calculation puzzle - A sequence defined by a single fraction

I have a sequence of numbers here, which follow a very simple rule that can be expressed using a single fraction:

1, 2, 10, 11, 13, 15, 20, 22, 25, 28, ...

What is this rule (in particular, what is the fraction), and what are the next few numbers in the sequence?

---

Since nobody's gotten it yet, I'll add a few hints over the next little while.

(06-16 13:22) 1. The numbers in the sequence never get more than 2 digits long.

Answer

The items in the sequence should be numbered from 2 onwards, and then item $b$ is

the base-$b$ representation of $\displaystyle \left \lfloor \frac{b^2}{4} \right \rfloor$, which can also be written as just $\displaystyle \left \lfloor \frac{100}{4} \right \rfloor$, since $b$ in base $b$ is always $10$.

So the fraction that defines the sequence is

100/4

and the next few entries in the sequence are:

30, 33, 37, 3B, 40, 44, 49, 4E, 50.

thought experiment - How is Q superposition in Q information different from a classical dice?

Let's consider a dice with 1 or -1 on its faces, opposites faces adding up to 0.

The point of this post is to relate the quantum coherence of a single qubit to something "weird", i.e., incapable of being reproduced by classical probability.

All the faces of the dice are covered, and we are repeatedly going to uncover and read the top face, and then cover it again. A $\sigma_z$ measurement will do simply this, while a $\sigma_x$ measurement turns the dice 90 degrees to the right, uncovers the top face, reads and covers it, and finally rotates the dice back, but assume that a device (maybe yourself) MUST roll the dice by keeping its vertical axis fix right after covering the face, similar to spinning a top. Note that this protocol tries to reensemble the incompatibility of the two measurements in QM.

The state preparation is what we do to the dice before giving it to someone else.

By simply rolling the dice, performing a $\sigma_z$ or $\sigma_x$ measurement will give a -1 or 1 with half probability, as for the quantum state {{1/2,0},{0,1/2}}.

By rolling the dice and then making sure that the left face has a "1", you will get the same output as if I gave you the quantum state {{1/2,1/2},{1/2,1/2}} for both $\sigma_z$ and $\sigma_x$ measurements!

After this analysis, I see that it is possible to reproduce the quantum superposition in Q information by classical means, which then makes it not "weird" at all. The only "special" feature I see is that nature itself automatically does the covering of the faces and the spinning at the measurements for you, but this is not my point, so...

...without going into non-locality, Bell inequalities and so on (which I understand are truly quantum and needed for Q cryptography), is there anything "weird" in the coherence alone of Q states in Q information that I can use to say "this is not classical" after looking at the measurements (which is all I can look at)? Am I missing something? If not, then I don't understand how a single qubit can be somehow better than a single bit in the same sense as a quantum superposition state can be more powerful than a single (maybe tricked and in a state not prepared by you) dice.

Answer

So what you are describing is a hidden variable theory of quantum mechanics, and I believe the answer is ultimately no, there is nothing fundamentally non-classical in a single qubit. You can construct hidden variable theories of quantum mechanics as long as they are non-local, and since there is no notion of locality in a single qubit, that last caveat should not be a concern.

Bell wrote a paper in 1966 on this topic:

The question at issue is whether the quantum mechanical states can be regarded as ensembles of states further specified by additional variables, such that given values of these variables together with the state vector determine precisely the results of individual measurements.

He concluded that indeed such models can be constructed, but that they have a non-local nature:

The curious feature is that the trajectory equations for the hidden variables have in general a grossly nonlocal character.

With regard to a qubit being more powerful than a bit, we must be careful about the comparisons we are making. A hidden variable model of a qubit will have more states than a classical bit, so in that sense already a qubit is greater than a bit, although that is not particularly compelling. Additionally, I do not know of any useful quantum computations that can be performed with a single qubit.

The more interesting question is why are collections of qubits, which presumably as a group should have some non-local but classical hidden variable description, able to perform computations which a classical computer can not? I do not have a comprehensive answer to that question, but there is a paper written by Scott Aaronson which claims that for any hidden variable theory, if you had access to the hidden variables you would have even more computational power than a quantum computer. So it seems in some sense that computational power does not come from non-classicality exclusively. We must remember that computers are physical systems, and when you lose non-locality you change the physical rules under which they operate, and thus what they are capable of doing.

quantum mechanics - Product of exponential of operators

in the context of non-relativistic quantum mechanics I want to show that, for any $A$ and $B$ operators

$$e^{A}e^{B}=e^{A+B} $$

if and only if

$$[A,B]=0$$

I remember my professor told use about looking for a differential equation but I don't remember the details, and I want to be able to prove it. Brute force the series doesn't seem to be a good idea.

Any hint will be appreciated thanks.

Answer

There is no 'only if' because it is not true: \begin{align} e^{A+B} = e^A e^B \end{align} does not necessarily imply $[A,B] = 0$.

One can easily find an example of this using matrices. Here's one: \begin{align} A= \begin{pmatrix} 0 & 0 \\ 0 & 2\pi i \end{pmatrix}, B=\begin{pmatrix} 0 & 1 \\ 0 & 2 \pi i \end{pmatrix}. \end{align} $[A,B] \neq 0$ but $e^{A+ B} = e^A e^B = I$.

Edit: Let me help with the if part, using a differential equation as OP desires. Compute \begin{align} \frac{d}{dt}(e^{t(A+B)}e^{-tA}e^{-tB}), \end{align} and show that it is $0$ if $[A,B] = 0$.

That implies that $e^{t(A+B)}e^{-tA}e^{-tB}$ is independent of $t$. In particular, plugging in $t = 0$ gives $e^{t(A+B)}e^{-tA}e^{-tB} = I$ for all $t$. Then plug in $t = 1$ to get $e^{(A+B)}e^{-A}e^{-B} = I$.

QED.

electromagnetism - The energy stored in the electromagnetic field of an electron

Due to Wikipedia the total energy per unit volume stored in an electromagnetic field is

$$u_{EM}=\frac{\varepsilon}{2}|\mathbb E|^2+\frac{1}{2\mu}|\mathbb B|^2$$

How does the energy stored in the electric field of the electron relates to its rest mass? How large part of the rest mass comes from this field?

And related, how does the energy stored in the magnetic field induced by a moving electron relates to its kinetic energy?

A Rigorous Derivation of Electromagnetic Self-force seems to give relevant information related to this question.

general relativity - Why doesn't gravity speed up light?

We know that gravity speeds up a body; for instance, a meteor which enters the earth gets constantly accelerated up by earth's gravity. And from relativity we know that light bends near a massive body, because Newton's law of gravitation is just an approximation and actually gravity depends on energy and momentum. So my question is: If a ray of light is aimed exactly at the center of a body, then will it get accelerated like a meteor? And if does get accelerated, then won't it surpass the universal speed limit of 3,00,000 km/s (approx.)?

Answer

If a ray of light is aimed exactly at the center of a body, then will it get accelerated like a meteor?

Short answer: no. However, when falling in a gravity field, the momentum of light increases.

Some background...

In Newtonian mechanics, the rate of change of momentum of a (massive) particle is proportional to the acceleration:

$$\frac{d\vec p}{dt} = m \vec a $$

In Relativistic mechanics, these quantities are not proportional. In fact, an accelerating massive particle can never reach speed $c$ but the momentum can reach arbitrarily large values.

This is because relativistic momentum is a non-linear function of velocity

$$\vec p = \frac{m \vec v}{\sqrt{1 - \frac{v^2}{c^2}}} $$

which diverges as $v \rightarrow c$.

In the special case of a massless particle, which must travel at speed $c$ in all frames, the numerator and denominator in the above are zero so, by this formula, the momentum of a massless particle is indeterminate.

However, the relativistic energy-momentum relation

$$E^2 = (pc)^2 + (mc^2)^2 $$

gives the momentum of a massless particle:

$$p = \frac{E}{c} $$

Thus, the momentum can change even though the speed does not. In falling from a higher potential to a lower potential, the massless particle gains energy and thus momentum but not additional speed.

