mathematics - Divisible by seventeen

Determine the smallest integer $n \geq 0$ for which

• the decimal digit sum of n is a multiple of 17


• the decimal digit sum of $n+1$ is a multiple of 17.

No computers! The puzzle has a nice direct solution.

Answer

We know that the digit sum of $n$ is a multiple of $17$, let us write that as ($d_1$ being the least significant digit): $$d_m + d_{m-1} + ... + d_2 + d_1 = x * 17$$ If $d_1$ would be smaller than $9$ then the digit sum of $n+1$ would be $$x * 17 + 1$$ which is obviously not divisible by $17$, so $d_1$ must be $9$. If $d_2$ would be smaller than $9$, then the digit sum of $n + 1$ wold be $$x * 17 - 9 + 1 = x * 17 - 8$$ which is not divisible by $17$ again, so $d_2$ is must be $9$ well. If $d_3$ would be smaller than $9$, then the digit sum of $n + 1$ would be $$x * 17 - 9 - 9 + 1 = x*17 - 17$$ which is divisible by $17$. This means we look for the smallest number with digit sum divisible by $17$ with the last 2 digits equal $9$ and the third last digit lower than $9$. This is obviously: $$8899$$ .

mathematics - Winnie-the-Pooh and the 27 honey pots

Winnie-the-Pooh keeps his $27$ honey pots in the larder. Each pot contains up to $1$ kilogram of honey, and different pots contain different quantities of honey. All $27$ pots together contain $17$ kilogram of honey.

Every day, Winnie-the-Pooh selects $7$ pots, picks a real number $x$, and then eats exactly $x$ kilogram of honey from each of the selected pots.

Question: Is it always (independently of the initial distribution of honey) possible for Winnie to empty all $27$ honey pots in a finite number of days?

Answer

Here is a rigorous proof that Pooh can always succeed.

Let $P_i$ be the amount of honey in the $i^{th}$ pot when sorted in decreasing honey contents, so $1\ge P_1\ge P_2\ge \dots\ge P_{27}\ge0$ and $P_1+\dots+P_{27}=17$.

Using the beginning of Sleafar's solution, we now need only show how to equalize the seven heaviest pots. At any point, let $m$ be the number of pots which have the same amount of honey as the heaviest, and let $n$ be the number of nonempty pots (so initially, $m=1$ and $n=27$). We show that, as long as $m<7$ and $n\ge 14$, it is possible to either increase $m$ or decrease $n$. We then argue why $n$ can't drop below $14$ before $m$ reaches $7$, so that eventually we equalize the top $7$ pots.

Here is the method. Let $x$ be the gap in honey contents between the heaviest $m$ pots and the next strictly lighter pot. Each day for the next $n-7$ days, Pooh eats $\frac{x}{n-7}$ kg from each of the $m$ heaviest pots. Each of these days, he must eat the same amount from some $7-m$ other pots. He does this in such a way that the $n-7$ pots $P_8,P_9,\dots,P_n$ are eaten from equally over this $n-7$ day period (for example, by arranging the nonempty pots in a circle, and eating from each consecutive segment of length $7-m$).

As long as each of the nonempty pots contain at least $\frac{x}{n-7}\cdot (7-m)$ kg of honey, then this will work, causing the $m$ heaviest pots to decrease to the next highest weight, thus increasing the number $m$. If the nonempty pots do not contain enough honey to do this, then adjust $x$ so that the smallest pot contains exactly $\frac{x}{n-7}\cdot (7-m)$. Then doing the procedure will empty that pot, thus decreasing $n$.

Now, why can't $n$ drop below $14$ before $m$ reaches $7$? First off, the amount of honey that needs to be removed from pots 8 through $n$ in order to equalize pots 1 through 7 is equal to $$6(P_1-P_2)+5(P_2-P_3)+4(P_3-P_4)+3(P_4-P_5)+2(P_5-P_6)+(P_6-P_7)$$ This is because, during the phase when pots $P_1,\dots,P_m$ are equal, we have to remove a total of $P_{m}-P_{m+1}$ from each of the first $m$ pots, and $(7-m)$ times that from the last pots 8 through $n$. A different way of writing this is $$6P_1-P_2-P_3-\dots -P_7\le 6 - 6P_7< 3$$ The last inequality follows since $P_7> \frac12$. Why is this true? If not, then the first 6 pots would initially contain at most $6$ kg, and the last 21 would contain at most $\frac12$, so that the total amount of honey was at most $6+\frac12\cdot 21=16.5$, which is impossible since we started with $17$ kg of honey.

So, the procedure will remove less than $3$ kg from the last 20 pots. However, the first 14 pots contain at most 14 kg, meaning the lightest pots together contain at least $17-14=3$ kg. Since our procedure eliminates the lightest pots first, the first 14 pots will not be emptied, as claimed.

astronomy - Why don't we see solar and lunar eclipses often?

Since we see the new moon at least once in a month when the Moon gets in between of the Sun and the Moon at the night and as far as I know if this happens during the day, you'll get to see a solar eclipse. Why don't we get to see this often or in the day?

Does it mean that in some part of world there's a solar eclipse when we are seeing a new moon? I'm looking for a diagram or interactive way to understand this if possible as I'm not a native English speaker, but I'll try my best to do so.

Answer

If the Moon's orbit around the Earth were in exactly the same plane as the Earth's orbit around the Sun, we'd have a total solar eclipse every month (but 100% totality would be seen only from the tropics).

But in fact they're not in the same plane. The Earth's spin axis is tilted by about 23 degrees relative to the Earth's orbit around the Sun, and the Moon's orbit is closely aligned with the Earth's spin. As a result, the Sun and the Moon do not follow the same path in the sky.

We get a solar eclipse only when (a) there's a new Moon, so the Sun and Moon are in the same position east-to-west, and (b) the new Moon happens when the Sun and Moon happen to be closely aligned north-to-south.

Since the Sun and Moon are both about half a degree wide (as we see them in the sky), the 23-degree offset of their paths makes solar eclipses relatively rare events.

Lunar eclipses, which occur during the full Moon when Moon passes into the Earth's shadow, are more common because the Earth is bigger than the Moon, and so has a much wider shadow.

speed of light - How soon that a force affect another object?

Imagine this scenario: I have 2 objects in vacuum without any force exerted upon them not even a possible gravitational force between them. Now if one of them gets a gravitational or magnetic force, how long from the moment the object gained this force to the other object to be affected by it? Is it instantaneous? Is it the same as speed of light?

Answer

In your comment you ask:

That question only addresses gravitational force. Is it the same for all kind of forces?

Generally speaking a force transmitted by massless particles like the photon and graviton obeys an inverse square law while a force transmitted by massive particles falls off exponentially with distance. This means that only the forces transmitted by massless particles are long range and therefore that any long range force propagates at the speed of light.

So for any macroscopic objects the interaction between them is going to travel at the speed of light.

Trying to assign a speed to the weak and strong nuclear forces is a bit complicated. At longish range, i.e. within a few proton diameters, the strong force is transmitted by massive mesons and therefor propagates at less then the speed of light. However the underlying force is transmitted by massless gluons. This doesn't give rise to a long range inverse square law because of confinement. The weak force is transmitted by massive W and Z bosons at low energies, but above the electroweak transition these become massless and in principle the electroweak force is long range.

cipher - Yet there is method in this Morse code

Sometimes a nonsense telegram makes much more sense after a methodical maneuver.
Here’s one for fun:

 - . -.. / ... .... . . - ... / ... . -. - / .... .. ... / .- -. - /
 - --- / - . . -. / ... .... . .. -.- / .. / ... .... .- -. - / .... . -..- /
 ... .. ... - . .-. / ... .- .-.. / .- ... / ... .... . / .. ... ... ..- . -.. /
 ..-. . . / ... .... .. ..-. - / .. ... / .-- . - / ... ..- . - /
 . ...- . .-. / - .. -. / ... ..- .. - / . . -.- / ... .. .-. . -.


 TED SHEETS SENT HIS ANT
 TO TEEN SHEIK I SHANT HEX
 SISTER SAL AS SHE ISSUED
 FEE SHIFT IS WET SUET
 EVER TIN SUIT EEK SIREN

 20 5 4 0 19 8 5 5 20 19 0 19 5 14 20 0 8 9 19 0 1 14 20 0
 20 15 0 20 5 5 14 0 19 8 5 9 11 0 9 0 19 8 1 14 20 0 8 5 24 0
 19 9 19 20 5 18 0 19 1 12 0 1 19 0 19 8 5 0 9 19 19 21 5 4 0
 6 5 5 0 19 8 9 6 20 0 9 19 0 23 5 20 0 19 21 5 20 0

 5 22 5 18 0 20 9 14 0 19 21 9 20 0 5 5 11 0 19 9 18 5 14

 10100 101 100 0 10011 1000 101 101 10100 10011 0 10011 101 1110 10100 0 1000 1001 10011 0 1 1110 10100 0
 10100 1111 0 10100 101 101 1110 0 10011 1000 101 1001 1011 0 1001 0 10011 1000 1 1110 10100 0 1000 101 11000 0
 10011 1001 10011 10100 101 10010 0 10011 1 1100 0 1 10011 0 10011 1000 101 0 1001 10011 10011 10101 101 100 0
 110 101 101 0 10011 1000 1001 110 10100 0 1001 10011 0 10111 101 10100 0 10011 10101 101 10100 0
 101 10110 101 10010 0 10100 1001 1110 0 10011 10101 1001 10100 0 101 101 1011 0 10011 1001 10010 101 1110

What’s that telegram really trying to tell?

Everything you need to know is below and most of the work to be done is above.

