astronomy - How is distance measured to far away stars and galaxies?

What I need is an accurate description of the methods used to determine the distance to Andromeda. The Parallax method is for nearby objects as I presume. The red shift method applies, but how do you really determine a red shift? Maybe the radiation emanated from that object already has the wave lengths reach Earth? Or maybe it's travelling at a large speed from us? Or maybe its mass is greater than we think?

Answer

The parallax method is shorter than required for Andromeda and the red shift method is longer, than required for it.

The distance to Andromeda galaxy is measured by the method of standard candles: http://en.wikipedia.org/wiki/Standard_candle#Standard_candles

I think Cepheid variable starts are appropriate for Andromeda galaxy. The period of pulsing of those stars is related with it's maximal brightness, which allows to calculate the distance.

http://www.physlink.com/education/askexperts/ae589.cfm

quantum mechanics - Delocalization in the square root version of Klein-Gordon equation

In this Wikipedia article a relativistic wave equation is derived using the Hamiltonian $$H=\sqrt{\textbf{p}^2 c^2 + m^2 c^4}$$ Substituting this into the Schrödinger equation gives the square root version of the Klein-Gordon equation: $$\left( \sqrt{ (-i \hbar \nabla)^2 c^2 + m^2 c^4 } \right) \psi = i\hbar\frac{\partial}{\partial t} \psi$$ Then the article says:

Another problem, less obvious and more severe, is that it can be shown to be nonlocal and can even violate causality: if the particle is initially localized at a point $r_0$ so that $\psi(r_0 ,t=0)$ is finite and zero elsewhere, then at any later time the equation predicts delocalization $\psi(r,t)\neq 0$ everywhere, even for $r>ct$ which means the particle could arrive at a point before a pulse of light could.

What is this solution explicitly? I have read also this Phys.SE question but there is no clue for my question.

special relativity - Is the clock hand off version of the twin paradox real or fake?

The clock hand off version of the twin paradox is when clock info is transferred at the point an outgoing ship meets an incoming ship. For example an outgoing ship at .6c, clock sync'd to earth on takeoff, will sync an incoming ship's clock 3 ly from earth so that when he meets up with the earth clock at .6c, the time on the ship's clock is 2 yrs less than the earth clock at earth in agreement with the results from a normal twin paradox scenario.

Why do I ask if this is fake?

The ships and earth are all engaged in constant relative velocity so they are all ageing at the same proper time rate relative to each other as if they were all stationary. The only thing subject to a frame jump, which is what causes age difference in the twin paradox, is the clock info. The incoming ship itself has not experienced a frame jump so the clock in the ship will not reflect the ageing the rest of the ship does. The captain will not end up ageing a different total time from a valid start than the earth clock as he would after a frame jump in a true twin paradox scenario. (A valid start is determined by syncing the two clocks at co-location and working backward to see when they would have both been zero.)

The info is not subject to the Rindler metric that a real clock would experience in a frame jump. The info instantaneously changes the clock whereas the Rindler metric's effect on time itself takes time to affect a physical clock having frame jumped.

dielectric - Dependence of capacitance on its build material

We know that the capacitance of a capacitor depends on the dielectric material in between the plates as $$C=\dfrac{K\epsilon_0 A}{d}\,.$$

But what if the electrode material is changed? For example suppose that the electrodes were built of Iron (Fe) and had a capacitance of $C$. Now I change it and make the electrodes with another material, say Copper (Cu). How will its capacitance change? What will be its new value?

Please do not assume that the capacitor is ideal. I am asking for a Real capacitor.

Answer

Since many metals form a thin oxide layer of relatively high resistivity, they can be used to affect the dielectric properties of a capacitor made from that material.

For example, aluminium, tantalum and even copper all form oxide layers with high resistivity (compared to the respective metals) and high relative permittivity (dielectric constant). Since each of these oxides has a different dielectric constant and dielectric strength (withstand voltage), a capacitor made from one material would have different characteristics to a capacitor of similar geometry made from another material.

This effect is already used in practice in the fabrication of electrolyte capacitors, tantalum capacitors, where one or both of the plates is made from aluminium or tantalum in order to exploit the very high dielectric constants of the respective oxide.

It is also the basis of forming capacitors in semiconductor microchips as well as metal oxide semiconductor field effect transistors (MOSFETs).

More recent advances in fabrication of multilayer nanosheets has allowed the development of supercapacitors, made from very thin layers of copper oxide formed from heating the copper metal in a controlled atmosphere, resulting in extremely high dielectric constants and high devices with capacitance.

An ideal capacitor is a device which can store electrical charge indefinitely. The relationship between voltage $v(t)$ and current $i(t) through a capacitor is given by:

$$i=C\frac{dv}{dt}$$

As you have already stated, the capacitance ($C$ in Farads, F) of a parallel plate capacitor is given by:

$$C=K\epsilon_0 \frac{A}{d}$$

where $K$ is the of the dielectric constant of the material (also called the relative permittivity), $\epsilon_0$ is the permittivity of a vacuum ($\epsilon_0 \approx 8.854 \times 10^{-12}F/m$ in free space), A is the area of overlap of the two plates [in $m^2$] and $d$ is the distance between the plates [in $m$].

So by utilising the fact that a very thin (small $d$) oxide layer with high dielectric constant (large $K$) can be formed on a metal plate used for a capacitor, capacitors with very large capacitances can be formed.

In practice, a 'real' capacitor a finite withstand voltage (dielectric strength, $D$), as well as some (non-zero) resistance due to connections, leads and plate material (total $R$), as well as resistance of the anodic oxide film ($r$) and inductance of the plate foils ($L$). A 'real' capacitor can therefore typically be represented as:

Practical values for L are typically very small at low frequencies (50Hz-1kHz) so inductance can sometimes be ignored, although not so at higher frequencies such as RF and microwaves. The dielectric strength (or withstand voltage) is represented by an ideal zenner diode with breakdown voltage $D$. That is, if the applied voltage exceeds $D$, the dielectric will 'break down' (conduct).

Some practical values for dielectrics are given below:

Since most metals are good electrical conductors, they have very low electrical resistivity), so the additional resistance offered by one metal over another when selecting the plate material of a capacitor usually has negligible affect on the capacity of the capacitor.

The reason for this is that the total resistance of a capacitor is the sum of the electrical resistance of the connections, leads and plates of the capacitor. The capacitance of the capacitor however is a function of the properties of the dielectric between the plates of the capacitor.

Tantalum and magnesium oxide capacitors are useful for their high capacity and small size, however, since they are typically fabricated from sintered metal powders, the effective series resistance of these capacitors can be significant.

Making a grid deduction puzzle

When making a grid-deduction puzzle, what are the strategies in making it?

For most grid-deduction puzzles, simply putting numbers/dots/whatever in the grid simply doesn't cut it. However, sometimes you need to fit a certain constraint, such as the answer looking like a certain letter.

How do you go about it?

Please post different strategies as different answers, with the strategy name at the top and an explanation beneath. Try to include an example if you can. No need to spoilerise!

I am interested in both general strategies and those for specific puzzles, so post both!

statistical mechanics - Mathematical form of distribution function with high energetic beam

Maxwell Boltzmann distribution function (MBDF) has the form $$f(v)=n(\frac{m}{2\pi k_BT})^{\frac{3}{2}} exp(-\frac{mv^2}{2k_BT})$$ [Basic Space Plasma Physics by Rudolf A. Treumann & Wolfgang Baumjohann]. The shifted MBDF has the form $$f(v)=n(\frac{m}{2\pi k_BT})^{\frac{3}{2}} exp(-\frac{m(v-v0)^2}{2k_BT}).$$

This is shown in the below figure for ms= 1.660539040 $10^{-27}$, kb=1.380 $10^{-23}$, T=100, v0=2000, wherein thick lines give the MBDF and dashed one gives the shifted MBDF. .

As far as the beams in plasma are concerned, I believe that one should see a figure of the form: Kindly correct me if I am wrong.

What is the mathematical expression that describes the above figure?

Answer

Since the only plasmas where that bump-on-tail velocity distribution function (VDF) can exist are those that are either weakly collisional or collisionless, it is perfectly okay to add VDFs. That is, those plasmas are neither in thermodynamic or thermal equilibrium, so there is nothing wrong with adding two VDFs as they are not a single temperature (e.g., see https://physics.stackexchange.com/a/268594/59023 or https://physics.stackexchange.com/a/375611/59023).

The general bi-Maxwellian VDF of species $s$ is given by: $$ f_{s}\left( v_{\parallel}, v_{\perp} \right) = \frac{ n_{s} }{ \pi^{3/2} \ V_{T \parallel, s} \ V_{T \perp, s}^{2} } \ exp\left[ - \left( \frac{ v_{\parallel} - v_{o, \parallel, s} }{ V_{T \parallel, s} } \right)^{2} - \left( \frac{ v_{\perp} - v_{o, \perp, s} }{ V_{T \perp, s} } \right)^{2} \right] \tag{0} $$ where $\parallel$($\perp$) refer to directions parallel(perpendicular) with respect to a quasi-static magnetic field, $\mathbf{B}_{o}$, $V_{T_{j, s}}$ is the $j^{th}$ thermal speed (actually the most probable speed), $v_{o, j, s}$ is the $j^{th}$ component of the bulk drift velocity of the distribution (i.e., from the 1st velocity moment), and $n_{s}$ is the number density or zeroth velocity moment of species $s$.