For light, the momentum and frequency are proportional:

$$p = \frac{h\nu}{c} $$

so, while the speed of light does not increase as it falls, the frequency of light increases. From the Wikipedia article "Blueshift":

Photons climbing out of a gravitating object become less energetic. This loss of energy is known as a "redshifting", as photons in the visible spectrum would appear more red. Similarly, photons falling into a gravitational field become more energetic and exhibit a blueshifting

mathematics - Variation of 100 prisoners light problem

There is a classic problem below for which Dr. Yisong Song wrote a very well written and thought out series of solutions. The variation here I've never seen anywhere but am curious. I want to work it out myself but think others may also want to give it a go.

One hundred prisoners have been newly ushered into prison. The warden tells them that starting tomorrow, each of them will be placed in an isolated cell, unable to communicate amongst each other. Each day, the warden will choose one of the prisoners uniformly at random with replacement, and place him in a central interrogation room containing only a light bulb with a toggle switch. The prisoner will be able to observe the current state of the light bulb. If he wishes, he can toggle the light bulb. He also has the option of announcing that he believes all prisoners have visited the interrogation room at some point in time. If this announcement is true, then all prisoners are set free, but if it is false, all prisoners are executed. The warden leaves, and the prisoners huddle together to discuss their fate. Can they agree on a protocol that will guarantee their freedom.

For this variation, there is a set of such 100 prisoners who are all exactly 50 years old and reasonably expect to live to 80. The light starts off and they will be taken into room on average roughly once a day. The warden may, however, send an extra prisoner of his choosing to the cell on the same day (but they still never see each other) to disrupt strategies. This excludes the use of binanary token strategy. Because the warden is not completely sadistic, a fair random number generator picks the first prisoners each day.

In their meeting, they initially agree to a simple one counter solution when one prisoner remembers (having once been a puzzle fan) that this should take about 30 years if they are lucky. He argues that each year outside of prison has an equal value but each year in prison is no better or worse than being dead. Being released on his 80th birthday is therefore worth about the same as never being released at all. He would gladly accept a 50% chance of death to be released with 20 years left of life remaining compared to a 100% chance of survival but being 79 when that chance came.

So if the value of making a guess is the number of years remaining of life multiplied by the probability of surviving the guess, what strategy should they employ to maximize this value?

Answer

I still believe there is a better answer than this but that this is a significant improvement over the current best answer. I have used only the same math used by Mr. Millikan as I like how understandable it made the problem.

On day 917, according to the

$\displaystyle \left(1-\left(\frac {99}{100}\right)^d\right)^{100}=p$

used by Mr. Ross Milikan the value has a maximum with only a roughly 1% chance of everyone being executed horribly. This does not use the light bulb at all and the likelyhood of more than 1 person never having been in the room is less than .001%.

Strategize that every time a prisoner (assumed to be all male) walks into the room the first time, he switches the light. After the 100th person goes in the room for the first time the light will always be off. On day 917, if the light is on: they know the 1% chance has happened and they must wait for it to go off before they say everyone has been there. The likelyhood that there are 2 people who have not been there so the light was off but they die is very small.

If this strategy is used on day 495 (for instance). The probability that everyone has been there is under 50%. The probability that 2 people have not been there, however, is less than 1%. Therefore, on day 495 if the light is off they can guess with >98% certainty that they can be freed; if the light is on, the person who has not been there yet, turns off the light and guesses that everyone is there and they win on the first day it is true.

Unless my calculations are too approximated, this strategy could yield a larger likelyhood of survival and will likely release prisonners a year ealier than Mr. Millikan's (depending on they day they choose). This strategy has a large likelyhood of releasing the prisonners on the first day possible (but of course still has a non-zero chance of causing horrible death).

lagrangian formalism - The derivation of fractional equations

Recently I saw some physical problems that can be modeled by equations with fractional derivatives, and I had some doubts: is it possible to write an action that results in an equation with fractional derivatives? For example, consider a hypothetical physical system with the principle of least action. Is there a "wave equation" with the time-derivative $3/2$? Does such a question make sense?

Answer

When I've seen fractional derivatives I've assumed that one place where they would naturally arise is in physical situations where there's a fractional dependency on time.

For example, random walks typically result in movement proportional to $\sqrt{t}$. Googling for "fractional+derivative+random+walk" on arxiv.org finds some papers that explore this:

http://www.google.com/search?q=fractional+derivative+random+walk+site%3Aarxiv.org

So I'm wondering if there's a way of relating some of the diffusion versions of QM with fractional derivatives.

Simple grid deduction puzzle

I will let you figure the rules of this one yourselves.

Here is the empty grid to assist solvers.

Here is a text version of the grid:

 _______________________________________________________________________________________________________________
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|       |       | C G P |       | C G W |       |       |       |       |       |       |       |       |       |
| C G P |  C G  |       |  G Y  |       | C G P | C G Y | G P Y | G P Y |  G Y  | G P Y | G O Y |  G O  | G O Y |

|       |       |  W Y  |       |   Y   |       |       |       |       |       |       |       |       |       |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|       |       | C G P |       | C G W |       | C G O |       | G O P |       | G P R | C G O |       | C G O |
| C G P | G P R |       | G R W |       | G P W |       | G O P |       | G P R |       |       | C G Y |       |
|       |       | R W Y |       |   Y   |       | P W Y |       |  R Y  |       |   Y   |   Y   |       |   Y   |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|       | C G O | C G O | G O R | C G W |       | C G O |       | G O R |       | B G R | B C G |       |       |
|  C G  |       |       |       |       | C G R |       | G R Y |       | B G Y |       |       | G O W | C G O |

|       |   R   |   Y   |   Y   |   Y   |       | R W Y |       |   Y   |       |  W Y  |  O W  |       |       |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|   G   | G O R |  G R  | G O R |  G W  | G R W | G O W | G O R | G O R | B G R | G R W | B C G | C G W |  C G  |
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|       |       |       |       |       |       |       |       |       |       | B C G | B C G | B C G |       |       
| G O W | G O W | G O W |  G O  |  B G  | B G R | B G R |  G R  |  G W  | B G W |       |       |       |  B G  |

|       |       |       |       |       |       |       |       |       |       |   W   |   W   |   W   |       |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|       | G O P |       |       |       |       |       |       |       |       | C G P |       | B C G |       |
| G R W |       | G P W | G O P | B G O | C G O | B C G | B C G | B G W |  B G  |       | G P W |       |  G W  | 
|       |   R   |       |       |       |       |       |       |       |       |   W   |       |  P W  |       |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
| C G O | C G O |       | G O P |       | B C G |       | B C G |       | B G W | C G O | C G O | B C G | B G R |
|       |       | C G W |       | B G W |       | B G R |       | G W Y |       |       |       |       |       |

|  W Y  |P R W Y|       |   W   |       | O R W |       |  R Y  |       |   Y   |  P W  |  R W  | O P R |   W   |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
| B C G | B G P | C G P | G O P |       | C G O |       | B C G |       | B G R | G O P | G P R | C G O | C G R |
|       |       |       |       | G O P |       | C G P |       | B G R |       |       |       |       |       |        
|   R   |  R Y  |   Y   |  W Y  |       | P R W |       |  R Y  |       |   Y   |  R Y  |  W Y  | P W Y |   W   |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|
|       |       |       |       |       |       |       |       |       |       |       |       |       |       |
|       | B C G |       |       |       |       |       |       |       |       | G O R |       | C G O |       |
| C G Y |       |  C G  | G W Y | G P W | G R W | G P R | G R Y | G R Y |  G Y  |       | G O R |       |  G R  |

|       |   Y   |       |       |       |       |       |       |       |       |   Y   |       |  R Y  |       |
|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|_______|

Answer

This is more of a continuation of rand al'thor's answer.

As Rand al'thor surmised the letters indicate that that cell has a cell with the corresponding colour as one of its neighbours. For instance the cells which include a Y are:

From which we can readily deduce the yellow cells are just the ones indicated. We repeat for B(rown), O(range), C(yan), R(ed), P(urple), W(hite).

Now every cell has a 'G' (which might be an error, since for all the other cases a cell doesn't include its own colour indicator) so we can skip that since it only tells us all the remaining cells are green (I have used gray though).

You end up with the following (but here mapping G to gray, rather than green).

Spelling out the message "YOU ROCK".

• Note for the text version of the grid we need to change the bottom element of the second column to B C G Y to match the image.

fluid dynamics - Can there be a pressure gradient in a large enough closed container?