     Morse     Binary           Morse      Binary            Morse      Binary

    A  .-     1      1         J  .---   10    1010         S  ...    19   10011
    B  -...   2     10         K  -.-    11    1011         T  -      20   10100
    C  -.-.   3     11         L  .-..   12    1100         U  ..-    21   10101
    D  -..    4    100         M  --     13    1101         V  ...-   22   10110
    E  .      5    101         N  -.     14    1110         W  .--    23   10111
    F  ..-.   6    110         O  ---    15    1111         X  -..-   24   11000
    G  --.    7    111         P  .--.   16   10000         Y  -.--   25   11001

    H  ....   8   1000         Q  --.-   17   10001         Z  --..   26   11010
    I  ..     9   1001         R  .-.    18   10010      space         0     0

Answer

I can see how David Thomas got his answer as I worked it out and got the same answer, but as he hasn't added an explanation (he has now), this is how you get the answer:

The telegram is telling you

THE MORE MIRTH THE MERRIER

TL;DR

The morse code gives binary for the phrase

The morse code represents binary, each dash is a 1 and each dot is a 0.

Converting gives:

1 0 100 / 000 0000 0 0 1 000 / 000 0 10 1 / 0000 00 000 / 01 10 1 / 1 111 / 1 0 0 10 / 000 0000 0 00 101 / 00 / 000 0000 01 10 1 / 0000 0 1001 / 000 00 000 1 0 010 / 000 01 0100 / 01 000 / 000 0000 0 / 00 000 000 001 0 100 / 0010 0 0 / 000 0000 00 0010 1 / 00 000 / 011 0 1 / 000 001 0 1 / 0 0001 0 010 / 1 00 10 / 000 001 00 1 / 0 0 101 / 000 00 010 0 10

Squishing that together gives:

10100 / 0000000001000 / 0000101 / 000000000 / 01101 / 1111 / 10010 / 0000000000101 / 00 / 000000001101 / 000001001 / 0000000010010 / 000010100 / 01000 / 00000000 / 000000000010100 / 001000 / 00000000000101 / 00000 / 01101 / 00000101 / 000010010 / 10010 / 000001001 / 00101 / 00000010010

Removing the slashes and leading zeroes, and also for any word which is comprised of only 0s, replacing it with a / to make it easier to read gives:

10100 1000 101 / 1101 1111 10010 101 / 1101 1001 10010 10100 1000 / 10100 100 101 / 1101 101 10010 10010 1001 101 10010

Now converting from binary using the table, and the / as a space gives the final phrase:

THE MORE MIRTH THE MERRIER

Red Herrings

The red herring that threw me and Gareth in chat off to start:

There are 26 words in the question, which we thought might correspond to the alphabet.

And another red herring which initially convinced Gareth to post an answer is that

The first part of the morse is -.-.. which matches the first piece of binary 10100, however this doesn't work for all the parts.

But the biggest red herring:

Everything in the telegram but the morse code is unneeded!

lateral thinking - Connect 3 houses with 3 wells

Connect every house with every well without the lines intersecting.

I am not sure if this puzzle has a solution. I have been puzzled by it for a long long time. An old man from my village mentioned this puzzle to me 12 years ago. I have never been able to solve it, I'm hoping someone here can.

thermodynamics - Is a plasma a distinct phase of matter?

Long ago I learned that a plasma was a distinct state of matter after solid, liquid and gas, and also that it was achieved by imparting heat to a the matter. But most references describe a plasma as an ionized gas. So I'm having trouble understanding, what then, does it mean to be a distinct phase of matter? Is ionization, as opposed to heat, all that's required to make a gas a plasma? If so, what makes a plasma more distinguished than, say, an ionized liquid?

Answer

For clarity, there is a common misconception about plasma here. Plasma when being introduced for the first time to someone who doesn't know what it is, it is called "The fourth state of matter" which is an inaccurate description of it. Since this term is used for introducing some one to plasma, it is no big deal.

When a material changes from a distinct phase to another, it goes through a physical process called phase transition. When gas becomes plasma, it doesn't go through the standard phase transition. Hence plasma-in a general sense-can't be regarded as a distinct phase as solid, liquid and gas phases. It is a phase of the gaseous state. In certain rare cases however, transition from gas to plasma can be described as phase transition.

Plasma by definition is a mixture of free electrons and their ions (possibly negative ions). You need enough energy to liberate electrons from atoms. Roughly speaking, When you put that energy in a solid, energy might be dissipated as heat. If you put that energy in a liqued, energy might be dissipated in vaporization. If you put it in a gas it goes into breaking atoms and molecules (creating plasma). The following figure makes it clearer

Hopefully that was useful

general relativity - Why does time slow down the closer you are to a mass?

When ever i look this up all I get is sites saying how its because general relativity says "-" why does it do it though? it is because there is more motion near gravity than further away? Or is it something completely different?

quantum mechanics - Electron Double Slit Experiment-de broglie wavelength relation to distance btw slits

In the 2 slit experiment with electrons, is the distance between the slits related to the individual electron's de broglie wavelength?

In other words, if the slits are too far apart which would prevent the electron's matter wave from passing through both slits, does the interference pattern then fail?

A broader question is what is the relationship between the size of the slits, the distance between the slits, and the observed interference pattern?

Answer

You ask:

A broader question is what is the relationship between the size of the slits, the distance between the slits, and the observed interference pattern?

The answer here covers your question.

The de Broglie wavelength describes the effective wavelength that a particle would have when it was behaving as a wave.

to decide what wavelength an electron should have so as to be able to see the interference pattern one has to see separation and and distance to the screen that the slits should have.

The interference pattern is dictated by the distance from one bright line (coherence) to the next:

$$\Delta y = \frac{\lambda D}{d}$$

where D is the distance from the slit to the screen (or detector), little d is the spacing between the slits, and λ is going to be our de Broglie wavelength.

Let's assume we want to use electrons for our experiment. We build a setup with the screen placed 1 meter from the slits, and the two slits 1 millimeter apart (maybe we found this equipment in a storage closet in the physics department...). This setup will make the distance between the bright spots on our screen 1000 times what the de Broglie wavelength of our incoming electron is. We want to be able to actually see the interference pattern in our detectors, so perhaps we should request that the spacing of the bright spots be about 1 millimeter (this would depend on the detectors, of course). This means the de Broglie wavelength of our electron has to be about one meter. Now we go back to the equation for de Broglie wavelength, and see that we know h and we now know λ, so we can calculate what p should be. Since we know the mass of the electron, calculating the momentum is essentially the same as calculating the speed; for our experiment, we find the electron needs to be going about 0.0007 m/s! That's a tiny speed... about 2 inches a minute (kind of like pouring ketchup)!

So experiments are not easy with electrons.

For the buckyball experiment , the researchers used slits about 100 nanometers apart (a nanometer is one millionth of a millimeter), and shot the buckyballs through the slits at about 200 meters per second (roughly 500 mph), much slower than the speed of light.

gravity - Why the galaxies form 2D planes (or spiral-like) instead of 3D balls (or spherical-like)?

Question: As we know, (1) the macroscopic spatial dimension of our universe is 3 dimension, and (2) gravity attracts massive objects together and the gravitational force is isotropic without directional preferences. Why do we have the spiral 2D plane-like Galaxy(galaxies), instead of spherical or elliptic-like galaxies?

Input: Gravity is (at least, seems to be) isotropic from its force law (Newtonian gravity). It should show no directional preferences from the form of force vector $\vec{F}=\frac{GM(r_1)m(r_2)}{(\vec{r_1}-\vec{r_2})^2} \hat{r_{12}}$. The Einstein gravity also does not show directional dependence at least microscopically.

If the gravity attracts massive objects together isotropically, and the macroscopic space dimension is 3-dimensional, it seems to be natural to have a spherical shape of massive objects gather together. Such as the globular clusters, or GC, are roughly spherical groupings Star cluster, as shown in the Wiki picture:

However, my impression is that, even if we have observed some more spherical or more-ball-like Elliptical galaxy, it is more common to find more-planar Spiral galaxy such as our Milky Way? (Is this statement correct? Let me know if I am wrong.)

Also, such have a look at this more-planar Spiral "galaxy" as this NGC 4414 galaxy:

Is there some physics or math theory explains why the Galaxy turns out to be planar-like (or spiral-like) instead of spherical-like?

See also a somehow related question in a smaller scale: Can I produce a true 3D orbit?

p.s. Other than the classical stability of a 2D plane perpendicular to a classical angular momentum, is there an interpretation in terms of the quantum theory of vortices in a macroscopic manner (just my personal speculation)?

Thank you for your comments/answers!

pattern - Two lists of favorite words

My great aunt Edith keeps this list of her favorite words tacked on the refrigerator:

Atlas
Chance

Detail
Frost
Hand
Magnet
Orange
Ring
Sand
Tiger
Winter
Wolf

My great uncle Mark has this list of his favorites:

Angel
Art
Bad
Boot
Gift
Hall
Kind

Mutter
Rat
Stern

Which one of them would like the word Listen?

Answer

Your

great uncle Mark

likes the word "Listen". Here's why:

All words can be English or German words. The words in Aunt Edith's list have the same meaning in German and in English. (Or there is at least one common meaning. "Ring" in German means a round band of metal, but not a call on the phone.) The words in Uncle Mark's list have a different meaning in English and German. "Gift", for example, means poison. And "Listen" means lists and has nothing to do with hearing.

cosmology - Is large-scale "time reversal" (Poincaré recurrence) possible given infinite time?

The following are some assumptions I'm basing my question on, from what (little) I understand of physics. I list them so an expert can (kindly) tell me where I'm going wrong.

• There is a probability assigned to every possible sequence of events.


• Besides sequences that violate the "rules", every sequence seems to have a nonzero probability.


• This applies to large scale systems as well, except the odds usually seem to tend toward nearly 100% in favor of certain sequences (i.e., paths of highest entropy).