Typically beams (i.e., the little bump in your merged VDF) are not isotropic, so it is common to have a VDF for the core and a VDF for the beam. It is okay to use a form like Equation 0 for both, each with different densities and thermal speeds. The non-equilibrium nature of the plasma allows two such VDFs to effectively stream past each other. They tend to excite instabilities that lead to fluctuations like Langmuir waves.

You could get a little fancier and use a bi-kappa VDF. The bi-kappa distribution function is given by: $$ f_{s}\left( v_{\parallel}, v_{\perp} \right) = A_{s} \left[ 1 + \left( \frac{ v_{\parallel} - v_{o, \parallel, s} }{ \sqrt{ \kappa_{s} - 3/2 } \ \theta_{\parallel, s} } \right)^{2} + \left( \frac{ v_{\perp} - v_{o, \perp, s} }{ \sqrt{ \kappa - 3/2 } \ \theta_{\perp, s} } \right)^{2} \right]^{- \left( \kappa_{s} + 1 \right) } \tag{1} $$ where the amplitude is given by: $$ A_{s} = \left( \frac{ n_{s} \ \Gamma\left( \kappa_{s} + 1 \right) }{ \left( \pi \left( \kappa_{s} - 3/2 \right) \right)^{3/2} \ \theta_{\parallel, s} \ \theta_{\perp, s}^{2} \ \Gamma\left( \kappa_{s} - 1/2 \right) } \right) \tag{2} $$ and where $\theta_{j, s}$ is the $j^{th}$ thermal speed (also the most probable speed), $\Gamma(x)$ is the complete gamma function, and $\kappa_{s}$ is the kappa index and can be any value larger than 3/2.

Further we can show that the average temperature is just given by: $$ T = \frac{ 1 }{ 3 } \left( T_{\parallel} + 2 \ T_{\perp} \right) \tag{3} $$ if we assume a gyrotropic distribution (i.e., shows symmetry about $\mathbf{B}_{o}$ so that the two perpendicular components of a diagonalized pressure tensor are equal).

special relativity - Do relativistically-contracted electron states have the same energy and angular momentum values?

I've been reading that electron bound states are defined by four quantum numbers, $n$, $l$, $m_l$, and $m_s$, respectively the principal quantum number, the azimuthal quantum number, the magnetic quantum number, and the spin quantum number. Together these determine the total energy of the electron (proportional to the negative inverse square of $n$) and the angular momentum.

However, they are derived in a non-relativistic setting. When I read about relativistic quantum chemistry I have a hard time visualizing the relativistic contraction of various orbitals because I don't understand how the quantum numbers relate to the energy and angular momentum values in the relativistic setting. When the "relativistic mass" of an electron bound state becomes significantly increased in heavy atoms, does the total energy and angular momentum for that electron stay the same as it would be otherwise?

See also Why does an electron's orbital contract as its relativistic speed increases?. (I don't quite understand the answer to that question)

quantum field theory - Are group representations possible when the solution space is not a vector space?

As far as I understand, the motivation for using representation theory in high energy physics is as follows. Assume that a theory has some (internal or external) symmetry group which acts on a vector space. Then fields satisfying the theory will have to transform under some representation of that symmetry group, by construction.

What happens if we have some internal or external symmetry structure that is no longer acting on a vector space? The gauge group diffeomorphisms of general relativity spring to mind. Is there some more general 'representation' type theory which comes to our aid? And are there any examples of internal symmetries where this viewpoint is needed?

Apologies if this question is imprecise or flawed - I'm just starting to get my head around the foundations of the subject! Many thanks in advance!

Answer

Let $G$ be a group, e.g. a finite group or a Lie group.

Then there exists the notion of a group action $G\times X\to X$, where $X$ is a set. The set $X$ does not necessarily have to be a vector space. It could e.g. be a manifold. And even if $X$ has vector-space structure, the group action could be non-linearly realized, i.e., a group element $g\in G$ is represented by a non-linear operator $T_g:X\to X$.

Non-linear realizations pop up all over the place in modern physics. For instance, in nonlinear realization of supersymmetry, or in nonlinear realization of the conformal group.

Example: Let the Lie group $G=GL(2,\mathbb{C})$ of invertible $2\times2$ matrices

$$\tag{1} A~=~\begin{pmatrix}a & b\\c & d \end{pmatrix}, \qquad \det(A)\neq 0, $$

act on the complex plane $\mathbb{C}$ (which, by the way, is a vector space) as

$$\tag{2} A.z ~:=~\frac{az+b}{cz+d}, \qquad (AB).z ~=~ A.(B.z)~. $$

In this way, matrices get non-linearly represented as meromorphic functions. The subgroup $SL(2,C)$ is the global conformal group in two space-time dimensions, which e.g. plays a fundamental role in the world-sheet description of string theory.

Finally, let us mention that in mathematics there exists a generalization of the notion of a $\mathbb{F}$-vector space, where the field $\mathbb{F}$ is replaced by a ring $R$. It is known as an $R$-module.

general relativity - Jacobi equation in the book The Large scale structure of space-time

On pp. 79, it is obvious that equation (4.2) \begin{equation} \frac{D}{\partial s}Z^a = {V^a}_{;\ b}Z^b \end{equation} holds, where $Z$ is the deviation vector and $V$ is the unit tangent vector along the timelike curves.

The author then project the deviation vector to ${}_\bot Z^a = {h^a}_bZ^b$, where ${h^a}_b = {g^a}_b+V^aV_b$ (the author used ${\delta^a}_b$, but I think it should be the metric?) is the tensor which projects a vector into its component in the subspace orthogonal to $V$.

My question is on equation (4.3): \begin{equation} {}_\bot \frac{D}{\partial s}({}_\bot Z^a) = {V^a}_{;\ b} {}_\bot Z^b \end{equation} In components it should be \begin{equation} {h^a}_c({h^c}_d Z^d)_{;\ b} V^b = {V^a}_{;\ b}{h^b}_c Z^c \end{equation} Using equation (4.2), which in component \begin{equation} {Z^a}_{;\ b}V^b = {V^a}_{;\ b}Z^b \end{equation} I just can make equation (4.3) equal on both side. Since the equation looks so simple, the derivation should be rather intuitive?

Also, there is no derivation to equation (4.4). It looks quite formidable though. Any hint on the derivation would be appreciated. There are a lot of detail derivation on geodesic deviation, but they did not project the deviation vector to the orthogonal ones though.

Thanks!

Answer

We first show that $h=g+V\otimes V$ is a projection operator into the subspace $H_p\mathcal{M}$ orthogonal to $V\in T_p\mathcal{M}$.

Idempotence ($h\circ h=h$). A simple calculation gives: $$h^a{}_bh^b{}_c=(\delta^a{}_b+V^aV_b)(\delta^b{}_c+V^bV_c)=\delta^a{}_b\delta^b{}_c+V^aV_b\delta^b{}_c+\delta^a{}_bV^bV_c+V^aV_bV^bV_c=\delta^a{}_c+V^aV_c+V^aV_c-V^aV_c=h^a{}_c.$$ Identity on orthogonal subspace ($h|_{H_p\mathcal{M}}=\operatorname{id}_{H_p\mathcal{M}}$). This is also not hard. Let $X\in H_p\mathcal{M}$, then $$h^a{}_bX^b=\delta^a{}_bX^b+V^aV_bX^b=\delta^a{}_bX^b=X^a.$$ We note also that $h(V)=0$, since $$h^a{}_bV^b=\delta^a{}_bV^b+V^aV_bV^b=V^a-V^a=0$$

So $h$ is a proper projection operator.

We will now explain the derivation of Eq. (4.3), since there is a little trick involved. The object $${}_\bot \frac{\mathrm{D}}{\partial s}{}_\bot Z^a$$ is understood as follows: project $Z^a\in T_p\mathcal{M}$ into $H_p\mathcal{M}$ and take the covariant derivative of the result along the integral curves of $V$. Then project this into $H_p\mathcal{M}$ again. Thus $${}_\bot \frac{\mathrm{D}}{\partial s}{}_\bot Z^a=h^a{}_b\frac{\mathrm{D}}{\partial s}(h^b{}_cZ^c)$$ Now we begin a long calculation: $$h^a{}_b\frac{\mathrm{D}}{\partial s}(h^b{}_cZ^c)=h^a{}_b\frac{\mathrm{D}h^b{}_c}{\partial s}Z^c+h^a{}_bh^b{}_c\frac{\mathrm{D} Z^c}{\partial s}=h^a{}_bZ^c\frac{\mathrm{D}}{\partial s}(\delta^b{}_c+V^bV_c)+h^a{}_cV^c{}_{;b}Z^b$$ Note that $\delta^a{}_b$ is covariantly constant (page 32), so we have $$\tag{1} h^a{}_cV^c{}_{;b}Z^b+h^a{}_bZ^cV^dV^b{}_{;d}V_c+h^a{}_bZ^cV^bV_{c;d}V^d$$ and the last term vanishes because $h(V)=0$.