If we had an airtight container that was large enough, say extended out to the Karman line, would there still be a pressure gradient? If we assume that the pressure at the bottom was $1\ \mathrm{atm}$, would it have the same exact pressure gradient at 6 miles up on the outside? This is assuming that the gravity is still $9.8\ \mathrm{m/s^2}$.

differential geometry - Gauge theory for mathematicians?

I'm looking for a textbook or set of lecture notes on gauge theory for mathematicians that assumes only minimal background in physics. I'd prefer a text that uses more sophisticated mathematical concepts like principal bundles and connections, and categorical language whenever convenient.

Answer

I have been writing something in this direction in section 1 of the book Differential cohomology in a Cohesive topos (pdf). Have a look, just focus on section 1 and ignore the remaining sections on first reading.

The survey-part is presently also appearing as a series on PhysicsForums. See at Higher prequantum geometry I, II, III, IV, V and Examples of Prequantum Field Theories I -- Gauge fields, II -- Higher gauge fields.

logical deduction - Knights and knaves in a foreign language

You die and ascend to heaven, there is a knight (truth-teller), a knave (pure liar) and a joker (random) sitting on a cloud - they all look the same. In order to gain entry you must determine their identities. You have 3 yes/no questions (each directed to only one of them). They will only respond to questions speaking their own heavenly language's words for "yes" and "no" which you do not know - "pluh" and "plit" (unknown which means what).

Are you going to be allowed into heaven?

Answer

This is so called The Hardest Logic Puzzle Ever. Wikipedia has a thorough description of it and its solutions, including different versions of formulations of how the joker functions (which was not defined in the OP's question).
I can cite the very basics here:

Formulation:

Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.

Solution:

Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination.

Does quantum mechanics violate the equivalence principle?

I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1} $$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$ m_I \frac{ d^2 x}{ dt^2} = m_g g \tag{2}$$ we can choose a transformation $x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see)

newtonian mechanics - Rigid (planar) Multibody system: How are these equations of motion linked to the model?

This a second question I have on the same subject as my previous question, so I will make a link here (Motion equations for Woodpecker toy (multibody system)) to that question (and a link from that question to here). I believe this deserves to be a question of its own since the modeling is different and I will provide more information here about the Woodpecker's motion. The question is how the below stated equation of motion is derived from the pictures of the model and motion of the Woodpecker. Let's get to it!

Below is a picture of a rigid multibody system consisting of a sleeve (ring) and Woodpeckerbody joined by a spring. The sleeve can travel down a rod with some play, and both the sleeve's and the Woodpecker's position can be described by their absolute rotational angles (see the figure). The whole system's position and motion is thus described by three generalized coordinates: $(y,\varphi_M,\varphi_S)$.

Now, the motion of the Woodpecker is non-smooth and the equation of motion is given as

$\textbf{M}(t,\textbf{q})\ddot{\textbf{q}}-\textbf{h}(t,\textbf{q},\dot{\textbf{q}})-\Sigma_{i\in I_S}(\textbf{w}_N\lambda_N+\textbf{w}_T\lambda_T)_i=0$

where the third term are constraints, but let's not mind them right now, since I am only intrested in the motion "inbetween" constraints. Besides, the constraints are all equal to zero (right? or does that only apply for holonomic constraints?).

The massmatrix $\textbf{M}$ and the force-vector $\textbf{h}$ are given as

$\textbf{M}(t,\textbf{q})= \begin{bmatrix} (m_S+m_M) & m_Sl_M & m_Sl_G \\ m_Sl_M & (J_S+m_Sl_M^2) & m_Sl_Ml_G \\ m_Sl_G & m_Sl_Ml_G & (J_S+m_Sl_G^2) \end{bmatrix}$

$\textbf{h}(t,\textbf{q},\dot{\textbf{q}})= \begin{bmatrix} -(m_S+m_M) \\ -c_{\varphi}(\varphi_M-\varphi_S)-m_Sgl_M \\ -c_{\varphi}(\varphi_M-\varphi_S)-m_Sgl_G \end{bmatrix}$

Since this system is non-smooth with friction (not a conservative system), I presume that there is no expression for the kinetic and potential energy, so we cannot derive this equation using the Lagrangian. If we write these equations out (there are three; one for each degree of freedom maybe?) we get:

$(m_S+m_M)\ddot{y}+m_Sl_M\ddot{\varphi_M}+m_Sl_G\ddot{\varphi_S}+(m_S+m_M)g=0\quad (1)$

$m_Sl_M\ddot{y}+(J_S+m_Sl_M^2)\ddot{\varphi_M}+m_Sl_Ml_G\ddot{\varphi_S}+c_{\varphi}(\varphi_M-\varphi_S)+m_Sgl_M=0\quad (2)$

$m_Sl_G\ddot{y}+m_Sl_Ml_G\ddot{\varphi_M}+(J_S+m_Sl_G^2)\ddot{\varphi_S}+c_{\varphi}(\varphi_M-\varphi_S)+m_Sgl_G=0\quad (3)$

According to the report (http://www.dct.tue.nl/New/Leine/ASME21608.pdf) I got this problem from, it the equations "follow in a straight way manner" from the figure. How exactly are these equations describing the motion of the Woodpecker?

I know this is might be a lot to ask, but what is so straight forward about this?

Charge conjugation and chirality

while I was reading about charge conjugation I found some (apparently) contradictory facts. For example Itzykson & Zuber says (page 153) "Up to a phase, $\cal C$ interchanges particles and antiparticlas with the same momentum, energy and helicity" while Zee (pag. 101) "You can easily convince yourself that the charged conjugated of a left handed field is a right handed field and viceversa" How can this be possible (in the case with $m=0$ when chirality coincides with helicity)?

In conclusion what is the handedness of the charge conjugated of a left handed field?

In order to convince myself I worked out two contradictory proofs: Let $U$ be the charge conjugation operator in Fock space, and $C$ the matrix that realizes charge conjugation on spinors: $\Psi^c = U^\dagger\Psi U = C\bar{\Psi}^t$, then:

1) $U^\dagger P_L\Psi U = P_L U^\dagger\Psi U = P_L C\bar{\Psi}^t$, (because $P_L$ is acting only upon creation and annihilation operators) and so here we proved that the charge conjugated of a left handed field is still left handed;

2) $U^\dagger P_L\Psi U = C\overline{(P_L\Psi)}^t = C\gamma^0 P_L\Psi^* = P_RC\bar{\Psi}^t$, (using Pauli-Dirac as well as Weyl representation of gamma matrices) and so here we proved that the charge conjugated of al left handed field is instead right handed.

Can you help me?

Note added: does a Majorana massless fermion exist?

quantum electrodynamics - Deriving the Coulomb force equation from the idea of photon exchange?

Since Newton's law of gravitation can be gotten out of Einstein's field equatons as an approximation, I was wondering whether the same applies for the electromagnetic force being the exchange of photons. Is there an equation governing the force from the exchange of photons? Are there any links which would show how the Coulomb force comes out of the equations for photon exchange? I know that my question is somewhat similar to the one posted here The exchange of photons gives rise to the electromagnetic force, but it doesn't really have an answer to my question specifically.

Answer

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles.

Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is

$$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p - E_{p'})(-\mathrm{i})\int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r$$

This is to be compared to the amplitude obtained from the Feynman diagram:

$$ \int \mathrm{e}^{\mathrm{i}k r_0}\langle p',k \rvert S \lvert p,k \rangle \frac{\mathrm{d}^3k}{(2\pi)^3}$$

where we look at the (connected) S-matrix entry for two electrons scattering off each other, treating one with "fixed" momentum as the source of the potential, and the other scattering off that potential. Using the Feynman rules to compute the S-matrix element, we obtain in the non-relativistic limit with $m_0 \gg \lvert \vec p \rvert$

$$ \langle p',k \rvert S \lvert p,k \rangle \rvert_{conn} = -\mathrm{i}\frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}(2m)^2\delta(E_{p,k} - E_{p',k})(2\pi)^4\delta(\vec p - \vec p')$$

Comparing with the QM scattering, we have to discard the $(2m)^2$ as they arise due to differing normalizations of momentum eigenstate in QFT compared to QM and obtain:

$$ \int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r = \frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}$$

where Fourier transforming both sides, solving the integral and taking $\epsilon \to 0$ at the end will yield

$$ V(r) = \frac{e^2}{4\pi r}$$

as the Coulomb potential.

cipher - This is an opportunity. Can you take it?