• There is therefore a probability assigned to the following sequence of events: a system (large or small scale) reversing everything it's done since a certain point in time (Poincaré recurrence), and ending up back "almost" where it started; perhaps not an exact copy.


• The universe appears to be expanding, and for all we know will do so forever, possibly leading to the heat death of the universe.

Here's my question:

Assuming the universe will continue on forever (and therefore that we have infinite time in our equations), isn't there a nonzero probability that everything in it can "reverse" itself and end up back at the big bang, perhaps in a somewhat slightly altered path?

Is dark energy is the only thing preventing this? If so, (1) would this be possible without dark energy? (2) could the force carriers, assuming they exist, behind dark energy obey quantum mechanics, and therefore be "time reversable"?

newtonian mechanics - Force acting on an object?

You have a bar of metal in an environment with no gravity. A force is applied on one end of it. How does it rotate?

There is a non-zero torque on any random point selected on the bar. For example, on point A the torque is $T_A = F*y$, but on point B the torque is $T_B = F*x$. So the torque on any point on the bar except the point where the force is applied, is non-zero. If there is a non-zero torque on one point, the object must rotate around that point. But since there is a non-zero torque on every point, and the object can't rotate around every point, how does it rotate and around what point?

Answer

See here for something about rotation and torque.
As is explained there, the IAOR is defined as the point about which every point is in pure rotational motion i.e. every point has a velocity perpendicular to the position vector to that point from IAOR. This point's position can vary in space and with time and is usually difficult to treat in most problems.
Theoretically, you can find the torque about any point from where you define the rotational variables ($\theta, \omega$) but a particularly useful point is the centre of mass.

Treating the rotation to be about the COM, you eliminate pseudo forces in the COM frame and that allows you to treat that frame as an inertial frame. Therefore, for any arbitrary unconstrained body, motion can be divided into two parts:-

1)The translational motion of the COM ($dv_{com}/dt=a$,$a=F_{net}/m$)
2)Rotation of the body about the COM ($d\omega/dt=\alpha$,$\alpha=\tau_{com}/I_{com}$)

Tell me if you want any more details.

quantum mechanics - Can two different Schrödinger equations have the same wavefunction?

Is it possible for two Schrödinger equations describing different systems to have the same wavefunction? And if that is the case, why or why not?

Answer

The question is very broadly posed, so there's a bunch of different ways to interpret it, and each of them can give a different answer.

• 
  

  Can two different Hamiltonians share an eigenfunction?

  
  

  Yes. The answer by ComptonScattering gives an example in finite dimensions; if you want something closer to the usual three-dimensional quantum mechanics of a massive particle, there's even more examples (because the Hilbert space is much bigger).

  
  
  

  For something concrete, you can try the hydrogenic Hamiltonian $\hat H_0$ and $$\hat H = \hat H_0 + f(\hat r) \, \hat L{}^2,$$ where $f(r)\geq0$ is a non-negative function of the radius. Here the ground state will be the same for both (as will all the $\ell=0$ states) but $\ell\neq 0$ states will differ.

  
  


• 
  

  Can two different Hamiltonians share all their eigenfunctions?

  
  

  Yes. For a simple example, take any Hamiltonian $\hat H_1 = \hat H$ that is not the identity operator, and compare it with $\hat H_2 = \hat H{}^2$. Then every eigenfunction of $\hat H_1$ will be an eigenfunction of $\hat H_2$ (though the converse of that is not necessarily true), but the eigenvalues will in general differ.

  
  


• 
  

  Can two different Hamiltonians share all their eigenfunctions and the eigenvalues?

  
  

  No. This is because you can express the Hamiltonian as a function of the eigenvalues and eigenfunctions. In Dirac notation, that reads $$\hat H = \sum_n E_n |n⟩⟨n|.$$

  
  



• 
  

  Can two different Hamiltonians share at least one solution of the time-dependent Schrödinger equation, for all times?

  
  

  Yes. The eigenfunction examples of the first point above are a suitable example.

  
  


• 
  

  Can two different Hamiltonians share an arbitrary wavefunction in their solutions of the Schrödinger equation, at least for a single time $t_0$?

  
  

  Yes. Easy: set up some arbitrary wavefunction $\psi(t_0)$, and let it run under your two arbitrary Hamiltonians $\hat H_1$ and $\hat H_2$. They'll typically take $\psi(t_0)$ in arbitrary, different directions, but hey, the solutions matched at time $t_0$.

  
  


• 
  

  Can two different Hamiltonians share solutions of the time-dependent Schrödinger equation for all times and for arbitrary initial conditions?

  
  
  

  No. Take an arbitrary wavefunction $\psi_0$, and use that as the initial condition $\psi(t_0) = \psi_0$ for the Schrödinger equation under arbitrary $\hat H_1$ and $\hat H_2$. By assumption, both solutions are equal, and in particular their time derivatives at $t_0$ are also equal. (In fact, that is all you need: agreement of the wavefunction and its time derivative, for arbitrary initial conditions.) This then implies that $$\hat H_1 \psi_0 = i\hbar \partial_t \psi(t_0) = \hat H_2 \psi_0,$$ i.e. that $\hat H_1$ and $\hat H_2$ agree on your arbitrary initial condition $\psi_0$. This then means that they must agree as operators.

  
  

As a final note, please note the huge number of questions that folded into your original query, because of its imprecise phrasing, and use it to learn the importance of providing sharp, well-defined questions.

quantum mechanics - Intuitive meaning of Hilbert Space formalism

I am totally confused about the Hilbert Space formalism of Quantum Mechanics. Can somebody please elaborate on the following points:

1. 
   

   The observables are given by self-adjoint operators on the Hilbert Space.

   
   
   


2. 
   

   Gelfand-Naimark Theorem implies a duality between states and observables

   
   


3. 
   

   What's the significance of spectral decomposition theorem in this context?

   
   


4. 
   

   What do the Hilbert Space itself corresponds to and why are states given as functionals on the Hilbert space.

   
   

I need a real picture of this. I posted in Math.SE but got no answer. So I am posting it here.

How does the intensity not change when the width of the slit changes in single slit diffraction

The picture in the book by Resnick and Halliday shows as the width of the slit increases the width of the central maximum decreases and becomes flat as the slit width decreases,

my concern is when the slit width decreases and the curve becomes flat in the first picture,the area under the central maximum graph increases keeping the intensity same,from where the extra energy is coming to cover the whole screen by the central maximum,

in this case when the width decreases the amount of light entering the slit actually decreases,so when the area under the central maximum increases,the pattern on the screen should have absolutely low intensity light making the diffraction pattern invisible,but in the diagram it shows the intensity remains the same at 1.0,

the figure 10.17 in the other attachment shows the intensity is I(0) meaning similar graphs come for both intensity and relative intensity verses 0,

if the same graph in figure 10.17 is done for a = λ i.e if the slit width is very small,will the value of intensity I(0) decrease,will the relative intensity remain the same in all cases.

electromagnetic radiation - Why do prisms work (why is refraction frequency dependent)?

It is well known that a prism can "split light" by separating different frequencies of light:

Many sources state that the reason this happens is that the index of refraction is different for different frequencies. This is known as dispersion.

My question is about why dispersion exists. Is frequency dependence for refraction a property fundamental to all waves? Is the effect the result of some sort of non-linearity in response by the refracting material to electromagnetic fields? Are there (theoretically) any materials that have an essentially constant, non-unity index of refraction (at least for the visible spectrum)?

Answer

Lorentz came with a nice model for light matter interaction that describes dispersion quite effectively. If we assume that an electron oscillates around some equilibrium position and is driven by an external electric field $\mathbf{E}$ (i.e., light), its movement can be described by the equation $$m\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2}+m\gamma\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}+k\mathbf{x} = e\mathbf{E}.$$ The first and third terms on the LHS describe a classical harmonic oscillator, the second term adds damping, and the RHS gives the driving force.

If we assume that the incoming light is monochromatic, $\mathbf{E} = \mathbf{E}_0e^{-i\omega t}$ and we assume a similar response $\xi$, we get $$\xi = \frac{e}{m}\mathbf{E}_0\frac{e^{-i\omega t}}{\Omega^2-\omega^2-i\gamma\omega},$$ where $\Omega^2 = k/m$. Now we can play with this a bit, using the fact that for dielectric polarization we have $\mathbf{P} = \epsilon_0\chi\mathbf{E} = Ne\xi$ and for index of refraction we have $n^2 = 1+\chi$ to find out that $$n^2 = 1+\frac{Ne^2}{\epsilon_0 m}\frac{\Omega^2-\omega^2+i\gamma\omega}{(\Omega^2-\omega^2)^2+\gamma^2\omega^2}.$$ Clearly, the refractive index is frequency dependent. Moreover, this dependence comes from the friction in the electron movement; if we assumed that there is no damping of the electron movement, $\gamma = 0$, there would be no frequency dependence.

There is another possible approach to this, using impulse method, that assumes that the dielectric polarization is given by convolution $$\mathbf{P}(t) = \epsilon_0\int_{-\infty}^t\chi(t-t')\mathbf{E}(t')\mathrm{d}t'.$$ Using Fourier transform, we have $\mathbf{P}(\omega) = \epsilon_0\chi(\omega)\mathbf{E}(\omega)$. If the susceptibility $\chi$ is given by a Dirac-$\delta$-function, its Fourier transform is constant and does not depend on frequency. In reality, however, the medium has a finite response time and the susceptibility has a finite width. Therefore, its Fourier transform is not a constant but depends on frequency.

thermodynamics - Thermodynamically possible to hide a Dyson sphere?

You build a Dyson sphere around a star to capture all its energy. The outer surface of the Dyson sphere still radiates heat at much higher temperature than the cold space background, so you're easy to detect.

