lagrangian formalism - Semiclassical limit of Quantum Mechanics

I find myself often puzzled with the different definitions one gives to "semiclassical limits" in the context of quantum mechanics, in other words limits that eventually turn quantum mechanics into classical mechanics.

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In a hand-wavy manner

• Classical or semiclassical limit corresponds to the limit of taking $\hbar \to 0.$


• Often when talking about the correspondence principle, the semiclassical limit is obtained in the limit of large quantum numbers (large orbits and energies).

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More precisely

• Exemplary source of confusion: One way to show why $\hbar \to 0$ describes a classical limit, is as follows:

Take the $1D$ Schrödinger equation for a particle of mass $m$ in a potential $V(\vec x)$:

\begin{equation} i\hbar \frac{\partial}{\partial t}\psi(\vec x,t) = \left[-\frac{-\hbar^2}{2m}\vec \nabla^2+V(\vec x)\right]\psi(\vec x,t) \end{equation}

By inserting $\psi(\vec x,t)=e^{iS(\vec x,t)/\hbar}$ in the Schrödinger equation above, and simplifying for $\psi$, we obtain:

$$ -\frac{\partial S}{\partial t}=\frac{1}{2m}(\vec\nabla S)^2-\frac{i\hbar}{2m}(\vec \nabla^2S)+V $$ Now taking $\hbar \to 0$, the above just becomes the classically well known Hamilton-Jacobi equation, where $S$ describes Hamilton's principal function or the action:

$$ -\frac{\partial S}{\partial t}=\frac{1}{2m}(\vec\nabla S)^2+V $$ Using such result, then we can use an $\hbar$ expansion of $S$ in the second equation. Unfortunately I fail to see why reaching the Hamilton-Jacobi equation necessarily implies a classical behavior!

1. 
   

   Alternatively one talks about classical limits of QM by saying: When Planck's quantum $\hbar$ becomes very small compared with values of the Lagrangian action integral (Feynman's path integral formalism). I probably shouldn't ask this (since the discussion is rather vague here), but is there any neat way of demonstrating the above idea mathematically? (e.g. by showing whether such limit necessarily leads to quantum decoherence and hence the classical trajectories become dominant.)

   
   


2. 
   

   Finally, are the two statements of $\hbar \to 0$ and taking the limit of high quantum numbers somehow equivalent? (i.e. a reformulation of one another?)

   
   
   

Of course any other ways (whether physical or mathematical) of thinking about and understanding semiclassical limits of quantum mechanics are also welcomed as answer.

Answer

First, the classical and semiclassical adjectives are not quite synonyma. "Semiclassical" means a treatment of a quantum system whose part is described classically, and another part quantum mechanically. Fields may be classical, particle positions' inside the fields quantum mechanical; metric field may be classical and other matter fields are quantum mechanical, and so on.

Also, we often treat the "quantum part" of the semiclassical treatment in another approximation – where we take the leading classical behavior plus the first quantum correction only. For this part of the system, the "semiclassical" therefore means "one-loop approximation" (like in the WKB approximation).

Now, the laws of quantum mechanics may be shown to imply the laws of classical physics for all the "classical questions" whenever $\hbar\to 0$. More properly, $\hbar\to 0$ indeed means $J / \hbar \to \infty$ for all the normal "angular momenta" $J$, actions $S$ (instead of $J$), and everything else with the same units. So yes, indeed, the $\hbar\to 0$ classical limit and the limit of the large quantum numbers is the same thing. It is not kosher to ask whether a dimensionful quantity such as $\hbar$ is much smaller than one; whether the numerical value is small depends on the units. So we must make these claims about "very small" or "very large" dimensionless, and that's why we need not just $\hbar$ but also $J$ or $S$ of the actual problem, and that's why all the inequalities dictating the classical limit that you mentioned are equivalent.

In this limit, the spectra become so dense that the observables (such as the energy of the hydrogen atom) are effectively continuous even though they are discrete in the exact quantum treatment. The Heisenberg equations of motion for the operators reduce to the classical equations of motion. Decoherence guarantees that with some environment, the diagonal entries of the density matrix may be interpreted as classical probabilities, and the off-diagonal ones quickly go to zero. We may always imagine that the wave functions in this limit are "narrow packets" whose width is negligible and whose center moves according to the classical equations. It just works.

One should understand all aspects of this proof that "classical physics is a limit of quantum mechanics", what it assumes, how we should ask the questions and translate them from one formalism to another, and so on. But at the end, the fact that this statement holds is more important than some technical details of the proof.

Historically, the Hamilton-Jacobi equation is a way to describe classical physics because it was discovered and shown equivalent to classical physics long before the quantum theory was first encountered. Mathematically, you can see that the Hamilton-Jacobi equation only contains the quantities we may actually measure with classical apparatuses such as $S,t,V,m$ etc. and it doesn't depend on $\hbar$ at all – even if you use the SI units, for example – which proves that the equation is independent of quantum mechanics.

There are lots of things to say about the classical limit of quantum mechanics and some more specific classes of quantum mechanical theories, see e.g.

http://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html?m=1

mathematics - Professor Halfbrain and the 99x99 chessboard (Part 1)

Professor Halfbrain has spent the last weekend with filling the squares of a $99\times99$ chessboard with real numbers from the interval $[-1,+1]$. Whenever four squares form the corners of a rectangle (with sides parallel to the sides of the chessboard), then the four numbers in these squares had to add up to zero.

Professor Halfbrain has proved two extremely deep theorems on this.

Professor Halfbrain's first theorem: It is possible to fill the chessboard subject to the above rules, so that the sum of all numbers in all the squares is $0$.

Professor Halfbrain's second theorem: If we fill the chessboard subject to the above rules, then the sum of all numbers in all the squares is at most $9801$.

This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper. Find an integer $x$, so that "the sum in all the squares is $0$" in the first theorem may be replaced by "the sum in all the squares is $x$", and so that "the sum is at most $9801$" in the second theorem may be replaced by "the sum is at most $x$" (again yielding true statements, of course).

Answer

This is a sketchy proof of what Ivo suspected in the comments.

Take an arbitrary selection as below.

$A+B+E+F = 0$

$A+B+I+K = 0$

$=> E+F=I+K$

$E+F+I+K = 0$

$=>E +F = -I-K$

$E = -F$

$I = -K$

Similarily, we can obtain

$F+K+E+I = 0$

$G+L+F+K = G+L + E+ I = 0 => F+K-E-I = 0$

$F = -K$

$E = -I$

$F = I$

$E = K$

So, for an arbitrary selection, we know that the diagonal elements are equal. Taking an arbitrary selection with odd sides (e.g. $A+C+I+L = 0$), we know that $A,C,I,L$ must be equal and hence 0. Since this can be done for all squares, they must all be 0.

electromagnetism - How is Gauss' Law (integral form) arrived at from Coulomb's Law, and how is the differential form arrived at from that?

On a similar note: when using Gauss' Law, do you even begin with Coulomb's law, or does one take it as given that flux is the surface integral of the Electric field in the direction of the normal to the surface at a point?

Answer

Let us for simplicity consider $n$ point charges $q_1$, $\ldots$, $q_n$, at positions $\vec{r}_1$, $\ldots$, $\vec{r}_n$, in the electrostatic limit, with vacuum permittivity $\epsilon_0$.

Now let us sketch one possible strategy to prove Gauss' law from Coulomb's law:

1. 
   

   Deduce from Coulomb's law that the electric field at position $\vec{r}$ is $$\tag{1} \vec{E}(\vec{r})~=~ \sum_{i=1}^n\frac{q_i }{4\pi\epsilon_0}\frac{\vec{r}-\vec{r}_i}{|\vec{r}-\vec{r}_i|^3} . $$

   
   



2. 
   

   Deduce the charge density $$\tag{2} \rho(\vec{r})~=~\sum_{i=1}^n q_i\delta^3(\vec{r}-\vec{r}_i). $$

   
   


3. 
   

   Recall the following mathematical identity $$\tag{3}\vec{\nabla}\cdot \frac{\vec{r}}{|\vec{r}|^3}~=~4\pi\delta^3(\vec{r}) .$$ (This Phys.SE answer may be useful in proving eq.(3), which may also be written as $\nabla^2\frac{1}{|\vec{r}|}=-4\pi\delta^3(\vec{r})$).

   
   


4. 
   

   Use eqs. (1)-(3) to prove Gauss' law in differential form $$\tag{4} \vec{\nabla}\cdot \vec{E}~=~\frac{\rho}{\epsilon_0} .$$

   
   


5. 
   
   

   Deduce Gauss' law in integral form via the divergence theorem.

   
   

quantum field theory - How to count and 'see' the symmetry factor of Feynman diagrams?

Could somebody explain how one can derive the symmetry factor both by counting possible contractions and by looking at the symmetry of a diagram. Consider for example this diagram

in $\phi^4$-theory with $\mathcal{L}_\text{int} = -\frac{\lambda}{4!}\phi^4$.

special relativity - Why does the (relativistic) mass of an object increase when its speed approaches that of light?

I'm reading Nano: The Essentials by T. Pradeep and I came upon this statement in the section explaining the basics of scanning electron microscopy.

However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. At such velocities, one needs to do relativistic correction to the mass so that it becomes[...]