Note that since $V^aV_a=-1$, we have $V^a{}_{;b}V_a=0$. This implies $V^a{}_{;b}{}_\bot Z^b\in H_p\mathcal{M}$. Thus $V^a{}_{;b}{}_\bot Z^b=h^a{}_bV^b{}_{;c}{}_\bot Z^c$, and one can show this equals (1) by simply expanding the definitions.

knowledge - The flying colours

Here's a $36\times36$ nonogram with two colours. Name all 12 things depicted in it.

Note that cells filled with different colours do not need to have a gap between them.

Answer

Pictured in the nonogram are ...

... the flags of Norway, the United States, Costa Rica, Japan, the Netherlands (or perhaps Luxembourg, given the shade of blue), Iceland, Cuba, France, Switzerland, Russia, the Dominican Republic and Czechia. (The title is a hint at flags.)

The nonogram:

quantum mechanics - Pair production at high laser intensity?

1. 
   

   Using high laser intensity to produce electron-positron pair, is it still required interaction with nucleus as is the case when gamma rays are used?

   
   


2. 
   

   What causes the pair creation ?

   
   

Answer

I had not realized this had been happening, very interesting.

Hand waving: when the intensity of the laser is high enough electrons can be accelerated to relativistic energies and create e+e- pairs interacting with the collective electric and magnetic fields of the laser pulse.

Here is a link to search for calculations , (first in list) .

Electrons have to be supplied from the atoms of a gas on which the laser is shining. The interaction is a collective field one with the electron, in these calculations, not with the nucleus.

And another using quantum electrodynamics:

Production of electron-positron pairs from vacuum in the combined electromagnetic fields of a high-intensity laser pulse and an atomic nucleus is studied within the framework of laser-dressed quantum electrodynamics. The focus lies on the influence exerted by a finite laser pulse length on the energy spectra of created electrons and positrons, which is examined in a broad range of field frequencies and intensities. The results for an isolated short laser pulse are also compared with corresponding calculations for an infinite train of laser pulses. It is shown that the laser pulse length and its carrier-envelope phase have a substantial effect on the pair creation process, leading to both quantitative and qualitative differences in the particle spectra.

It is behind a pay wall so I cannot see what "atomic nucleus" means in this case. Certainly the diagrams will be different than the gamma-nucleus diagrams.

cipher - A letter telling my great-great-grandfather's death-day

Yesterday, while I was tidying my attic, I found a letter dedicated to my great-great-grandfather. (Note that the letter was inside a box which my great-great-grandfather transmitted to my great-grandfather, and so on...) It is supposed to tell him his death-day. I actually know he got murdered, and I also know the exact date when it happened. Now I just want to know if this letter is actually accurate, but I can't decrypt/solve it. Can someone please help me? This is what it says (except the part about how he will be killed):

... (5.3-912-83-19-24) + (2.3-12-931-453-1) = mm

(119.9-12-32-183-873) + (300-200-193-930-39.2) = dd

(4.2-342-12-123-847) + (293-1-2-93-8.7) + (59-21-329-12-23-987) + (4.2-181-192-712-9995) = yyyy

Have fun decrypting it. Meanwhile I will wait. You should start learning how to make SOS-signs, because no-one will hear you.

P.S: Don't even think of solving it the "regular" way, like you've learned in school, you silly!

Answer

It was on

19 November 1901

because (short answer)

Morse code

or because (long answer)

Ignore the numbers in the puzzle; they're just a red herring. Concentrate instead on the dots and dashes between the numbers. We get:

mm = (.----)(.----) = 11
dd = (.----)(----.) = 19
yyyy = (.----)(----.)(-----)(.----) = 1901.

quantum mechanics - Why doesn't De Broglie's wave equation work for photons?

Well, as I am learning about quantum physics, one of the first topics I came across was De Broglie's wave equation. $$\frac{h}{mc} = \lambda$$ As is obvious, it relates the wavelength to the mass of an object. However, what came to my mind is the photon. Doesn't the photon have zero mass? Therefore, won't the wavelength be infinity and the particle nature of the particle non existent? Pretty sure there is a flaw in my thinking, please point it out to me!

Answer

What you have there isn't actually de Broglie's equation for wavelength. The equation you should be using is

$$\lambda = \frac{h}{p}$$

And although photons have zero mass, they do have nonzero momentum $p = E/c$. So the wavelength relation works for photons too, you just have to use their momentum. As a side effect you can derive that $\lambda = hc/E$ for photons.

The equation you included in your question is something different: it gives the Compton wavelength of a particle, which is the wavelength of a photon that has the same electromagnetic energy as the particle's mass energy. In other words, a particle of mass $m$ has mass energy $mc^2$, and according to the formulas in my first paragraph, a photon of energy $mc^2$ will have a wavelength $\lambda = hc/mc^2 = h/mc$. The Compton wavelength is not the actual wavelength of the particle; it just shows up in the math of scattering calculations.

homework and exercises - Work done by Electric Field vs work done by outside force

I'm confused as to the signage of the equation: W=qv, W=-U, W=-qv? When is work positive? When is it negative? Why is this different for the work done by the electric field vs the work done by an outside force? What is the relationship between electric potential energy and work?

riddle - Giving Gifts to Twins

If you give the first twin a gift, he will attach two letters on one side and give it back to you.

If you give the second twin a gift, he will ignore it and show you his ID.

Despite being twins, these two are literally opposites.

Who are the twins?

---

Hint #1:

One of the letters attached is not an English letter.

Hint #2:

ID means "identity," not the literal letters "ID" as some answers have tried.

Hint #3:

This is not wordplay, although the concept underlying this question has applications in linguistics (or so I'm told).

Hint #4:

When you give someone a gift, they will always return a gift. Also, there is no distinction between gifts and giftees.

Can there be general relativity without special relativity?

Can General Relativity be correct if Special Relativity is incorrect?

homework and exercises - Derviation of group velocity

I am working thru a derivation of the group velocity formula and I get to this stage: $$y=2A\cos(x\frac{\Delta K}{2} -t\frac{\Delta \omega}{2})\sin( \bar k x-\bar \omega t)$$ Then all the derivations I have seen say that $\frac{\Delta \omega}{\Delta K} $ is the group velocity. I know mathematically why this is a velocity but what I don't get is why do we know that this is the group velocity rather then the phase velocity and that $\frac{\bar \omega}{\bar k}$ is the phase velocity and not the group velocity?

logical deduction - Enlightened, Part I: Finding the Temple

Some time in the future, humanity discover a planet which once held an alien civilization. This civilization is long gone now, and only the empty structures remain, a testament to the achievements of their society. By working out the language system the aliens have used, researchers find several references to a temple built in the rocky wastelands, that contains the most sacred doctrines of their people. This is a story about recovering those doctrines.

Ricky: Chris! We have a serious problem!
Chris: What's up?
Ricky: Our map of the rocky wastelands are shredded, and the logs are ruined!

Chris: What map? What logs?
Ricky: (sighing) You don't know about the map, do you? Fine, I'll explain.
Ricky: We have a very strict protocol for exploring the wasteland. We send out groups of men, distinguished by a color, and get them to write logs about where they go, and what landmarks they find. Then, back at base, because we know where they started, we can construct a map showing their starting position and the landmarks they found. Here, I've got an example in my pocket- (rumbles around)

Orange team log:
Travelled 4km north, found a landmark.
Travelled 3km east, found a landmark.
Travelled 5km south, found a landmark.
Travelled 2km east, found a landmark.
Travelled 6km north, found a landmark.

Travelled 3km west, found a landmark.
Travelled 4km south, found a landmark.

Ricky: And sometimes we abbreviate this like so... (scribbles on a piece of paper)

Orange: N4 E3 S5 E2 N6 W3 S4

Ricky: This then allows us to draw a map back at base:

Chris: Ok, I get that. How is it relevant, though? What happened?

Ricky: Well, yesterday we sent out four teams to survey the area where we thought the temple might be. They all came back, we collected their logs and made a square map. It was late, and I hadn't had much sleep, so I put both the logs and the map in a folder and went to sleep. However, when I woke up - (pauses) - the maps and the logs had been sabotaged.
Chris: It is salvageable?
Ricky: Well, I'm not sure. That's why I came to ask you. You see, the map had been cut in a most mysterious fashion - into twelve identical L shaped pieces. I'll show you - just wait a bit. (dashes out, and comes back with a whole lot of paper)
Ricky: Have a look at this.

Chris: I don't see how those could fit into a square.
Ricky: Some of them have been rotated.
Chris: (slowly) ...oh.
Chris: But surely you can just reconstruct the map using the logs?
Ricky: That's just it. Some of the text off the logs has been removed. I'll show you what's left.

Red: N?,W6,N5,E?,S10,W11,N?,E?,S?,E6
Green: W3,N9,W11,S?,W?,S7,E11,N?,W6,S11
Blue: S2,W?,N2,W8,S?,E4,S7,W?,S3,E?
Purple: E?,N3,E1,S6,E?,N7,W5,S?,W?,N4

Ricky: Those question marks could be anything from 1 to 17, as far as I'm concerned. I don't remember what the numbers were before they were erased!
Chris: It still looks possible to reconstruct it... I think.
Ricky: That would be good, because the green team told us that the last landmark they visited was the temple we were looking for. Oh, you should know one more thing... (trails off)
Chris: What's that?

