How can you work out the FWHM of the gain profile for a longitudinal laser cavity?

Consider a longitudinal cavity with a gain medium.

Assuming knowledge of the geometry of the cavity (i.e. its length) and the free spectral range of the modes, as well as the frequency of the light that it emits, is there a quick and easy way to calculate the FWHM (full width half max) of the gain profile?

Answer

assuming knowledge of the geometry of the cavity (i.e. its length) and the free spectral range of the modes, as well as the frequency of the light that it emits, Is there a quick and easy way to calculate the FWHM of the gain profile?

No.

First, because normally, gain profile refers to a property of the gain medium only, independent of the cavity geometry. So the parameters you specified have nothing to do with the gain profile.

The gain profile will depend on the composition of the gain medium and on the pump power applied. Normally the gain profile is much wider than the spectral width of the resonant modes of the cavity. Because of this, the "FWHM" of the gain profile isn't something that's even normally calculated.

Second, if you actually meant the spectral width of the output beam, you can't easily calculate that either, because it depends not only on the cavity length, but also on the mirror reflectivities, optical loss in the cavity, and phase noise introduced by spontaneous emission in the gain medium. It may also depend on engineering details such as "mechanical vibrations, temperature fluctuations, and pump power fluctuations" (source).

riddle - Are you clever enough?

I'm always in your thoughts, technically twice
If displayed correctly, just to be precise

If you see my left you will call me a con
and it would tear me apart if you called me John

Take down a note and drop your cup
All you need is what keeps up

Please don't apply pressure or I will be off
and try to solve my riddle if you can take the scoff

---

Don't you dare look at the solution below.

Shame upon you! Try harder.

The solution is a single, short word. This won't help you for now but "meow"

Answer

Perhaps the single short word is the word "on"

It appears in "your" and "thoughts" if thoughts is viewed upside down and reflected.

To make the word John, you have to split the o and the n, thus tearing him up.

If you "c" the left, you will now be "con", a con.

If a switch is "on" and you apply pressure, it is now "off".

classical mechanics - Can a ship float in a (big) bathtub?

I am confused.

Some sources say it is possible at least theoretically ( http://www.wiskit.com/marilyn/battleship.jpeg ) and some say it is not true ( http://blog.knowinghumans.net/2012/09/a-battleship-would-not-float-in-bathtub.html )

Is it necessary or not that there exists an amount of water around the ship that weights at least the same as the weight of the ship?

Answer

Yes a ship can float in a big bath tub with very little water. No you do not need as much water as the weight of the ship. In theory, you can use less than a cupful!

Image from B5-2. How much water is needed to float a wood block?

Explanation

Suppose the ship is floating in the ocean, far from the sea bed or any shore. The layer of water in contact with the ship (let's say a layer of thickness 1mm all round the submerged surface) is providing enough force ("upthrust") to support the whole weight of the ship. This layer of water is in equilibrium vertically : it is pulled down by gravity (its own weight), pushed down by the weight of the ship above it, and pushed up by the water below it. It is in equilibrium horizontally : it is pushed outwards by the ship, and pushed inwards by the surrounding water.

Suppose the surrounding water beyond the contact layer is replaced by a thick concrete wall like a dam, many meters thick. The concrete wall is pushing both horizontally and vertically on the contact layer of water next to the ship. It makes no difference to the contact layer whether it is being pushed by other water or by the concrete wall. In both cases the contact layer remains in equilibrium and does not move. It does not, for example, get squeezed upwards and outwards between the concrete wall and the ship. That process of adjustment has already taken place, when the ship was launched or loaded.

The pressure in the contact layer of water varies only with its depth, not with its thickness. If there is more or less water between the ship and the wall, the pressure at that depth does not change.

The concrete wall or dam can be replaced by an enormous bath tub, providing that it is strong enough to exert the same force which the wall exerted.

Comment on Explanations by "Marilyn" and Brian Holtz

My explanation is essentially the same as Marilyn's (wiskit.com) - without the benefit of her excellent diagrams. The only difference is that Marilyn starts with water in the dock and replaces most of it with the battleship. I start with the ship floating in the ocean and replace most of the ocean water with dry dock.

Brian Holtz (Knowing Humans blog) is incorrect. His reasoning (accessed 12 January 2017) is not very clear to me, so I apologise to him if I have misunderstood it. His arguments are as follows :

1. The bathtub must initially contain sufficient water to be displaced by the ship when it floats.

eg If the battleship (USS Missouri, for example) weighs 45,000 tonnes then the bathtub must initially contain at least 45,000 tonnes of water.

The volume of the bath tub does need to be at least equal to the volume of 45,000 tonnes of water. However, it is not necessary for water to be actually displaced from the bathtub and to overflow from it. A small amount of water displaced upwards is just as good as an ocean displaced sideways.

If the battleship is lowered gradually into the bathtub and fits snugly into it, the cupful of water at the bottom will be squeezed up into the gap, increasing the depth of water. With only a cupful of water this will happen very quickly when the ship is almost in place. As this water moves upwards the upthrust which it provides increases. Eventually the upthrust is sufficient to support the whole weight of the ship.

As Deep says in his answer, and Jim in his comment, the volume of water displaced in Archimedes' Principle refers to the volume of the ship which is below the final waterline, not below the initial waterline. We cannot, of course, float a battleship in a bathtub which is only 1m deep, however wide it is. The draught of the USS Missouri is 8.8m, so our bathtub must be at least this deep. It must also be at least as wide and long as the ship at this height above the keel.

2. The battleship could not be floated by pouring an arbitrarily small amount of water into the gap between the ship and the bathtub. "You can't do the enormous work of lifting a massive ship merely by balancing it against a small mass of water."

Not correct. Hardly any work needs to be done to float the ship. Floating it is just a question of redistributing the load from direct contact with the bathtub to indirect contact via the layer of water. The ship does not need to be lifted up any further than say 1 micron - just enough to ensure that it is not pressing directly on the bath tub at any point.

To create the narrow gap the ship could rest with its whole weight on the bottom of the bathtub (both the ship and the tub must be enormously strong to do this) while narrow supports, perhaps 1 mm thick, at the sides prevent it from making contact with the sides of the bathtub. Water could easily be poured into this gap under its own gravity - there is nothing to prevent it falling down, until it reaches the level of the water already poured in.

As the water gets deeper it gradually exerts more pressure on the ship, so there is less pressure on keel of the ship (directly from the bathtub) and more on the rest of the underneath surface (from the water, which is in turn pushing on the rest of the bathtub). When the water is deep enough (at least 8.8m for USS Missouri) the pressure from it will be enough to support the whole weight of the ship and it will no longer exert any contact force directly on the bathtub.

To float the battleship any finite distance higher up in the bathtub (say 5mm higher) it is only necessary to add more water to the tub. However, the amount of water required to do this could be very large because raising the ship will increase enormously the volume of the gap which needs to be filled. A gap which is initially "snug" does not remain "snug" as the ship rises vertically.

The work done will of course also be enormous : the weight of the battleship times the distance moved upwards. However, this work is done by gravity, acting on the extra water. If the extra water is already in a reservoir above the current water level, gravity will carry it down into the tub. But if it is necessary to pump this extra water up to the current water level from below the level of the keel, the energy required to do so will be at least equal to the work done in raising the 45,000 tonne battleship by 5mm.

So as Brian Holtz says, There is no free lunch. Lifting the battleship even by 5mm requires enormous energy. But this is not the same as getting it to float, which is just a question of transferring the weight from the keel to the contact layer of water.

3. In canal locks there is sufficient clearance all round (behind and in front as well as at the sides) to hold a volume of water equal to the weight of the ship above the original water level.

Not necessarily true. There is no reason why a rectangular ship (eg a barge) cannot "dock" with clearance of say only 6" on either side and below. The water it displaced has been pushed aside and behind it. A dam can then be erected behind it, again with only 6" clearance. When the dam is strong enough, the water behind it can be pumped away, leaving the barge afloat in an isolated "dock" which contains much less than its own weight of water.

The same thing happens in canal locks when the ship occupies almost the total volume of the lock. After the lower gate is closed, water is allowed to fall in from the lock above, raising the height of the ship. This requires enormous energy, but it is all done by gravity, courtesy of rainwater and reservoirs.