We all know about the famous theory of relativity, but I couldn't quite grasp the "why" of its concepts yet. This might shed new light on what I already know about time slowing down for me if I move faster.

Why does the (relativistic) mass of an object increase when its speed approaches that of light?

Answer

The complete relevant text in the book is

The de Broglie wave equation relates the velocity of the electron with its wavelength, $\lambda = h/mv$ ... However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. ...

Actually, the de Broglie wavelength should be $$ \lambda = \frac hp, $$ where $p$ is the momentum. While $p = mv$ in classical mechanics, in special relativity the actual relation is $$ \mathbf p = \gamma m \mathbf v = \frac{m\mathbf v}{\sqrt{1-\frac{v^2}{c^2}}} $$ where $m$ is the rest mass. If we still need to make the equation $p = mv$ correct, we introduce the concept of "relativistic mass" $M = \gamma m$ which increases with $v$.

electromagnetism - Magnetic force on a charged particle viewed from 2 inertial frames

Suppose we have a magnetic field directed into the computer screen. Assume that there is no gravitational field. Now suppose we have a stationary positively charged particle in the magnetic field. The magnetic force is given by $$\mathbf{F}=q\mathbf{v}\times\mathbf{B}$$

Since $\mathbf{v}=0, \mathbf{F}=0$.

Now pick a reference frame moving to the left with constant velocity. In this frame, the particle is moving to the right with constant velocity. So, the magnetic force is directed upwards.

From what I've learnt in physics, forces are independent of choice of reference frames as long as the frames are inertial. But in the 2 inertial reference frames I have chosen, I get 2 different magnetic forces. How do we reconcile this?

Answer

The resolution to this paradox is that the electric and magnetic fields need to be transformed when you change from one inertial frame to another, through rules fixed by special relativity. This is explained in this Wikipedia page or in your favourite EM textbook's relativity section, but the short of it is that in the frame transformation you describe, the magnetic field is tranformed into an electric field given by $$ \mathbf E'= \gamma \, \mathbf u \times\mathbf B, $$ where $\mathbf u$ is the velocity of the new frame with respect to the old one and $\gamma = 1/\sqrt{1-u^2/c^2}$. This new electric field acts on the particle to fully cancel out the effect of the magnetic force it feels on the new frame - electric and magnetic fields change from one inertial frame to the next, but if the total force is zero in one frame it must be zero in all frames.

It can feel pretty funny that one needs to invoke relativity to deal with what was originally a galilean reference-frame transformation, but that's just the way the cookie crumbles: electromagnetism was a fully relativistic theory from the time Maxwell formulated it, and in fact it was precisely these sorts of symmetries in electromagnetic analyses from different reference frames that motivated Einstein to develop special relativity.

geometry - How do you walk across a 10' x 10' hole using two 9' boards?

Given a 10' x 10' hole which is very deep, how do you walk across the hole using two boards that are each 9' long?

You can't jump, pole vault across, nail the boards together, or go around the hole.

You can start on any side you like, but once you start to cross, you must stay on the two boards. No stepping off on the sides.

Answer

You put the first board

at a 45 degree angle across one corner

of the hole. Then you put the other one

with one end on the first board and the other end on the side of the hole.

Thanks to Julian Rosen and Geobits for their insight on how my drawing was bad. I updated it to be more accurate:

poetry - I will be uttering this with a moan

What line is hidden behind the stars?

I will be uttering this with a moan
someplace aeons and aeons therefore:

$*****$ $*****$ $*****$ $*****$
and I picked up the one with inferior migration,
And this has caused complete disagreement

Answer

Thanks for this.

The Path Not Taken by Robert Frost.

I shall be telling this with a sigh
Somewhere, ages and ages hence:
Two roads diverged in a wood and I,

I took the one less traveled by
And that has made all the difference.

Much as I hate to add to or subtract from the verse above, @BigBlackBox and @LightnessRacesInOrbit have a point and an obscured line ought really to be part of an accepted answer. Any upvotes therefore properly belong to him. Technically, there should also be an explanation of the reasoning as well but there is pretty much a one-to-one mapping between the original and the obscured version. So:

A thoroughfare bifurcated in a copse and, myself-

pattern - What is a BEN Number™?

_{This is in the spirit of the What is a Word/Phrase™ series started by JLee with Number version puzzles.}

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If a number conforms to a special rule, I call it a BEN Number™.
Use the following examples below to find the rule.

$$ % set Title text. (spaces around the text ARE important; do not remove.) % increase Pad value only if your entries are longer than the title bar. % \def\Pad{\P{0.0}} \def\Title{\textbf{ BEN }} % \def\S#1#2{\Space{#1}{20px}{#2px}}\def\P#1{\V{#1em}}\def\V#1{\S{#1}{9}} \def\T{\Title\textbf{Numbers}^{\;\!™}\Pad}\def\NT{\Pad\textbf{Not}\T\ }\displaystyle \smash{\lower{29px}\bbox[yellow]{\phantom{\rlap{rubio.2019.05.15}\S{6px}{0} \begin{array}{cc}\Pad\T&\NT\\\end{array}}}}\atop\def\V#1{\S{#1}{5}} \begin{array}{|c|c|}\hline\Pad\T&\NT\\\hline % \text{ 17 }&\text{ 57 }\\ \hline \text{ 20 }&\text{ 60 }\\ \hline \text{ 24 }&\text{ 64 }\\ \hline \text{ 49 }&\text{ 89 }\\ \hline \text{ 121 }&\text{ 161 }\\ \hline \text{ 158 }&\text{ 198 }\\ \hline \text{ 538 }&\text{ 578 }\\ \hline \text{ 830 }&\text{ 870 }\\ \hline \text{ 1,059 }&\text{ 1,099 }\\ \hline \text{ 1,144 }&\text{ 1,184 }\\ \hline \text{ 1,622 }&\text{ 1,662 }\\ \hline \text{ 2,345 }&\text{ 2,385 }\\ \hline \end{array}$$

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

BEN Numbers™,Not BEN Numbers™
17,57
20,60
24,64
49,89
121,161
158,198
538,578

830,870
1059,1099
1144,1184
1622,1662
2345,2385

Bonus Question:
It's true that the cardinality of BEN Numbers' set is finite. What is the largest BEN Number?

Hint 1:

Hint 2:

Hint 3:

Hint 4:

Hint 5:

Hint 6:

Hint 7:

Answer

Ok, so a ben number is a number that

when split like a clock number, and put on a clock, make an acute angle (< 90º)

Proof:

• 17 -> 01:07 -> 12º


• 20 -> 02:00 -> 60º


• 24 -> 02:04 -> 36º


• 49 -> 04:09 -> 66º


• 121 -> 01:21 -> 66º


• 158 -> 01:58 -> 42º


• 538 -> 05:38 -> 78º



• 830 -> 08:30 -> 60º


• 1059 -> 10:59 -> 54º


• 1144 -> 11:44 -> 66º


• 1622 -> 16:22 -> 12º


• 2345 -> 23:45 -> 60º

(To calculate just multiply the hour by 5, subtract the minutes, and then the multiply by 6)

All numbers in the Not Ben Column end up with a number bigger than 59 (except the first 57 -> 05:07 -> 108º)

The hints helped so:

1-

It's the Sun Moon Reservoir, sun and moon indicating time (thanks to @JS1)

2-

It's represanting day

3-

A triangle, that contains 3 acute angles

4-

Degree

5-

They are cute -> Acute

6-

BEN -> Big Ben, the clock

7- What helped me

From the movie 'SPLIT', this is what helped me because I was splitting the numbers differently (I was making 49 as 00:49, not 04:09)

The largest BEN number is then

2359 -> 23:59 -> 24º

quantum field theory - Regarding a small step in the derivation of the LSZ formula

I'd like to prove the LSZ formula, but there is a specific step that is bugging me a lot. I know there are many subtleties in its derivation, but I'm not worrying about this right now: I'm trying to understand the naive proof, so to speak.

You can find an example of the usual proof here: http://isites.harvard.edu/fs/docs/icb.topic473482.files/06-LSZ.pdf

My question is: how to get eqs. (24-25) from eq. (22-23). In (22-23) the time-ordering symbol is to the left of $(\partial^2+m^2)\phi$ and in (24-25) it is to the right of the differential operator.

I'm asking how to go from $T(\partial^2+m^2)\phi\cdots$ to $(\partial^2+m^2) T\phi\cdots$. I feel that this cannot be done in general, because the symbol $T$ will introduce some $\Theta(x_0-y_0)$ functions, which when differentiated will give rise to some deltas.

This step makes no sense to me... Is my question legitimate? Is there anything from the proof I'm missing?

If I think of specific examples, I find different results depending on whether the $T$ symbol is to the right or to the left of the Klein-Gordon differential operator. This means that the order is important... so, which is the right order? Should the $T$ symbol be placed to the right or to the left of the KG operator? In the begining of the proof the $T$ symbol is always to the left, and in the end, it is always to the right.

This same problem appears on many proofs online, such as

http://isites.harvard.edu/fs/docs/icb.topic473482.files/06-LSZ.pdf (eqs. 22-25)

http://www2.ph.ed.ac.uk/~egardi/MQFT_2013/MQFT_2013_lecture_2.pdf (eqs. 32-33)

Srednicki's book, page 51 (eqs. 5.14-5.15) online: http://web.physics.ucsb.edu/~mark/qft.html

etc.