Ricky: I didn't write the compass directions on there. So I don't know which way is north.
Chris: ...

Can you help Chris reconstruct the map?

Answer

The solution is below:

The first tile to place is the one with the blue flag. It must be oriented like an "r" so that it can go south 2 and hit a landmark. It must be in the upper right as it manages to go extreme distances east and south without the possibility of going extreme distances west or north in between.

Simularly one can determine that the tile with the green flag must be on the far right and somewhat down. It must be oriented and positioned where it is or else the additional tiles around it cannot be placed to make the square.

The next major step was to find a pairing of tiles which could allow for landmarks in one of the 8th through 12th columns from left and the 3rd and 5th rows from the top. This was mostly done by process of elimination based on the green and blue paths. Which ever tile bordered the top could not have any landmarks in the 1st row. It could not be positioned as an "L" or there would be space a tile could not fill. It could not be positioned as a "r" or backwards "L" as no tiles fit the bill and did not lead to later complications. When the 1 tile shaped as a backwards "r" with an empty top row was tried, the top left tile was identified. From there, the final tile shapes were pieced together.

When I figured out the final tile arrangement, I assumed the blue path touched the landmark in column 5, row 8 as that was the only way I could see for it to go down 10 more squares after hitting a landmark 4 to the east. The tile with the red flag was the only tile where a landmark was 7 south of the previous blue landmark. This meant I had placed all 4 starting positions. The remaining tiles were added in as the paths were drawn, I can elaborate further if need be but the remaining placements were trivial.

geometry - Can you arrange 36 trees so that there are 9 rows of 8 trees?

How could you arrange 36 trees into 9 rows of 8? (Note that a row is a straight line that can go in any direction.)

Answer

Draw nine lines so that no two are parallel and no three meet in the same place, then place a tree at each intersection point. Every pair of lines intersects, so there are $\binom{9}{2}=36$ trees, each row being one of the nine lines.

Illustration:

special relativity - Doubts concerning Wigner's classification

Wigner classified particles in function of the eigenvalues of $P_\mu P^\mu$ and $W_\mu W^\mu$. Then, it can be proved that for massless particles spin values can be only $\pm s_{max}$. But for a particle with mass could have intermediate spin values.

1. 
   

   If we think that a massless particle is the limit where $m\rightarrow0$ (very small mass), how can we have this sudden change of the spin values (I think this is just an intuition error)?

   
   


2. 
   
   

   What is the difference between polarization and spin (or helicity)?

   
   


3. 
   

   Is it reasonable to say that massless particles have no spin but just helicity (I've read that this is because that don't have a center of mass and also because spin could point in any direction)?

   
   


4. 
   

   Should we consider photons of different helicity different particles?

   
   


5. 
   

   Is there any nice demonstration of why $W^2$ eigenvalues are $-m^2s(s+1)$, most books just refer to Wigner (1939).

   
   
   

Comment: I'm not sure whether I should divide this into several questions.

Answer

I want to refer you to Weinberg QFT1, if you have not read it yet. Below is my attempt to answer your question in the formalism addressed there.

You are classifying particles by representations of the Poincaré group. A one particle state has to be transformed under an element of this group. Part of this transformation just changes the momentum, the other part makes a particle-specific 'intrinsic' transformation of the state (i.e. rotates the spinor indices). Spin and helicity characterize the latter part and in this sense are rather similar.

It happens so that this intrinsic transformation can be understood in terms of what is called the small group. For any momentum $p_\mu$ you can go to a standard reference frame where we will have a standard momentum $k_\mu$ which lies in the same orbit of the Lorenz group as $p$. For example (I am usign signature $+---$), for massive particles, where you have $p^2=M^2$, you can choose the rest frame, and the standard $k_\mu=(M,0,0,0)$. For $p^2=0$ you can go to a reference frame where the momentum becomes $k^\mu=(\kappa,0,0,\kappa)$.

Now the small group is the subgroup of the Lorenz group which leaves your standard $k$ invariant. It happens (and for proof of this it is better to read Weinberg, Wigner, etc) that for every Lorenz transformation there is a corresponding element of the small group that determines the 'intrinsic' transformation.

Lets talk about the massive case first. We have $k^\mu=(M,0,0,0)$, and the small group is clearly just the rotations $SO(3)$. Irreducible representations of this group are well-known and they are parametrized by spin $s$ which can be integer of half-integer. (Formally, the latter are projective representations, but in QM a phase does not matter.) We can also pick an axis and look at the rotation generator $J_3$, and its eigenvalues, this is the 'spin projection'. As far as we have other rotations at our disposal, we can rotate the particle to change this projection. This is the intuitive reason why we have all spin projections between $-s$ and $s$ (with integer steps).

Now, what is different in the case of massless particles. The standard momentum is $k^\mu=(\kappa,0,0,\kappa)$ that is a, say, photon travelling along $z$ direction. Now we still have our $J_3$ rotation, but there is some problem with $J_1$ and $J_2$ -- they will change our vector. However, it is possible to save the situation by adding some boosts to these generators. This fact changes the group in this case, and it becomes isomorphic to $ISO(2)$ -- the group of isometries of 2D euclidian plane. In practical terms, we how have three generators in the small group -- $A,B,J_3$, where $A,B$ is what is left of $J_1$ and $J_2$ and correspond to translations of this plane, while $J_3$ is the rotation along the momentum direction and corresponds to rotations in the plane. Of course, this plane picture is just a mathematical abstraction.

It is more or less obvious that $A,B$ correspond to some 2D momentum, and so if they have a non-zero eigenvalue, they have continious spectrum (just rotate it with $J_3$). We do not observe any intrinsic continious degrees of freedom for massless patricles, so we conclude that $A,B$ have zero eigenvalues in the representations of interest. We now only have to think about $J_3$ eigenvalue, which is the projection of spin to the momentum direction. Here I say 'spin projection' because it determines the transformation of wavefunction under rotations around some axis. However, this eigenvalue is actually called helicity. For now, no restrictions on its value.

You can see that in the masseless case, the relevant irreducible representation of the Lorenz algebra is in fact one-dimensional (just one state with $A\psi=B\psi=0$, $J_3\psi=\lambda\psi$) and is parametrized by some parameter $\lambda$. Now, it is the topology of the Lorenz group that requires $\lambda$ to be integer or half-integer (in massive case, this requirement grows already from the algebra). Still, no requirements like "there should be also at least $-\lambda$" etc.

The physical reason for this is that in massive case, if we knew 'helicity' - the projection of spin to the momentum direction, we could go to the rest frame, rotate the spin as we want, and go back, thus changing the projection. In massless case, there is no rest frame -- when we try to rotate the 'spin', the momentum rotates as well. The helicity is Lorenz-invariant.

Why we have, as you say, '$\pm s_{max}$' in massless case, it is because spatial inversion changes the sign of helicity. So, for particles that 'respect $P$-inversion' we have also particles with opposite sign of helicity, and call them the same names. For neutrino (assume them massless), however, we dont have such a nice thing, so there are neutrinos and antinetrinos with different helicities.

So, if a mircale happened and I explained it in a correct as well as understandable way, I think this should clarify points 1,2,3.

4. It is the same as should we call spin-up electron and spin-down electron the same particle? If you believe that rotations are symmetries of our world -- yes. Now, helicity? If you believe that spatial inversion ($P$) is a symmetry -- yes. Well, we know it is violated, but, if we take QED alone, it is a symmetry, so it is reasonable to call it so. Also, the mathematical formalism ($A_\mu$-field) highly suggests this..

5. Go to the rest frame. (The square is a casimir, so it commutes with boosts) Then this is, up to a coefficient, $m^2\vec{J}^2$. Where $J_i$ is the generator of rotations.

1. I would say that you are expecting some continuity that is not here..

heisenberg uncertainty principle - Simultaneously measurement in quantum mechanics?

In quantum mechanics $A$ and $B$ can be simultaneous measured if mathematically $\hat{A}\hat{B}=\hat{B}\hat{A}$. But how do we actually measure thing simultaneously. $\hat{A}\hat{B}$ is not simultaneous measurement because because here we measure $B$ and then $A$. So the above mathematical equation only tells us that $\hat{A}\hat{B}$ ($B$ first, $A$ later) = $\hat{B}\hat{A}$ ( $A$ first, $B$ later) - how does this relate to simultaneous measurement?

Answer

simultaneous measurement doesn't mean that commuting operators are measured simultaneously (in time, as you asked in question) and not one by one. It means that when the two operators commute, then they are diagonal in the same ket basis, which are referred to as simultaneous eigenkets, and you can now measure them one by one without disturbing the collapsed state vector. The fact that they are diagonal in the basis of same eigenkets allows them to be measured simultaneously without disturbing the state vector. Now coming to your example of two commuting operators, it does not matter in which order you measure the operators, since the two operators commute with each other. You can measure them like $ XY$ or $YX$.

And simultaneous measurement is not prohibited by quantum mechanics. It is prohibited for those operators that don't obey commutation relations and thus are not diagonal in the same ket basis, and therefore the state vector is disturbed each time they are measured. A famous example is that of position and momentum, they have a non-zero commutator and thus cannot be measured simultaneously.

quantum field theory - What's the current status of the Casimir force $leftrightarrow$ vacuum energy controversy?