The first thing you hear on awakening is the clock. It sounds just like the one you have in your room. But what you see defies what you hear.

You've never seen this room before. You don't remember how you got here. The last thing you can remember is that your mission went horribly wrong...

Wincing, you get up and survey the scene in front of you. As you reach up, and feel a painfully large lump on the back of your head, a crumpled piece of paper falls out your top. You tuck into your pocket, thinking it will be important later. You look at the door and don't need to go over there to know that it's locked. Here's the room you see (click for a better view)

Can you escape?

This will be important later:

QNW

NOTE:

I created all the images from scratch though I occasionally used an online image in the making of my images which I don't claim as my own. If you find something odd in an image which doesn't appear to be a clue, it's probably an unintentional error which (probably) won't affect the puzzle.

Answer

Solution:

On the enlarged picture, there is an open book on the shelf with something that looks like an imgur link.
I made a screenshot about that gif, and typed the letters, so here they are for copy-paste:

fkx wpwax iggqkl mg zighva gsfhhixxdu
pshc wvsnfz erw ldmrd scemg'qky jhmjh whea jshv plsnyd
5gugo.lrk

fol'zf zvcbme wu tyqt fvod qu kplc jmcd ug apcu
i'd i ecbbcm bulnk
q'ws sewb zcb cccfg ao vadowe
wqor
ahv xbgz cflf hv lzdf
tpnu bjal ow mydsojqpb
zo pwv yuon ppk tutp
uwte pwv vhvv tfta

I tried decoding this as a Vigenere cipher with the key given in the hint ('QNW'), but it fails. Maybe that comes to use in a later step.

@Strawberry suggested using 'sweet' as a key. (I suppose it comes from the crossword to which there was another link on the table on the enlarged image).
That one gives:

not sweet enough to decode completely
look around and think again'you found some food though
5cqnw.png
mwp'vb gdgxil ey puxb jrkk yy glsk niyk ck wljc
m'z e lkfxyt jyhjr
y'ao olef vyi kgybn is rwkwaa
sxwv
wdc ffcv jnpb dc tdzb
axrq xqip ks tghokqytx
vv xar ubwr llr bypl

bexa ldd zdrc bjpw

The second key for the second half of the message is 'haribo'. That must be sweet enough indeed. I kind of bruteforced it, supposing the message starts with either 'you've' or 'you're'. It reads:

you're locked in this room it will blow up soon
i'm a double agent
i've left you clues to escape
find
the pass code to live
find time of explosion
so you know how much
time you have left

Maybe the time of explosion is 10:30 as written on the calendar. So we've around 20 minutes left. We have to hurry, but also, need to think with a calm head. What might the pass code be?

Ok, so in the first decoded version of the message, '5cqnw.png' suggests imgur again. Changing 'qnw' into 'QNW', we get a valid imgur link - thanks for the observation, @Ankoganit. The new image contains a carton of a cereal titled "NEW SWEENEES" which seems to be compass-directions. As the crumpled piece of paper had a compass on it, this should probably used on that. Starting from the specially marked box, we get: II, III, II, V, IV, V, IV, III, I, IV, III,
which are all roman numerals, the ones for which the directions were underlined read 3413.

---

EDIT FROM OP

This answer is almost completely correct, but I'd just like to add a few things that were missed out, or not clearly stated.

First of all in the first room, the newspaper headline says Another Bomb Murder. This is another clue that the room will explode.

When the image is clicked on new things appear. The newspaper changes to a Crossword with ucBTN.jpg written. Changing the end of the imgur link gives us:

The first clue is 'grain turned sugary'. Grarin is wheat. Wheat turned sugary is sweet
The second clue is 'A 3rd exists'. To be is to exist. Is is 3rd person of be.
The third clue is 'Grants power of entrance', which is a key
The last clue is 'Common determiner'. The

Rearranged this gives us Sweet is the key

The other thing that changes is that the book on the shelf changes to give 2yRkA.gif. Changing the link gives us a GIF of a book. A page flips over to give the text that @elias kindly wrote out.

Using 'sweet' as the key gives

'Not sweet enough to decode completely, look around and think again. You found some food though. 5cqnw.png.

So sweet isn't the key. But in the far left cupboard in the enlarged image is this:

It's hard to make out in the shadow of the cupboard but you can just make out the 'H' in Haribo. So 'Haribo' is the sweet that is the key.

Using Haribo to decode the cipher you learn that you're locked in the room which will explode, but a double agent has left clues to escape.

Changing the imgur adress to 5cqnw.png gives nothing but by using QNW, we get a new room. This has a cereal packet:

This has another clue that the room is going to explode with the explosion around new. So 'New Explosion'. The name of the cereal is NEW SWEENEES.

Interestingly =this only contains the letters NSEW which are all part of the compass. Under the packet is says 'Cereal that directs you'.

This tells us to use the NEWSWEENEES as directions. On the map which fell out or top.

There are four letters underlined in the cereal packet, the second, fifth, ninth and eleventh.
On the map those underlined letters give us 3413

On the table is an envelope this is an unsubtle clue to the postbox in the door. In the enlarged version we get this:

By typing in 3413 you escape! The explosion happens twenty minutes later (the calendar) after the area has been evacuated after you tell the police.

Air friction at supersonic velocities

I know that if an object moves in the air, it can experience two types of drag, laminar and turbulent. For instance, I have a meteor of ideal spherical shape falling from edge of space, say, 100 km up at the Earth surface with initial velocity of 1 km per second. I would consider turbulent drag, but is it still applicable at supersonic velocities? How do I estimate meteors velocity when it hits the ground?

quantum mechanics - How to choose proper measurement operator?

Let's assume I have two states inside the Bloch sphere, at radial vectors $r_1$ and $r_2$ respectively $(r_1

general relativity - Gravitational potential, effective refractive index, and vacuum charge density

In an earlier question, I asked about how to explain gravitational lensing to a layman in terms of propagating wave fronts, in a way analogous to the way an optical lens can be explained: the propagation velocity is inversely proportional to the refractive index of the glass and in the direction normal to the wavefront, which results in an initially plane wavefront being converted to a concave, converging wavefront.

Subsequently, I found several papers and presentations that described the vacuum as having an “effective refractive index” determined by the gravitational potential: (1), (2), (3), (4). The effective refractive index can then be used to explain gravitational lensing from either the ray perspective or the wave perspective.

This leads to a confusing inference, though, because dielectric constant is equal to the square of the refractive index (assuming that magnetic permeability is equal to 1). So, it seems that the vacuum around a gravitating mass must have an "effective dielectric constant" that varies with distance from the mass.

So far, so good. But Gauss' law

says that the effective charge of a charged particle depends on the dielectric constant of the vacuum surrounding it. If the dielectric constant is a function of distance $r$, then the effective charge is a function of distance.

This suggests, then, that radial dependence of the gravitational potential will cause the effective charge of a massive charged particle to vary with distance from the particle. (By the way, vacuum polarization [https://quantummechanics.ucsd.edu/ph130a/130_notes/node507.html] which is presumably an unrelated phenomenon, has a similar effect.)

However, answers and comments in response to this PhysicsSE question state emphatically that the divergence of E is zero in vacuum, regardless of the gravitational potential, according to the Einstein-Maxwell equations.

I suspect that the disconnect is due to the fact that derivation of the “effective refractive index” in (1), (2), (3), (4) assumes that space is flat. However, my confusion remains: it seems that a distant observer, ignorant of general relativity, would see a distance-dependent variation in the (effective) charge of a massive charged particle.

Is there a way out of this apparent contradiction?

Answer

The effective refractive index idea is not at all about electrodynamics, or dielectric constant, or anything like that. It is simply a nice way to calculate null geodesics. That is what it is, and that is all it is.