Edit: I was asked to post a self-contained question, so I'll write the delails here:

Take the in state to be $|i\rangle\propto a^\dagger_1(-\infty)a^\dagger_2(-\infty)|0\rangle$ and the out state to be $|f\rangle\propto a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle$. Then the transition amplitude is $$ \langle i|f\rangle \propto \langle 0|a_2(-\infty)a_1(-\infty)a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle= $$ $$ =\langle 0|Ta_2(-\infty)a_1(-\infty)a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle $$

Now write $a(+\infty)-a(-\infty)\propto \int \mathrm d x\ \mathrm e^{\cdots}(\partial^2+m^2)\phi$; all terms with $a$'s or $a^\dagger$'s annihilate the vacuum, so that the only remaining term is $$ \langle i|f\rangle \propto \langle 0|T \int \mathrm dx_1\mathrm dx_2\mathrm dx_3 \mathrm dx_4\ \mathrm e^{\cdots}(\partial_1^2+m^2)(\partial_2^2+m^2)(\partial_3^2+m^2)(\partial_4^2+m^2) \phi_1\phi_2\phi_3\phi_4 |0\rangle $$

This is usually writen as $$ \langle i|f\rangle \propto \int \mathrm dx_1\mathrm dx_2\mathrm dx_3 \mathrm dx_4\ \mathrm e^{\cdots}(\partial_1^2+m^2)(\partial_2^2+m^2)(\partial_3^2+m^2)(\partial_4^2+m^2) \langle 0|T\phi_1\phi_2\phi_3\phi_4 |0\rangle $$ but this step is problematic, because the $T$ symbol and the KG operator don't commute, right?

Answer

Comments to the question (v4):

1. 
   

   OP is wondering about the contact terms from commuting time-derivatives and time-ordering symbol $T$, cf. e.g. this and this Phys.SE posts.

   
   



2. 
   

   Consider the on-shell $S$-matrix side of the LSZ reduction formula. The time-differentiation from the boundary terms $$ T\left[\prod_{i=1}^n \left\{ a_{{\bf p}_i}^{\#}(t_i\!=\!\infty) -a_{{\bf p}_i}^{\#}(t_i\!=\!-\infty)\right\}\right] $$ $$~=~T\left[\prod_{i=1}^n \int_{\mathbb{R}}\! \mathrm{d}t_i~\frac{d}{dt_i}a_{{\bf p}_i}^{\#}(t_i)\right] ~=~\left[\prod_{i=1}^n \int_{\mathbb{R}}\! \mathrm{d}t_i~\frac{d}{dt_i}\right]T\left[\prod_{j=1}^n a_{{\bf p}_j}^{\#}(t_j)\right]\tag{A}$$ can be moved outside the time-ordering symbol $T$ because the contact terms vanish $$\delta(t_i-t_j) \left[a_{{\bf p}_i}^{\#}(t_i) ,a_{{\bf p}_j}^{\#}(t_j) \right]~=~0\tag{B} $$ for generic$^1$ 3-momenta ${\bf p}_i\neq {\bf p}_j$. Eq. (B) follows from locality, i.e. spatially separated operators commute. [Here $\#$ refers to creation/annihilation operators, i.e. with or without Hermitian conjugate.]

   
   


3. 
   

   In the Hamiltonian formulation with only first-order time-derivatives, the above shows that contact terms vanish.

   
   


4. 
   

   In the Lagrangian formulation with second-order time-derivatives (i.e. one more time differentiation, which is the case OP is asking about), one may show using similar arguments, that contact terms do not contribute to the S-matrix. See also e.g. Ref. 1.

   
   

References:

1. M.D. Schwartz, QFT and the Standard Model; Section 6.1, p.72, footnote 2.

--

$^1$ Also note that we are usually not interested in disconnected parts of the $S$-matrix, which implies more momentum conservation laws, and hence more special values of the momenta, such as, e.g., ${\bf p}_i={\bf p}_j$.

particle physics - Does the gravity affect voltage in a circuit?

The electric current is a flow of electrons, which have mass (small, but it is still a mass). So, considering a planar circuit, do the properties of the electric current (voltage, intensity) change with the orientation of the circuit with respect to the ground? I would expect some difference between the cases where the circuit is parallel or perpendicular to the ground.

mathematics - Creating 2020 in the fewest number of steps

You start with the number 1. You can create a new number by applying an operation on two existing numbers (can be the same). The operations are +, - and *. What is the fewest number of steps needed to reach the number 2020? Bonus question: can you find multiple solutions?

Good luck!

Answer

7 is the minimum number of operations

This should be all of the shortest length solutions, some of these have already been answered, and I'll leave credit to those that found them.

I'm also including the brute force python code that I used to exhaust all the combinations. That's how I was able to arrive at the the answer to the minimum length to be what it is.

Solution 1

Found first by @hexomino

1 + 1 = 2
2 + 1 = 3
3 + 2 = 5
5 * 3 = 15
15 * 3 = 45
45 * 45 = 2025
2025 - 5 = 2020

Solution 2

1 + 1 = 2
2 + 2 = 4
4 + 1 = 5
5 + 4 = 9
9 * 5 = 45
45 * 45 = 2025
2025 - 5 = 2020

Solution 3

First found by @Jens

1 + 1 = 2
2 + 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 5 = 100
100 + 1 = 101
101 * 20 = 2020

Solution 4

Found first by @Benoit Esnard

1 + 1 = 2
2 + 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 5 = 100
100 * 20 = 2000
2000 + 20 = 2020

Solution 5

1 + 1 = 2
2 + 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 20 = 400
400 + 4 = 404
404 * 5 = 2020

Solution 6

First found by @hexomino

1 + 1 = 2
2 + 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 20 = 400
400 * 5 = 2000
2000 + 20 = 2020

Solution 7

First found by @sudhackar

1 + 1 = 2
2 * 2 = 4
4 + 1 = 5
5 + 4 = 9
9 * 5 = 45
45 * 45 = 2025

2025 - 5 = 2020

Solution 8

1 + 1 = 2
2 * 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 5 = 100
100 + 1 = 101

101 * 20 = 2020

Solution 9

First found by @Teejay

1 + 1 = 2
2 * 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 5 = 100

100 * 20 = 2000
2000 + 20 = 2020

Solution 10

1 + 1 = 2
2 * 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 20 = 400

400 + 4 = 404
404 * 5 = 2020

Solution 11

1 + 1 = 2
2 * 2 = 4
4 + 1 = 5
5 * 4 = 20
20 * 20 = 400

400 * 5 = 2000
2000 + 20 = 2020

Brute force search python code

def mdFormat(nums, ops, ans, sol_no):
    #Formatting the solutions for markdown
    subheader="Solution  %s"%sol_no
    subheader_lines='-'*len(subheader)
    steps = []
    val = nums[0]

    ans = ans[1:]
    for i, num in enumerate(nums[1:]):
        steps.append('>! %s %s %s = %s 
'%(val, ops[i], num, ans[i]))
        val = ans[i]
    s = [subheader, subheader_lines]
    s.extend(steps)
    s.append('\n')
    return '\n'.join(s)

def apply_operations(numbers, operations):

    #Gives us the new list of number choices
    if len(numbers) == 1:
        return [numbers[0]]

    n_seq = (numbers[0], )
    n = numbers[0]

    for i, num in enumerate(numbers[1:]):
        if operations[i] == '+':
            n += num

        elif operations[i] == '-':
            n -= num
        elif operations[i] == '*':
            n *= num

        n_seq += (n, )

    return n_seq

solutions_found = 0


def search_n_operations(n, last_numbers=(1,), last_operations=None, choices=(1, )):
    global solutions_found

    if n == 0: #we're done with the recursion
        return

    if last_operations is None:
        op_combos = (next_op for next_op in ('+', '-', '*'))
    else:

        op_combos = (last_operations + (next_op,) for next_op in ('+', '-', '*'))

    for operation_seq in op_combos:
        num_combos = (last_numbers + (next_val,) for next_val in set(choices))
        for number_seq in num_combos:
            new_choices = apply_operations(number_seq, operation_seq)
            if new_choices[-1] == 2020: #This is an answer!
                solutions_found += 1
                print mdFormat(number_seq, operation_seq, new_choices, solutions_found)


            if last_operations is None:
                operation_seq = (operation_seq, )

            search_n_operations(n - 1, number_seq, operation_seq, new_choices)


n = 10
search_n_operations(n)
print "A total of %s solutions were found for %s operations"%(solutions_found, n)

Varying the n should illustrate where the minimum bound is.

Outputs for n < 7:
A total of 0 solutions were found for 1 operations
A total of 0 solutions were found for 2 operations
A total of 0 solutions were found for 3 operations
A total of 0 solutions were found for 4 operations
A total of 0 solutions were found for 5 operations
A total of 0 solutions were found for 6 operations

quantum mechanics - Invariance of a maximally entangled state under unitary operation $U otimes U^dagger$

Apparently the (d-dimensional) maximally entangled state, $|E \rangle = \sum_{i} |ii\rangle /\sqrt{d}$ is invariant under operations of the form $U \otimes U^\dagger$. I want to prove this result, which amounts to showing that

$$\left(\sum_i U | i \rangle \otimes V |i \rangle = \sum_i | ii \rangle \right)\Rightarrow (V = U^\dagger )$$

I have no idea how to even start. I suppose it's some simple linear algebra result, but I don't see it. A hint would be appreciated.