Most textbooks explain the Casimir force by using the infinitely large ground state energy associated with quantum fields. Since the total energy between two conducting plates is different from the total outside of the plates, there is a net energy difference and thus a force acting on them. This force is known as Casimir force and was observed in experiments. The effect therefore seems to confirm that the ground state energy of quantum fields is nonzero and leads to measurable effects.

On the other hand, it's sometimes argued that the Casimir force "is just the van der Waals force derived in a different language." In particular:

Conceptually, saying that "the plates force the electric field to vanish" is the same as saying "electronic fluctuations in the two plates become synchronized", which is how we describe the origin of the van der Waals force.

Moreover, this highly cited paper concludes that

Still, no known phenomenon, including the Casimir effect, demonstrates that zero point energies are “real”.

---

What's the exact relationship between the two explanations? Is one of them wrong or is the standard derivation in terms of ground state energies really just a derivation of the van der Waals force in disguise?

electrostatics - What's the shape of the flame for a candle in parallel plate capacitor?

Let me show a figure first about the shape of flame for a candle without any external electric field in the dark night:

---

---

Now consider the following physical configuration: If you put this lighted candle into the parallel plate capacitor, in which the electric field produced by it can be turned up or turn down arbitrarily.

So my question is: what's the shape of the flame of the candle when we keep the electric field stationary and how is it changed as we increased the field or decrease the field yielded by the parallel plate electrical capacitor?

Answer

According to http://archive.iypt.org/iypt_book/2011_3_Bouncing_flame_Iran_RMN_HA_RA_v3.pdf , in a typical scenario, electrons and positive ions in the flame plasma move to opposite plates of the capacitor, electrons leave the flame area and form negative ions with oxygen atoms, the flame gets positively charged and attracted to the negatively charged plate, as you can see in numerous videos. In some videos, though, the flame is "divided" in two parts attracted to opposite plates, so in such cases the negative ions probably remain within the flame.

thermodynamics - Maximum theoretical efficiency of internal combustion engine

This Wikipedia article states that:

Most steel engines have a thermodynamic limit of 37%.

Is it correct? If so then it is not clear where this number comes from. Is it derived somehow from Carnot's theorem? I tried to google it but found nothing usefull.

If this limit of 37% is wrong then what is correct limit?

Answer

I basically agree with Chase's answer - with some additions. But let's see thoroughly what is behind that sentence in Wikipedia ("Most steel engines have a thermodynamic limit of 37%").

Internal combustion engines are modeled by the Otto cycle rather than by the Carnot cycle. Looking at this link for the Otto cycle (or this great link from MIT), you may observe that in this cycle there are four temperatures involved, and the efficiency is given by

$$ \epsilon_O = 1-\frac{T_4- T_1}{T_3 - T_2}. $$

However, in an ideal Otto cycle we have that $T_4/T_1 = T_3/T_2$, and we get an "effective Carnot form" for the efficiency:

$$ \epsilon_O = 1-\frac{T_1}{T_2}. $$

Now, using the isentropic equations of ideal gases, we obtain the following simple expression

$$ \epsilon_O = 1- \frac{1}{(V_1/V_2)^{\gamma -1}}, $$

where $\gamma$ is the specific heat ratio ($c_p/c_v$)

At the end of the Wikipedia sentence that you quote, there is a reference to a course by the University of Washington (see link here). There it is stated that most current auto engines have compression ratio $V_1/V_2=10$ and the mixture of air, gasoline vapor, CO$_2$, CO and H$_2$O has an effective specific heat ratio of $\gamma=1.27$. Plugging this into our formula we get $\eta_O=0.46=46\% $, which is pretty close to what Chase said. I would call this the theoretical limit for the efficiency of the internal combustion engine!

But how do they get $35-37\%$ instead? Here I have to quote the Uni of Washington link directly:

(...) you get h = 0.46. Multiply this by about 0.75 to account for real cycle effects (such as the time it takes to burn, heat losses to the coolant, and exhaust valves that open before the piston fully reaches bottom position) and you have h = 0.35.

So in conclusion, after a careful look through the different links, I have to agree with Chase (with the above described additions), the efficiency limit dictated by thermodynamics is ~46%, and the 35% limit arrives when additional real-life considerations are taken into account. Although admittedly this "let's multiply by 0.75" approach does not sound very well grounded on its own. But at least it answers your question "where this number comes from".

mathematics - Martin Gardner - Persistence

A number's persistence is :

• 
  

  The number of steps required to reduce it to a single digit by multiplying all its digits to obtain a second number

  
  


• 
  
  

  Then multiplying all the digits of that number to obtain a third number, and so on until a one-digit number is obtained.

  
  

For example : 77 has a persistence of four because it requires four steps to reduce it to one digit: 77-49-36-18-8.

The smallest number of persistence one is 10

The smallest of persistence two is 25

The smallest of persistence three is 39

The smaller of persistence four is 77

What is the smallest number of persistence five?

Answer

Brute forcing with Lua gave me

679

This is also confirmed by Wikipedia and OEIS:

http://en.m.wikipedia.org/wiki/Persistence_of_a_number

http://oeis.org/A003001

lateral thinking - Puzzling about a cube

I do not claim credit for this puzzle; I saw it online years ago and realised it hadn't been asked here yet (as far as I could tell), so I thought I'd share it with the community here. I cannot recall the text verbatim, so I sincerely hope that I do not end up being misleading. Here goes:

You start with small and equal cubes, 27 of which are put together to form a large cube. How does one then remove 1 from the middle of the large cube to leave only 26, while not touching any of the sides of the large cube?

Answer

Maybe this is a large cube and if you take away the middle one you get 26.

The $216$ is $6$ cubed ($6^3$) made of $27$ small cubes.  Taking away the middle '1' leaves "2 6".

Edited by asker:

Cubes in the puzzle are used in a figurative sense. Start with 27 small cubes: 2^3 = 8. Put them together to form a large cube: 8*27 = 216 = 6^3. You are now free to remove 1 from the middle and leave 26 behind, all without touching any of the sides!

general relativity - How long would it take to travel through a wormhole?

Assuming wormholes exist and you put some matter into one, how long would it take to reach the other end versus how far apart the two ends are? Basically, by how much does a wormhole stretch spacetime?

Answer

There's a recent popularization of wormhole physics that nicely lists the properties of the four wormhole examples that Morris and Thorne considered in their 1988 paper. These properties include traversal times. Here's a summary

1. 
   

   Infinite-Exotic-Region Wormhole (exotic matter distributed throughout space) ~ 1 hour

   
   



2. 
   

   Large-Exotic-Region Wormhole (exotic matter confined to large finite radius) >= 7 days

   
   


3. 
   

   Medium-Exotic-Region Wormhole (exotic matter loosely restricted to throat) ~ 200 days

   
   


4. 
   

   Small-Exotic-Region Wormhole (exotic matter closely restricted to throat) >= 0.7 seconds

   
   

Morris and Thorne referred to the last example as "absurdly benign". It is not dissimilar to the thin-shell wormholes considered by Visser.

These times are completely independent of the distances between the mouths of each wormhole in normal space.

Source: The Physics of Stargates -- Parallel Universes, Time Travel, and the Enigma of Wormhole Physics, by Enrico Rodrigo (2010), Chapter 5

Relation between Black-Scholes equation and quantum mechanics

I am interested in the link between the Black & Scholes equation and quantum mechanics.

I start from the Black & Scholes PDE $$ \frac{\partial C}{\partial t} = -\frac{1}{2}\sigma^2 S^2 \frac{\partial^2C}{\partial S^2} -rS\frac{\partial C}{\partial S}+rC $$ with the boundary condition $$C(T,S(T)) = \left(S(T)-K\right)_+.$$ Performing the change of variables $q=\ln(S)$ this equation rewrites $$ \frac{\partial C}{\partial t} = H_{BS}C $$ with the Black & Scholes Hamiltonian given by $$H_{BS} = -\frac{\sigma^2}{2} \frac{\partial^2}{\partial q^2}+\left(\frac{1}{2}\sigma^2 - r\right)\frac{\partial}{\partial q} +r.$$

Now I compare this equation with the Schrödinger equation for the free particle of mass $m$ : $$i\hbar \frac{d\psi(t)}{dt} = H_0\psi(t),\quad \psi(0)=\psi$$ with the Hamiltonian (in the coordinate representation) given by $$H_0 = -\frac{\hbar^2}{2m} \frac{d^2}{dq^2}.$$

My problem comes from the fact that the various references I am reading for the moment explain that the two models are equivalent up to some changes of variables (namely $\hbar=1$, $m=1/\sigma^2$ and the physical time $t$ replaced by the Euclidean time $-it$). However, their justifications for the presence of the terms $$\left(\frac{1}{2}\sigma^2 - r\right)\frac{\partial}{\partial q} +r$$ in the Hamiltonian seem very suspicious to me. One of them tells that these terms are "a (velocity-dependent) potential". Another one tells this term is not a problem since it can be easily removed is we choose a frame moving with the particle.

I have actually some difficulties to justify why, even with this term, we can say that the Black & Scholes system is equivalent to the one coming from the quantum free particle. I don't like this potential argument, since (for me) a potential should be a function of $q$ (so it would be ok for example for the $+r$ term) but not depending on a derivative.