4. In the no-overlow scenario, there is enough room at the top of the bath tub to hold the mass of water which balances the floating object.

It is not clear what this means. If it means that the volume of the empty bath tub must be big enough to hold a volume of water equal to the ship's weight, then this is not in dispute. eg See Marilyn's diagrams. The bath tub needs to be at least as deep as the draught of the battleship. But the final amount of water in it can be very small. It is the weight of the missing ("displaced") water below the waterline which is crucial, not the weight of the water which remains.

If it means that the weight of water in the gap must be at least as much as the weight of the ship, this is false, as argued in #1 and #3.

5. When large machines like telescopes "float" on a thin film of lubricating oil, the oil is kept pressurized in a sealed system, and the pool of oil is not open to the atmosphere.

True, but this is a matter of convenience, not necessity. The telescope could float just as well on a much "deeper" film of oil which is open to the atmosphere. Sealing the vessel enables high pressure to be achieved fairly uniformly and with the minimum amount of oil. When open to the atmosphere, the balancing depth (not mass) of oil provides the required pressure.

Quantum entanglement and spooky action at a distance

When quantum entanglement is explained in "layman's terms", it seems (to me) that the first premise, that we have to accept on faith, is that a particle doesn't have a certain property (the particle is not IN one state or another) until that property is measured.

For example, I have read that a particle may have "spin up" or "spin down" but -- here's where I get lost -- it's not that we don't KNOW which spin the particle has until we measure it, but actually the particle has NEITHER spin until we measure it.

Now, if we accept that premise, then we can see how spooky it is, when two particles are created with correlated spins -- if one has spin up, the other has spin down, but neither particle "has" either spin until one of them is measured. At that instant, the other particle "assumes" the other spin.

At least that's how I interpret things I have read.

Certainly, if we accept that a particle doesn't actually HAVE the property until it is measured, but rather the particle exists in a superposition of states until the measurement is made, then we "regular citizens" can understand what is meant by "spooky action at a distance". (As soon as one particle is measured, the other particle assumes the "other" spin, no matter how far away it is. How did the second particle know the first particle was measured? That, I think, is the spooky action at a distance (and the information transfer can be faster than light).)

However, from a layman's perspective, I want to cry out "but the particle does have a definite spin, we just don't KNOW what it is, until it is measured! Duh!"

Now, I can't believe that all of the physics community hasn't thought of that objection -- but -- here's my point -- although the spooky action at a distance can be explained to us, once we accept the premise of superposition -- Why isn't the fact of superposition explainable also, in terms mortals can understand? I have looked, but I haven't found a layman's explanation of why we should accept the fact that a particle doesn't have a particular spin, or other property, until it's measured. It sure seems that the particles "have" the properties, even though we haven't measured them yet.

I know that much of quantum mechanics is not "intuitive". If anyone can explain why particles don't have a definite property, even before we measure them, I would be grateful.

Answer

The assumption (if you cry it or not) "but the particle does have a definite spin, we just don't KNOW what it is, until it is measured! Duh!" is called realism, or in mathier speak, a theory of hidden variables.

Bell's inequalities now say that no theory that fulfills local realism (equivalently that has local hidden variables) can ever predict the correct results of a quantum mechanical experiment.

So we are faced with a problem: Do we give up locality or realism?

Most people choose realism, since giving up locality would totally destroy our conceptions of causality. It is possible that there is a non-local theory that assigns a definite value to every property at all times, but due to its non-locality, it would be even more unintuitive than "particles do not have definite properties".

There is no intuitive explanation for the non-realism of reality (there has to be a way to phrase that better...) because our intuitions have been forged in the macroscopic world which is, to good approximation, classical. But the non-realism is an effect that has no classical analogon, so we cannot understand it in pretty simple pictures or beautiful just-so stories.

Sometimes, we just have to take the world the way it is. (I have assumed that you do not want the whole QM story of non-commuting observables and eigenbases and so on to explain why we, formally from QM principles, expect realism to be false. If I have erred in that respect, just tell me)

Give a description of Loop Quantum Gravity your grandmother could understand

Of course, assuming your grandmother is not a theoretical physicist.

I'd like to hear the basics concepts that make LQG tick and the way it relates to the GR. I heard about spin-networks where one assigns Lie groups representations to the edges and intertwining operators to the nodes of the graph but at the moment I have no idea why this concept should be useful (except for a possible similarity with gauge theories and Wilson loops; but I guess this is purely accidental). I also heard that this spin-graph can evolve by means of a spin-foam which, I guess, should be a generalization of a graph to the simplicial complexes but that's where my knowledge ends.

I have also read the wikipedia article but I don't find it very enlightening. It gives some motivation for quantizing gravity and lists some problems of LQG but (unless I am blind) it never says what LQG actually is.

So, my questions:

1. Try to give a simple description of fundamentals of Loop Quantum Gravity.



2. Give some basic results of the theory. Not necessary physical, I just want to know what are implications of the fundamentals I ask for in 1.


3. Why is this theory interesting physically? In particular, what does it tell us about General Relativity (both about the way it is quantized and the way it is recovered from LQG).

Answer

@Marek your question is very broad. Replace "lqg" with "string theory" and you can imagine that the answer would be too long to fit here ;>). So if this answer seems short on details, I hope you will understand.

The program of Loop Quantum Gravity is as follows:

1. 
   

   The notion of diffeomorphism invariance background independence, which is central to General Relativity, is considered sacrosanct. In other words this rules out the String Theory based approaches where the target manifold, in which the string is embedded, is generically taken to be flat [Please correct me if I'm wrong.] I'm sure that that is not the only background geometry that has been looked at, but the point is that String Theory is not written in a manifestly background independent manner. LQG aims to fill this gap.

   
   
   


2. 
   

   The usual quantization of LQG begins with Dirac's recipe for quantizing systems with constraints. This is because General Relativity is a theory whose Hamiltonian density ($\mathcal{H}_{eh}$), obtained after performing a $3+1$ split of the Einstein-Hilbert action via the ADM procedure [1,2], is composed only of constraints, i.e.

   
   

   $$ \mathcal{H}_{eh} = N^a \mathcal{V}_a + N \mathcal{H} $$

   
   

   where $N^a$ and $N$ are the lapse and shift vectors respectively which determine the choice of foliation for the $3+1$ split. $\mathcal{V}_a$ and $\mathcal{H}$ are referred to as the vector (or diffeomorphism) constraint and the scalar (or "hamiltonian") constraint. In the resulting phase space the configuration and momentum variables are identified with the intrinsic metric ($h_{ab}$) of our 3-manifold $M$ and its extrinsic curvature ($k_{ab}$) w.r.t its embedding in the full $3+1$ spacetime, i.e.

   
   

   $$ {p,q} \rightarrow \{\pi_{ab},q^{ab}\} := \{k_{ab},h^{ab}\} $$

   
   

   This procedure is generally referred to as canonical quantization. It can also be shown that $ k_{ab} = \mathcal{L}_t h_{ab} $, where $ \mathcal{L}_t $ is the Lie derivative along the time-like vector normal to $M$. This is just a fancy way of saying that $ k_{ab} = \dot{h}_{ab} $

   
   

   This is where, in olden days, our progress would come to a halt, because after applying the ADM procedure to the usual EH form of the action, the resulting constraints are complicated non-polynomial expressions in terms of the co-ordinates and momenta. There was little progress in this line until in 1986 $\sim$ 88, Abhay Ashtekar put forth a form of General Relativity where the phase space variables were a canonically transformed version of $ \{k_{ab},h^{ab}\} $ This change is facilitated by writing GR in terms of connection and vielbien (tetrads) $ \{A_{a}^i,e^{a}_i\}$ where $a,b,\cdots$ are our usual spacetime indices and $i,j,\cdots$ take values in a Lie Algebra. The resulting connection is referred to as the "Ashtekar" or sometimes "Ashtekar-Barbero" connection. The metric is given in terms of the tetrad by :

   
   

   $$ h_{ab} = e_a^i e_b^j \eta_{ij} $$

   
   

   where $\eta_{ij}$ is the Minkowski metric $\textrm{diag}(-1,+1,+1,+1)$. After jumping through lots of hoops we obtain a form for the constraints which is polynomial in the co-ordinates and momenta and thus amenable to usual methods of quantization:

   
   
   

   $$ \mathcal{H}_{eha} = N^a_i \mathcal{V}_a^i + N \mathcal{H} + T^i \mathcal{G_i} $$

   
   

   where, once again, $ \mathcal{V}_a^i $ and $\mathcal{H}$ are the vector and scalar constraints. The explanation of the new, third term is postponed for now.