Answer

Write the terms $U |i \rangle$ and $V | i \rangle$ explicitly in the matrix form, e.g., $U | i \rangle = \sum_{k} | k \rangle U_{ki}$ and analogously. Then play with exchanging the three summations of indices, you should get it proved.

quantum mechanics - Are all scattering states un-normalizable?

I am an undergraduate studying quantum physics with the book of Griffiths. in 1-D problems, it said a free particle has un-normalizable states but normalizable states can be obtained by sum up the solutions to independent Schrodinger equations. in my view the book also suggests that a scattering state with E>V(infinity) is to be un-normalizable. Is it true in 1-D situations? If so can it be generalized to 3-D situations? And why?

mathematics - How do I convince my grandmother?

How do I convince my grandmother (who really hates mathematics) that there exist three positive integers $x,y,z$ that satisfy the equation $28x+30y+31z=3650$?

Answer

You tell your Grandmother:

Take ten ordinary years. They contain exactly 3650 days, and these days are split among several months with 28, 30, and 31 days.

states of matter - Why did my windshield freeze instantaneously?

It was very cold outside, this morning, when I took the car that slept in the snow, with a simple cloth on the windshield. I entered the vehicle, drove a kilometer or so. The air inside was so cold I could see my breath (or maybe I forgot to brush my teeth). Warm air was blowing on the windshield from the inside and suddenly…

The glass in front of my became opaque, starting from the bottom (the place where the hot air was blowing) and freezing up, up, up until the whole screen was filled in about 5 seconds. It reminded my a little bit of how some baterias spread.

When I tried to remove the mist, I realised it was ice in the car. My windshield instant froze.

I know the theory behind supercooling: a very cold and still liquid can freeze when moved. I'm not sure what happened here, but there were some pretty big turns in the road before it froze. It really looked like the hot air froze the place and not the movement.

Any idea about what happened ?

Answer

Maybe that was just window frost (http://www.its.caltech.edu/~atomic/snowcrystals/frost/frost.htm - "forms when a pane of glass is exposed to below-freezing temperatures on the outside and moist air on the inside")?

quantum chromodynamics - Permissible combinations of colour states for gluons

My lecturer has said that there are 8 types of gluons (I'm assuming that the repetition of $r\bar{b}$ is a typo that is meant to be $r\bar{g}$)

$$r\bar{b}, b\bar{r}, r\bar{g}, g\bar{r}, g\bar{b}, b\bar{g}$$ which I'm fine with. However what I don't understand is that

$$\frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g})\text{ and }\frac{1}{\sqrt{6}}(r\bar{r}+g\bar{g}-2b\bar{b})$$

are allowed, as she states later that "we are not allowed colour-neutral combinations e.g. $r\bar{r}$"

How is $\frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g})$ not colour neutral? If you were to measure it surely it would collapse into either $r\bar{r}$ or $g\bar{g}$ state, which are colour neutral?

Also what forbids the state $\frac{1}{\sqrt{2}}(b\bar{b}+g\bar{g})$ and other similar combinations if $\frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g})$ is allowed?

Answer

The color language is not really well-suited to understand why there are eight gluons. Here's why, however:

The gluon field transforms in the adjoint representation of the color gauge group $\mathrm{SU}(3)$. The adjoint representation is a representation on the vector space of the generators of $\mathrm{SU}(3)$, the Lie algebra $\mathfrak{su}(3)$. An explicit realization of $\mathfrak{su}(3)$ is given by the Gell-Mann matrices $\lambda_i$, and it is simply a fact that $\mathfrak{su}(3)$ is an eight-dimensional vector space, and that there are hence eight independent gluons/Gell-Mann matrices.

To assign now colors to the gluon fields $G_i$, which are the real coefficients of the expansion of the $\mathfrak{su}(3)$-valued gluon field $G$ as $G = \sum_i G_i \lambda_i$, we look at the interaction term with the quarks, which is proportional to $$ \bar q G q $$ and known to be color-neutral (more formally, invariant under an $\mathrm{SU}(3)$ gauge transformation). With the chosen realization of $\mathfrak{su}(3)$ as the Gell-Mann matrices, we choose the basis of the corresponding (three-dimensional) fundamental representation for the quarks to be the basis of "colors", i.e. $q = (q_r,q_b,q_g)^T$, where $q_c$ is the $c$-colored part of the quark state. In particular, a pure red state is given be $(1,0,0)^T$, for example.

Now, insert $q = (q_r,q_b,q_g)$ into the interaction term and look only at what $\lambda_1$ does to this, i.e. explicitly compute $\bar q G_1\lambda_1 q$. The result is $\bar r G_1 b + r G_1 \bar b$ (up to normalization). From this we infer that $G_1$, in this realization is a gluon field corresponding to an equal mixture of red-antiblue and blue-antired, i.e. $\bar r b + \bar b r$. We would also find a $\bar r b - \bar b r$ gluon, allowing us to change into the choice of color basis of the question, where we have $\bar r b$ and $\bar b r$, by forming linear combinations like $(\bar r b - \bar b r) + (\bar r b + \bar b r)$.

The statement "color-neutral gluons are not allowed" is technically correct in the sense that the gluons together from the adjoint representation rather than a trivial representation. However, it does not translate into the naive meaning of "writing down combinations human color intuition would describe as colorless is not allowed", and that's why we also get a gluon $\bar r r - \bar g g$.

tl;dr: The color charges and color combinations of the strongly charged particles are not literally governed by human color intuition, but arise from assigning color rather arbitrarily to the representations of $\mathrm{SU}(3)$ the particles transform under.

Also, note that the "color" of a state is not measureable, as a "red state" is changed by a $\mathrm{SU}(3)$ gauge transformation into a "blue state", and states which are related by gauge transformation are not physically distinguishable, see also this answer of mine.

riddle - Simply a relationship of symbols

Mathematicians love me
Chemistry students study me
Even lovers of language use me.
I am
a relationship built of symbols,
a procedure or a hard-and-fast rule,
Or the things something’s made of,
feeding newborns when the mother’s not able.

What am I?

Answer

You are a...

FORMULA

Mathematicians love me

Refers to a mathematical formula or equation.

Chemistry students study me

Refers to a chemical formula that denotes chemical composition.

Even lovers of language use me.

Maybe refers to formulaic language in writing, or a formula in formal language in mathematics.

I am a relationship built of symbols, a procedure or a hard-and-fast rule,

Definition of a mathematical formula. Same meaning as the puzzle title.

Or the things something’s made of,

Definition of a chemical formula

feeding newborns when the mother’s not able.

Refers to infant formula, which is used for feeding young infants.

particle physics - Why is a nearby nucleus required for Pair Creation?

I was recently studying Pair Production and Annihilation. The author mentions that a nearby nucleus is required when the photon materialises into a particle and an anti-particle. The explanation given is that the momentum and the energy must be conserved. However, there is no calculation given that shows the violation of energy. The reason is just blankly stated. Is there more to this concept. Please explain? As far as I know that from the knowledge of Particle Physics, virtual photons can violate the conservation laws if the time scales are very small due to the Heisenberg Uncertainty Principle. Then why can't we apply the same idea here?

PS: I've read the other answers but none of them include the contribution of Nucleus' energy/momentum to conserve momentum or energy.

Answer

The argument depends on a fact about photons that you may or may not have encountered yet: if you have a photon with energy $E$, that photon must carry momentum $p = E/c$ in some direction. If you change your opinion about the photon's momentum by observing it from some other reference frame, you also must change your opinion about its energy. For instance if you run away from your light source your photons will exchange less momentum when they interact with you than if you and the light source were relatively at rest; you'll see these photons as having less energy, or "red shifted."

Matter particles (or collections of matter particles, like electron-positron pairs) have a property that photons do not: a "rest frame" where the momentum is zero.

So suppose you have a single photon which transforms into an electron-positron pair with equal and opposite momentum --- that is, you're in the electron-positron rest frame. In order to create these two particles, your single photon must have had a little more than 1 MeV of energy (more if the e-p pair have kinetic energy). But in order to conserve momentum your photon must have had zero momentum, and therefore zero energy. This contradiction is why a single photon cannot transform into an electron-positron pair.

The flaw goes away if the photon can steal momentum from its environment. The inside of an atom has a strong electric field, which is made of (in quantum electrodynamics) "virtual" photons. Your real photon can exchange momentum with the atom by scattering from one of these virtual photons; that's where the pair creation happens. Theoretically it should be possible to generate e-p pairs by sending gamma rays into a very intense laboratory electric field or a microwave cavity, but I don't think those environments are experimentally accessible.

word - Piece de Resistance - Five Flowers with One Missing

Five - Flowers with One Missing

_{This puzzle is part of the "Piece de Resistance" series. Go back to Part 1 (Ace) for the story.
Ace Two Three Four Five ...}

Removing the silver matter, you see this handwritten text:

To the West is of conquer, of victory,
To the North is for leather laundry.
To the East is a great river, alloy,

To the South is big and tall, ahoy!
I myself am the glorious one who brings in heat,
A four-letter word is what you need!

Hints

_{usefulness level 0:}

This is just another Four Directions puzzle...