Could you give me your thoughts on this problem?

quantum mechanics - Help on applying a Hadamard gate and CNOT to two single q-bits

I am stuck on a few issues in this video. (Note: It is at the frame concerning this question.)

In it, from what I understand (which could be wrong) we first apply the Hadamard gate to a qbit in the $\lvert 0 \rangle$ state. Then on that we use a controlled not on the output value, together with another input from another qbit set at $\lvert 0 \rangle$. I don't understand how to work this out and how the $\lvert 00 \rangle + \lvert 10 \rangle$ is derived for the output of combination of the two gates.

Here is what I am thinking (neglecting the $1/\sqrt{2}$ multiplier):

$$ \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \lvert 0 \rangle + \lvert 1 \rangle. $$

There however is another $\lvert 0 \rangle$ coming in from the second (bottom) wire and it interacts with the output from the Hadamard gate. I don't understand how to manipulate the values - have no idea if I can do something such as the following by distribution: $$ (\lvert 0 \rangle + \lvert 1 \rangle) \lvert 0 \rangle = \lvert 0 \rangle \lvert 0 \rangle + \lvert 1 \rangle \lvert 0\rangle. $$

I have no idea why we can do this (if we can even do so), but this is the only way I can see that we could get $\lvert 0 \rangle \lvert 0 \rangle + \lvert 1\rangle \lvert 0 \rangle = \lvert 00 \rangle + \lvert 10 \rangle$, as the author of the video has on the particular time in the video.

Any clarification would be extremely appreciated.

Answer

The basis you have for the two qubits in this example is the product basis, where all states describing the two qubits are products of states describing each individual qubit.

Some notation: $|01 \rangle$ is the same as $|0\rangle |1 \rangle$. The first state is for the first qubit, the second state is for the second qubit.

Thus, the most general 2-qubit state would look something like $( a|0\rangle + b|1\rangle)(c|0\rangle + d|1\rangle)$, taking normalization into account of course.

Some basic linear algebra would then tell you that you can indeed use the distributive property, so $$(|0\rangle + |1\rangle) |0\rangle = |00\rangle + |10\rangle$$

If you want to go the long way to convince yourself that you can do these things, write down the Hadamard gate in the 2-qubit basis: You have your four basis states, $|00\rangle, |01\rangle, |10\rangle$ and $|11\rangle$ and you know how the Hadamard gate acts on the first qubit and that it does not change the second qubit, so you could write down the corresponding 4x4 matrix.

newtonian mechanics - Effect of the tail of the cat in the falling cat problem

To explain why a falling cat can turn by 180 degree without external torque and without violation of the conservation of angular momentum, one usually models the cat as two cylinders as in

http://en.wikipedia.org/wiki/Falling_cat_problem

This may explain the turn. However I often heard contrary to this that she can rotate her body simply because she rotates her tail very fast into the opposite direction (and essentially keeps the rest of the body rigid).

So, what effect does the tail have in reality? Is there any detailed model, which takes the tail rotation into account and calculates how large its effect is?

quantum mechanics - Does measuring destroy entanglement

Before measuring a quantum particle(photon) it exists in a superposition state, once we observe(measure) it, it settles in one of the possible states(destroying superposition). For entangled particles, does measuring destroy entanglement as well?

Answer

Yes: If you measure an entangled property, you destroy the entanglement, always.

I think @OndřejČernotík has already already nicely answered this one, but I think your question included the assumption that the measurement and the entanglement were talking about the same property, e.g. spin polarization. And if so, the answer is just a simple yes.

Superposition and entanglement slight variants of the same phenomenon, that being the ability of a quantum system to contain more than one possible state at the same time. In superposition the multiple states mostly share a single relatively small region in xyz space, such as an atom, whereas in entanglement the states may be quite large.

Here's as more specific example of that commonality, first given by Einstein: If you have a very large wave function and then find the particle at one distant part of that wavefunction, how is it that that probability of finding that same particle at some other large region of the wave function a light second "instantly" drops to zero? How did that distant part of the wave function "know" at an apparently superluminal velocity that the particle had already been found?

Einstein's example was just another form of entanglement, not of conserved angular momentum but of mass-energy. The universe insists on absolute conservation of both, so in both cases finding the property (angular momentum or mass-energy) at one well-defined location requires that that property be balanced out or removed from the rest of the universe, no matter how large the wave function has become.

In terms of superposition, such examples show that any quantum wave function contains entanglement, even if it is "just" a smooth, simple wave packet for the particle location. It's just that the entanglement issue does not show up clearly until you make the wave packet so large that its curious conflicts with the speed of light become apparent.

fluid dynamics - Why does your car lurch toward an oncoming truck as it passes you?

I notice that the larger the truck the greater the magnitude of the lurch. Can anyone give a physical explanation to this?

Here's A Clean Riddle

I kinda like Riley riddles, so here goes nothing.

My beginning is a fake.
My ending, faeces it means.
My middle is a unit of electron flow rates.
However, as a whole, I keep you clean.

Please comment on how to improve on it! Thank you!

Answer

You are

SHAMPOO

Explanation:

SHAM - fake
AMP - unit of current (ampere)
POO - feces

special relativity - Why can't a spaceship accelerate for ever? Since there is no friction in space

I have seen many answers like: because we don't have infinite energy, because of gravity, because it is impossible, because of physics.

But they don't really answer my question.. I mean if there is no friction and at some points, even if just in theory, you could be so far away from other objects or at the right distance between many so that the total force vector is null... why is it still considered impossible?

pattern - What is a Hopeful Word™?

_{In the spirit of the What is a Word™/Phrase™ series started by JLee, a special brand of Phrase™ and Word™ puzzles.}

---

If a word conforms to a special rule, I call it a Hopeful Word™.
Use the examples below to find the rule.

$$ % set Title text. (spaces around the text ARE important; do not remove.) % increase Pad value only if your entries are longer than the title bar. % \def\Pad{\P{0.0}} \def\Title{\textbf{ Hopeful }} % \def\S#1#2{\Space{#1}{20px}{#2px}}\def\P#1{\V{#1em}}\ \def\V#1{\S{#1}{9}} \def\T{\Title\textbf{Words }^™\Pad}\def\NT{\Pad\textbf{Not}\T\ }\displaystyle \smash{\lower{29px}\bbox[yellow]{\phantom{\rlap{rubio.2017.02.04}\S{6px}{0} \begin{array}{cc}\Pad\T&\NT\\\end{array}}}}\atop\def\V#1{\S{#1}{5}} \begin{array}{|c|c|}\hline\Pad\T&\NT\\\hline % \text{ AGAINST }&\text{ FOR }\\ \hline \text{ AGENTS }&\text{ SPY }\\ \hline \text{ BASE }&\text{ LOFTY }\\ \hline \text{ BATTLE }&\text{ ACCORD }\\ \hline \text{ CIVIL }&\text{ CHURLISH }\\ \hline \text{ DEATH }&\text{ LIFE }\\ \hline \text{ DESTROY }&\text{ REBUILD }\\ \hline \text{ EVIL }&\text{ GOOD }\\ \hline \text{ POWER }&\text{ RESPONSIBILITY }\\ \hline \text{ REBEL }&\text{ COMPLY }\\ \hline \text{ SINISTER }&\text{ HOPEFUL }\\ \hline \text{ STRIKING }&\text{ CARESSING }\\ \hline \text{ VICTORY }&\text{ DEFEAT }\\ \hline \text{ WAR }&\text{ PEACE }\\ \hline \text{ WEAPON }&\text{ PLOWSHARE }\\ \hline \end{array}$$

And, if you want to analyze, here is a CSV version:

Hopeful Words™,Not Hopeful Words™
AGAINST,FOR
AGENTS,SPY

BASE,LOFTY
BATTLE,ACCORD
CIVIL,CHURLISH
DEATH,LIFE
DESTROY,REBUILD
EVIL,GOOD
POWER,RESPONSIBILITY
REBEL,COMPLY
SINISTER,HOPEFUL
STRIKING,CARESSING

VICTORY,DEFEAT
WAR,PEACE
WEAPON,PLOWSHARE

The puzzle satisfies the series' inbuilt assumption, that each word can be tested for whether it is a Hopeful Word™ without relying on the other words.
These are not the only examples of Hopeful Words™; many more exist.

What is the special rule these words conform to?

Answer

These are...

Words that appear in Star Wars: A New Hope's opening crawl:

Episode IV, A NEW HOPE: It is a period of civil war. Rebel spaceships, striking from a hidden base, have won their first victory against the evil Galactic Empire. During the battle, Rebel spies managed to steal secret plans to the Empire’s ultimate weapon, the DEATH STAR, an armored space station with enough power to destroy an entire planet. Pursued by the Empire’s sinister agents, Princess Leia races home aboard her starship, custodian of the stolen plans that can save her people and restore freedom to the galaxy...

And the non-Hopeful words are nowhere in sight. ;)

geometry - Tiling rectangles with Hexomino plus rectangle #1

Inspired by Polyomino T hexomino and rectangle packing into rectangle

See also series Tiling rectangles with F pentomino plus rectangles and Tiling rectangles with Hexomino plus rectangle #1

Next puzzle in the series: Tiling rectangles with Hexomino plus rectangle #2

The goal is to tile rectangles as small as possible with the given hexomino, in this case number 1 of the 25 hexominoes which cannot tile a rectangle alone. We allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one of the given hexomino will tile.