   
   

   Nb: Thus far we have made no modifications to the theoretical structure of GR. The Ashtekar formalism describes the exact same physics as the ADM version. However, the ARS (Ashtekar-Rovelli-Smolin) framework exposes a new symmetry of the metric. The introduction of spinors in quantum mechanics (and the corresponding Dirac equation) allows us to express a scalar field $\phi(x)$ as the "square" of a spinor $ \phi = \Psi^i \Psi_i $. In a similar manner the use of the vierbien allows us to write the metric as a square $ g_{ab} = e_a^i e_b^j \eta_{ij} $. The transition from the metric to connection variables in GR is analogous to the transition from the Klein-Gordon to the Dirac equation in field theory.

   
   


3. 
   

   The application of the Dirac quantization procedure for constrained systems shows us that the kinematical Hilbert space, consisting of those states which are annihilated by the quantum version of the constraints, has spin-networks as its elements. All of this is very rigorous and several mathematical technicalities have gradually been resolved over the past two decades.

   
   

---

This answer is already pretty long. It only gives you a taste of things to come. The explanation of the Dirac quantization procedure and spin-networks would be separate answers in themselves. One can give an algorithm for this approach:

1. 
   

   Write GR in connection and tetrad variables (in first order form).

   
   


2. 
   

   Perform $3+1$ decomposition to obtain the Einstein-Hilbert-Ashtekar Hamiltonian $\mathcal{H}_{eha}$ which turns out to be a sum of constraints. Therefore, the action of the quantized version of this Hamiltonian on elements of the physical space of states yields $ \mathcal{H}_{eha} \mid \Psi \rangle = 0 $. (After a great deal of investigation) we find that these states are represented by graphs whose edges are labeled by representations of the gauge group (for GR this is $SU(2)$).

   
   


3. 
   

   Spin-foams correspond to histories which connect two spin-networks states. On a given spin-network one can perform certain operations on edges and vertices which leave the state in the kinematical Hilbert space. These involve moves which split or join edges and vertices and those which change the connectivity (as in the "star-triangle transformation"). One can formally view a spin-foam as a succession of states $\{ \mid \Psi(t_i) \rangle \}$ obtained by the repeated action of the scalar constraint $ \mid \Psi(t_1) \rangle \sim \exp{}^{-i\mathcal{H}_{eha}\delta t} \mid \Psi (t_0); \mid \Psi(t_2) \rangle \sim \exp{}^{-i\mathcal{H}_{eha}\delta t} \mid \Psi (t_1) \cdots \rangle $ [3].

   
   



4. 
   

   The graviton propogator has a robust quantum version in these models. Its long-distance limit yields the $1/r^2$ behavior expected for gravity and an effective coarse-grained action given by the usual one consisting of the Ricci scalar plus terms containing quantum corrections.

   
   

There is a great deal of literature to back up everything I've said here, but this is already pretty exhausting so you'll have to take me on my word. Let me know what your Grandma thinks of this answer ;).

---

homework and exercises - Questions About Quantum Delta Function Potentials

1. 
   

   I didn't think that it would be possible for a wave function to get through the delta function because there is no "leakage" of the wave function through an infinite potential barrier. I can understand why a particle could get through a non infinite barrier because of this exponential leakage but I don't understand how a particle could get through an infinite delta potential. My thinking right now is that maybe there is some imaginary (as apposed to real) leakage, similar to how a quadratic that doesn't cross the $x$ axis in the real plane may cross the $x$ axis in the imaginary plane. Is this the reason?

   
   
   


2. 
   

   I tried to solve for the reflection and transmission coefficients by solving for the two solutions outside the delta function and you obviously get 4 different exponentials with imaginary exponents. I know that these two functions have to touch so you equate them at $x=0$ and get $A+B=C+D$. Then I did the finding the change in slope trick and I get another relation between these 4 coefficients. The problem is that this gives me 2 equations and 4 coefficients to solve for, so I'm not sure what to do now (also a k value so 5 variables I guess). For the bound solution the coefficients were much nicer and it was easy to solve, but I'm not sure how to solve this one. My initial thinking is that I can get rid of the D coefficient since I only want the transmitted wave to be right-moving, but that only cuts me down to 3 variables when I need either 2 or another equation.

   
   

Answer

Even with a delta function potential, continuity of the wave function is still required. (Please see comment from ACuriousMind below on this).

The derivative of the wavefunction is obviously not continuous, however. You can find the discontinuity by integration about the delta function from +s to -s, where s is a small parameter.

You then let s go to 0 and check the behaviour of the derivative. You don't state the wavefunction(s) you are using, but here is a plot of a typical one.

You can find the actual calculations here:

Delta Function Wikipedia

Mass and Energy

Would the mass of burnt firewood be equal to the mass of firewood before burning?

Then where does that heat come from?

According to Einstein's equation, $E=mc^2$ Shouldn't there be some mass going out of the Earth which contradicts the law of mass conservaton?

Answer

Would the mass of burnt firewood be equal to the mass of firewood before burning?

You won't get a good answer by simply looking at the "burnt firewood". The combustion is using oxygen from the air, and it is creating carbon dioxide and many volatilized materials that will disperse in the air. But we can imagine combustion happening in a box that is sealed to retain any materials (such as smoke and gases), but which allows energy (perhaps in the form of heat and light) to leave. If done to a very high precision, there would be a tiny difference found in the measured mass. The difference would be equal to the energy liberated during the combustion.

A generous value for the amount of energy liberated due to combustion would be $20MJ/kg$. If we were to combust a $1kg$ log in the presence of sufficient oxygen, then the mass deficit would be on the order of $$m = \frac{E}{c^2} $$ $$m = \frac{2\times 10^7 J}{9.0\times 10^{16}\frac{m^2}{s^2}}$$ $$m = 2.2\times10^{-10}kg$$ This difference in a $1kg$ mass is not measurable.

With such small differences, we can assume that mass is conserved during chemical reactions. When nuclear reactions are considered, the amount of mass converted becomes large enough to be measured and the law of mass conservation has to be modified.

quantum field theory - How to derive the form of the parity operator acting on Lorentz spinors?

I'm reading Berestetskii (Volume 4 of Landau & Lifshitz) section 19 on inversion of spinors. Berestetskii says parity $P$ maps undotted spinors into dotted spinors and vice-versa as $\xi^{A}\rightarrow i\eta_{\dot{A}}$ and $\eta_{\dot{A}}\rightarrow i\xi^{A}$. I think he means there are parity tensors $P_{\dot{A}B}$ and $P^{A\dot{B}}$ which act as, $$ \eta_{\dot{A}}=P_{\dot{A}B}\xi^{B} \\ \xi^{A}=P^{A\dot{B}}\eta_{\dot{B}} $$ and which satisy $P^{2}=-1$, $$ P^{A\dot{C}}P_{\dot{C}B}=-\delta^{A}_{B} \ . $$ Berestetskii's map then says, $P_{\dot{A}B}=P^{A\dot{B}}=i\delta_{AB}$ which is only true in (presumably) the rest frame of the fermions. I've tried to derive the components of the parity tensors by imposing various mathematical and physically motivated restrictions, but I can only get $P^{A\dot{B}}$ diagonal with entries $a,1/a^{*}$ and $P_{\dot{A}B}$ diagonal with entries $-1/a,-a^{*}$. Berestetskii's parity is recovered by taking $a=i$ and so my question is, "What are the extra restrictions needed to be able to set $a=i$?"

quantum field theory - Why is the functional integral of a functional derivative zero?

I'm reading Quantum Field Theory and Critical Phenomena, 4th ed., by Zinn-Justin and on page 154 I came across the statement that the functional integral of a functional derivative is zero, i.e. $$\int [d\phi ]\frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} = 0$$ for any functional $F[\phi ]$.

I would be most thankful if you could provide a mathematical proof for this identity.