What you do is first pick coordinates where the metric is static and with the spatial part 'conformally flat' (i.e. equal to some function of $r$ multiplied by a Euclidean metric). It won't always be possible to do that, but if it possible, as for example with the exterior Schwarzschild metric, then you can proceed. Next you notice that (i.e. prove that) the null geodesics lie exactly where you would calculate light rays to go if you were to pretend that spacetime is flat and light moves at the coordinate speed of light in your chosen coordinates, i.e. at $c$ divided by a factor given by the metric. So you call that factor $n$, an 'effective refractive index', and away you go. The method is useful because it gives a very good intuition about the effects you are calculating, and because it allows you to draw on known results and methods in ray optics.

But, I repeat, this whole calculation is simply a way to find null geodesics in whatever spacetime you have; it does not have anything to do with electromagnetic fields or Maxwell's equations or permittivity. Having said that, this does not rule out that it might be possible to employ a similar style of reasoning to get insight into electromagnetic phenomena in curved spacetime in some situations. The idea is to spot when the GR equations happen to fall out in a way that can be interpreted as a field theory or particle theory in flat spacetime.

special relativity - Direct Product vs Tensor Product

I am confused in the notation on page 67 and page 70 a text (http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf), whether it's talking about a direct product or an outer product:

1. 
   
   

   On page 67, it mentioned that

   
   

   
   

   "you can take a direct product of two $j = 1/2$ representations"

   
   

   
   

   and build representations of higher j.

   
   


2. 
   

   On page 70, it mentioned

   
   

   
   

   "we can think of [the Lorentz Group] as the direct product $SU(2) \times SU(2)$."

   
   
   

   
   

In each of the above, does the author mean Direct Product or Tensor Product?

electromagnetism - Magnetic field insulators

I was wondering if there is any way to stop the magnetic field, without having the insulator turned into magnet. Let me present this as a simple case, there is a magnet to the left and a piece of iron to the right, is there anything that I can put in between to stop the magnet from attracting the iron piece, provided that the insulator won't turn into a magnet as well?

I am open to all possibilities, any materials, electric fields or anything that can stop the magnet field without harming the magnet or the iron piece are welcome.

Answer

Mu-metals are your best bet, though superconductors are useful in some situations.

You can think of mu-metals as redirecting the magnetic field, rather than neutralizing it. So if you just have a flat piece mu-metal between the magnet and your piece of iron, you'll still get a reasonable amount of magnetic field -- it just had to "go farther" to get to the iron, so it will be weaker. The usual design for shielding is a box enclosing whatever you want to shield.

Bringing a superconductor close to a magnet, on the other hand, will simply cancel out changes to the magnetic field inside the superconductor itself. This is typically less useful for your problem, but it might be applicable for other applications you may need. In particular, if you need fields eliminated from a cavity inside a material, this could function better than the mu-metal.

special relativity - How do I know which observer is running the time faster or slower?

Ok, I'm not a physicist, so I don't know if my question is silly (probably yes), but there is something in special relativity that I can't understand and I would really like it to be clarified.

If two observers are experiencing different velocities, then they will experience time differently. Ok. But, since there is no absolute referential, ever, how can I tell which one is moving "away" from the other in order to know which one is experiencing time faster than the other? I'll try to be more specific. Let's get the classic experiment of the light clock where a beam of light is moving up and down between two mirrors, and counting time at each passage. Now let's get an exact copy of that mirror and put it side by side with the first one. Now let's start moving the second clock horizontally to the right, away from the first one. In order for the speed of light to be the same to all observers, the moving light clock should experience time slower than an observer attached to the first clock. Ok, got it. Very logical. But here's my doubt: if the entire universe is composed of just these two watches, how can I tell that the second mirror is the one that is moving away from the first to the right, and not the first mirror that is moving away from the second to the left? It's impossible to tell. So, how can I know which one of the clocks is the one "moving away" in order to define which one is experiencing time dilation?

I know the answer must be silly because I simply can't find it. Any help?

Critical Dimension of Bosonic Strings and Regularization of $sum_{n=1}^infty n$

If $D$ is critical dimension of Bosonic strings, a particular derivation goes like the following, where we arrive finally at $$ \frac{D-2}{2}\sum_{n=1}^\infty n + 1 = 0. $$ Now mathematically this is clearly a divergent series, but using zeta function regularization here we are taking $$ \sum_{n=1}^\infty n = \zeta(-1) = -\frac{1}{12}. $$ And obtain $ D = 26 $ where $\zeta $ is the analytic continuation of the zeta function we know. But it makes no sense in putting $ s = -1 $ in the formulae $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}. $$ As the above is only valid for $ Re(s) > 1 $. So what is going on in here? Can anyone give me a reasonable explanation about obtaining $ -1/12 $?

Answer

I know some derivations in which one can track the emergence of the concrete value, without having to buy that the second order contribution in the Euler-MacLaurin formula (see other answer) is $-\frac{1}{2!}$ times the second Bernoulli number $B_2$.

---

The limit $\lim_{z\to 1}$ of the sum

$0+1\,z^1+2\,z^2+3\,z^3+\dots$

diverges, because of the pole in

$\sum_{k=0}^\infty k\,z^k=z\frac{{\mathrm d}}{{\mathrm d}z}\sum_{k=0}^\infty z^k=z\frac{{\mathrm d}}{{\mathrm d}z}\frac{1}{1-z}=\frac{z}{(z-1)^2}, \hspace{1cm} z\in(0,1)$

We are instead going to consider the sum of smooth deviations of the above, using the local mean

$\langle f(k)\rangle:=\int_{k}^{k+1}f(k')\,{\mathrm d}k'$.

for which $\langle k\,z^k\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\langle z^k\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\langle {\mathrm e}^{k \log(z)}\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\frac{z^{k'}}{\log(z)}\left|_{k}^{k+1}\right.$.

Because of canceling upper and lower bounds, the sum $\sum_{k=0}^n\langle k\,z^k\rangle$ is $\frac{z^0}{\log(z)^2}$ plus terms suppressed by $z^n$. Finally, using the expansion

$\dfrac{1}{\left(\log(1+r)\,/\,r\right)^2}=\dfrac{1}{1-r+\left(1-\frac{1}{1!\,2!\,3!}\right)r^2+{\mathrm{O}}(r^3)}=1+r+\dfrac{1}{1!\,2!\,3!}r^2+{\mathrm{O}}(r^3),$

we find

$\sum_{k=0}^\infty \left(k\,z^k-\langle k\,z^k\rangle\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\log(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right).$

---

The picture shows the two functions $\dfrac{z}{(z-1)^2}$ and $\dfrac{1}{\log(z)^2}$, as well as their difference (blue, red, yellow). While the functions themselves clearly have a pole at $z=1$, their difference converges against

$$-\frac{1}{1!\,2!\,3!}=-\frac{1}{12}=-0.08{\dot 3}.$$

general relativity - A question about the physics involved in tracking satellites such as those used in the GPS system

I know that besides the effects of Newton's theory of Gravitation on the satellite's motion, one has to take account of the retardation of the satellite's clocks when compared to earth-fixed clocks. Is this just an effect of Special Relativity? Or are there other effects that must be taken into account, which would not be predicted by Special Relativity alone, but which only occur in General Relativity?

Answer

The short answer is yes. An object moving at non-relativistic velocity $\vec{v}$ in a weak gravitational field will have a proper time $\Delta \tau$ elapse that is related to the time $\Delta t$ on distant clocks (far from the Earth) by the equation $$ \frac{\Delta \tau}{\Delta t} \approx 1 - \left( \frac{1}{2} \frac{v^2}{c^2} + \frac{G M_E}{r c^2} \right) $$ where $r$ is the distance from the center of the Earth and $M_E$ is the Earth's mass. The first term in parentheses is just the change in the Lorentz factor $\gamma = (1 - v^2/c^2)^{-1/2}$ to lowest non-vanishing order in $v$. The second term, however, can be seen to have something to do with gravity (in fact, the general expression would be $\Phi/c^2$, where $\Phi$ is the Newtonian gravitational potential.) For a satellite in circular orbit, it's not too hard to show that $v^2 = G M_E/r$; thus, you can see that the correction due to special relativity is one-half as large as the correction due to gravity.