Example with the $1\times 1$ (or the $1\times 2$) you can tile a $2\times 5$ as follows:

Now we don't need to consider $1\times 1$ (or $1\times 2$) further as we have found the smallest rectangle tilable with copies of the hexomino plus copies of $1\times 1$ (or $1\times 2$).

Currently known rectangles exist for the hexomino plus $1\times 3$, $1\times 4$, $1\times 5$, $1\times 6$, $1\times 7$, $1\times 8$, $1\times 9$, $1\times 10$, $1\times 11$, $1\times 12$, $1\times 13$, $1\times 14$, $1\times 15$, $1\times 16$, $1\times 18$, $1\times 19$, $1\times 20$, $1\times 26$, $2\times 2$, $2\times 3$, $2\times 5$, $2\times 7$, $3\times 4$, $3\times 5$

Most of them could be tiled by hand fairly easily.

Answer

Well, the first one, for $2 \times 2$ is easy:

4x5=20

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It turns out to be rather unique; almost all other solutions I've found so far are part of three families of generalizable solutions, all of which are characterized by the fact that

the hexominos are lying next to each other, filling a narrow band of only two rows.

A summary of solution sizes is shown at the bottom.

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Type A

Let's start with type A, which is the minimal solution for $1 \times 3$ and $1 \times 4$ (shown):

This works by

putting $2k$ alternating hexominos in the indicated pattern, giving a bottom length of $6k$ which can be covered with $n-1$ rows of horizontal $1 \times n$ rectangles when $n$ divides $6k$. We'll need one horizontal rectangle on the top right, and $n$ vertical rectangles, giving a total box of $(n+1) \times (6k+n)$.

The minimum value of $k$ for which $n$ divides $6k$ is $\frac{\text{lcm}(6,n)}{6}$, so the total box is $(n+1) \times (\text{lcm}(6,n)+n)$.

A generalizable solution for $2 \times n$, $n$ odd uses the same hexomino pattern. It's the minimal solution for $2 \times 3$:

The total box here is $2n \times (6k+2)$. The minimum value of $k$ for which $n$ divides $6k$ is $\frac{\text{lcm}(6,n)}{6}$, so the total box is $2n \times (\text{lcm}(6,n)+2)$.

Type B

The minimal solution for $2 \times 7$ (and, by subdividing, $1 \times 7$) is

This leads to a generalizable solution (for both $1 \times n$ and $2 \times \text{odd} n$) by

putting $2k+1$ alternating hexominos in the indicated pattern, giving a bottom length of $6k+1$ which can be covered with $n-1$ rows of horizontal $1 \times n$ rectangles when $n$ divides $6k+1$. The total box size is $(n+1) \times (6k+5)$.

Type C

The last type, C, uses the same hexomino pattern as B but 'flipped'. Here is the minimal solution for $1 \times 11$:

This leads to a generalizable solution (only for $1 \times n$) by

putting $2k+1$ alternating hexominos in the indicated pattern, giving a bottom length of $6k+5$ which can be covered with $n-1$ rows of horizontal $1 \times n$ rectangles when $n$ divides $6k+5$. The total box size is $(n+1) \times (6k+1)+2n$.

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Let's make a list of minimal solutions. For $1 \times n$:

axb Type k Size axb Type k Size
1x3 A 1 4x 9 = 36 2x3 A 1 6x 8 = 48
1x4 A 2 5x16 = 80 2x4 -
1x5 C 0 6x11 = 66 2x5 B 4 6x29 = 174

1x6 A 1 7x12 = 84 2x6 -
1x7 B 1 8x11 = 88 2x7 B 1 8x11 = 88
1x8 A 4 9x32 = 288 2x8 -
1x9 A 3 10x27 = 270 2x9 A 3 18x20 = 360
1x10 A 5 11x40 = 440 2x10 -
1x11 C 1 12x29 = 348 2x11 B 9 12x59 = 708
1x12 A 2 13x24 = 312 2x12 -
1x13 B 2 14x17 = 238 2x13 B 2 14x17 = 238
1x14 (not minimal, see below)
1x15 A 5 16x45 = 720 2x15 A 5 30x32 = 960

1x16 A 8 17x64 = 1088
1x17 C 2 18x47 = 846
1x18 A 3 19x36 = 684
1x19 B 3 20x23 = 460

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The minimal $1 \times 14$ is an exception and does not belong to the generalizable solutions mentioned above:

16x50=800

The solutions for $3 \times 4$ and $3 \times 5$ use the same hexomino pattern but take more 'padding' to form a rectangle:

13x24 = 312

16x45 = 720

Finally, the $4 \times 5$ solution breaks with the pattern of generalizable solutions:

32x45 = 1440

optics - Does total internal reflection really reflect every single photon?

In certain cases of refraction, light can be totally internally reflected (TIR) instead of being transmitted. One learns that literally 100% of the light is reflected back in such an interaction.

My question is simple: To what degree is this true? Certainly, geometric imperfections could effectively lead to different geometry outside the TIR regime, but are there other effects? Tunneling? Photon-photon interaction? Interference? If so, about what strength are they ($1:10^3$, $1:10^6$, $1:10^{18}$, ...)?

Is there a maximum amount of photons that can exist in a certain amount of space?

If you have a set amount of space, lets say 10 cubic centimeters, and you would be able to trap photons in there.

If you would then add more and more photons to that space, could you then go on infinitely or would you eventually run into a maximum amount of photons that can be in that space?

Answer

1. 
   

   Photons are bosons, and they can exist with the same quantum numbers with an indefinite number of photons.

   
   


2. 
   

   Photons have zero mass, but they have energy $E=h\times \nu$; increasing their number and frequency increases the energy per cubic centimeter.

   
   


3. 
   
   

   Photons move with the velocity of light, and trapping them presupposes reflectors of one kind or another.

   
   

From 1., the answer is “no limit”.

From 2. and 3., the limit would come from the melting of the reflectors due to the high energy density. The number would be large and would depend on the frequencies present.

black holes - Do wormhole exits require matter? If not, would they change the edge of the universe?

As a layperson who watches way too many theoretical physics and astronomy documentaries, I've seen the folded-piece-of-paper analogy of wormholes a bajillion times. Usually they explain that black holes might connect to other parts of the universe, but they don't really explain what would have to be on the other end for any kind of information to travel through it. When they do, they usually suggest white holes -- but I'm not entirely convinced.

I don't have too much trouble understanding how a black hole forms, and to some extent, how that could lead to a distortion of space-time sufficient to causing the "fold". But it's hard to imagine what would cause a white hole to form. Even if there were such thing as negative mass, there doesn't seem to be any way that an object can come to have a negative density sufficient to becoming a white hole, so there couldn't possibly be a process that causes them to form.

So if black holes could be wormholes at all, it doesn't seem to be the case that the exit would require matter to be present before it forms. If anything, a black hole wormhole would have to "perforate" a white hole or some other kind of exit into existence -- somewhere.

But there doesn't seem to be anything that determines where that exit would form, so one could exist outside the boundaries of the matter that defines the edges of the universe. That would mean that the universe is infinite in size, since white holes could (and probably would, if their position were completely random) appear very, very far away.

So then either black holes can't be wormholes, the universe is infinite in size, or I'm missing something. Neither of the former two conclusions have been reached to the best of my knowledge -- so what am I missing?

Answer

The sort of wormholes you'll have seen on documentaries are not related to black holes.

The Scharzschild metric that describes a black hole contains a wormhole. This type of wormhole is known as a Schwarzschild wormhole, or there is a variant called an Einstein-Rosen Bridge, but nothing can traverse these wormholes. Even if it could, we'd need to watch for an infinite time to see anything emerge from the other end.

The sort of wormholes you've probably seen on TV are traversable wormholes, and these do allow matter to pass through them. However they probably don't exist. It's not hard to find wormhole solutions to the Einstein equations. You start with the wormhole geometry and work backwards to find out what stress-energy tensor you need to create the wormhole. The problem is that the result is always that you need negative matter i.e. matter that gravitationally repels other matter. It would take a brave scientist to declare absolutely that negative matter doesn't exist, but we've never seen any trace of it.

Incidentally, the universe is probably infinite or at least it doesn't make any sense to consider it having an edge. It could be finite but unbounded, like a sphere, but if so the length scale must be vastly larger than the observable universe because the observable universe is flat to within experimental error.

Rotating and charged black holes can act as links between different regions of space, but these aren't wormholes in the sense that you could use them as a short cut. If you pass through a rotating or charged black hole into a different region of space there is no way back to where you started from, or at least not without travelling faster than light. Actually the same is true of the Schwartzschild/Einsten-Rosen bridge wormhole. It links different regions of spacetime, but doesn't link two points in our spacetime.