A Riley Rebus As Suggested By Duck

This is gonna be a Riley rebus suggested by @Duck (Thanks)
Be sure to check out some of his great riddles and puzzles like this and this :D

Note :

I will probably (Definitely) post another one of these. It's here :D

Will also add hints if nobody gets if after a while

Small Hint :

Don't take things too literally

Another Small Hint :

The Suffix picture is a game called agario

Specific Hint for Prefix :

It's a way of measuring something that is also related to a circle

Answer

Could it be something like

Radio?

Prefix

Rad (for radian)

Infix

di

Suffix

Io

computer puzzle - ‘‘Loopy’’ C loop

What is the smallest number of non-space characters that can be added in order to change the following code into a complete C program that prints  LoopyLoopyLoopyLoopy?

#include
main(){
<-- ^
| @ |
v --> printf("Loopy");}

Starts as: 5 lines with no leading spaces, totaling 52 non-space characters and 5 within-line spaces.

Possible edits (additions that don't move the original characters):
•   Single non-space characters may replace any of the 5 original spaces.
•   Spaces and non-space characters may be added to the ends of lines.

No-nos and pedantry:
•   No commenting.
•   No additional " quotation marks.
•   No new lines. (Intended, but not made explicit, in the original puzzle statement.)
•   All syntax should abide by The C Programming Language, 2nd Edition, Kernighan & Ritchie.
•   If any variables are used before they are initialized, the program should be successful with any initial values for those variables.

_{Notes about the original puzzle statement: The word visible was used instead of instead of non-space. Adding lines was unintentionally allowed as a newline would count as a character added to the end of an existing line. The line #include could have been left out for a more streamlined version of essentially the same puzzle, as mentioned by Arkku.}

Answer

In accordance with the updated rules:

#include
main(){int v=4;for('\
<-- ^\
| @ |';v;)
v --> printf("Loopy");}

The original rules (in effect at the time of posting this) unintentionally permitted adding a newline character to the end of a line, which enabled:

#include
v;
main(){for(;-8
<--v^0;'\
| @ |')
v --> printf("Loopy");}

(16 if adding a newline character to end of line counts as space, 17 if it counts as a non-space.)

Note: The type of the global v defaults to int and is implicitly initialized to zero in ANSI C (the standard at the time of K&R 2nd edition) due to static duration.

If it were permitted to insert a single non-space character between two non-space characters:

#include
main(v){for(;-8
<--v^0;'\
| @ |')

v --> printf("Loopy");}

This with the caveat:

The program must be run without arguments in an environment that passes 1 as the initial argument of main in that case. (As discussed in comments, this is permitted.)

homework and exercises - Based on newtons third law

If an elephant can apply 250N force and a rabbit can apply 25N force , and if they are pulling each other , then elephant will pull with 250N force and according to newtons third law the rabbit should also pull the elephant with 250N force.then from where do the rabbit got the extra 225N i.e.(250-25=225) force??? can anyone please explain.

Answer

To check your data image that a spring balance was connected to a wall and the elephant and the rabbit were asked to pull as hard as they could so that you could take a reading of their maximum pulling force.
What that reading will show is the maximum horizontal force that the elephant and the rabbit can exert on the ground with their feet.
This is because when the elephant pulls on the spring balance the spring balance pulls on her with an equal force. (N3L)
For the elephant not to move the ground must push the elephant with an equal force so that the net force on the elephant is zero. (Not N3L)
So the maximum pull is the maximum horizontal push that the feet can exert on the ground which is equal and opposite to the maximum push the ground can exert on the feet. (N3L)

So now we come to the tug of war.

The rabbit pulls with a force of 10 N and the elephant also pulls with a force of 10 N. There is not a lot of movement. The rabbit then pulls with a maximum force for him of 25 N and the elephant counters with a force of 25 N. Stalemate again.

The elephant now pulls with a force greater than 25 N, a force which is greater than that which the ground can exert on the rabbit.
There is thus a net force on the rabbit.
So the rabbit accelerates towards the elephant.

classical mechanics - Is *every* planar/2D system integrable?

Consider the generic following planar/2D system:

$$\begin{cases} \frac{dx}{dt} = A(x,y)\\ \\ \frac{dy}{dt} = B(x,y), \end{cases}$$

where $A,B$ are two functions. Reading Classical Mechanics by Joseph L. McCauley I found the following statements:

Every two dimensional flow, $$dx/dt = A(x,y), \qquad dy/dt = B(x,y),$$ whether dissipative or conservative, has a conservation law,

and, if we rewrite the system equations as $dt=dx/A=dy/B$,

every differential form $B(x,y)dx-A(x,y)dy=0$ in two variables either is closed or else has an integrating factor $M(x,y)$ that makes it integrable.

So is really every planar system integrable, or have I missed some detail?

Answer

A global integrability statement for general 2D systems does not hold, but a local integrability statement is true. Let us reformulate OP question as follows.

Suppose that we are given a two-dimensional first-order problem $$ \dot{x}~=~f(x,y), \qquad \dot{y}~=~g(x,y), \tag{1}$$ where $f$ and $g$ are two given smooth functions. Is eq. (1) a Hamiltonian system $$ \dot{x}~=~\{x,H\}, \qquad \dot{y}~=~\{y,H\}, \tag{2}$$ with a symplectic structure $\{\cdot,\cdot\}$ and Hamiltonian $H(x,y)$?

The answer is, perhaps surprisingly: Yes, always, at least locally. The Hamiltonian $H$ is the sought-for integral of motion/first integral.

1. 
   

   Proof: In two dimensions, a Poisson bracket is completely specified by the fundamental Poisson bracket relations $$ \{x,y\} ~=~B(x,y)~=~-\{y,x\}, \qquad \{x,x\}~=~0~=~\{y,y\}, \tag{3} $$ where $B$ is some function that doesn't take the value zero. [Exercise: Check that eqs. (3) automatically satisfy the Jacobi identity.] The Hamilton's eqs. (2) become $$ \dot{x}~=~B\frac{\partial H}{\partial y}, \qquad \dot{y}~=~-B\frac{\partial H}{\partial x}.\tag{4} $$ Next consider the one-form $$ \eta ~:=~ f{\rm d}y -g{\rm d}x, \tag{5}$$ which is possibly an inexact differential. However, it is known from the theory of PDE's, that there locally exists an integrating factor $\frac{1}{B}$, so that the one-form $$ \frac{1}{B}\eta~=~{\rm d}H \tag{6} $$ is locally an exact differential given by some function $H$. It is straightforward to check that one can use $B$ as the Poisson structure (3) and $H$ as the Hamiltonian. $\Box$

   
   



2. 
   

   Remark. The existence of a pair of canonical variables $q(x,y)$ and $p(x,y)$, with $\{q,p\}=1$, are, in turn, guaranteed locally by Darboux' Theorem.

   
   


3. 
   

   Example: 1D system with friction force: Eqs. of motion: $$m\dot{v}~=~-kv-V^{\prime}(x), \qquad \dot{x}~=~v. \tag{7} $$ According to the theorem there in principle exists locally a Hamiltonian formulation without explicit time dependence, however it is not possible to give a general formula. If we are allowed to have explicit time dependence (which is outside the main topic of this answer), then there is a simple solution: Define $e(t):=\exp(\frac{kt}{m})$. Lagrangian: $L=e(t)L_0$, where $L_0=\frac{m}{2}v^2-V(x)$. Momentum: $p=e(t)mv$. Hamiltonian: $H=\frac{p^2}{2me(t)}+e(t)V(x)$.

   
   


4. 
   

   Example: Math.SE q1577274.

   
   


5. 
   
   

   Counterexample. Solutions to diff. eqs. exist in general only locally. Consider $$f(q,p)~=~\frac{q}{q^2+p^2}\quad\text{and}\quad g(q,p)~=~\frac{p}{q^2+p^2}\tag{8}$$ in the domain $D=\mathbb{R}^2\backslash\{(0,0)\},$ which is not contractible. It is relatively straightforward to check that $$\eta~=~f\mathrm{d}p-g\mathrm{d}q ~=~\frac{q\mathrm{d}p-p\mathrm{d}q}{q^2+p^2} \tag{9}$$ is a closed $1$-form, and there doesn't exist a globally defined Hamiltonian $H$ on $D$ such that eqs. (2) are satisfied. The best one can do is to put $H$ equal to a single-valued branch of ${\rm arg}(q+ip)$, which is not globally defined.

