tycho-2 catalog made from data observed by the ESA Hipparcos satellite in the late 1980s and early 1990s.

Parallax is the only direct distance measurement we have on astronomical scales. (There is another method, the moving cluster method, but it has very limited applicability.) Beyond that everything else is based on data calibrated using stars for which we can determine parallax. And they all rely on some application of the distance-luminosity relationship

$m - M = 5log_{10}\left(\frac{d}{10pc}\right)$

where

• m = apparent magnitude (brightness) of the object


• M = Absolute magnitude of the object (brightness at 10 parsecs)


• d = distance in parsecs

Given two of the three you can find the third. For the closer objects, for which we know the distance, we can measure the apparent magnitude and thus compute the absolute magnitude. Once we know the absolute magnitude for a given type of object, we can measure the apparent magnitudes of these objects in more distant locations, and since we now have the apparent and absolute magnitudes, we can compute the distance to these objects.

It is this relationship that allows us to define a series of "standard candles" that serve as ever more distant rungs on our distance ladder stretching back to the edge of the visible universe.

The closest of these standard candles are the Cepheid variable stars. For these stars, the period of their variability is directly related to the absolute magnitude. The longer the period, the brighter the star. These stars can be seen in both our galaxy and in many of the closer galaxies as well. In fact, observing Cepheid variable stars in distant galaxies, was one of the original primary mission of the Hubble Space Telescope (named after Edwin Hubble who measured Cepheids in M31, the Andromeda Galaxy, thus proving that it was an “island universe” itself and not part of the Milky Way).

Beyond the Cepheid variables, other standard candles, such as planetary nebula, the Tully-Fisher relation and especially Type 1a supernova allow us to measure the distance to even more distant galaxies and out to the edge of the visible universe. All of these later methods are based on calibrations of distances made using Cepheid variable stars (hence the importance of the Hubble mission to really nail down those observations.

homework and exercises - If an astronaut had stationed in International Space Station for the duration of mission, 17 years, would he be older?

Today the NASA International Space Station started the 100000 orbit after 17 years in the space.
I just wonder if there were a team of astronauts which were in the Lab for all the duration of last 17 years, how much older they would be due to the combination of Einstein's especial theory of relativity and general theory of relativity?
Here is the link to NASA's youtube video. http://youtu.be/g5chOA-WEuw

Answer

The first answer has all the results, but I will try to show some calculations, cause I have been writing them since there was no answer.

It is known from General Theory of Relativity (GTR) that the closer you are to a massive object - the slower the time goes. On the other hand Special Theory of Relativity (STR) gives us the next statement: the faster you move - the slower the time goes.

I can give here only a half of an answer due to my low knowledge of GTR.

Let's compare relative speeds of the ISS and a person on the surface of the Earth (on the equator):

We know that the orbit of ISS is approximatele 400 km. Knowing this we can calculate its speed: $$\frac{mV_1^2}{r+400000}=G\frac{mM}{(r+400000)^2}$$ where $r$ is a radius of the Earth, $M$-its Mass, $m$-mass of the ISS, $G$-Gravitational constant. From this we get $V_1=\sqrt{\frac{GM}{r+400000}}$

Also we can approximate the speed of a person on the surface of the Earth:$$V_2=r\frac{2\pi}{T}$$

So from theese two equations we can calculate the time dilation per second relatively to the center of the Earth for each case: $$\tau_1=\frac{1}{\sqrt{1-\frac{V_1^2}{c^2}}}, \tau_2=\frac{1}{\sqrt{1-\frac{V_2^2}{c^2}}}$$

Then we can calculate the difference and multuply it on 17 years:$$|\Delta \tau|*17*365*24*60*60=0.174634$$

For better understanding we can look at the problem from different angle i.e. Lorentz contraction (effect of STR) or contraction of the distances in the direction of speed:

Let the orbit length be $l=2 \pi (r+400000)$ With every turn the ISS will cover the distance equal to $l_1=l \sqrt{1-\frac{V_1^2}{c^2}}$. Then we just calculate the differenc between 2 times $$T_1=\frac{100000l}{V_1}, T_2=\frac{100000l_1}{V_1}$$ We get that $\Delta T \approx 0.17$

Note:we have only paid respect to time dilation due to STR. The GTR requires more complex calculations but it will shift our answer a bit to the direction of zero, because it will slow the time more for those who are closer to the center of the Earth.

mathematics - The three utilities puzzle

Here is a variant of the three utilities puzzle. Famously, It is impossible to solve it on a plane paper (Euler's theorem can prove that). But can you solve it on a coffee mug?