Of course, we don't measure the time difference relative to clocks at infinity; we measure the time difference relative to clocks on the ground, which are themselves moving (due to rotation of the Earth) and in a gravitational field. To find the true correction effects, you could combine the correction factor for orbital clocks relative to infinity with the correction for Earth-based clocks relative to infinity. This would lead to a fractional correction relative to Earth of something like $$ \frac{\Delta \tau_s}{\Delta \tau_g} \approx 1 - \left( \frac{1}{2} \frac{v_s^2}{c^2} + \frac{G M_E}{r_s c^2} \right) + \left( \frac{1}{2} \frac{v_g^2}{c^2} + \frac{G M_E}{r_g c^2} \right) $$ where the subscripts $g$ and $s$ stand for "ground" and "satellite" respectively. If we assume a station at the Earth's equator, then we have $v_s \approx 464$ m/s and $r_s \approx 6380$ km; for GPS satellites, we have $r_s = 26 580$ km (an altitude of approximately 20,200 km) and, since they are in approximately circular orbits, $v_s = 3 872$ m/s (using the formula above.) Thus, the overall correction factor due to special-relativistic effects would be $$ \frac{1}{2} \left( \frac{v_g^2}{c^2} - \frac{v_s^2}{c^2} \right) \approx -8.2 \times 10^{-11} \approx - 7 \, \mu\text{s / day} $$ while the correction factor due to gravitational effects would be $$ \frac{G M_E}{r_g c^2} - \frac{G M_E}{r_s c^2} \approx 5.2 \times 10^{-10} \approx 46\, \mu \text{s / day}. $$ Thus, the correction effects work in opposite directions, but you can see that the gravitational effects cannot be neglected compared to the special-relativistic effects.

quantum field theory - Intuition for parameter $mu$ in dimensional regularization

In dimensional regularization, a dimensionless coupling $g$ is replaced by $\mu^{4-d}g$ so that it can remain dimensionless. $\mu$ is unphysical, though its choice affects the values of counterterms. By setting the derivatives of observables with respect to $\mu$ equal to zero, we get renormalization group equations.

I'm fine with this, but there's an additional claim slipped in: the renormalized couplings you get from using $\mu$ "describe physics at the energy scale $\mu$".

But when I first learned DR, $\mu$ was presented as a parameter that was just thrown in for convenience; it was totally meaningless, like the ghost particle mass in Pauli-Villars. Where does this interpretation of $\mu$ come from?

Answer

Dim. reg. is not very intuitive. You could say MS is not a very physical renormalization scheme. There are however several ways in which $\mu$ is connected to an actual physical energy scale in applications:

• 
  

  $\mu$ is arbitrary in general, however, in calculations you usually get logarithms of the form $log \left( \frac{\mu}{M}\right)$ where $M$ is some energy scale in your problem. Could be a momentum transfer for example. If you want your perturbative corrections to be small, you better choose $\mu \sim M$ otherwise the logs would be large and your perturbative correction would not be small. This is mainly the reason why $\mu$ is usually tought of as an energy scale in the problem, even though in principle it is arbitrary. If you follow this prescription for choosing $\mu$, you will find that it really is true that at high momentum transfer the QCD 2->2 scattering becomes weaker.

  
  


• 
  

  In problems with several interesting scales this leads to a problem, as you get several logarithms, say $log \left( \frac{\mu}{m} \right)$ and $log \left( \frac{\mu}{M} \right)$, with say $m \ll M$. In this case you can not choose a $\mu$ such that all logarithms are small. To solve problems of this kind with dim. reg. Effective Field Theory techniques are needed. That is you first construct an EFT valid for momenta smaller than $M$ and matching small momentum S matrix elements between the theories. For definiteness lets say $M$ is some heavy particle mass. In this case you would match the S matrix elements for the light particle between the full theory and the EFT witout heavy fields at some scale, say $\mu \sim M$, implementing the decoupling of the heavy particle BY HAND. The MS scheme does not satisfy the decoupling theorem, but you can put it in by hand. Similarly as with the former case, you match the theories at $\mu$ of order $M$ to avoid large logs in the matching.

  
  

In both cases, you put in the "interpretation" of $\mu$ by hand, to make your life easier, and make perturbative correction actually small. In this sense $\mu$ in applications is usually connected to some physical scale, even though in principle it could be arbitrary.

weighing - Two heaps each with a heavy ball, and a 3-pan balance

The three-pan balance

Imagine a balance with not two, but three pans. Weighings using the balance follow these rules:

• If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)


• If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)

Let's call this the "lightest-pan-detection-rule" (LPDR).

The problem

You are given, one after the other, three heaps of n identical looking balls. (3n balls in total.) The first two heaps each contain one heavier ball. The third heap only contains normal balls. Your task is to find the two heavy balls using a 3-pan balance. How many weighings are required to identify the two heavy balls without needing to guess?

The normal (non-heavy) balls are all of the same weight. The heavy balls may or may not have the same weight. You should not assume that they are only slightly heavier, i.e. your solution should work even if one heavy ball is twice as heavy as a normal ball.

You may weigh only the given balls, and you may put an arbitrary and possibly different number of balls on each pan. Please present a method to identify the heavy balls, and explain why the number of weighings is minimal.

Note

This problem arises when you try to identify two heavy balls among 3n balls, and put n balls on each pan, and find that one pan is the lightest.

As we saw in a previous puzzle, when you want to find the one heavy ball in one of our n-ball heaps alone, for n > 3, you can do it in no more than $\lceil\log_3(n+2)+1\rceil$ weighings. So naively, you can solve the problem in twice that many weighings. Can you do better, by utilizing balls from both heaps, and the extra normal-only heap?

Answer

The number of weighings required is $\lceil{\log_2 n}\rceil$.

Label the heaps as red, blue, and green, with the green heap being the normal ones. Put $k = \lceil {n\over 2}\rceil$ red balls on pan A, and the rest of the red on pan B. Put $k$ blue on pan B, the rest on pan C. Put $k$ green balls on pan C, the rest on pan A. All the pans have the same number of balls on them, so the only weight differences will be due to the heavy balls.

If no pan rises, both heavy balls are on the same pan. The two heavy balls are one red, one blue, and only pan B holds balls of both those colors. If pan A rises, then the red heavy ball must be on pan B and the blue heavy ball must be on pan C. No matter what happens, you have divided the heaps in half (possibly rounding up if $n$ was odd).

Repeat the process with the smaller heaps until each heap is size 1. This takes at most $\lceil \log_2 n\rceil$ weighings.

Note that for $n=1$ we get $\log_2 1 = 0$, which is correct; we don't need to do any weighing to identify which ball in each heap is heavy, since each heap has only one ball.

Proof of minimality The number of different results of a weighing is 4, so the number that can be differentiated in $k$ weighings is $4^k$. If $n = 2^k$, then the number of possible starting states is $n^2 = {(2^k)}^2 = {(2^2)}^k = 4^k$. The algorithm above meets that limit.

electrostatics - How should one interpret $vec{f}=0$ in an ideal battery?

In a circuit there are two forces that act on the charges to keep the current uniform through out,$\vec{f}=\vec{E}+\vec{f_s}$, where $\vec{E}$ is the electrostatic field and $\vec{f_s}$ is the electric field produced by chemical reactions. Inside an ideal battery, $\vec{E}$ and $\vec{f_s}$, oppose each other such that $\vec{f}$, is zero. What exactly would be the correct way of thinking of $\vec{f}=0$? I mean obviously the charges are moving, but it looks like in the battery they are not.

cosmology - Did spacetime start with the Big bang?

Did spacetime start with the Big Bang? I mean, was there any presence of this spacetime we are experiencing now before big bang? And could there be a presence/existence of any other space-time before the big bang?

faster than light - Is the GW150914 signal consistent with a superluminal gravitational wave burst?

I've been following the news about the detection of gravitational waves with interest and had a question for those with a physics background. I've gathered that the preferred explanation for GW150914 (Fig 1) is a binary black hole inspiral because it is characterized by an increasing frequency and amplitude over time (Fig 2). However, the other types of signal expected (termed "bursts") have the opposite properties, ie a decreasing amplitude and frequency with time (Fig 3), something like what you get out of a bass drum.

As I understand it, the speed at which this signal propagated is constrained to between the speed of light (c) and 1.7*c. [ref 1] Consistent with a superluminal velocity, the only other observation that has been proposed to have some relation was by the Fermi Gamma-ray Burst Monitor. That team reported a gamma ray burst 0.4 s after GW150914. [ref 2]

I searched the literature a bit and found that apparently what we would expect from a superluminal signal is that it appears in chronologically reversed order from a subluminal signal. Here is a quote [ref 3]:

Therefore, if a macroscopic phenomenon is known to produce a radioemission obeying a certain chronological law, and one happens to detect the reversed radioemission, the observed source should be considered as a Superluminal, approaching object.