   
   


6. 
   

   Counterexample: Contractible domain without global solution: Math.SE q2710698.

   
   

electromagnetic radiation - Can light exist in $2+1$ or $1+1$ spacetime dimensions?

Spacetime of special relativity is frequently illustrated with its spatial part reduced to one or two spatial dimension (with light sector or cone, respectively). Taken literally, is it possible for $2+1$ or $1+1$ (flat) spacetime dimensions to accommodate Maxwell's equations and their particular solution - electromagnetic radiation (light)?

Answer

No, because the polarization of the electromagnetic field must be perpendicular to the direction of motion of the light, and there aren't enough directions to enforce this condition. So in 1d, a gauge theory becomes nonpropagating, there are no photons, you just get a long range Coulomb force that is constant with distance.

In the 1960s, Schwinger analyzed QED in 1+1 d (Schwinger model) and showed that electrons are confined with positrons to make positronium mesons. A much more elaborate model was solved by t'Hooft (the t'Hooft model, the nonabelian Schwinger model) which is a model of a confining meson spectrum.

EDIT: 2+1 Dimensions

Yes, light exists in 2+1 dimensions, and there is no major qualitative difference with 3+1 dimensions. I thought you wanted 1+1, where it's interesting.

logical deduction - The final showdown

I am amazed at my adviser's brilliance once again. Blackbeard's vanguard fleet was quickly sunk. And as luck would have it, the moment we finished cleaning up the mess, the scout returned with the lost scout from the first engagement. He seemed extremely excited and kept pointing to a piece of paper. Upon closer look, it is another map the scout drew.

$$ \begin{array}{|c|c|c|c|c|c|c|c|l} \hline&&&&&&&&4\\ \hline&&&&&&&&2\\ \hline&&&&&&&&5\\ \hline&&&&&&&&2\\ \hline&&&&&&&&1\\ \hline&&&&&&&&6\\ \hline&&&&&&&&1\\ \hline&&&&&&&&4\\ \hline3&1&5&2&5&2&2&5 \end{array} \quad\quad \bbox[4pt,border: 1px solid black]{ \begin{array}{llc} \bf{Legend:}&\\ \hline \text{Man-of-War:}& \blacktriangleleft\blacksquare\blacksquare\blacksquare\blacktriangleright \text{or} &\blacktriangle\\ &&\blacksquare\\ &&\blacksquare\\ &&\blacksquare\\ &&\blacktriangledown\\ \hline \text{Queen Anne's Revenge:}& \blacktriangleleft\blacksquare\blacksquare\blacktriangleright \text{or} &\blacktriangle\\ &&\blacksquare\\ &&\blacksquare\\ &&\blacktriangledown\\ \hline \text{Frigate:}& \blacktriangleleft\blacksquare\blacktriangleright \text{or} &\blacktriangle\\ &&\blacksquare\\ &&\blacktriangledown\\ \hline \text{Corvette:}& \blacktriangleleft\blacktriangleright \text{or} &\blacktriangle\\ &&\blacktriangledown\\ \hline \text{Patrol Boat:}& \blacklozenge \end{array}} $$

After he finally calmed down, he told me the following

I have spotted Blackbeard himself on his Queen Anne's Revenge. However, in addition to his standard setup (1 flagship, 2 frigates, 3 corvettes, and 4 patrol boats), there is a man-of-war with 74 cannons. The size of man-of-war is bigger than even Blackbeard's flagship. Their formation is slightly different though, and they are closer together, but interestingly, none of the ships touch each other, not even diagonally. Unfortunately, since it was still dark, I wasn't able to pinpoint their exactly location, but I have drawn the map with the number of ship pieces on each column and row.

My adviser pondered at the map for a while, and then smiled.

Men, it seems like Blackbeard himself isn't aware of our presence yet, we must strike them now before dawn or we will face Blackbeard's full wrath. I have figured out their exact formation, so follow my orders and I will lead you into victory!

With a cheer from the rest of the sailors, everyone frantically prepare to sail again, leaving me behind pondering just how he managed to find out the enemy fleet formation, and why the scout keeps adding a legend but not using it.

Answer

It can't go in Column 3 or 5

It can't go in Column 8

It can't go in Row 6

That leaves the Man-of-War in Row 3.

We know that in column three and five there has to be a vertical segment of at least length two, so we can mark off a couple sea segments.

Fill in a couple segments:

Now column 8 needs to have a vertical segment of length at least two, so we can mark off a couple sea segments.

Extending with known info takes us here:

The two blanks adjacent to the Man-of-War will be a ship piece and a sea piece, leaving column eight, row three as a ship piece. Filling in knowns takes us here:

If column six, row eight is a ship then forced moves leave us with too many ships of size two or larger.

Row one cannot contain the Queen Anne's Revenge. If it did then column 8 would hold the frigate and columns one, three, and five would have to contain a frigate, two corvettes and two patrol boats. This will make the two in row 7 impossible.

Column eight cannot contain the Queen Anne's revenge. If it did then two of columns one, three, and five would contain frigates. Therefore column five, row one would be a sea segment forcing column five to contain a ship of length four, which is not possible.

Therefore column 8, row 1 is sea and forced moves lead us here:

Of the two spaces left, one must contain a frigate, and the other two patrol boats. Column three needs the frigate. This solves the puzzle.

As an aside, there are some steps that I would not have been able to do easily on paper. The hunt and check to mark off row six as impossible for the Man-of-War was a little deep. Curious to see if anyone else has any tricks.

calculus - Rigorous underpinnings of infinitesimals in physics

Just as background, I should say I am a mathematics grad student who is trying to learn some physics. I've been reading "The Theoretical Minimum" by Susskind and Hrabovsky and on page 134, they introduce infinitesimal transformations. Here's the first example they use:

Consider a particle moving in the x,y plane under the influence of a potential, $V$, which depends only on the radius, with Lagrangian:

$L=\frac{m}{2}(\dot{x}^2+\dot{y}^2)-V(x^2+y^2)$

This is clearly invariant under rotations:

$x \rightarrow x\cos \theta + y\sin \theta$

$y \rightarrow -x\sin \theta + y\cos \theta$

All well and good. Now they say "consider what happens ... when the angle $\theta$ is replaced by an infinitesimal angle $\delta$." Already I could say "What the heck is $\delta$ really?", but I'm willing to play along with my intuition. Since $\delta$ is infinitesimal, we work to first order and say

$\cos \delta=1$ and $\sin \delta= \delta$.

Plugging this into our rotation formulas above, we obtain:

$x \rightarrow x+y\delta$

$y \rightarrow y-x\delta$

By differentiating, we see that:

$\dot{x} \rightarrow \dot{x}+\dot{y}\delta$

$\dot{y} \rightarrow \dot{y}-\dot{x}\delta$

Plugging these into the Lagrangian and ignoring terms higher than first order, we see that the Lagrangian is invariant under this transformation.

My main problem with all of this is that I don't understand what the physical nature of an infinitesimal transformation actually is. All I got from the above was that if you do this formal calculation by following rules like "only work to first order in $\delta$," then the Lagrangian is invariant. This is in contrast to the case where we have an actual transformation, like a rotation, where there is no question about what is going on physically.

I would also like to know how all of this relates to rigorous mathematics. In mathematics, I can't recall ever using infinitesimals in an argument or calculation, so it would be useful if there was some way of formulating the above in terms of limits/ derivatives/ differential forms (for example). I sense a connection to Lie Algebras, as the infinitesimal version of the rotation is $(I+A)$ where $I$ is the identity matrix and $A$ is an element of the Lie Algebra of $SO(2)$.

Here are some questions whose answers I believe may be useful to me (feel free to answer some or all):

-What is an infinitesimal quantity like $\delta$ to the physicist?

-Why do physicists argue using infinitesimals rather than "standard" calculus?

-What is the physical meaning of infinitesimal transformation? How does it relate to Lie Algebras?

-Is there a rigorous theoretical apparatus for justifying the computations shown above?

-What is meant by the Lagrangian being invariant under infinitesimal transformations?

If any of the questions seem too vague, please say so. Thanks in advance for your insights!