But you'd like to stay hidden. So you cool the outer surface of the Dyson sphere to near cold space background. Of course you still need to radiate your excess heat somewhere, so you plan to radiate it off in directed beams, away from the directions of the nearby solar systems, to stay hidden from your neighbors at least.

Questions:

1. Is such directed radiation of excess heat allowed by known laws of physics?


2. Would the energy of the star be sufficient for running the cooling system?

What Am I Riddle (Part 6)

Am I real? It's up to you..
I am real, if you flip me..
But, once you flip, I may bite you...

If I spit a 'POLO', I will become a little good..
If I split a 'POLO', I will become a little more good..
If I swallow a 'POLO', I will become fully good...

Can you guess who I am???

Hint:

A Polo is approximately 1.9 cm in diameter, 0.4 cm deep and has a 0.8 cm wide hole. The original Polo is white in colour with a hole in the middle, and the word 'POLO' embossed twice on one side around the ring, hence the popular slogan The Mint with the Hole.

---

Note: Please place your answers in spoiler tags. Also mention how you get to the answer.

Answer

GOD.
Whether [a] God is real is up to personal belief. If flipped, it becomes "dog", which is something that is definitely real and can bite you.
A Polo looks like the letter "O". I don't know what the lines about spitting or splitting a 'POLO' refer to, but if "god" "swallows" an O, it becomes "good", which is certainly fully good.

newtonian mechanics - With Newton's third law, why are things capable of moving?

I've got a rather humiliating question considering newton's third law

"If an object A exterts a force on object B, then object B exerts an equal but opposite force on object A" -> $F_1=-F_2$

Considering that, why is there motion at all? Should not all forces even themselves out, so nothing moves at all?

When I push a table using my finger, the table applies the same force onto my finger like my finger does on the table just with an opposing direction, nothing happens except that I feel the opposing force.

But why can I push a box on a table by applying force ($F=ma$) on one side, obviously outbalancing the force the box has on my finger and at the same time outbalancing the friction the box has on the table?

I obviously have the greater mass and acceleration as for example the matchbox on the table and thusly I can move it, but shouldn't the third law prevent that from even happening? Shouldn't the matchbox just accommodate to said force and applying same force to me in opposing direction?

I've found a lot of answers considering that question but none was satisfying to an extend that I had an epiphany solving my fundamental problem I've got understanding it.

Answer

I think it's a great question, and enjoyed it very much when I grappled with it myself.

Here's a picture of some of the forces in this scenario.$^\dagger$ The ones that are the same colour as each other are pairs of equal magnitude, opposite direction forces from Newton's third law. (W and R are of equal magnitude in opposite directions, but they're acting on the same object - that's Newton's first law in action.)

While $F_{matchbox}$ does press back on my finger with an equal magnitude to $F_{finger}$, it's no match for $F_{muscles}$ (even though I've not been to the gym in years).

At the matchbox, the forward force from my finger overcomes the friction force from the table. Each object has an imbalance of forces giving rise to acceleration leftwards.

The point of the diagram is to make clear that the third law makes matched pairs of forces that act on different objects. Equilibrium from Newton's first or second law is about the resultant force at a single object.

$\dagger$ (Sorry that the finger doesn't actually touch the matchbox in the diagram. If it had, I wouldn't have had space for the important safety notice on the matches. I wouldn't want any children to be harmed because of a misplaced force arrow. Come to think of it, the dagger on this footnote looks a bit sharp.)

Why does water reflect light?

Why does water reflect light? What is actually happening when light is reflected by water? We know why metals reflect light; water, however, is not metal, but it still reflects light and we can see our image reflected on calm water.

probability - The Ravages of Purple Spots Disease

While browsing the Internet for medical lore, you happen on a medical encyclopedia article about purple spots disease.

Instances of purple spots disease have been documented in 150 countries. The disease affects one in every 100,000 individuals and there is no known cure. Individuals who have the disease remain completely asymptomatic until one day they very suddenly break out in purple spots that never go away.

Concerned that you may have purple spots disease and not know it, you scour the Internet and locate a reputable medical testing company that provides a screening test for the disease. The test is state of the art. It guarantees 99.5% accuracy, meaning that whether you have the disease or not, the result returned by the test will be correct 99.5% of the time. The test costs $300.00 USD to administer.

Figuring that your peace of mind is worth $300.00, you undergo the test for the disease.

Three weeks later, the results come back. The test is positive! You're advised to come in for a retest to be absolutely sure.

Based on the information above, should you be deeply worried? Why or why not?

Puzzlers are politely encouraged to place answers in spoiler blocks to avoid inadvertently spoiling the fun for other readers. :)

Answer

No you shouldn't be too worried. The probability you have the disease is 0.199%, or in other words about a 1/500 chance.

First, let us assume there are 10,000,000 people in the society (the number you assume here is irrelevant, and you could even just let x be the number if you were mathematiaclly inclined). Now given that 1/100,000 people have the disease, therefore, 100 people will have it in this society. Using this information a 2 by 2 table can be constructed as seen below:

             Has Disease        No Disease         Total
+ve test        99.5              49999.5          50099       

-ve test         0.5            9949900.5        9949901

Total            100              9999900       10000000

From the above table, given a +ve test result, the probability of having the disease can be evaluated by computing by 99.5/50099 = 1.99 x 10^-3, or about a 1/500 chance.

quantum field theory - Matching Dirac/Majorana/Weyl Spinor Degrees of Freedom in Minkowski signature

Question: How do we match the real degrees of freedom (DOF) of Dirac/Majorana/Weyl Spinor in terms of their quantum numbers (spin, momentum, etc) in any dimensions [1+1 to 9+1] in Minkowski signature?

(p.s. This question is partially inspired by this one, but I am asking something more generally in any dimensions from 1+1 to 9+1 in Minkowski signature)

My question concerns that how to match each component of spinors in physical degrees of freedom (real degrees of freedom of Dirac/Majorana/Weyl Spinor) reflecting into their quantum numbers.

For example in 3+1d, in Weyl basis,

For 3+1d Dirac spinor,

we have 4 component complex spinor thus we have 8 real degrees of freedom. For the massless particles, in the boost limit along the $z$ direction, $p_z \to E$ or $p_z \to -E$, we can match

$$8=2 \times 2 \times 2,$$ as $$8 \text{ real DOF}= 2 \text{ (spin up/down)} \times 2 \text{ (momentum up/down)} \times 2 \text{ (particle/ anti-particle)}$$ More precisely, the 8 real DOF becomes the following:

$\psi_{par}(x)=u(p) e^{-ip \cdot x}=u(p) e^{-i (Et -\vec{p} \cdot \vec{x})}= \begin{pmatrix} \sqrt{p \cdot \sigma} \zeta^s \\ \sqrt{p \cdot \bar{\sigma}} \zeta^s \end{pmatrix} e^{-i (Et -\vec{p} \cdot \vec{x})}$, $s=1,2$

$\psi_{anti}(x)=v(p) e^{ip \cdot x}=v(p) e^{i (Et -\vec{p} \cdot \vec{x})} = \begin{pmatrix} \sqrt{p \cdot \sigma} \eta^s \\ -\sqrt{p \cdot \bar{\sigma}} \eta^s \end{pmatrix} e^{i (Et -\vec{p} \cdot \vec{x})}$, $s=1,2$

particle, $p_3=p_z=E$, spin up,

$\psi_{par}(x)= \sqrt{2E} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} e^{-ip \cdot x}$

particle, $p_3=p_z=E$, spin down,

$\psi_{par}(x)= \sqrt{2E} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} e^{-ip \cdot x}$

particle, $p_3=p_z=-E$, spin up,

$\psi_{par}(x)= \sqrt{2E} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} e^{-ip \cdot x}$

particle, $p_3=p_z=-E$, spin down,

$\psi_{par}(x)= \sqrt{2E} \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} e^{-ip \cdot x}$

anti-particle, $p_3=p_z=E$, spin up,

$\psi_{anti}(x)= -\sqrt{2E} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} e^{ip \cdot x}$

anti-particle, $p_3=p_z=E$, spin down,

$\psi_{anti}(x)= \sqrt{2E} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} e^{ip \cdot x}$

anti-particle, $p_3=p_z=-E$, spin up,

$\psi_{anti}(x)= \sqrt{2E} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} e^{ip \cdot x}$

anti-particle, $p_3=p_z=-E$, spin down,

$\psi_{anti}(x)= -\sqrt{2E} \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} e^{ip \cdot x}$

---

For 3+1d Weyl spinor,

we separate Dirac into left-hand and right-hand Weyl spinors, thus

We have the particle is the same as the anti-particle, which is the real representation. Thus we have

for left-handed ($P_L$)

$$4=(2) \times 2,$$ as $$4 \text{ real DOF}= (2 \text{( $-1/2$ helicity : spin up/down lock momentum up/down)} \times 2 \text{ (particle / anti-particle)}$$

similarly for right-handed ($P_R$)

$$4=(2) \times 2,$$ as $$4 \text{ real DOF}= 2 \text{($+1/2$ helicity : spin up/down lock momentum up/down)} \times 2 \text{ (particle / anti-particle)}$$

---

For 3+1d Majorana spinor,

We have the particle is the same as the anti-particle, which is the real representation. Thus we have

$$4=2 \times 2 \times 1,$$ as $$4 \text{ real DOF}= 2 \text{ (spin up/down)} \times 2 \text{ (momentum up/down)} \times 1 \text{ (particle = anti-particle)}$$

How about other dimensions?

There will be certainly an even-odd dimensions subtlty. Also the the helicity like degrees of freedom (the spin projected along the boost direction) may NOT be enough for other dimensions higher than 3+1d.

rebus - Good rebuses come in nines

A teacher might do this if a student is making rebuses during class.