My questions are:

1. Has anyone been searching the LIGO data for such "reverse" signals?


2. Would this be considered a useful line of inquiry? If not, why not?

Figure 1. GW150914 Signal https://losc.ligo.org/events/GW150914/

Figure 2. A typical expected inspiral signal http://www.ligo.org/science/GW-Inspiral.php

Figure 3. A typical expected burst signal http://www.ligo.org/science/GW-Burst.php

References:

1. Blas et al. On constraining the speed of gravitational waves following GW150914. 12 Feb 2016. http://arxiv.org/abs/1602.04188


2. Connaughton et al. Fermi GBM Observations of LIGO Gravitational Wave event GW150914. 16 Feb 2016. http://arxiv.org/abs/1602.03920


3. R. Mignani, E. Recami. Astrophysics and tachyons. Il Nuovo Cimento B (1971-1996) May 1974, Volume 21, Issue 1, pp 210-226. http://link.springer.com/article/10.1007%2FBF02737452

Answer

Is someone searching for a superluminal signal? I'm sure someone is, in the same way there are crackpots like apparently Mignani & Recami publishing papers on the topic.

However this isn't a particularly credible line of research. The thing is, superluminal anything violates causality in special relativity. This is not something to be taken lightly. Once you allow for causality violations, the most cherished and fundamental princples of science go right out the window. One can no longer make predictions about the future based on the past, since time travelers from distant times and places are free to affect your experiments whenever it is convenient. With causality violations you could write down your observations in your lab notebook and then the next day go back in time and tamper with the experiment, changing what really happened... and altering what you wrote down? This sort of thing hardly even makes sense, and our usual notions of causality, not to mention time and space, fail to adequately describe such a universe.

One might argue that superluminal physics could do less violence to science as a whole, that we might reformulate some meta-relativity that handles it in a logically consistent manner. But this is speculation well beyond science fiction. Even if such were possible, it would still destroy the foundations of the theory that underpins gravitational waves in the first place.

If you destroy special relativity as we know it, general relativity must also fall, since general relativity has at its core the assumption that the universe is locally governed by special relativity. But general relativity is what gives us merging black holes and gravitational waves and predictions for what we see in our interferometers. If you claim to have detected violations of special relativity with LIGO, you have to explain what exactly you are seeing, because you've just undermined all of physics, gravitational waves and lasers included. Indeed you would have to explain most of the last century of physics with an unfathomably different theory that somehow matches all of the observations done to date but also includes the possibility for superluminal signaling.

---

More philosophically, consider Quine's epistemology and his web of belief. All our experiences (or at least all our science) forms an interconnected web, with various propositions affecting and deriving from others. Some such propositions are predictions: "LIGO will see waveforms that look like..." These predictions can be tested empirically.

But what do we do with conflicting data? What if a prediction turns out to not match reality? Something must be wrong, but what? It turns out we have (too many) choices. Perhaps the prediction was made in error, or perhaps our apparatus doesn't work as we thought, or perhaps our data was poorly analyzed, or, just maybe, our core principles at the heart of the web of belief are wrong.

The point is, you'd hesitate to throw out the core principles that clearly work so well in other cases. If indeed you detected a "time-reversed gravitational wave in your data" you'd sooner look to the more error-prone parts of the web of belief in the area. For example, a new astrophysical phenomenon might be the cause, one that proceeds in conformity with relativity.

special relativity - How do simple two-component Fierz identities follow from a property of the Pauli matrices?

On page 51 Peskin and Schroeder are beginning to derive basic Fierz interchange relations using two-component right-handed spinors. They start by stating the trivial (but tedious) Pauli sigma identity $$(\sigma^\mu)_{\alpha\beta}(\sigma_{\mu})_{\gamma\delta}=2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta}.\tag{3.77}$$ They then claim to "sandwich" this identity in between the right-handed part of four Dirac spinors $u_1,u_2,u_3,u_4$: $$(\bar u_{1R}\sigma^\mu u_{2R})(\bar u_{3R}\sigma_{\mu}u_{4R})=2\epsilon_{\alpha\gamma}\bar u_{1R\alpha}\bar u_{3R\gamma}\epsilon_{\beta\delta}u_{2R\beta}u_{4R\delta}.$$ I understand the first identity with elements of the contraction of the Pauli vector perfectly fine, but this one completely mystifies me. The next step in their calculation swaps indices in the Levi-Citiva symbols and uses essentially the same equation in the other direction to get the expected Fierz identity, so if I understood the first equality I would also know the second. However, I just can't see how it easily follows from the Pauli matrix equation. I thus have two questions.

a) Is there an elegant way to actually "sandwich" the identity into four right-handed Weyl spinors, or do I have to manually expand the bilinears?

b) Does right-handedness actually play a role here? That is, it seems to me like this derivation would work just as well with any left-handed spinors, but is this true?

wordplay - Why will my friend have to grow his beard out?

Knowledge required for this riddle: English-language proverbs.

My friend Leonard came to me with a problem the other day. He said, “Tanner, I want to have a clean-shaven face. But every time I use a razor, I end up with this nasty painful rash.”

“Razor burn,” I said.

“Yeah, that's right,” said Leonard. “Do you have any advice for me?”

“Unfortunately, there's no good solution,” I told him. “You're going to have to either grow your beard out, or just cope with those painful rashes.”

Why will Leonard have to grow his beard out?

Answer

Maybe it is because

A Lenny shaved is a Lenny burned.
(A penny saved is a penny earned.)

visual - We interrupt your scheduled series for a pretzel rebus

What does this rebus, drawn with pretzels, say? (This is what happens when I get bored with pretzels on the table... ;P)

_{Note: The 'scheduled series' mentioned is this. It still needs to be solved! Finally solved by @randal'thor! :D}

Overall picture:

Individual pictures:

Note: That black dot in the second one is not important.

Answer

The first panel shows

a god and a small human, therefore the panel stands for myth.

The second panel shows

a running human, therefore the panel stands for run. However the backwards pointing indicates that the word should be in past tense, so the panel ultimately stands for ran. (Thanks to @CWoods)

The third panel shows

a deer

Now,

say the three words out loud and you'll have the final solution: Mithrandir

astronomy - What are the prerequisites for considering any other planet to be habitable?

Well, there is a measure of how a planet could be considered like Earth, called Planetary habitability. Based on this measure, what are the prerequisites needed to consider a planet to be a habitable one?

Answer

Habitable by whom? There are conditions that are uninhabitable by humans, however, many "extremophiles" survive perfectly happy.

Although, if you are talking about humans, here is a small list (all the rest are probably more "nice to have" requisites):

• Approximately 20% oxygen (more or less depending on the pressure)


• Temperatures that allow for liquid water


• Adequate access to water (and food)



• Adequate protection from radiation.


• Then a whole host of conditions that wouldn't end human life. Such as deadly pathogens on chemicals in the atmosphere.

All this said, since we only have a sample size of one currently for planetary life, we really don't know what is possible, or how to bound the problem. ANYTHING is just speculation drawn from this one sample. That said, we have found life on our own planet where we never suspected it to be. Life has proven to be nearly unstoppable in propagating throughout every niche on this planet. So, the better question I think would be what are the requisites for abiogenesis? I think once life manages to start on any planet, it will adapt to whatever conditions the planet presents (to within a reasonable degree).

grid deduction - Double feature: In concert

_{This puzzle is part 3 of the Double feature series (first part here). The series will continue in "Double feature: Hot stuff".}

---

Rules of Ripple Effect¹

• Fill the grid so that each cell contains one number.


• The bold lines divide the grid into rooms. Every room contains consecutive numbers starting from 1. For example, a room of size three will have the numbers 1, 2 and 3 in some order.


• If there are more than one of the same number on the same row or column, the numbers must have at least as many cells between them as the number indicates. For example, two cells with the number 3 on the same row or column must be separated by at least three cells.