Answer

When I asked my undergrad analytic mechanics professor "what does it mean for a rotation to be infinitesimal?" after he hand-wavily presented this topic in class, he answered "it means it's really small." At that point, I just walked away. Later that day I emailed my TA who set me straight by pointing me to a book on Lie theory.

Fortunately, I don't intend to write an answer like my professor's.

In general, whenever you see the term "infinitesimal BLANK" in physics, you can be relatively certain that this is merely a placeholder for "first order (aka linear) approximation to BLANK."

Let's look at one of the most important examples.

Infinitesimal transformations.

To be more rigorous about this, let's consider the special case of "infinitesimal transformations." If my general terminological prescription above is to be accurate, we have to demonstrate that we can make the concept of a "first order approximation to a transformation" rigorous, and indeed we can.

For concreteness, let's restrict the discussion to tranformations on normed vector spaces. Let an open interval $I=(a,b)$ containing $0$ be given, and suppose that $T_\epsilon$ is a transformation on some normed vector space $X$ such that $T_0(x)$ is the identity. Let $T_\epsilon$ depend smoothly on $\epsilon$, then we define the infinitesimal version $\widehat T$ of $T_\epsilon$ as follows. For each point $x\in X$, we have $$ \widehat T_\epsilon(x) = x + \epsilon\frac{\partial}{\partial\epsilon}T_{\epsilon}(x)\bigg|_{\epsilon=0} $$ The intuition here is that we can imagine expanding $T_\epsilon(x)$ as a power series in $\epsilon$; $$ T_\epsilon(x) = x + \epsilon T_1(x) + \mathcal O(\epsilon^2) $$ in which case the above expression for the infinitesimal version of $T_\epsilon$ gives $$ \widehat {T}_\epsilon(x) = x+\epsilon T_1(x) $$ so the transformation $\widehat T$ encodes the behavior of the transformation $T_\epsilon$ to first order in $\epsilon$. Physicists often call the transformation $T_1$ the infinitesimal generator of $T_\epsilon$.

Example. Infinitesimal rotations in 2D

Consider the following rotation of the 2D Euclidean plane: $$ T_\epsilon = \begin{pmatrix} \cos\epsilon& -\sin\epsilon\\ \sin\epsilon& \cos\epsilon\\ \end{pmatrix} $$ This transformation has all of the desired properties outlined above, and its infinitesimal version is $$ \widehat T_\epsilon = \begin{pmatrix} 1& 0\\ 0& 1\\ \end{pmatrix} + \begin{pmatrix} 0& -\epsilon\\ \epsilon& 0\\ \end{pmatrix} $$ If we act on a point in 2D with this infinitesimal transformation, then we get a good approximation to what the full rotation does for small values of $\epsilon$ because we have made a linear approximation. But independent of this statement, notice that the infinitesimal version of the transformation is rigorously defined.

Relation to Lie groups and Lie algebras.

Consider a Lie group $G$. This is essentially a group $G$ that can also be thought of as a smooth manifold in such a way that the group multiplication and inverse maps are also smooth. Each element of this group can be thought of as a transformation, and we can consider a smooth, one-parameter family of group elements $g_\epsilon$ with the property that $g_0 = \mathrm{id}$, the identity in the group. Then as above, we can define an infinitesimal version of this one-parameter family of transformations; $$ \widehat g_\epsilon = \mathrm{id} + \epsilon v $$ The coefficient $v$ of $\epsilon$ in this first order approximation is basically (this is exactly true for matrix Lie groups) an element of the Lie algebra of this Lie group. In other words, Lie algebra elements are infinitesimal generators of smooth, one-parameter families of Lie group elements that start at the identity of the group. For the rotation example above, the matrix $$ \begin{pmatrix} 0& -1\\ 1& 0\\ \end{pmatrix} $$ is therefore an element of the Lie algebra $\mathfrak{so}(2)$ of the Lie group $\mathrm{SO}(2)$ of rotations of the Euclidean plane. As it turns out, transformations associated with Lie groups are all over the place in physics (particularly in elementary particle physics and field theory), so studying these objects becomes very powerful.

Invariance of a lagrangian.

Suppose we have a Lagrangian $L(q,\dot q)$ defined on the space (tangent bundle of the configuration manfiold of a classical system) of generalized positions $q$ and velocities $\dot q$. Suppose further that we have a transformation $T_\epsilon$ defined on this space, then we say that the Lagrangian is invariant under this transformation provided $$ L(T_\epsilon(q,\dot q)) = L(q, \dot q) $$ The Lagrangian is said to be infinitesimally invariant under $T_\epsilon$ provided $$ L(T_\epsilon(q,\dot q)) = L(q, \dot q) + \mathcal O(\epsilon^2) $$ In other words, it is invariant to first order in $\epsilon$. As you can readily see, infinitesimal invariance is weaker than invariance.

Interestingly, only infinitesimal invariance of the lagrangian is required for certain results (most notably Noether's theorem) to hold. This is one reason why infinitesimal transformations, and therefore Lie groups and Lie algebras, are useful in physics.

Application: Noether's theorem.

Let a Lagrangian $L:\mathscr C\times\mathbb R$ be given where $\mathscr C$ is some sufficiently well-behaved space of paths on configuration space $Q$. Given a one-parameter family of transformations $T_\epsilon:\mathscr C\to\mathscr C$ starting at the identity. The first order change in the Lagrangian under this transformation is $$ \delta L(q,t) = \frac{\partial}{\partial\epsilon}L(T_\epsilon(q),t)\Big |_{\epsilon=0} $$ One (not the strongest) version of Noether's theorem says that if $L$ is local in $c$ and its first derivatives, namely if there is a function $\ell$ such that (in local coordinates on $Q$) $L(q,t) = \ell(q(t), \dot q(t), t)$ and if $$ \delta L(q,t) = 0 $$ for all $c\in\mathscr C$ that satisfy the equation of motion, namely if the Lagrangian exhibits infinitesimal invariance, then the quantity $$ G = \frac{\partial \ell}{\partial \dot q^i}\delta q^i, \qquad \delta q^i(t) = \frac{\partial}{\partial\epsilon}T_\epsilon(q)^i(t)\bigg|_{\epsilon=0} $$ is conserved along solutions to the equations of motion. The proof is a couple lines; just differentiate $G$ evaluated on a solution with respect to time and use the chain rule and the Euler-Lagrange equations to show it's zero.

general relativity - Boundary term in Einstein-Hilbert action

Why is the boundary term in the Einstein-Hilbert action, the Gibbons-Hawking-York term, generally "missing" in General Relativity courses, IMPORTANT from the variational viewpoint, geometrical setting and the needs of Black Hole Thermodynamics? Shouldn't it be also included in modern courses of General Relativity despite its global effect on the equations of motion is irrelevant (at least in the classical theory of relativistic gravity)?

astronomy - How did Halley calculate the distance to the Sun by measuring the transit of Venus?

What numbers did Halley, Cook, et al. have? What was the strategy by which they calculated the AU?

Answer

Halley's method requires one to measure the timing of the beginning of the transit and the end of the transit; both pieces of data have to be measured at two places of the Earth's globe whose locations must be known.

The picture by Vermeer, Duckysmokton, Ilia shows that the two places on Earth have differing locations in two different directions (the differences in the distance from the Sun and Venus are too small to be measurable): one of them is parallel to the direction of the transit of Venus and will be reflected in the overall shift of the timing; the other component is transverse to it and it will actually shift the line along which Venus moves and crosses the Sun in the up/down direction i.e. it will make the duration of the transit longer.

Each of these pieces of data – overall shift in the timing arising from one coordinate's difference between the two terrestrial locations – and the difference between the length of the transit – due to the other coordinate – are in principle enough to determine the solar parallax. Because synchronization of clocks at very different locations was difficult centuries ago, I suppose that the latter – the difference between $\Delta t_1$ and $\Delta t_2$ – was probably more useful historically. But we're talking about O(10) minutes differences in both quantities.

The calculation of the parallax from $\Delta t_1$ and $\Delta t_2$ and their difference is a simple exercise in geometry but I want to avoid trigonometric functions here.