_{usefulness level 1:}

All sites in a nation

_{usefulness level 2:}

Above hint: [tag:???ost??]

_{usefulness level 3:}

The oval is 11 meter.

_{usefulness level 3.5:}

Above hint: cryptic-clues

_{usefulness level 4:}

Above hint: It's a special type of clue... Enumeration is (5).

_{usefulness level 5:}

BIG giveaway

Answer

After sussing that we were supposed to interpret the titular 'flowers' as:

A cryptic clue for 'RIVERS' (see my comment on the main question above),

I had gone on to wonder whether the area of interest was indeed:

INDIA, as confirmed by @hexomino via the printer's devilry of Hint 3 (The oval is 11 in diameter), although I had not solved that particular clue. My own flawed reasoning had been based on 'putra' in 'Brahmaputra' (a river in East India) sounding like the alloy 'pewter'.

This logic inadvertently led me to the answer for the missing 'flower' (I myself am the glorious one who brings in heat / A four-letter word is what you need!) being:

TAPI - a river of central India named after the goddess Tapati, the daughter of Surya the sun god and "mother-goddess of the south, the home of the southern sun where she brings the heat to the earth" (source: Wikipedia).

After the OP's hints and prodding (thanks!) I have now worked out that the directions might not strictly indicate the part of the country they are located in but rather their position in relation to the solution.

So WEST (To the West is of conquer, of victory,) is:

Shetrunjaya (a river west of the Tapi), whose name literally means 'victorious; one who overcomes enemies'.

NORTH (To the North is for leather laundry) is:

Chambal (a river north of the Tapi) since its ancient name was Charmanvati, meaning 'the river on whose banks leather is dried' (Source: Wikipedia).

EAST (To the East is a great river, alloy) is:

Mahanadi (a river east of the Tapi) since its name is a compound ('alloy') of two Sanskrit words: maha meaning "great", and nadi meaning "river".

SOUTH (To the South is big and tall, ahoy!) is:

Bhima (a river south of the Tapi) since its name in Sanskrit literally means 'big' or 'tall' (Source: Wikipedia).

And thus the riddle is (definitely!) complete!

quantum mechanics - What is the physical interpretation of the density matrix in a double continuous basis $|alpharangle$, $|betarangle$?

(a) Any textbook gives the interpretation of the density matrix in a single continuous basis $|\alpha\rangle$:

• 
  

  The diagonal elements $\rho(\alpha, \alpha) = \langle \alpha |\hat{\rho}| \alpha \rangle$ give the populations.

  
  



• 
  

  The off-diagonal elements $\rho(\alpha, \alpha') = \langle \alpha |\hat{\rho}| \alpha' \rangle$ give the coherences.

  
  

(b) But what is the physical interpretation (if any) of the density matrix $\rho(\alpha, \beta) = \langle \alpha |\hat{\rho}| \beta \rangle$ for a double continuous basis $|\alpha\rangle$, $|\beta\rangle$?

I know that when the double basis are position and momentum then $\rho(p, x)$ is interpreted as a pseudo-probability. I may confess that I have never completely understood the concept of pseudo-probability [*], but I would like to know if this physical interpretation as pseudo-probability can be extended to arbitrary continuous basis $|\alpha\rangle$, $|\beta\rangle$ for non-commuting operators $\hat{\alpha}$, $\hat{\beta}$ and as probability for commuting ones.

[*] Specially because $\rho(p, x)$ is bounded and cannot be 'spike'.

---

EDIT: To avoid further misunderstandings I am adding some background. Quantum averages can be obtained in a continuous basis $| \alpha \rangle$ as

$$\langle A \rangle = \int \mathrm{d} \alpha \; \langle \alpha | \hat{\rho} \hat{A} | \alpha \rangle$$

(a) Introducing closure in the same basis $| \alpha \rangle$

$$\langle A \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \alpha' \; \langle \alpha | \hat{\rho} | \alpha' \rangle \langle \alpha' | \hat{A} | \alpha \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \alpha' \; \rho(\alpha,\alpha') A(\alpha',\alpha)$$

with the usual physical interpretation for the density matrix $\rho(\alpha,\alpha')$ as discussed above.

(b) Introducing closure in a second basis $| \beta \rangle$, we obtain the alternative representation

$$\langle A \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \beta \; \langle \alpha | \hat{\rho} | \beta \rangle \langle \beta | \hat{A} | \alpha \rangle = \int \mathrm{d} \alpha \int \mathrm{d} \beta \; \rho(\alpha,\beta) A(\beta,\alpha)$$

When the two basis are momentum $| p \rangle$ and position $| x \rangle$ the density $\rho(p,x)$ is the well-known Wigner function whose physical interpretation is that of a pseudo-probability. My question is about the physical interpretation of $\rho(\alpha,\beta)$ in two arbitrary basis $| \alpha \rangle$, $ | \beta \rangle$.

Energy conservation and quantum measurement

Consider a particle in a potential well. Let’s assume it’s a simple harmonic oscillator potential and the particle is in its ground state with energy E₀ = (1/2) ℏω₀. We measure its position (measurement-1) with a high degree of accuracy which localises the particle, corresponding to a superposition of momentum (and therefore energy) states.

Now we measure the particle’s energy (measurement-2) and happen to find that it’s E₁₀ = (21/2) ℏω₀. Where did the extra energy come from?

In the textbooks it’s claimed that the extra energy comes from the act of observation but I wonder how that could work. Measurement-1 which probed the position of the particle can’t have delivered to it a precise amount of energy, while measurement-2 might just have been passive. No doubt there is entanglement here between the particle state and the measuring device but where, and which measurement?

Answer

For a high degree of accuracy you would have to probe the particle with a high energy (short wavelength) photon so there is plenty of energy that can go into vibrational excitation. After such hard hit the particle will be smeared across a wide range of states $$\Psi=a_0*\Psi_0+a_1*\Psi_1+...$$This is not an entanglement but a simple superposition of eigenstates. The expectation value of the particle's energy $\bar{E}=\sum_{i}a_i^2E_i$ should not be equal to the energy of any particular state and the second measurement will yield $E_i$ with $a_i^2$ probability.

So, the extra energy comes from an interaction with a probe particle and it doesn't have to be precisely equal to the energy of a certain vibrational state.

Logical English Riddles

Someone's mother has four sons. name of three sons are north,east,west, what is the name of the fourth son

can anyone tell?

Answer

i believe that the answer is

What

as

whenever this question is asked, it’s actually asked this way: “Someone's mom has four sons, North, East and West. What is the name of the fourth son. Can you guess the name of the fourth son?”

moreover

If you read it again, the second sentence ends with a full stop. It’s a statement. It tells you the name of the fourth son is, “What”.

logical deduction - How Many Queens to Threat Themselves?

How many queens can be placed in a standard $8*8$ chessboard to threat each of them exactly once by other queens?

Answer

I found a 10 queen answer. I realized that d'alar'cop's reasoning was slightly off, because threatening is part of the puzzle, so 8 wasn't necessarily a limit. With the 8 inner queens in this position, you can put the final 2 queens in any 2 corners.

Update: I have a proof for a maximum of 10 queens.

Given that 2 queens in a row or a column use 1x2 and 2 queens diagonally use 2x2, it is advantageous to avoid the diagonals.

Put 2 queens in a row. You have now used up 1 row and 2 columns.
Do this again. 2 rows, 4 columns.
Again. 3 rows, 6 columns.

If you do it a 4th time, you will have used all 8 columns, so don't.
Put 2 queens in a column. 5 rows, 7 columns.
Do this again. 7 rows, 8 columns.

You can no longer add any queens, as your columns are used up.
Repeat swapping columns and rows and you end up with 8 rows and 7 columns, so you still cannot add any queens.

Any attempts at using diagonals simply make the problem end faster (exception: a single diagonal, like my displayed answer, gets you to 8 rows, 8 columns).

Therefore, 10 queens is the maximum.

Now as for allowing other pieces, I noticed that there might be extra space from d'alar'cop's 24 queens answer, and I was able to get 2 more queens in.

newtonian mechanics - Satellite in Elliptical orbit

When finding the period of a satellite orbiting the earth we equate the centripetal force to the gravitational force

$$\frac{mv^2}{r} = \frac{-GMm}{r^2}$$ If I understood well the $r$ cancels into the $r^2$ because the distance from the earth at any point on the orbit equals the radius of the orbit. Now what if we have an ellipse instead of a circle?

And by the way why does Kepler's $1^{st}$ law say In elliptical orbits

mathematics - Make 5 5 5 5 = 19

Can you find a way to make:

$5\ 5 \ 5 \ 5 = 19$

by adding any operations or symbols? You can use only these symbols:

$+,\ -,\ *,\ !,\ /,\ \hat\, ,\ ()$.

It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.

Answer

Here's one way I found:

$\dfrac{5!-(5\times 5)}{5}=19$

Or, using just the characters explicitly allowed in the question:

(5! - (5 * 5)) / 5 = 19

models - Why are continuum fluid mechanics accurate when constituents are discrete objects of finite size?

Suppose we view fluids classically, i.e., as a collection of molecules (with some finite size) interacting via e&m and gravitational forces. Presumably we model fluids as continuous objects that satisfy some differential equation. What mathematical result says that modeling fluids as continuous objects can accurately predict the discrete behavior of the particles? I don't know anything about fluid mechanics, so my initial assumption may in itself be wrong.