Answer

CASTIGATE

Explanation:

cast + gate

computer puzzle - Debugging with printf()

Bob, a C programmer and a believer of printf-debugging, gets hired by a failing software company. The company is not doing well because they adopt an archaic source code management system which does not allow any modification to any lines of code once they are checked in.

As you can imagine, poor Bob is too excited at resolving a bug with his printf debugging technique and accidentally checked in that line of C code:

printf("OK, bug gone - this should make my nasty boss happy");

Now, he desperately wants to silence it. But the stupid system won't allow him to edit that line. He can only add new lines of code around it to cancel its effect. There are many ways to achieve that, but can you find the one which adds the fewest lines, with the fewest characters?

Answer

Well, he can do it by adding

if(0)

on the line before. So that's

one extra line, five extra characters.

Or, slightly better, exploit the fact that

printf returns a value

and make that first line

0&&

for

one line, three characters.

quantum mechanics - Expectation values of the position operator is equal to zero in case of even potentials?

Assuming the eigenvalue of position operator $\hat x$ equal to $k$, can I not write:

$$\begin{align} \langle\psi_n|x|\psi_m\rangle &= \langle x\psi_n|\psi_m\rangle \\ &=\langle k\psi_n|\psi_m\rangle \\ &=k\langle\psi_n|\psi_m\rangle \\ &=k\delta_{nm} \end{align}$$

But I know that $\langle x \rangle =0$ in case of even potentials (I don't know how that happens) and what I have written above is wrong, at least in case of even potentials.

Taking the example of infinite 1D square well, the states are : $$\psi_{n} \left(x\right)=A\sin\left(\frac{nx\pi}{L}\right)dx$$ then, $$\langle\psi_n|x|\psi_m\rangle =A\int_{-\infty}^{\infty}\sin\left(\frac{mx\pi}{L}\right)x\sin\left(\frac{nx\pi}{L}\right)dx$$ If m=n=1, $$ =A\int_{-\infty}^{\infty}x\sin^{2}\left(\frac{x\pi}{L}\right)dx $$if we apply an even potential then the equation gets reduced to$$ =\frac{1}{L}\int_{-L}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=0 $$while in case of a potential(neither even nor odd), the equation leads to$$ =\frac{2}{L}\int_{0}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=L/2 ~? $$ Here $n=m=1$ but $ =0 $for even potentials, which is confusing me! It should be$k$ right?

homework and exercises - How are the Lagrange points determined?

According to Hyper Physics, there are 5 equilibrium, or Lagrange points of the Earth-Moon system and only 2 of them are said to represent stable equilibrium points.

This made me think if there is an equation that describes this system, and from which Physics Laws was it derived?

quantum field theory - Perturbation series in QED

The coupling constant in the QED lagrangian is clearly the electric charge $e$. However, one often hears the statement that the expansion parameter in QED is the fine structure constant $\alpha = e^2/4\pi$, not $e$.

Unfortunately, I've never seen a formal proof that the sum of all contributions to S-matrix which are proportional to the given odd power of the electric charge $e$ must vanish.

My question is whether this is really true?

E.g. why the combined third order contributions to the proces of the scattering of an electron and a photon into an electron and two photons vanish?

Maybe what is meant by the statement that the expansion parameter in QED is $\alpha$ is simply that the perturbation series has the form $\sum_{L=0}^\infty e^{E-2+2L} a_L = e^{E-2}\sum_{L=0}^\infty \alpha^L a_L$ ($L$ - number of loops, $E$ - number of external lines), which is not difficult to prove.

Answer

Your latter option is what is meant - the perturbation series is an expansion in loop orders, and the power of $\alpha$ is what counts the loop order.

quantum field theory - when is the stationary phase approximation exact?

I am thinking about some topological field theories, and I am wondering when one can say that the stationary phase approximation (ie. a sum of the first-order variations about each vacuum) is exact.

I am looking perhaps for conditions on how the space of vacua is embedded into the space of all field configurations. I suspect that when the action is a Morse function (and I suppose the space of field configurations is finite dimensional) that the exactness of the stationary phase approximation implies some very strict topological constraints on the configuration space... torsion-free and so on.

Anyone have a good reference or some wisdom?

Also I'd like to dedicate this question to the memory of theoreticalphysics.stackexchange

Answer

In general, the situation where the stationary phase approximation is exact is described by the Duistermaat Heckman theorem, which states (not in its most general form) that if $M$ is a compact symplectic manifold and $H$ is a Hamiltonian generationg a torus action on $M$, then for the "partition" function

$Z = \int_M e^{it H} d_L(M)$

the stationary phase approximation is exact ($d_L(M)$ is the Liouville measure) and the integral can be computed by summing the contributions from the extrema of $H$ (fixed points of the torus action).

An equivalent characterization of the hamiltonian $H$ is that it is a perfect Morse function.

Two very known examples are the Gaussian integral and the spin partition function in a magnetic field where the classical and the quantum partition functions are exactly the same.

This theorem was applied and generalized to more complicated situations (e.g., when the fixed points are not isolated), to path integrals of certain theories (coherent state path integrals), loop spaces and to topological field theories.

Further reserach of the Duistermaat-Heckman theorem and its generalizations led to the discovery of a general phenomenon leading to this type of exactness, now called "equivariant localization".

Please see the following review article"Equivariant Localization of Path Integrals" by Richard J. Szabo, where numerous applications are described.

riddle - I'm purple and small, what am I?

I'm purple and small,
not hard to make at all.
I take you to new places,
Where you will see some new faces.

I'm part of the world, yet take you away,
When you step through me, you see mostly grey.
You must walk through me to complete the game,
And when you see me, you'll know my name.

Sorry for the lame poetry, but

HINT (big hint, so use only as a last resort):

I am related to https://www.youtube.com/watch?v=gMoS2FvkyLE

Answer

Sounds like you are a...

Minecraft Nether Portal

I'm purple and small,
not hard to make at all.

Made up of only an obsidian frame, then lit on fire.

I take you to new places,
Where you will see some new faces.

A portal ... well, takes you somewhere else. This one in particular opens up a new dimension.

I'm part of the world, yet take you away,
When you step through me, you see mostly grey.

Stepping through, your view is obscured by a purplish hue, and you will be whisked away from the Overworld.

You must walk through me to complete the game,
And when you see me, you'll know my name.

A portal is pretty clearly identified. You must complete tasks and gather materials in the Nether to reach The End.

thermodynamics - In counting degrees of freedom of a linear molecule, why is rotation about the axis not counted?

I was reading about the equipartition theorem and I got the following quotations from my books:

A diatomic molecule like oxygen can rotate about two different axes. But rotation about the axis down the length of the molecule doesn't count. - Daniel V. Schröder's Thermal Physics.

A diatomic molecule can rotate like a top only about axes perpendicular to the line connecting the atoms but not about that line itself. - Resnick, Halliday, Walker s' Fundamentals of Physics.

Why is it so? Doesn't the rotation take place that way?

Answer

The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is:

$$B = \frac{\hbar^2}{2I}$$

Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This means the moment of inertia around an axis along the bond is going to be about $10^{10}$ smaller than the moment of inertia around an axis normal to the bond. Therefore the energy level spacings will be around $10^{10}$ times bigger along the bond than normal to it.

In principle we can still excite rotations about the axis along the bond, but you'd need huge energies to do it.

reverse puzzling - What are they trying to solve?

I was on the the train once, bored as hell (because trains in Romania are really slow) and at one point I hear 2 guys behind me, probably as bored as I was, talking about puzzles.
And I thought:

"Oh! I like puzzles. I should join them. I'm sure they won't mind. But wait...what if I have no idea what's it about and I look like a fool? I better eavesdrop a bit and but in only if I know something".
But a window was open and the noise from outside covered part of their conversation.
All I could understand was this:

Guy1: Look at this (mmmmm). It has numbers and letters.
Guy2: it has to be a crossword.
Guy1: That's not possible because (mmmm). But they can be the moves for a Rubik cube.
Guy2: Pff...you're an (mmmm). There are only 6 letters in the Rubik cube moves. Maybe it's a matrix of something. Letters on (mmmm), numbers on (mmmm).
Guy1: Or a cipher? Man, I hate ciphers. Or a pattern?
Guy2: There are (mmmm), so this is probably not a rebus.

I this point I realized I have no idea what's this about and decided to shut up. Also, I'm a shy guy and I didn't ask them what's this about.
But I am still puzzled (pun intended). What was that about?

[Edit] In the light of the 2 answers I feel obligated to mention that the fact that letters and numbers are involved is not important.
And I remembered one of the guys saying at one point that they were trying to solve a puzzle from puzzling.stackexchange.com.
[/Edit]

Note: (mmmm) marks the parts I couldn't hear, but they are not really important. You can solve this without knowing what I missed. They are there only for dramatic effect.
Note 2: (not part of the puzzle, just "fun" fact): The trains in Romania are really, really slow.

Answer

Just from the OP's comment this is maybe too literal but

I suppose a possibility is that two guys in a train could be reading this puzzle now and making the same statements as in the puzzle. This may be too self-referential and hope ok to post.

I guess this could only happen after the puzzle has been posted unless the OP was present when it was being created, or that Romania time is ahead of mine, or being on a train reminds me of special relativity and the fact they are going slow may be be an observer's effect of near light speed travel - I've stretched too far now. I was set up to answer this by Marius :)

cosmology - On Flatness problem, Inflation etc

I have a couple of naive questions from the topic of the title.