At any rate, Halley didn't live to see a proper measurement (the transit occurs about twice a century and the two events are clumped together with a 8-year break in between). The best he could get was 45 angular seconds for the parallax; the right answer is about 8.8 seconds. He knew that his result was very inaccurate. Note that the solar parallax is the angle at which the Earth's radius is seen from the Sun, i.e. the difference in the rays needed to observe the Sun from the Earth's center and/or a point on the Earth disk's surface.

When you convert 8.8 angular seconds to radians, i.e. multiply by $1/3,600\times \pi/180$, you get $4.3\times 10^{-5}$. Now, divide 6378 km by this small number to get about 150 million km for the AU.

Some orders-of-magnitude estimates for the numbers. Venus orbits at 0.7 AU so it's actually closer to the Earth than it is to the Sun during the transit. It means that a shift by 6,000 km up/down on the Earth's side corresponds to about 12,000 km up/down on the Sun's side. So the two horizontal lines crossing the Sun on the picture (places on the solar surface where Venus gets "projected") may be separated by about 12,000 km. Compare it with the solar radius near 700,000 km: you may see that we're shifting the horizontal lines by about 1% of the Sun's radius and the relative difference between $\Delta t_1$ and $\Delta t_2$ will be comparable to 1%, too. The last transit in 2004 took about 6 hours so the difference in the duration at various places is of order 10 minutes.

The 2012 transit of Venus on Tuesday night UTC will take over 6 hours, too; the timing and duration differs by about 7 minutes depending on the location, too.

If you've been dreaming about observing the transit of Venus, don't forget about Tuesday 22:49 night UTC; the following transits will occur in 2117 and 2125. There is a blog version of this answer, too.

charge - Why do the free electrons in N-type want to diffuse?

I'm trying to understand how a diode works and for this I've used(among other resources) the book written by Albert Malvino, Electronic Principles.

Everywhere I read about this topic, it says that when the N-type and P-type semiconductors are joined together, the free electrons from the N-type diffuse to the P-type.

I don't understand why is the diffusion happening.

The book Electronic Principles contains an attempt to explain this and states that the electrons diffuse because they have the same charge and they repel each other, but in my understanding the P-type and the N-type semiconductors have a neutral charge, because the number of positive charges (protons in the nuclei) is equal to the number of negative charges (free electrons and covalent bonds electrons), so the electrons repeling each other can't actually be the cause of the diffusion.

Answer

The answer lies in a thermodynamic argument. The diffusion is a spontaneous process that occurs when particles with random motions are not uniformly distributed. The electrons in the conduction band in the n region are much more numerous that in the p junction. They will naturally tend to balance the concentration from the n junction to the p junction since the equilibrium tends to be the state with maximum disorder ie maximum entropy.

A good visual picture of how the diffusion process occurs is in this video at 4'50. https://www.youtube.com/watch?v=JBtEckh3L9Q

So the answer to what causes diffusion is thermal motion of quasi free electrons.

optics - Why does the refractive index depend on wavelength?

Why do different wavelength get impeded more or less when in different materials? Moving with the same speed, but a longer physical distance would imply that the fields oscillate less times in the material, but I don't know why a difference in the number of oscillations would impede the wave- I don't even know why things slow down in general. Why some electromagnetic wave would slow down just because it's entering other electromagnetic fields... It would seem to me that the only factor would be time taken to physically move some electron or something in the direction of the fields... But that seems to simple of an explanation to me.

optics - Optical absorption -- what are the common ranges and mechanisms?

So let's say you do some reflection/transmission spectroscopy of a material. It's clear that it's absorbing in some range.

What would be your first step in identifying the source of the absorption? The two main clues I would look at would be

1. The range it occurs in (if your incident light is in the x-ray range, it's clearly not phonon absorption, for example)


2. The shape of the absorption (is it a peak? Or an absorption 'edge', where it isn't absorbing below some energy, and absorbs everything above, for example)

But I'd like to know if there are other ways as well.

Additionally, does anyone know of a 'map' of the ranges you'll typically find different types of absorption in? I saw one on wikipedia once, but it only had three types, and now I can't find it of course. For example, a few types of absorption I can think of off the top of my head are

• Interband electron transitions


• Intersubband electron transitions


• Impurities


• Phonons



• Vibrational modes


• Atomic transitions

I understand that a lot of these will be material-dependent, but often the same mechanism in different materials will yield absorption in the same order of magnitude range. What ranges would one typically see these (and other) absorption mechanisms in?

Thank you!

pattern - Hard questions from an IQ test

I have edited the post to reduce the number of questions (I only kept the ones that were already answered) and I am creating a new post

The questions are from the test taken from the website: http://www.iq-brain.com/example/1

1:

2:

5:

Answer

2.

Bottom left. When the first two shapes are superimposed, regions that are only occupied by one shape are red, by both are white.

5.

Bottom left. Purple\Blue is widen, turn tip black. Green is rotate, Red is turn body yellow.

special relativity - Relevance of spinor in relativistic (classical) electrodynamics

I'm following a course about relativity and electrodynamics (not the "quantum" one), and the lecture notes introduces the concept of spinor by a map between an orthogonal basis in Minkowski spacetime and the pauli matrices with the identical matrix as the temporal part, like

$$ T: M^4 \rightarrow H(2,C) $$ $$ T(e_0) = I,\,\,\,\, T(e_i) = \sigma_i, i = 1,2,3 $$

It introduces some formulas like the direct mapping and its inverse as $$ v_i = \frac{1}{2}Tr(\hat{v}\sigma_i) $$ and the norm as the determinant of the resulting hermitian matrix. It explain that SL(2,C) is a double covering of the Lorentz group. And that's all. I mean, the notes don't speak about the spinor ever again.

So I wonder, why are they introduced in a course about classical relativistic electrodynamics? can some phenomena being explained using spinors that can't be explained using old good 4-vectors? or should I take it as a mathematical curiosity until I get introduced to quantum electrodynamics and dirac spinor (which I know is the direct sum of a Weyl left and right spinor)?

Answer

Transformations of the Faraday tensor by the spinor map and Lorentz transformation calculations are often less messy than their $S0(1,\,3)$ counterparts.

Even the most everyday and mundane EM calculations (without needing to do Lorentz transformations) can benefit from this: I have in my time written many lines of code to simulate electromagnetic propagation, and I use the following representation almost exclusively in my codes: code becomes VASTLY less complicated and much more readable - an more importantly maintainable - in this notation (you do need to be writing in an OO language (to allow tensor, quaternion, matrix etc classes) with operator overloading implemented for these comments to hold true. C++, Ada and I believe Python and the modernest version of Fortran also qualify (I only have direct experience of the former two)). There is also a modest code speedup for some spinor operations, but that's becoming less important these days.

We can write the Faraday tensor in the space of $2\times2$ Hermitian matrices as follows:

$$\begin{array}{lcl}F_\pm &=& \left(\begin{array}{cc}E_z & E_x - i E_y\\E_x + i E_y & -E_z\end{array}\right) \pm i \,c\,\left(\begin{array}{cc}B_z & B_x - i B_y\\B_x + i B_y & -B_z\end{array}\right)\\ & =& E_x \sigma_1 + E_y \sigma_2+E_z\sigma_3 + i\,c\,\left(B_x \sigma_1 + B_y \sigma_2+B_z\sigma_3\right)\end{array}\tag 1$$

where one often keeps both the $+$ and $-$ form but one throws away the negative frequency part. In this notation, $F_+$ is the left hand circularly polarized part of the field and $F_-$ the right.

The Pauli spin matrices are simply Hamilton's imaginary quaternion units reordered and where $i=\sigma_1\,\sigma_2\,\sigma_3$ so that $i^2 = -1$. When inertial reference frames are shifted by a proper Lorentz transformation:

$$\Lambda = \exp\left(\frac{1}{2}W\right)\tag 2$$

where:

$$W = \left(\eta^1 + i\theta \chi^1\right) \sigma_1 + \left(\eta^2 + i\theta \chi^2\right) \sigma_2 + \left(\eta^3 + i\theta \chi^3\right) \sigma_3\tag3$$

encodes the transformation's rotation angle $\theta$, the direction cosines of $\chi^j$ of its rotation axes and its rapidities $\eta^j$, the entities $F_\pm$ undergo the spinor map:

$$F_\pm \mapsto \Lambda \,F_\pm \Lambda^\dagger\tag 4$$

We can write Maxwell's equations in their quaternion form:

$$\begin{array}{lcl} \left(c^{-1}\partial_t + \sigma_1 \partial_x + \sigma_2 \partial_y + \sigma_3 \partial_z\right) \,F_+ &=& 0\\ \left(c^{-1}\partial_t - \sigma_1 \partial_x - \sigma_2 \partial_y - \sigma_3 \partial_z\right) \,F_- &=& 0\end{array}\tag 5$$

which are actually the Dirac equation (when Weyl spinors instead of Dirac ones are used) for a massless particle.