Answer

There are many physical intuitions often presented in various texts on fluid dynamics. I won't mention those here. I will, however, mention that mathematically the passage from a particle point of view to a continuum point of view is still a largely un-resolved problem. (With suitable interpretation, this problem was already posed by Hilbert as his 6th of 23 problems.)

We can interpret the problem as one of starting from "a Newtonian description of particles interacting through collisions" and try to end up with "an approximation of the physical system by a continuum obeying certain laws of fluid dynamics (Euler, Navier-Stokes, etc.)"

Most work up through now takes an intermediate step through the Boltzmann equation: in this kinetic theory model, instead of individual particles we consider distributions of particles, where the "density" of particles is given based on both position and velocity. So it makes one level of continuum approximation. But it still keeps the facet of Newtonian theory where particles interact through direct collisions. Under an assumption known as molecular chaos (more on this later), that Boltzmann's equation follows from Newtonian laws of motion have been demonstrated, to various degrees of rigour, by Boltzmann himself, as well as Grad, Cercignani, and Lanford, building on the work of Bogoliubov, Born, Green, Kirkwood, and Yvon. For a mathematically sophisticated, but more or less self-contained description one can refer to Uchiyama's write-up. There are a few issues with this derivation.

1. The problem of potentials. The derivations listed above made assumption that the particles are hard spheres: that the only interaction between two particles is when they actually collide (so no inter-molecular forces mediated by electromagnetism, like hydrogen bonds and such), and that the particles are spherical. This is satisfactory for monatomic gases, but less so for diatomic molecules or ones with even stranger shapes. Most people don't think of this as a big problem though.


2. The derivation is only valid under the so-called Grad limit assumption. To take the continuum limit, generally the assumption is made that the particle diameter decreases to zero, while the number of particles (per unit volume) increases to infinity. Exactly how these two limits balance out affects what the physical laws look like in the continuum limit. The Grad limit assumes that the square of the particle diameter scales like the inverse of the number density. This means that the actual volume occupied by the particles themselves (as opposed to the free space between the particles) decreases to zero in this limit. So in the Grad limit one actually obtains an infinitely dilute gas. This is somewhat of a problem.


3. The derivation also makes use of what is called molecular chaos: it assumes that, basically speaking, the only type of collision that matters is that between two particles, and that the particle, after its collision, "forgets" about its previous zig-zag among its cousins in the dilute gas. In particular, we completely ignore the case of three or more particles colliding simultaneously, and sort of ignore the billiards-trick-shot like multiple bounces. While both of this can be somewhat justified based on physical intuition (the first by the fact that if you have a lot of small particles spaced far apart, the chances that three of them hit at the same time is much much much smaller than two of them colliding; the second by the fact that you assume some sort of local thermodynamic equilibrium [hence the name molecular chaos]), one should be aware that they are taken as assumptions in the Boltzmann picture.

Starting from the Boltzmann equation, one can arrive at the Euler and Navier-Stokes equations with quite a lot of work. There has been a lot of recent mathematical literature devoted to this problem, and under different assumptions (basically how the Reynolds and Knudsen numbers behave in the limit) one gets different versions of the fluid equations. A decent survey of the literature was written by F. Golse, while a heavily mathematical discussion of the state-of-the-art can be found in Laure Saint-Raymond's Hydrodynamic limits of the Boltzmann equation.

It is perhaps important to note that there are still regimes in which the connection between Boltzmann equation and the fluid limits are not completely understood. And more important is to note that even were the connections between the kinetic (Boltzmann) picture and the fluid limits, there is still the various assumptions made during the derivation of the Boltzmann equation. Thus we are still quite far from being able to rigorously justify the continuum picture of fluids from the particle picture of Newtonian dynamics.

string theory - T-duality approaches

The textbook approach to explaining T-dualities is to show that a type of T-duality transformation "inverts the radius of the circle, that is, it maps $R\rightarrow\tilde{R} = \alpha'/R$ and it leaves the mass formula for the string invariant provided that the string winding number is exchanged with the Kaluza-Klein excitation number." 1

However, the broader classification of T-duality appears imply that theories are T-daul if there is any type of transformation that changes on theory into the other. So the question is whether there are other approaches to showing T-duality, or is there a "canonical" approach for T-duality.

String Theory and M-Theory: A Modern Introduction [Hardcover] Katrin Becker (Author), Melanie Becker (Author), John H. Schwarz (Author)

Answer

"Any transformation that changes one theory into another" (or the same) theory is not called T-duality. It is just a "duality". A condition is that the two theories seemingly look different - otherwise the equivalence would be vacuous - but it must be true that their spectrum and the strength of interactions between their states must be totally isomorphic: physics has to be indistinguishable. A duality is therefore just a very fancy redefinition of coordinates that can't be quite done in the classical limit so you can't really write any "explicit" redefinition. But the impact is the same - the two theories behave in the same way. For each state/object on one side, you find a counterpart on the other side - whose geometrical interpretation may be very different but whose behavior is isomorphic.

T-duality in string theory

T-duality is a duality that changes the geometrical properties of the target spacetime but that holds order-by-order in the expansion in terms of the string coupling in perturbative string theory. In particular, a T-duality must preserve the notion of the "fundamental string" (or "world sheet") and its tension and it is a weak-weak duality when it comes to the interactions between strings. The world sheet of a given shape remains the world sheet of the same shape; however, the fields on it get transformed. That's T-duality in general.

Quite generally, T-dualities that we know may be interpreted as the duality you mentioned although it may also invert the radii of several circles (a torus) at the same moment, and the shape of these circles (and torus) may depend on other coordinates of spacetime (the so-called "Buscher procedure" is a standardized method to perform such a transformation). In particular, "mirror symmetry" relating two completely different Calabi-Yau shapes may be understood as a T-duality acting on a 3-dimensional toroidal fibers of the Calabi-Yau space (a special way to choose coordinates on the 6D shape so that 3 of them look like a torus whose shape depends on the remaining 3.) However, in recent years, people have also appreciated the importance of fermionic T-dualities that mostly act on fermionic fields on the world sheet but that are otherwise analogous.

Other dualities

There are several other important kinds of dualities. S-dualities invert the weak and strong coupling. In particular, for a very large value of $g$ much greater than one, they find another theory with a weak coupling $g'=1/g$ or $1/g^2$ - which is much smaller than one - that is equivalent to the previous one. So S-dualities usually invert a small and large couplings, much like T-dualities exchange small and large radii of circles. A fundamental string doesn't remain a fundamental string under an S-duality; it usually gets transformed into another string, e.g. a D-string (also known as a D1-brane), that used to be heavy and unimportant when the coupling was weak but becomes the lightest and most important object when the coupling is strong. Again, an S-duality exchanges fundamental strings and D-strings, and similar pairs of objects.

More generally, there are also U-dualities that combine the actions of S-dualities and T-dualities: they exchange a strong coupling with large radii with a weak coupling and small radii, among similar actions. In particular, M-theory (essentially 11-dimensional supergravity with the extra "stringy" stuff needed to make it consistent at all energies) compactified on a $k$-dimensional torus has a self-U-duality group which is something like $E_{k(k)}(Z)$, a discrete subgroup of a noncompact edition of the exceptional Lie groups (for $k\leq 8$: for $k\leq 5$, the exceptional group can be written in terms of non-exceptional groups). Such an exceptional U-duality group exchanges momenta in compact dimensions with wrapped membrane and fivebrane charges (and other discrete charges such as KK-monopole charges if there are at least 6 compact dimensions).

Several other equivalences between two seemingly different theories are also called "dualities" - and the AdS/CFT holographic correspondence itself is sometimes classified as a duality as well. It would triple this article if I wanted to cover all important kinds of dualities that appear in field theory and string theory.

Why the existence of dualities was surprising

Finally, it's useful to stress that dualities - a key notion of theoretical physics of the last 20 years or so - didn't have to exist. More precisely, people thought that a theory compactified on an extremely tiny circle, or a theory extrapolated to an insanely strong coupling, is a completely new mysterious beast filled with sea dragons that behaves like nothing that the people had previously encountered. However, it turned out that the behavior of those "mysterious limits" is almost always governed by another "mundane" theory that the people have essentially known, and that the two are equivalent. It's like sailing to India and America and expecting totally new creatures over there - but finding the same animals and plants we're used to.

Whenever some radius is taken extremely tiny, coupling is made very strong, or even the number of colors in a gauge theory is sent to infinity, so that the perturbative techniques of the original theory totally break down, there always seems to be a new description that is totally weakly coupled. That's one of the lessons of physics since the mid 1990s - something that we didn't previously know but nowadays, we do know it. The fact that we seem to understand the "opposite extreme" limits of the previously known theories places an upper bound on the amount of "new enigmas" that may still be hiding in field theory and string/M-theory. If there are totally new enigmas, they have to hide at the intermediate values - like Atlantis in the middle of the Atlantic Ocean, if you wish. ;-)

cosmology - What if the size of the Universe doubled?

My question has a silly formulation, but I want to know if there is some sensible physical question buried in it:

1. Suppose an exact copy of our Universe is made, but where spatial distances and sizes are twice as large relative to ours. Would this universe evolve and function just as ours?