1. We know $$\begin{eqnarray} \Omega-1=\frac{k}{a^2H^2}-\frac{\Lambda}{3H^2} \end{eqnarray}$$ Now I read that from the standard big bang (SBB) model $\frac{1}{aH}$ increases with time and in recent time $\Omega\approx 1$. So, $\Omega$ has to be fine tuned to stay close to 1. I guess its because as we go back in time $\frac{1}{aH}$ decreases in SBB, and for fixed $\Lambda$ this drives $\Omega$ away from 1. I guess that's o.k as $aH$ decreases faster than $H$ alone in the second term above. But why do we want $\Omega\approx 1$ in earlier times?

My confusion deepens after reading how inflation solves this problem. In "Inflation", it is argued that $\frac{1}{aH}$ decreases with time. So I thought to myself As before going back in time resulted in decrease in Hubble length (comoving), so in the above equation the 1st term was large that is effectively with non-zero $k$. But now as going back in time increases $\frac{1}{aH}$, the first term decreases and it acts as if $k$ is nearly zero which is the criterion for flat universe. So my confusion boils down to the question on Why do we want $\Omega\approx 1$ in earlier times? Also how does Inflationary era solve the whole flatness problem? What about when inflation ends and SBB begins? Moreover, the above equation is valid for every case (inflation and SBB). So, even if inflation kills the problem, SBB should revive it yielding non unity of $\Omega$ now!

1. Also I came to know that as long as $\dot{\phi}^2
   

Related to that, I learned in Hybrid Inflation, since $\phi$ is driven to zero for $\psi>1$ (Why??), the potential in $\psi$ direction is flat and satisfies slow-roll conditions, so that $\psi$ is considered inflation. How can we say potential in $\psi$ direction is flat?

Answer

the normal cosmological evolution dominated by matter or radiation - which was the case for billions of years after the Big Bang - makes $|\Omega-1|$ increase with time because of changes captured by the Friedmann equations. However, WMAP and others show that $|\Omega-1|$ is not greater than 0.01 today.

It follows that $|\Omega-1|$ had to be even much smaller a minute - and a fraction of second - after the Big Bang - something like $10^{-{\rm dozens}}$. This is called the flatness problem because there's no reason for a generic Universe to have such a small number of a quantity such as $|\Omega-1|$ which can a priori be anything.

The cosmic inflation solves the flatness problem because it reverses the evolution $|\Omega-1|$: as time goes to the future, $|\Omega-1|$ is (more precisely, was) decreasing in this case. It's because the evolution is driven by the last, temporary cosmological constant term in your equation. So because of inflation, one may start with a generic value of $|\Omega-1|$ before the inflation, and one ends up with $|\Omega-1|$ close to zero at the end, anyway.

I hope that I don't have to explain that if a quantity increases/decreases with time, it will decrease/increase if you read the time in the opposite direction. Nevertheless, it seems that 1/3 of your question is focusing on this triviality.

$\dot\phi^2 < V (\phi)$ is just a condition saying that $V(\phi)$, the cosmological constant term, is the dominant term in the Friedmann equation during inflation. If that's so, you may neglect the kinetic term $\dot \phi^2$. I am not sure whether the coefficient in the equation is one - I doubt so. You should only view the inequality as an estimate. If the kinetic term is much smaller, it's guaranteed that you can neglect it and the exponential inflationary expansion dictated by $V(\phi)$, the temporary cosmological constant, follows.

Concerning hybrid inflation, obviously, you can't determine that the potential for $\psi$ is a slow-rolling one just from the assumptions you have shared with us: it's just an assumption that the potential has the property. In particular, the potential is supposed to resemble the potentials in the theory of second order phase transitions - which occurs very generically in string theory models etc. Imagine that $\phi$ is the temperature, more precisely $T-T_c$, and $\psi$ is the magnetic field. For a positive $\phi$, the minimum is at $\psi=0$, so only $\phi$ is changing. However, once $\psi$ reaches zero, you get a phase transition and the right magnetic field is at a nonzero value, so one rolls the channel at $\psi=\pm M$. Near the $\phi=0$ point, the potential is inevitably changing slowly because it's a stationary point. See e.g.

http://nedwww.ipac.caltech.edu/level5/Liddle/Liddle5_7_2.html

The purpose of the other fields in hybrid inflation is to stop the inflation.

quantum mechanics - Noether theorem, gauge symmetry and conservation of charge

I'm trying to understand Noether's theorem, and it's application to gauge symmetry. Below what I've done so far.

First, the global gauge symmetry. I'm starting with the Lagragian $$L_{1}=\partial^{\mu}\Psi\partial_{\mu}\Psi^{\ast}-m^{2}\left|\Psi\right|^{2}$$ with classical complex fields. This Lagragian is invariant with respect to the global gauge symmetry $\Psi\rightarrow\tilde{\Psi}=e^{\mathbf{i}\theta}\Psi$, ... such that I end up with $$\delta S=\int dv\left[\dfrac{\delta L_{1}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{1}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\mathbf{i}\left(\Psi\partial^{\mu}\Psi^{\ast}-\Psi^{\ast}\partial^{\mu}\Psi\right)\partial_{\mu}\delta\theta\right]=\int dv\left[\partial_{\mu}j^{\mu}\right]\delta\theta$$ provided the equations of motion ($\delta L / \delta \Psi = 0$, ...) are valid. All along I'm using that $$\dfrac{\delta L}{\delta\phi}=\dfrac{\partial L}{\partial\phi}-\partial_{\mu}\dfrac{\partial L}{\partial\left[\partial_{\mu}\phi\right]}$$ and that $\int dv=\int d^{3}xdt$ for short. The conserved current is of course $$j_{1}^{\mu}=\mathbf{i}\left(\Psi^{\ast}\partial^{\mu}\Psi-\Psi\partial^{\mu}\Psi^{\ast}\right)$$ since $\delta S / \delta \theta =0 \Rightarrow\partial_{\mu}j_{1}^{\mu}=0$.

Here is my first question: Is this really the demonstration for conservation of charge ? Up to now, it seems to me that I only demonstrated that the particle number is conserved, there is no charge for the moment...

Then, I switch to the local gauge symmetry. I'm starting with the following Lagrangian $$L_{2}=\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi\left(\partial_{\mu}-\mathbf{i}qA_{\mu}\right)\Psi^{\ast} -m^{2}\left|\Psi\right|^{2} -\dfrac{F_{\mu\nu}F^{\mu\nu}}{4}$$ with $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$. This Lagrangian is invariant with respect to the local gauge transformation $$L_{2}\left[\tilde{\Psi}=e^{\mathbf{i}q\varphi\left(x\right)}\Psi\left(x\right),\tilde{\Psi}^{\ast}=e^{-\mathbf{i}q\varphi\left(x\right)}\Psi^{\ast},\tilde{A}_{\mu}=A_{\mu}-\partial_{\mu}\varphi\right]=L_{2}\left[\Psi,\Psi^{\ast},A_{\mu}\right]$$

Then I have $$\delta S=\int dv\left[\dfrac{\delta L_{2}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{2}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\dfrac{\delta L_{2}}{\delta A_{\mu}}\delta A_{\mu}\right]$$ with $\delta\Psi=\mathbf{i}q\Psi\delta\varphi$, $\delta A_{\mu}=-\partial_{\mu}\delta\varphi$, ... such that I end up with $$\dfrac{\delta S}{\delta\varphi}=\int dv\left[\mathbf{i}q\Psi\dfrac{\delta L_{2}}{\delta\Psi}+c.c.+\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]\right]$$ with $j_{2}^{\mu}=\partial L_{2}/\partial A_{\mu}$ and $F^{\nu\mu}=\partial L_{2}/\partial\left[\partial_{\nu}A_{\mu}\right]$

Then, by application of the equations of motion, I have $$\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]=0\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$ since $\partial_{\mu}\partial_{\nu}F^{\nu\mu}=0$ by construction. Of course the new current is $$j_{2}^{\mu}=\mathbf{i}q\left(\Psi^{\ast}\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi-\Psi\left(\partial^{\mu}-\mathbf{i}qA^{\mu}\right)\Psi^{\ast}\right)$$ and is explicitly dependent on the charge. So it seems to me this one is a better candidate for the conservation of charge.

NB: As remarked in http://arxiv.org/abs/hep-th/0009058, Eq.(27) one can also suppose the Maxwell's equations to be valid ($j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu} = 0$, since they are also part of the equation of motion after all, I'll come later to this point, which sounds weird to me), and we end up with the same current, once again conserved.

Nevertheless, I still have some troubles. Indeed, if I abruptly calculate the equations of motions from the Lagrangian, I end up with (for the $A_{\mu}$ equation of motion) $$j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$ by definition of the $F^{\mu \nu}$ tensor.

So, my other questions: Is there a better way to show the conservation of EM charge ? Is there something wrong with what I did so far ? Why the Noether theorem does not seem to give me something which are not in the equations of motions ? said differently: Why should I use the Noether machinery for something which is intrinsically implemented in the Lagrangian, and thus in the equations of motion for the independent fields ? (Is it because my Lagrangian is too simple ? Is it due to the multiple boundary terms I cancel ?)

Thanks in advance.

PS: I've the feeling that part of the answer would be in the difference between what high-energy physicists call "on-shell" and "off-shell" structure. So far, I never understood the difference. That's should be my last question today :-)

Answer

Comments to the question (v1):

1. 
   

   Last thing first. On-shell means (in this context) that equations of motion (eom) are satisfied. Equations of motion means Euler-Lagrange equations. Off-shell means strictly speaking not on-shell, but in practice it is always used in the sense not necessarily on-shell. [Let us stress that every infinitesimal transformation is an on-shell symmetry of an action, so an on-shell symmetry is a vacuous notion. Therefore in physics, when we claim that an action has a symmetry, it is always implicitly understood that the symmetry is an off-shell symmetry.]

   
   


2. 
   