Across
3. Top removed from crooked talking tree (3)
6. A bush for wingless peacock having no energy left (4)
7. Drive without a stick (3)

8. Writer of Introductions to Pharaohs of Egypt (3)
9. Oxford educator identifying flower in Russia's south (3)
10. French city seen in anime series (5)
12. The finale is held in crescendo (3)
16. Innumerable? Calculate a smaller amount (9)
17. Philistine refusing to boil chicken (3)
18. Constructed language words that may be pronounced at the altar (3)

Down
1. Christian church group of two people rejecting Buddhist mantra (5)
2. Pear-shaped harvest (4)

4. Fashion tailor's fringe finishing (5)
5. A reservoir maker; he lost a rib (4)
9. An album of original 1970s music (5)
11. Part of TV series Wild Deep is receiving Oscar (7)
13. Wacky things that go with bolts (4)
14. Hybrid talisman discovered (4)
15. Periodically upbeat companion (3)

_{¹ Paraphrased from the original rules on Nikoli.}

Solve both puzzles to answer the question: What is "Clap" by Steve Howe?

Answer

What is "Clap" by Steve Howe?

Unaccompanied Solo

Grids

Across

3. Top removed from crooked talking tree (3)

ENT (talking tree, BENT - B)

6. A bush for wingless peacock having no energy left (4)

COCA (bush, (P)EACOC(K) - E, reversed)

7. Drive without a stick (3)

ROD (stick, ROAD - A)

8. Writer of Introductions to Pharaohs of Egypt (3)

POE (writer, initials)

9. Oxford educator identifying flower in Russia's south (3)

DON (ddef)

10. French city seen in anime series (5)

NIMES (french city, aNIME Series)

12. The finale is held in crescendo (3)

END (the finale, crescENDo)

16. Innumerable? Calculate a smaller amount (9)

COUNTLESS (innumerable, COUNT + LESS)

17. Philistine refusing to boil chicken (3) [Thanks M Oehm]

HEN (chicken, HEATHEN - HEAT)

18. Constructed language words that may be pronounced at the altar (3)

IDO (ddef)

Down

1. Christian church group of two people rejecting Buddhist mantra (5)

DUOMO (christian church, DU(OM)O)

2. Pear-shaped harvest (4)

REAP (harvest, PEAR anagrammed)

4. Fashion tailor's fringe finishing (5)

TREND (fashion, TailoR + END)

5. A reservoir maker; he lost a rib (4)

ADAM (he lost a rib, A + DAM)

9. An album of original 1970s music (5)

DISCO (1970s music, DISC + O)

11. Part of TV series Wild Deep is receiving Oscar

(7) EPISODE (part of tv series, EP(ISO)DE)

13. Wacky things that go with bolts (4)

NUTS (ddef)

14. Hybrid talisman discovered (4) [Thanks M Oehm]

MULE (hybrid, aMULEt)

15. Periodically upbeat companion (3) [Thanks M Oehm]

PET (companion, uPbEaT)

cipher - An encoded message for Nigel

Nigel came inside from his back garden carrying the peas he'd harvested, and dumped them onto the table to be sorted according to which were edible and which weren't. Just as he did so, he noticed a letter lying on the table, now mostly buried by the peas, in what looked like the handwriting of his colleague Richard. He scrabbled through the sour pods, picked up the cryptic missive, and read:

1, 2, 4, 8, 16, 32, 64, 128

I1, Y1, O8, A5, R1, U6, V3, M4, D2, A4, S4, U3, E3, J1, E11, A13, M2, I3, A2, F3, I6, O5, W1, S4, Y3, A14, U4, S1, P1, A8, E6

What does the message say?

Answer

The bit in the text about

scrabbling through the sour pods

is a hint that this has to do with

Scrabble, and the so-called SOWPODS word list.

The list of powers of two indicates

that we should consider two-letter words.

So we'll make use of

the list of 2-letter words in SOWPODS in alphabetical order as found e.g. here.

Now we interpret each item in the message as follows:

"I1" means "1st two-letter word beginning with I", "U6" means "6th two-letter word beginning with U". And then we take the second letter of each of these words.

So, e.g., we begin as follows:

I1 -> ID -> D
Y1 -> YA -> A
O8 -> ON -> N

and eventually we get

DANGER!MEETMEATTENBYTHETOWNHALL

where the "!" represents

a word that doesn't exist (there are no 2-letter words in SOWPODS containing a V).

---

User IAmInPLS cunningly notes that

Nigel and Richard are two names extracted from the full name Nigel Richards, who is the world's highest-rated Scrabble player.

paper folding - Create a 1 inch measurement

This is a follow up to the "Create a 3 inch measurement" puzzle, which got a lot of innovative solutions. Using the standard 8.5 x 11 inch paper, can you create a 1 inch measurement only by folding? Again no marking allowed. No ruler either.

One more thing. I realize that folding the 8.5 inch side three times can get you 1.06 inches. But can you do better than that? Maybe I was overdoing it, but I think I got it in more than 3 folds.

Answer

A solution in 4 folds:

1. Fold one corner down to the opposite side. We now have a triangle that is 8.5" by 8.5", and a flap below it that is 2.5" by 8".
2. Fold the flap up, and call this Flap A. Now unfold the paper and turn it upside down.
3-4. Repeat the same steps to get a Flap B.
Now, unfold the paper and just fold down Flap A and Flap B. the thin space between the two flaps should have length exactly 1".

This is because

The flaps are of length 2.5", and when folded up (or down) that is another 2.5" for a total of 5". If Flap A accounts for 5" and Flap B accounts for another 5" from either side of the 11" length of the paper, the space in between should be 1".

enigmatic puzzle - An overheard gang meeting, a double agent, and a murder

I was walking home late last night and heard raised voices coming from a boarded up old warehouse. I stopped, out of sight, to listen...

[When I started eavesdropping, the conversation was already under way...]

Gangster 1: -infiltrator in our midst, and I know who it is... [dramatic pause] ...It's you isn't it!
Gangster 2: Me?
G1: What? No... him.
Gangster 3: That's ridiculous! I'm no traitor! How do we know it's not you?!
G2: It's not me!
G3: No... him... [under breath] ...idiot.
G1: That doesn't even make sense. I'm the most prolific amongst us. When would I even have the chance to work for the opposition?
G3: What do you mean? I'm in every day!
Gangster 4: [to himself] Me too, but you don't hear me harping on...

G3: [continuing] Not to mention I saw you with your ex last week on my way out of here... And I won't mention who I saw him hanging out with again yesterday...
G2: We weren't hanging out! I was just stuck behind him in a queue!
Gangster 5: Hey, we all fraternize with the enemy now and then... It's necessary to meet our goals... But this is different. From what I hear, half the time you act like you are one of them!

[A street sweeper drove slowly by, so I missed a little of the conversation...]

G3: It's a set up! Think about it... Who would gain the most from seeing me dead? ...I would!
Gangster 6: 'strue, but 'e would too...
G1: Sure, he occasionally fills in for me, but he could practically replace you!
G2: No he couldn't!
G6/G1/G3: Would you shut up!
G4: Geez settle down guys, what's with all the yelling?

G1: Because no one was talking to you!
G4: Ahh... Gotcha...

[A crowd of drunken revellers stumbled by and as I slipped into an alley to remain unseen, I missed another few minutes of the argument...]

G1: Gimme one good reason why I shouldn't kill you right now!
G2: I didn't do anything wrong!
G3: Wait... Who are we talking about now?
G1: You!
G2: Oh...
G5: Why?
G1: Yes!

[From here things escalated rapidly, and when gunshots were fired, I made a run for it...]

I went to the police but when they visited the scene, the gang was gone, and everything had already been cleaned. They said there wasn't enough to go on unless I could identify the victim and the perpetrators.

Who was the gang?
Which member was killed?
Why were they killed?

Bonus: What was the name of each gangster?

Answer

The gang was the notorious

Vowel Syndicate!

The victim was

Y

because

he was a double agent for the Vowel Syndicate's chief rivals, the Consonant Guild!

The gangsters are, I believe:

1: e ("most prolific")
2: u (every time someone says "you" he thinks they're talking to him, also was stuck behind "Q")
3: y (is in every "day", is the one that seems to be the accused)
4: a (is also in every "day")
5: o (responds to gangster 2 saying "oh" near the end)
6: i (responds to gangster 3 saying "I" would gain the most from seeing "Y" dead)