Since mass is proportional to volume which is distance cubed, but strength of a rope is only distance squared, I think not everything would scale proportionally.

2. 
   

   Does this mean that a universe with our physical laws that evolves like ours can only have one particular size?

   
   


3. 
   

   Or would the physical laws scale proportionally so that it evolves in the same way?

   
   


4. 
   

   If it has a particular size, what is it relative to?

   
   
   


5. 
   

   Another variation: suppose another exact copy of our Universe is made, but where everything happened twice as fast relative to ours. Would it evolve in the same way as ours?

   
   

Answer

If you think for a moment about how lengths and speeds in our universe are set (that is, independent of how we choose to measure them, by meters or seconds or whatever), you'll see that these must ultimately come from different ratios of fundamental constants.

I don't know and would be very surprised if there's a way to change these ratios so that all lengthscales or all timescales would change, since so much of what we observe actually comes from very complicated interacting systems acting at different timescales and lengthscales.

So the answer is probably no, there is no sensible physical question behind what you've asked, at least the way you've asked it.

You could ask of course, what would happen if some (dimensionless) physical constant was doubled or halved, and then we could chat some more.

Why do I emphasize dimensionless here? What if you asked "What happens if the speed of light was doubled?" Well, because the speed of light is what sets our time and length scales, we would observe no changes at all! In other words, think about what you can compare the speed of light to that doesn't depend on the speed of light itself. Just saying the numerical quantity changes is meaningless because the units we use to measure it rely (via a possibly long chain of dependencies) on the speed of light itself!

higgs - Is it (theretically) possible to reduce inertial mass without reducing matter quantity?

Ok, please bear with me because I only have a very little (or no?) understanding of physics outside Newtonian laws... So I was playing a video game called "Mass Effect" in which a particular compound ("Element Zero") can create a field in which objects lose what I understand is inertial mass without being shrunk down or losing matter (with this I mean that they are still physically formed by the same particles), thus rendering possible to accelerate them with a smaller force. Is it theoretically possible for such a thing to happen? Like interfering with Higg's Field?

Answer

I don't think a big effect like the one in the game is possible, but a small effect is possible...

The mass of a proton plus an electron is greater than the mass of a hydrogen atom (by a little tiny bit) because $E=mc^2$, where $m$ is the inertial mass. The energy of the electron and proton in a hydrogen atom is lower than the energy of the hydrogen atom and the electron which are separated from each other. So the reduction in mass $\Delta m$ is given by $\Delta E=\Delta m c^2$ and $\Delta m =\Delta E / c^2$. The value of $\Delta E$ is $13.6 ~eV$, which gives a value for $\Delta m$ of $\sim 2.4 \times 10^{-35} ~kg $, which is about $1.5 \times 10^{-8}~m_H$ where $m_H$ is the mass of the hydrogen atom.

So the mass of a hydrogen atom is slightly lower than the combined mass of its components. This is the only example I can think of relevant to your question.

edit after good comment from - Francesco Bertolaccini

The release of energy in fusion and fission processes can also be calculated with $\Delta E=\Delta m c^2$ - the energies released are much larger than the energy mentioned above in the formation of the hydrogen atom from an electron and a proton. For more information the wikipedia page on fusion as a power source may be a good place to start.

newtonian mechanics - Are Newton's 1st and 3rd laws just consequences of the 2nd?

Can Newton's 1st and 3rd laws be assumed given just $F=ma$. I know that the argument would be, "No, then there would only be 1 law". But I can't think of any situation where 1 and 3 aren't superfluous.

If you just told me $F=ma$: I would assume nothing else causes an acceleration besides a force. So things not experiencing a force don't change velocity, even when velocity is 0. 1 ✔️

And, when two things that exist interact they use only their mass and acceleration to do so so they both must change in opposite ways. 3 ✔️

A tyo riddle...why not?

With two I make someone quiet,
With three I'm an algorithm,
With four I'm fake,
With all five I'm dishonor.

Inspired by this.

Answer

I think you are

Shame

With two I make someone quiet,

Sh

With three I'm an algorithm,

SHA

With four I'm fake,

Sham

With all five I'm dishonor.

Shame

classical mechanics - Does mass affect speed of orbit at a certain distance?

Does the mass of both the parent object, and the child object affect the speed at which the child object orbits the parent object?

I thought it didn't (something like $T^2 \approx R^3$) until I saw an planet on the iphone exoplanet app, that is closer to it's star than a planet in another system, yet takes longer to complete one orbit. Both of the planets were a similar mass, as were the stars.

fluid dynamics - Is it possible to change Pressure without changing Density?

I am still trying to understand Navier-Stokes equations. I understand the equations in general, BUT there is one aspect I still cannot digest: The equations assume that density is constant (okay, I understand why, because if it varies, then the derivation of the equations would be more difficult), HOWEVER, what I don't understand is : how it is possible to change pressure without changing density? meaning we use the negative gradient of P to represent forces acting on a blob of fluid, but is it even possible to change P without changing density? I understand that the ONLY way to increase pressure at a one point in space is via increasing the density of the smoke at this point?? Is there other way to change pressure without changing density? Maybe I am missing another definition of pressure that does not depend on density!

Just one last note, I understand if the fluid is liquid then changing its density is not "easy", but for gases, I "think" this is rather doable...

Thanks in advance :)

Answer

In the case of incompressible fluids such as liquids, pressure changes do not change the density of the fluid very much.

And in the case of gases, and if you can treat your fluid as an ideal gas it can be written as

$$\rho=\frac{P}{RT}$$

So you can see by properly adjusting $T$ to compensate for changes in $P$ you can hold density, $\rho$ constant

enigmatic puzzle - (10 of 11: Nurikabe) What is Pyramid Cult's Favorite Topic?

<sub>Dear PSE users and moderators,
I’m new here in PSE, but I really need your help. There was this person who gave me a black envelope consisting 10+1 pages of puzzles, and also a scribble saying: “Find our favorites and you will be accepted to join our ‘pyramid cult’. Feel free to ask for help from your beloved friends on PSE. They will surely guide you into all the truth.” I’m also a newbie on grid puzzles, so, could you please give me any hint to solve these? It’s getting harder and harder later on..
- athin</sub>

_{Jump to the first page: #1 Numberlink | Previous page: #9 Shikaku | Next page: #END Meta}

---

Rules:

1. Shade the cells under the following rules.


2. You cannot shade cells containing numbers.


3. A number tells the number of continuous white cells. Each area of white cells contains exactly one number in it and they are separated by shaded cells.


4. The shaded cells are linked to be a continuous wall.


5. Shaded cells cannot be linked to be 5-in-a-row cells.

---

^{Special thanks to chaotic_iak for testing this puzzle series!}

Answer

Answer is here:

Favourite topic:

Anagram of yellow area (wall) is FASHION

Welp... I'm too stupid and blind to the apparent... jafe has found out the same answer can be obtained by

reading the letters left-to-right, row by row. Thanks @jafe!

lateral thinking - Do I Work with an Orangutan?

This morning I found myself face-down in front of a large building. It was obvious that I had taken a great fall from one of the rooms above me, but now I apparently had amnesia.
I went in.
I could tell that the building had been built by an American architect, as the ground floor was designated as the first floor, and I guessed that there might be a basement below me. I started going into random rooms, trying to jog my memory a bit.

1st floor: I poked my head into room 130 and saw someone with a deck of cards. She was staring intently at it, then flipped the top card over with great conviction. She then sighed sadly and scribbled something down on a notepad.

3rd floor: There were a bunch of unsavory mobster-types hanging out by room 364, so I kept going up.

4th floor: I couldn't seem to understand what anyone was talking about, so I continued up.

6th floor: I stopped in room 633 and mellowed out for exactly 7 minutes.

9th (top) floor: I looked into room 910 and was reminded that I had not looked in the basement, so I went back down for a look-see.

Basement: Ahhh! This was finally familiar! I snuggled right into my desk in room 25 and took a nap.

When I woke up I was home, in bed. What a strange dream that had been! Well, no worries, everyone dreams about work sometimes.

What is my job and where do I work?

Answer

You are a:

Librarian working in a library

and the rooms are:

Dewey Decimal Classifications

1st floor: I poked my head into room 130 and saw someone with a deck of cards. She was staring intently at it, then flipped the top card over with great conviction. She then sighed sadly and scribbled something down on a notepad.

Dewey Decimal Classification 130: Parapsychology and occultism
The woman is attempting to divine the future, and is either attempting to predict the top card of a deck (as a self-test) or is using the deck as a divination tool (see further: Tarot cards)

3rd floor: There were a bunch of unsavory mobster-types hanging out by room 364, so I kept going up.

Dewey Decimal Classification 364: Criminology
This is where books that study the criminal mind go, especially the minds of career criminals or particularly famous criminals.

4th floor: I couldn't seem to understand what anyone was talking about, so I continued up.

Dewey Decimal Category 400-499: Language
Linguistics, phonology, etymology, grammar, etc. They're all here, across a variety of languages, including "Hellenic languages; classical Greek" (480) and "Austronesian and other languages" (499)

6th floor: I stopped in room 633 and mellowed out for exactly 7 minutes.

Dewey Decimal Classification 633: Field and Plantation Crops
7 minutes is exactly 420 seconds.

9th (top) floor: I looked into room 910 and was reminded that I had not looked in the basement, so I went back down for a look-see.