   OP wrote: Here is my first question: Is this really the demonstration for conservation of (electric) charge? For that particular action: Yes. More generally for QED: No, because the $4$-gauge-potential $A_{\mu}$, the Maxwell term $F_{\mu\nu}F^{\mu\nu}$, and the minimal coupling are missing in OP's action. It is in principle not enough to only look at the matter sector. On the other hand, global gauge symmetry for the full action $S[A,\Psi]$ leads to electric charge conservation, cf. Noether's first Theorem. [Two comments to drive home the point that it is necessary to also consider the gauge sector: (i) If we were doing scalar QED (rather than ordinary QED), it is known that the Noether current $j^{\mu}$ actually depends on the $4$-gauge-potential $A_{\mu}$, so the gauge sector is important, cf. this Phys.SE post. (ii) Another issue is that if we follow OP's method and are supposed to treat the $4$-gauge potential $A_{\mu}$ as a classical background (which OP puts to zero), then presumably we should also assume Maxwell's equations $d_{\mu}F^{\mu\nu}=-j^{\nu}$. Maxwell's equations imply by themselves the continuity equation $d_{\mu}J^{\mu}=0$ even before we apply Noether's Theorems.]

   
   


3. 
   

   There is no conserved quantity associated with local gauge symmetry per se, cf. Noether's second Theorem. (Its off-shell Noether identity is a triviality. See also this Phys.SE question.)

   
   



4. 
   

   Perhaps a helpful comparison. It is possible to consider an EM model of the form $$S[A]~=~\int\! d^4x~ \left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+J^{\mu}A_{\mu}\right),$$ where $J^{\mu}$ are treated as passive non-dynamical classical background matter sources. In other words, only the gauge fields $A_{\mu}$ are dynamical variables in this model. Before we even get started, we have to ensure local (off-shell) gauge symmetry of the action $S[A]$ up to boundary terms. This implies that the classical background sources $J^{\mu}$ must satisfy the continuity equation $d_{\mu}J^{\mu}=0$ off-shell. Thus a conservation law is forced upon us even before we apply Noether's Theorems. Note that global gauge symmetry is an empty statement in this model.

   
   

homework and exercises - Why doesn't a spinning object in the air fall?

Let's say I have a ball attached to a string and I'm spinning it above my head. If it's going fast enough, it doesn't fall. I know there's centripetal acceleration that's causing the ball to stay in a circle but this doesn't have to do with the force of gravity from what I understand. Shouldn't the object still be falling due to the force of gravity?

Answer

We have the ball orbiting at a distance $R$ from the centre of rotation and the string inclined at angle $\theta$ with respect to the horizontal.

Two main forces act on the ball: gravity $mg$ ($m$ is the mass of the ball, $g$ the Earth's gravitational acceleration) and $F_c$, the centripetal force needed to keep the ball spinning at constant rate. $F_c$ is given by:

$$F_c=\frac{mv^2}{R},$$

where $v$ is the orbital velocity, i.e. the speed of the ball on its circular trajectory.

Trigonometry also tells us that if $T$ is the tension in the string, then:

$$T\cos\theta=F_c.$$

Similarly, as the ball is not moving in the vertical direction, thus $F_{up}$:

$$T\sin\theta=F_{up}=mg.$$

From this relation we can infer:

$$T=\frac{mg}{\sin\theta}.$$

And so:

$$\frac{mg}{\tan\theta}=F_c=\frac{mv^2}{R}.$$

Or:

$$\tan\theta=\frac{gR}{v^2}.$$

From this follows that for small $\tan\theta$ and thus small $\theta$ we need large $v$. But at lower $v$, $\theta$ increases. Also note that $\theta$ is invariant to mass $m$.

special relativity - How time dilation happens by velocity?

i got many explanation and proving that relative velocity causes time dilation. Einstein's moving light experiment proves it. but that is a clock works with light detector. as detecting the light gets latter the clock works slowly. but in reality how a real watch can be slowed down? And won't the man feel that the time is moving slowly as he knows the general speed of the clock? Sources says that if a man goes to space with very high speed if he returns after few years as his clock moved slowly earth would be moved for many years forward. But my doubt is won't he feels the fact that his clock was slow and he spend many years in space? If he won't feel it how does it happens? and how is he not getting older?

logical deduction - Write digits 1 to 9 in the table, each pointing to next 3x3 grid

Write 1 to 9 digits in an empty Sudoku table with these rules (similar to 9-Queen problem):

• There are 9 Boxes indexed from 1 to 9 in the image below. and each box has a 3x3 grid child.


• Every two digits should NOT be in the same row, column or diagonal.



• Pointing rule: You have to place next number in an specific box. (e.g. if you start to place 1 in top-left box and middle-center child grid, you have to continue with 2 in middle-center box)


• You are free to begin from any box.


• Close the loop: The last number (9) should point to the box that you have placed the first number (1) in it.


• At the end there will be a Sudoku table with only 9 digits in it each in a separate 3x3 box. (other 72 cells will remain empty)

Write down the 9 digits with the rules, or if there is no such a solution, prove it!

See the following example that have some mistakes:

1. 8 and 9 are in the same row.


2. 8 and 5 are in a diagonal.


3. 5 and 2 are in the same row.

PS. I don't know whether it was a stupid question or not.

PS. PS. an grammatically edit from would be appreciated!

Answer

It is not possible.

Any solution to this puzzle must also be a solution to the 9x9 Queens puzzle. Luckily, that is a well-known puzzle. It has 352 solutions, but due to symmetry, those 352 solutions can be reduced to 46 solutions. After that, it is just a matter of checking against those 46 solutions.

I found a page showing the solutions at: http://stamm-wilbrandt.de/en/xsl-list/n-queens/n-queens.xsl.xml

I then looked at the first answer in each of the 46 columns.

Items that quickly ruled a solution out:

1. Queen in a corner. Any queen in a corner would mean that the box referred to itself.


2. Queen in the center of a side. Again, that box referred to itself.


3. Queen in center box. Same reason.


4. 2 Queens in same 3x3 box / empty 3x3 box.

This left me with very few actual solutions to check. From that page, #'s 19, 20, 31, 39, 42.

To check, I simply started in box 1 and followed the pattern until I returned to a box that I'd already seen. None of those took all 9 boxes.

electrostatics - How does one prove that Energy = Voltage x Charge?

We know $$E = q V$$ where $E$ is the energy (in Joules), $V$ is the potential difference (in Volts), and $q$ is the charge. Why is this equation true and how we prove it?

Answer

There are various ways to decide which of the assumptions are primary and which of them are their consequences but $E=VQ$ may be most naturally interpreted as the definition of the potential.

The potential energy is a form of energy and the potential (and therefore voltage, when differences are taken) is defined as the potential energy (or potential energy difference) per unit charge, $V = E/Q$. That's equivalent to your equation. The potential energy is proportional to the charge essentially because of the linearity of Maxwell's equations (the superposition principle). Once we know about the proportionality, we must just give a name to the proportionality factor between $E$ and $Q$ and we simply call it potential (or voltage).

quantum mechanics - What are the frameworks of physics?

Are there physical theories in use, which don't fit into the frameworks of either Thermodynamics, Classical Mechanics (including General Relativity and the notion of classical fields) or Quantum Mechanics (including Quantum Field Theory and friends)?

Answer

The proposed partition of physics into Thermodynamics, Classical Mechanics, and Quantum Mechanics is quite arbitrary. To take just one conspicuous example, statistical mechanics does not fit, as it is the discipline that mediates between these three areas of physics.

The Physics and Astronomy Classification Scheme (PACS) http://www.aip.org/pacs/pacs2010/individuals/pacs2010_regular_edition/index.html , ''an internationally adopted, hierarchical subject classification scheme, designed by the American Institute of Physics (AIP)'', partitions physics instead into

• The physics of elementary particles and fields


• nuclear physics


• atomic and molecular physics


• electromagnetism, optics, acoustics, heat transfer, classical mechanics, and fluid dynamics


• physics of gases, plasmas, and electric discharges


• condensed matter: structural, mechanical and thermal properties


• condensed matter: electronic structure, electrical, magnetical, and optical properties


• interdisciplinary physics and related areas of science and technology



• geophysics, astronomy, and astrophysics.
  

  It would be quite meaningless to put each of these general containers under the hood of either Thermodynamics, Classical Mechanics, or Quantum Mechanics. In many cases, there is an interplay between thermodynamical, classical, and/or quantum aspects that bear on a given physical problem.

  
  

  But let me respond to the challenge by proposing a systematic view of physics not by its phenomena but by classifying it in terms of 7 orthogonal criteria.

  
  

  The first criterion is methodological, and distinguishes between

  
  


• applied physics (AP), didactical physics (DP), experimental physics (EP), theoretical physics (TP), and mathematical physics (MP).
  

  The other six criteria are defined in terms of the six limits that play an important role in physics:

  
  


• the classical limit ($\hbar\to 0$) distinguishes between classical physics (Cl), in which $\hbar$ is negligible, and quantum physics (Qu) where it is not.


• the nonrelativistic limit ($c\to \infty$) distinguishes between nonrelativistic physics (Nr), in which $c^{-1}$ is negligible, and relativistic physics (Re) where it is not.



• the thermodynamic limit ($N\to\infty$) distinguishes between macroscopic physics (Ma), in which microscopic details are negligible, and microscopic physics (Mi) where they are not.


• the eternal limit ($t\to\infty$) distinguishes between stationary physics (St), in which time is negligible, and nonequilibrium physics (Ne) where it is not.


• the cold limit ($T\to 0$) distinguishes between conservative physics (Co), in which entropy is negligible, and thermal physics (Th) where it is not.


• the flat limit ($G\to 0$) distinguishes between physics in flat space-time (Fl), in which curvature is negligible, and general relativistic physics (Gr) where it is not.
  