You might be interested to have a look at my answer to Entering a black hole, jumping into another universe---with questions. This describes in more details hole black holes can link different areas of spacetime.

statistical mechanics - RG of the Gaussian Model: Finding the scaling factor

I'm studying how the Renormalization Group treatment of the simple Gaussian model, $$\beta H = \int d^d r \left[ \frac{t}{2} m^2(r) + \frac{K}{2}|\nabla m|^2 - hm(r)\right]$$

In momentum space, the Hamiltonian reads $$\beta H = \frac{1}{(2\pi)^d} \int d^d q \left[\frac{t + q^2 K}{2} |m(q)|^2\right] - hm(0)$$ and the $q$-integral runs from $0$ to some long-wavelength cut-off $\Lambda$.

The coarsening is done by splitting this integral into one from $0$ to $\Lambda/b$ and one from $\Lambda/b$ to $\Lambda$, and because the Gaussian model is so simple, the two integrals don't mix and decouple nicely. The high-momentum integral contributes just a constant additional term to the free energy, so we ignore it, and then we are left with $$\beta H = \frac{1}{(2\pi)^d} \int_0^{\Lambda/b} d^d q \left[\frac{t + q^2 K}{2} |m(q)|^2 \right] - hm(0).$$

The rescaling is done by introducing a new momentum $q' = bq$. For the order parameter $m(q)$, one makes the scaling assumption $m'(q') = m(q)/z$. Then I can rewrite $\beta H$ in terms of the new momentum variable $q'$, and then I demand that the rescaled Hamiltonian has the same functional form as the old Hamiltonian, which allows me to read off

$$t' = b^{-d} z^2 t$$ $$K' = b^{-d-2} z^2 K$$ $$h' = zh$$

Now we don't know $z$, and in the literature I found that one somehow demands that $K' = K$, so that $z = b^{d/2 + 1}$, and I don't really understand why we can make that demand, and if there are other possibilities. Could we also demand that $t' = t$ and read off a different $z$?

Answer

Yes. It is a merely a convenient choice to fix the term in front of the gradient. You may make some other choice and get some other equivalent RG flow. An interesting toy to play with is a field with kinetic term $K_1 q^2 + K_2 |q|^\alpha$ where $\alpha >0$ is some parameter. You can write an RG flow choosing to fix either $K_2$ or$K_1$. Look at the resulting fixed points and their stability in either flow.

electromagnetic radiation - How large or small can frequency in the EM spectrum get?

The largest frequency range is gamma rays, but does the EM spectrum 'stop' somewhere? Like is there a limit to how large a frequency can get? Or how small frequency can get? Is it one of those things that theoretically nothing is stopping it, but nothing in the universe can produce ways of such a frequency or beyond a certain limit?

Does light from galaxies redshift to the point where the wavelength is just insanely long? Long enough that we can't see them?

Answer

I'll start with the second of your questions. Yes, light from very distant galaxies gets redshifted to such long wavelengths that there practically isn't any light to see. The lower limit on frequency is zero. Obviously. Technically one could say there is no signal at $0\,Hz$, but that still put a lower boundary on the frequency. Objects on the edge of our cosmological horizon have their light redshifted almost infinitely by the time it reaches us.

As for upper bounds on frequency, there's two ways to think about this. Strictly speaking, there is no limit on how high a frequency can be. I could say the limit is that wavelength can't go below zero, but that would feel cheap and like cheating. A photon can have an arbitrarily high amount of energy. However, something else to consider is the limits on knowledge of frequency. If the wavelength goes below the Planck Length, we really don't have any measurement equipment that could theoretically ever measure it. Not only that, but the energy of a photon with this wavelength would be at the Planck Scale. At this scale, the Compton wavelength and the Schwarzschild radius of a Black Hole are about equal. That means any photon with this high a frequency has enough energy to spontaneously create a particle that immediately becomes a black hole and swallows the photon. This indicates two things: 1) We really need Quantum Gravity to adequately describe this energy scale and 2) we can be pretty sure that whatever is flying around with this energy is probably not photons.

So a lower limit on frequency is zero and an upper limit is that with a wavelength of one Planck Length. (At least, that's the upper limit until we find a nice theory of Quantum Gravity).

quantum mechanics - Free particle propagation amplitude (Peskin & Schroeder 2.1)

In section 2.1 on p. 14 of Peskin and Schroeder QFT, they calculate the amplitude of a free non-relativistic particle to propagate from $\mathbf{x_0}$ to $\mathbf{x}$ as $$U(t) = \left<\mathbf{x}|e^{-i\left(\mathbf{p}^2/2m\right)t}|\mathbf{x_0}\right>$$ $$ = \int \frac{d^3p}{\left( 2\pi\right)^3} \left<\mathbf{x}|e^{-i\left(\mathbf{p}^2/2m\right)t}|\mathbf{p}\right>\left<\mathbf{p}|\mathbf{x_0}\right>$$ $$ = \frac{1}{\left( 2\pi\right)^3} \int d^3p \ e^{-i(\mathbf{p}^2/2m)t} e^{i\mathbf{p}\cdot (\mathbf{x}-\mathbf{x_0})}$$ $$ = \ ...$$

using the nonrelativistic Hamiltonian $\mathbf{H} = \mathbf{p}^2/2m$. I think in going from the first to second line, they just inserted the identify operator as a sum over all the momentum projection operators, since the momentum eigenstates form a complete set. Could someone please explain the step from the second line to the third line?

Answer

The transition from the 2nd line to the 3rd simply uses the fact that

$$\langle \mathbf{p}|\mathbf{x} \rangle = e^{-i \mathbf{p}\cdot\mathbf{x}}.$$

The exponent of the Hamiltonian merely picks up the eigenvalue $\mathbf{p}$ from the state $|\mathbf{p}\rangle$, so you can take it out of the inner product, and what you have left is

$$\langle \mathbf{x}|\mathbf{p} \rangle \langle \mathbf{p}|\mathbf{x}_0 \rangle = e^{i \mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_0)},$$

since switching the order of states in the inner product gives you the complex conjugate, that is,

$$\langle \mathbf{x}|\mathbf{p} \rangle = \langle \mathbf{p}|\mathbf{x} \rangle^*.$$

Let me know if this is clear.

particle physics - What is the mass density distribution of an electron?

I am wondering if the mass density profile $\rho(\vec{r})$ has been characterized for atomic particles such as quarks and electrons. I am currently taking an intro class in quantum mechanics, and I have run this question by several professors. It is my understanding from the viewpoint of quantum physics a particle's position is given by a probability density function $\Psi(\vec{r},t)$. I also understand that when books quote the "radius" of an electron they are typically referring to some approximate range into which an electron is "likely" to fall, say, one standard deviation from the expectation value of its position or maybe $10^{-15}$ meters.

However it is my impression that, in this viewpoint, wherever the particle "is" or even whether or not the particle "had" any position to begin with (via the Bell Inequalities), it is assumed that if it were (somehow) found, it would be a point mass. This has been verified by my professors and GSIs. I am wondering if its really true.

If the particle was truly a point mass then wherever it is, it would presumably have an infinite mass density. Wouldn't that make electrons and quarks indistinguishable from very tiny black holes? Is there any practical difference between saying that subatomic particles are black holes and that they are point masses? I am aware of such problems as Hawking Radiation although at the scales of the Schwarchild radius of an electron (back of the envelope calculation yields $\tilde{}10^{-57}$ meters), would it really make any more sense to use quantum mechanics as opposed to general relativity?

If anyone knows of an upper bound on the volume over which an electron/quark/gluon/anything else is distributed I would be interested to know. A quick Google Search has yielded nothing but the "classical" electron radius, which is not what I am referring to.

Thanks in advance; look forward to the responses.

Answer

Let me start by saying nothing is known about any possible substructure of the electron. There have been many experiments done to try to determine this, and so far all results are consistent with the electron being a point particle. The best reference I can find is this 1988 paper by Hans Dehmelt (which I unfortunately can't access right now) which sets an upper bound on the radius of $10^{-22}\text{ m}$.

The canonical reference for this sort of thing is the Particle Data Group's list of searches for lepton and quark compositeness. What they actually list in that reference is not exactly a bound on the electron's size in any sense, but rather the bounds on the energy scales at which it might be possible to detect any substructure that may exist within the electron. Currently, the minimum is on the order of $10\text{ TeV}$, which means that for any process occurring up to roughly that energy scale (i.e. everything on Earth except high-energy cosmic rays), an electron is effectively a point. This corresponds to a length scale on the order of $10^{-20}\text{ m}$, so it's not as strong a bound as the Dehmelt result.

Now, most physicists (who care about such things) probably suspect that the electron can't really be a point particle, precisely because of this problem with infinite mass density and the analogous problem with infinite charge density. For example, if we take our current theories at face value and assume that general relativity extends down to microscopic scales, an point-particle electron would actually be a black hole with a radius of $10^{-57}\text{ m}$. However, as the Wikipedia article explains, the electron's charge is larger than the theoretical allowed maximum charge of a black hole of that mass. This would mean that either the electron would be a very exotic naked singularity (which would be theoretically problematic), or general relativity has to break at some point before you get down to that scale. It's commonly believed that the latter is true, which is why so many people are occupied by searching for a quantum theory of gravity.

However, as I've mentioned, we do know that whatever spatial extent the electron may have cannot be larger than $10^{-22}\text{ m}$, and we're still two orders of magnitude away from probing that with the most powerful particle accelerator in the world. So for at least the foreseeable future, the electron will effectively be a point.