Dewey Decimal Classification 910: Geography and Travel
Where did you come from? Where did you go?

Basement: Ahhh! This was finally familiar! I snuggled right into my desk in room 25 and took a nap.

Dewey Decimal Classification 025: Library operations
Get back to work!

Connection to title:

In Terry Pratchett's Discworld series of novels, the Librarian of the magical Unseen University is an orangutan. (Don't call him a monkey!)

Sphinx-related riddle

My 3rd grader brought an assignment home about the Giza plateau, and one of his questions was the following riddle:

What creature with one voice walks on four legs in the morning, on two at noon, and on three in the evening?

This totally puzzles me; the only thing I can think of is an unfortunate human being who's been through a lot the night before (silly joke on my part).

Can you solve the Sphinx's riddle?

general relativity - Equivalence of definitions of ADM Mass

ADM Mass is a useful measure of a system. It is often defined (Wald 293)

$$M_{ADM}=\frac{1}{16\pi} \lim_{r \to \infty} \oint_{s_r} (h_{\mu\nu,\mu}-h_{\mu\mu,\nu})N^{\nu} dA$$

Where $s_r$ is two sphere with a radius going to infinity.

However, I have also seen the following definition (EDIT: for example, Page 147 of "A Relativists Toolkit"):

$$M_{ADM} = - \frac{1}{8\pi} \lim_{r \to \infty} \oint_{s_r}(H-H_0)\sqrt{\sigma}d^2\theta$$

Where $H_0$ is the extrinsic curvature of $s_r$ embedded in flat space and $H$ is the extrinsic curvature of $s_t$ embedded on a hypersurface of the spacetime.

It is not overly clear to me how these two are related. Obvious we can get the same differential with the follow definition.

$$dA=\sqrt{\sigma}d^2\theta$$

But more so than that it doesn't seem clear why those extrinsic curvatures should be equal to those metric derivatives (times the constant term of $-1/2$). Presumably the fact that we are taking the limit out to $r \to \infty$ is important as we are assuming that the metric should be asymptotically flat (or at least asymptotically constant metric) and these quantities might be reduced to some similar form in the limit.

Any insight?

Answer

First, let's try and clean up the MTW expression a little bit by getting rid of the manifestly coordinate variant stuff, by introducing a proper contraction, a proper integral over the surface at infinity, and the unit normal $r^{a}$ to the surface at infinity:

$$M_{ADM}={16\pi}\oint\sqrt{\gamma}\,\,d^{2}x \gamma^{ab}r^{c}\left(\gamma_{ac,b}-\gamma_{ab,c}\right)$$

This is only valid, however, not just only for in asymptotically flat spaces, but actually only for asymptotically Cartesian coordinates. You can prove this to yourself by calculating the ``ADM Mass'' of the flat 3-metric in polar coordinates:

$\begin{align*} 16\pi M_{ADM}&=\lim_{r\rightarrow\infty}\oint r^{2}\sin\theta \gamma^{ab}\left(\gamma_{ra,b}-\gamma_{ab,r}\right)d\theta d\phi\\ &=\lim_{r\rightarrow\infty}4\pi r^{2}\left(\gamma^{rr}\gamma_{rr,r}-\gamma^{ab}\gamma_{ab,r}\right)\\ &=\lim_{r\rightarrow\infty}4\pi r^{2}\left(0-\frac{4}{r}\right)\\ &=-\infty \end{align*}$

Obviously, this is wrong, and just an artifact of the way that spherical coordinates behave at infinity. To fix this, we need to always subtract the divergence from the ADM mass of flat spacetime from the ADM mass of the coordinate system in question. So, that's the origin of the $H$ and $H_{0}$ terms. The $H$ term is the extrinsic curvature of the space in question, while $H_{0}$ is the extrinsic curvature of flat spacetime. Now, it's just a matter of showing that the expression inside the integral is equal to $H$.

Typically, I define the extrinsic curvature as $\gamma^{ab}\nabla_{a}r_{b}$. So, let's go on an adventure:

$\begin{align*} H&=\gamma^{ab}\nabla_{a}r_{b}\\ &=\gamma^{ab}\left(\partial_{a}r_{b}-\Gamma_{ab}{}^{c}r_{c}\right)\\ &=\gamma^{ab}\partial_{a}r_{b}-\frac{1}{2}\gamma^{ab}\gamma^{cd}\left(2\gamma_{ad,b}-\gamma_{ab,d}\right)r_{c}\\ &=\gamma^{ab}\partial_{a}\left(\gamma_{bc}r^{c}\right)-\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\frac{1}{2}\gamma_{ab,c}\right)\\ &=\delta^{a}{}_{c}\partial_{a}r^{c} + \gamma^{ab}r^{c}\gamma_{bc,a}-\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\frac{1}{2}\gamma_{ab,c}\right)\\ &=\partial_{a}r^{a}-\frac{1}{2}\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\gamma_{ab,c}\right) \end{align*}$

Now, we note that in any coodinate system adapted so that the coordinate $r=constant$ that determines the surface is chosen for one of the coordinates, we have, necessarily $r^{a}=(1,0,0)$, we note that $\partial_{a}r^{a}=0$, and we thus conclude that the two expressions for ADM mass are equivalent.

statistical mechanics - What happens for the spins around the phase transition

Suppose we now consider a lattice of spin, say Ising model, and the phase transition at the critical temperature $T_c$. There are few scaling laws describe the regime around the critical temperature such as correlation length $\xi$:

$\xi\sim|T-T_{c}|{}^{-\nu}$

where $\nu$ is the critical exponent. This equation describes the divergence of the correlation length as the temperature approaching $T_c$.

However, it gives no direct physical picture of the spins near the $T_c$ because we need to compute the summation over each pair of spins to find the correlation. So the question is: If we really look at the lattice, what is the most obvious feature and scaling that we can obverse?

Answer

Generalities on Conformal Invariance

In two dimensions, a lot is known / conjectured about statistical models at criticality. For instance, at $T_c$, the spin configuration that you see will not only be self-similar (what others here have been calling "fractal") but actually fully conformally invariant (in the continum limit); that is, the probability distributions ("ensembles") are invariant not only under scale transformations, but indeed, under any transformation which preserves angles. (I think conformal invariance is expected in models with "short-range" interactions at criticality in any dimension, but it's only well understood in 2 dimensions).

Let me go a bit more into what "conformal invariance" means. Consider the following "thought experiment", which you might try with a computer simulation. First, intersect a disk with a square lattice with side length $\epsilon$ with $\epsilon$ quite small. On this finite graph, draw a random spin configuration for the Ising model at criticality. Call this ensemble "A".

Now, perform a conformal mapping of a disk onto whatever (simply connected) shape you like, say a triangle or a square (this is guaranteed by the Riemann mapping theorem). Approximate this shape with a square lattice with side $\epsilon$ and draw a random critical Ising spin configuration for this. Call this ensemble "B".

The claim is that if you perform a conformal transformation of ensemble "A" to get A', the typical picture from A' will be identical to the typical picture from B (this is a statement that the distribution of spin configurations will be nearly identical, if $\epsilon$ is small).

Here's some illustrations from slides of Stanislav Smirnov (who recently won a Fields medal for related work!).

On the left is a critical Ising model configuration under a conformal transformation; on the right is a random walk under a conformal transformation. You have to kind of use your imagination to see that (other than finite lattice effects) the pictures would be similar to those if you just ran a Ising model or random walk in the funny shapes below.

This has recently become a hot topic in mathematics, due to the breakthroughs in Schramm-Loewner evolution, though it has been studied a bit longer in physics using the less rigorous techniques of conformal field theory.

A specific "geometrical observable"

I've been kind of vague above about what conformal invariance means for individual "pictures" of spins (well, most of the claims are about probability distributions). You seem to be dissatisfied with various claims about correlation functions; I also feel that these are often a bit hard to grasp geometrically. There is one solidly geometric thing that one can say about certain lattice models at criticality, and this is about the boundaries between spin clusters.

The boundaries between up and down spins in the Ising model at criticality form an ensemble of closed loops in the plane. This loop ensemble is believed to be related to SLE $\kappa=3$ curves; this is a very particular (conformally invariant) ensemble of random planar curves. SLE curves are parametrized by a real number $\kappa\geq0$, when $\kappa=0$ one just has straight lines, as $\kappa$ gets larger, the curve gets more "wriggly", the fractal dimension of these curves is equal to $1+\kappa/8$ for $\kappa<8$ (thus in the continuum limit, boundaries between Ising spin clusters are expected to be 11/8 = 1.375 dimensional!). For $\kappa\geq8$, the curves actually become plane-filling!

All of the things I talked about above do not hold except when you are exactly at $T=T_c$. For the Ising model, it is sometimes hard to generate spin configurations at $T=T_c$ due to critical slowing down of Markov chains which are used, but there are many other systems such as critical percolation which are easier to study.

References

One good place to start reading about conformal invariance in lattice models for a theoretical physicist is Malte Henkel's book "Conformal Invariance and Critical Phenomena". Of course if you're serious, you'll eventually turn to the big yellow book by DiFrancesco et al on conformal field theory...

A good intro to SLE for physicists is this one by Ilya Gruzberg. Probably it won't be long before some books combining the SLE approach with the CFT approach start coming out.