  A particular subfield is characterized by a signature consisting of choices of labels (or double arrows between labels) in some categories.

  
  

  A few examples:

  
  


• Thermodynamics: Ma ,Th


• Equilibrium thermodynamics: Ma, Th, St


• Classical Mechanics: Cl, Co



• Classical field theory: Cl, Co, Ma


• General relativity: Cl, Re, Ma, Gr


• Quantum mechanics: Qu, Nr


• Relativistic quantum field theory: TP, Qu, Re, Mi


• Statistical mechanics: TP, Mi$<->$Ma, Th


• Precision tests of the standard model: TP$<->$EP, Qu, Re, Mi, St, Co


• The empty signature is simply the field of physics itself.
  

  In each category, one can choose no label, a single label, or an arrow between two labels, giving $1+5+5*4/2=16$ cases for the first category, and $1+2+1=4$ cases in the six other categories. Thus the classification splits physics hierarchically into $16*4^6=65536$ potential subfields with different signatures, of which of course only the most important ones carry conventional names.

  
  

  Let me give what I think is a particularly useful subhierarchy of the complete hierarchy. This subhierarchy splits the whole physics recursively into quadrangles of subfields.

  
  

  On the highest first level, we split physics according to the cold limit and the flat limit. This gives a quadrangle of first level theories of

  
  
  


• thermal physics in curved spacetime (Th Cu)


• thermal physics in flat spacetime (Th Fl)


• conservative physics in curved spacetime (Co Cu)


• conservative physics in flat spacetime (Co Fl) together with two first level interface theories


• statistical physics (Th<->Co)


• geometrization of physics (Cu<->Fl)
  

  These first level theories describe very general principles on the theoretically most fundamental level of physics.

  
  

  On the second level, we split each first level theory according to the eternal limit and the thermodynamic limit. This gives in each case a quadrangle of theories of

  
  



• nonequilibrium particle physics (Ne Mi)


• nonequilibrium thermodynamics (Ne Ma)


• physics of bound states and scattering (St Mi)


• equilibrium thermodynamics (St Ma) together with two second level interface theories


• long time asymptotics (Ne<->St)


• thermodynamic limits (Ma<->Mi)
  

  These second level theories describe physics on a level already close to many applications, especially outside physics, though still lacking detail.

  
  

  On the third, lowest level, we split each second level theory according to the nonrelativistic limit and the classical limit. This gives in each case a quadrangle of theories of

  
  


• relativistic quantum physics (Re Qu)



• relativistic classical physics (Re Cl)


• nonrelativistic quantum physics (Nr Qu)


• nonrelativistic classical physics (Nr Cl) together with two third level interface theories


• nonrelativistic limit (Re<->Nr)


• quantization and classical limit; quantum-classical systems (Qu<->Cl)
  

  These third level theories describe physics on the usual textbook and research level.

  
  

  (Maybe someone who likes to do graphics can illustrate this hierarchy with appropriate diagrams.)

  
  

lagrangian formalism - Noether theorem with semigroup of symmetry instead of group

Suppose You have semigroup instead of typical group construction in Noether theorem. Is this interesting? In fact there is no time-reversal symmetry in the nature, right? At least not in the same meaning as with other symmetries ( rotation, translation etc). So why we construct energy as invariant of such kind of symmetry, not using semigroup instead? Is is even possible? There is a plenty of references about Lie semigroups - i looks like it is active field of mathematics. Is there any kind of Noether theory constructed within such theories? Could You give any references to, ideally introductory texts?

Answer

Here is an argument why "a Noether Theorem with Lie monoid symmetry" essentially wouldn't produce any new conservation laws. Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws together form a representation of the Lie algebra), then one may focus on a $1$-dimensional Abelian subgroup of symmetry. The corresponding Lie subalgebra then becomes just $u(1)\cong\mathbb{R}$. Now returning to the question, one may, of course, artificially truncate a Lie group into a Lie monoid, say, if $q$ is a cyclic variable for a Lagrangian $L$, then artificially declare that the symmetry monoid is $q \to q + a$ for only non-negative translations $a\geq 0$, while artificially denying all negative $a<0$. On the other hand, one needs at least access "from one side" because Noether's Theorem is about continuous symmetry. But in practice, one can then always extend, at least infinitesimally, to "the other side" as well, and then one is back to a standard $u(1)$ Lie algebra and a standard Noether Theorem.

quantum mechanics - Tensor Product of Hilbert spaces

This question is regarding a definition of Tensor product of Hilbert spaces that I found in Wald's book on QFT in curved space time. Let's first get some notation straight.

Let $(V,+,*)$ denote a set $V$, together with $+$ and $*$ being the addition and multiplication maps on $V$ that satisfy the vector space axioms. We define the complex conjugate multiplication ${\overline *}:{\mathbb C} \times V \to V$ as $$c {\overline *} \Psi = {\overline c} * \Psi,~~\forall~~\Psi \in V$$ The vector space formed by $(V,+,{\overline *})$ is called the complex conjugate vector space and is denoted by ${\overline V}$.

Given two Hilbert spaces ${\cal H}_1$ and ${\cal H}_2$ and a bounded linear map $A: {\cal H}_1 \to {\cal H}_2$, we define the adjoint of this map $A^\dagger: {\cal H}_2 \to {\cal H}_1$ as $$\left< \Psi_2, A \Psi_1 \right>_{{\cal H}_2} = \left< A^\dagger \Psi_2 , \Psi_1 \right>_{{\cal H}_1}$$ where $\left< ~, ~ \right>_{{\cal H}_1}$ is the inner product as defined on ${\cal H}_1$ (similarly for ${\cal H}_2$) and $\Psi_1 \in {\cal H}_1,~\Psi_2 \in {\cal H}_2$. That such map always exists can be proved using the Riesz lemma.

Here the word "bounded" simply means that there exists some $C \in {\mathbb R}$ such that $$\left\| A(\Psi_1) \right\|_{{\cal H}_2} \leq C \left\| \Psi_1 \right\|_{{\cal H}_1}$$ for all $\Psi_1 \in {\cal H}_1$ and where $\left\| ~~ \right\|_{{\cal H}_1}$ is the norm as defined on ${\cal H}_1$ (similarly for ${\cal H}_2$)

Great! Now for the statement. Here it is.

The tensor product, ${\cal H}_1 \otimes {\cal H}_2$, of two Hilbert spaces, ${\cal H}_1$ and ${\cal H}_2$, may be defined as follows. Let $V$ denote the set of linear maps $A: {\overline {\cal H}}_1 \to {\cal H}_2$, which have finite rank, i.e. such that the range of $A$ is a finite dimensional subspace of ${\cal H}_2$. The $V$ has a natural vector space structure. Define the inner product on $V$ by $$\left< A, B \right>_V = \text{tr}\left( A^\dagger B \right)$$ (The right side of the above equation is well defined, since $A^\dagger B: {\overline {\cal H}}_1 \to {\overline {\cal H}}_1$ has a finite rank). We define ${\cal H}_1 \otimes {\cal H}_2$ to be the Hilbert space completion of $V$. It follows that ${\cal H}_1 \otimes {\cal H}_2$ consists of all linear maps $A: {\overline {\cal H}}_1 \to {\cal H}_2$ that satisfy the Hilbert-Schmidt condition $\text{tr}\left( A^\dagger A \right) < \infty$.

My question is

1. How does this definition of the Tensor product of Hilbert spaces match up with the one we are familiar with when dealing with tensors in General relativity?

PS - I also have a similar problem with Wald's definition of a Direct Sum of Hilbert spaces. I have decided to put that into a separate question. If you could answer this one, please consider checking out that one too. It can be found here. Thanks!

Answer

I don't think Wald ever defines a tensor product for infinite dimensional space in his GR text, so I presume your question is about the finite dimensional case where we simply write the tensor product as the vector space over pairs $u_iv_j$ where $u$ and $v$ are a basis. I will show the equivalence in that case.

If we have two finite dimensional Hilbert spaces $H_1$, $H_2$ we can take the orthonormal bases $u_i\in H_1$ , $v_j\in H_2$. Since everything is finite dimensional, everything is finite rank, so the the vector space is just the the space of linear maps from $H_1$ to $H_2$. Take a linear map $A$ and define $a_{ij}= \langle A(u_i),v_j\rangle = \langle u_i,A^{\dagger}(v_j)\rangle$. Using the orthonormality of the bases that means $a_{ij}$ is simply the matrix presentation of $A$, and the vector space is simply the appropriate vector space of matrices. Then we can interpret $Tr(A^\dagger B)$ as the usual matrix trace which gives $\sum_{ij} a_{ij}^*b_{ij}$.

This is equivalent to the usual notation whereby we write tensor products as elements $\sum_{ij}a_{ij}u_i\otimes v_j$. Again the vector space is the appropriately sized matrices. The inner product is defined to be $\langle a\otimes b, c\otimes d\rangle = \langle a,b\rangle\cdot\langle c,d\rangle$. This gives the same result as above after plugging in the basis.

special relativity - How does the solar sailing concept work?

Wikipedia describes solar sailing as

a form of spacecraft propulsion using a combination of light and high speed ejected gasses from a star to push large ultra-thin mirrors to high speeds.

I understand the part where ejected gasses bump into the sail pushing the spacecraft. On the other hand, I don't understand how light can do this, since light has no mass.

How does that work? Does this mean that if I have a mirror balancing on a needle I would be able to push it over with my flashlight?

Answer

I think you are really asking "how can light deliver an impulse to the sail". The answer is that although light has no mass it does carry momentum. When light is reflected off the sail, conservation of momentum requires that the sail changes momentum by twice the momentum of the light. The extra kinetic energy of the sail comes from the red shift of the reflected light.

This question has several answers that discuss the momentum of light in some detail.