Connect all three utilities to each of the houses, such that no two lines intersect.

Don't draw on the inside, It's for drinking ofcourse.

Answer

When trying to connect all utilities $A$, $B$ and $C$ to the three houses one realizes that

one needs just 1 "crossing" of lines (see drawing). This can be done on a topology like a mug (or donut) by having one of the lines going on the handle of the cup and the other one through it.

And this is how it looks like on a cup:

soft question - Journal for high school physics papers

I am currently a senior in high school. I have spent the past four years participating in physics research at a local lab. Are there any journals in which I as a high schooler can publish my research?

word - Time For Another Teapot Riddle (No. 35)

RULES :

I have one word which has several (2 or more) meanings

Each of the meanings is a teapot (first, second ...)

Try to figure out the word with my hints.

HINT 1 :

My first teapot exists in only some places

The second one is what you will have to pay first

HINT 2

The first teapot is what everyone wants

The second one can be fun

HINT 3

My first teapot is actually legit

The second one is usually quite big

WILL ADD MORE HINTS IN THE FUTURE

Answer

I think it might be

Fair as in fair people and a fair or event

My first teapot exists in only some places

There aren't many fair people around

The second one is what you will have to pay first

Not completely sure might be that you have to pay to go to a fair or maybe referencing to paying a fair amount of money

The first teapot is what everyone wants

everyone wants to be a fair person

The second one can be fun

Some fairs can be fun to go and visit

My first teapot is actually legit

I would say that fair people are legit

The second one is usually quite big

most fairs are big and want to appeal to as much people as possible

quantum field theory - Lagrangians combining terms with 1 and 2 derivatives

How are field theory Langrangians treated when some terms have 2 derivatives but others have only 1? Because the number of derivatives in a Lagrangian term is more easily even than odd, the discussions for newcomers to physics of breaking up Lagrangians into free theories and perturbations often does not give clear instruction on how to conceptualize and handle terms with a single derivative.

How should a mathematician speaking to physicists refer to the role played by terms with only one derivative (e.g. Chern Simons like terms) in the presence of a Yang Mills kinetic term? Would it be a kinetic term of lower order? A velocity dependent potential term? Would the interpretation change if the second order Yang - Mills like term were discarded? Is an m D m -like term with 1-derivative of a field m treated as part of the perturbation in perturbation theory or as generating a lower order summand of the 2nd order 'free' operator to be inverted?

Please feel free to rephrase the question if you understand what is being asked and the confusion is complicating the inquiry. Thanks in advance.

Answer

The OP is asking several questions.

1) The kinetic terms are, roughly speaking, the terms with temporal derivatives, e.g., $\frac{m}{2} \dot{q}^2$; or $p_i\dot{q}^i$; or a symplectic potential term $\vartheta_I(z)\dot{z}^I$, where $z^{I}$ is a phase space variable. (I'm using notation from point mechanics for the recognition value, however it works in field theory as well.)

2) Physicists often prefer a Hamiltonian formulation (also known as a first order formulation) withtime-derivatives (=velocities) appearing at most linearly, cf. the symplectic potential term $\vartheta_I(z)\dot{z}^I$. It is in principle possible to achieve a Hamiltonian form of the Lagrangian density ${\cal L}$ by Legendre transformation in the pertinent sectors at the expense of sometimes introducing first and second constraints via a Dirac-Bergmann analysis. Similarly to the Darboux theorem in finite dimensions, it is often possible to locally achieve a symplectic potential term of the standard form $p_i\dot{q}^i$ by a change of variables. See also the Faddeev-Jackiw method.

3) Concerning theories with both quadratic and linear dependencies of the same velocity, it is a good idea to first look at the simple example of a charged particle in an electro-magnetic field. Yang-Mills action plus a Chern-Simons action is, a bit oversimplified, just a field-theoretic version of this.

4) A fourth question is related to how to divide a Lagrangian density $\cal L={\cal L}_0+{\cal V}$ into a free part $\cal L_0$ and an interaction part ${\cal V}$. As a rule of thumb, we put as much as we can solve exactly in the free part, which usually restrict us to quadratic (and linear) terms only. In particular quadratic terms that are linear in velocities belong to the free part. Finally, the Hessian of the free part has to be invertible so that we can find the propagator. Zero-modes may cause extra subtleties. We set up the perturbative expansion around a stationary value of the free part.