See my answer here for some more information on the physical meanings of the above.

---

Question from OP

...I would want to know what is the relation (mapping?) between the Faraday tensor in Minkowski and this one..

They are identical, in the senses that:

1. 
   

   If you solve (5) with appropriate boundary conditions and other forms of Maxwell's equations with the same boundary conditions, then the field components you read off (1) as functions of time and space will be exactly the same answers that you get from any other method of solution;

   
   


2. 
   

   If you transform between relatively boosted and rotated frames using (4), you'll get exactly the same answer that you get by acting on the left and right of the wonted form of the Faraday tensor with the relevant Lorentz transformation written in the matrix representation of $SO(1,\,3)$. Witness that the form of (4) looks exactly like the transformation of the rank 2 tensor (insofar that you need two Lorentz matrices for that one too).

   
   
   

The "differences" are:

1. 
   

   The difference between contravariant and covariant forms of $F$ are hard to spot and subtle in this notation. Multiplying the rank two Faraday tensor by the Minkowski metric $\eta$, i.e. raising or lowering indices is, by (1), taking the negative Hermitian conjugate of the $F_\pm$ quantities.

   
   


2. 
   

   Two boosts, both $+\Lambda\in SL(2,\,\mathbb{C})$ and $-\Lambda\in SL(2,\,\mathbb{C})$ in the form (2) have the same action as a single boost in $SO(1,\,2)$ because $SL(2,\,\mathbb{C})$ is the simply connected double cover of $SO(1,\,3)$ - the two groups are identical in the small, but have different global topology.

   
   

gravity - Does light have mass?

1. 
   

   Does light have mass?

   
   


2. 
   

   If yes, will it exert force?

   
   
   


3. 
   

   If no, then how are light particles are travelling at light speed?

   
   


4. 
   

   If light doesn't have mass how is it attracted by gravitational force (black holes)?

   
   


5. 
   

   And what actually is light?

   
   

Answer

Photons having zero rest mass is precisely the reason why they travel at light speed. A particle can have momentum without having mass, the only condition is $$(mc^2)^2=E^2-(pc)^2$$ in the case of light, left side is zero and $E=pc$ where $E$ is energy and $p$ is linear momentum.

Furthermore, the gravitational attraction isn't really attraction in the traditional sense. Gravitation curves the space-time so that the geodesic (the locally straightest and shortest path, which the light takes) is curved. So the light just goes along its way, as straight as it can get, and the curvature of the space makes this geodesic path deflect in angle when the ray passes a massive object.

Light is an electromagnetic wave - change in electric field induces magnetic field, and change in magnetic field induces electric field again... this cycle propagates in space with the speed of light.

newtonian mechanics - Coriolis force and Newton's third law

I would like to ask a stupid question here.

If a body 'b' moving downward with a velocity $v$ in a rotating frame of reference with angular velocity $\omega$, and $\omega$ and $v$ not being parallel and anti parallel. We all know that the body 'b' experiences a Coriolis force.

If it gets deflected in its trajectory then according to newtons third law shouldn't it push some body with equal force?

what if the frame of reference was moon where there is no atmosphere? will the body be deflected?

Answer

Nothing feels a reaction force to the Coriolis force because Newton's laws only apply in an inertial frame. If we imagine a frame with some sort of wild, random acceleration, then in that frame at a given moment it would look like, in a room with completely stationary tables and chairs and things, everything was simultaneously accelerating the same direction, and there would be huge "fictitious forces" needed to explain this. There are no third-law pairs going on, though. The fictitious forces were necessary to make $F=ma$ work out in this accelerating frame, but by introducing them we sacrificed the third law (at least in regards to those kinematic forces).

Imagine dropping a ball off a tower. It gets deflected east.

If we analyze this scenario from a frame rotating along with the Earth, there is no conflict with Newton's third law because it does not hold for the Coriolis force or other kinematic forces.

If we analyze the scenario from an inertial frame in which the Earth is rotating, the trajectory simply is not deflected. The ball lands east of where it was dropped because its velocity in a tangential direction, which comes from the rotation of the Earth, stayed the same, but as it fell its angular velocity $v/r$ increased because $r$ decreased. That led it to "deflect east" when it was really moving in a perfectly ordinary parabola.

nuclear physics - The "binding energy" of bonded particles adds mass?

This is a follow-up my previous question. Several of the answers indicated that the mass of a particle (atom, proton, etc.) increase with the "binding energy" of it's component particles - the energy needed to split them apart - which doesn't make sense to me. I'm convinced that, at least in an exothermic chemical reaction (where the product bond energies are larger) the product's particles will lose mass (proportionally to the heat dissipated) or at least have no change.

To use a larger-scale analogy, if an object, a "particle", is 100m above the Earth's surface, it has potential energy from gravity. As it falls, this energy is lost, converted into KE. Overall, the two "particles", the object and the Earth, end up with the total energy, and therefore the same total mass. There is no room for a "binding energy" that adds mass.

My reasoning is that this extends to particles, with electrostatic or nuclear forces taking the place of gravity. Potential energy of component particles becomes KE of the bonded particle, they end up with the same mass. Actually, if this KE is dissipated (as in a burning / nuclear fusion reaction) the particles should actually have more mass in their uncombined/unreacted state, thanks to their PE. Surely it isn't possible for the mass to increase without an external input of energy?

However, answerers of my energy in chemical reactions question said that:

the energy involved in the bonds is ... half of what we normally consider the "mass" of the proton - David Zaslavsky

and

potential energy of the chemical bonds do correspond to an increase of mass - Ben Hocking

So, how can this be, and where is my reasoning incorrect? What exactly is the binding energy (if not just the energy needed to break the bond), and where does it come from?

particle physics - Why are high energies equivalent to short distances?

Why are the regimes in high-energy collisions called short distances?

Qualitatively I only know that the higher the given energy of the colliding particles "the smaller are the pieces yielded by cracking the collision particles". Is there a quantitative derivation of the term relating the energy $E$ and the distance $x$?

Answer

Energy is related to frequency via Planck's relationship $E=h\nu$ and frequency is related to wavelength via the speed of light $\nu=c/\lambda$. Therefore, the higher the energy the smaller the associated wavelength. This wavelength represents the resolution with which one makes an observation.

hilbert space - Is non-linear quantum mechanics possible?

Say we have a state vector $|A\rangle$. Is it possible to have a theory where the evolution of $|A\rangle$ depends on the vector $|A\rangle$ itself? e.g.

$$ i\frac{\partial}{\partial t} \psi(t) = \hat{F}(\psi(t)) \psi(t) $$

On reason I was thinking about this was the idea that space-time is related to entanglement. But entanglement means knowledge of the state-vector. So if this were true, the state-vector would have to affect itself.

In other words is non-linear quantum mechanics possible?

word - Teapot Riddle no.44

Teapot riddle no.40:

Rules
I have one word which has several (2 or more) meanings
Each of the meanings is a teapot (first, second ...)
Try to figure out the word with my hints.

First Hint:

My first teapot is a group of people using my second teapot
My second teapot is a system using my third teapot
My third teapot is using force

Second Hint:

My first teapot shouldn't be dirty
My second teapot is partly dirty
My third teapot can be dirty in sports

Third Hint:

My first teapot reproduces everything
My second teapot reproduces data

My third teapot doesn't reproduce

Good luck and have fun :)
last riddle

Answer

Another try again... with the help of JGibbers!

May it be...

.. press? The press (a group of journalists), the press (printing machine), the press (urgency)

My first teapot is a group of people using my second teapot

untill here, it matches

My second teapot is a system using my third teapot

they are always under pressure because news must remain ... news :)

My third teapot is using force

Hmm.... first flaw, or I don't know how to formulate such as it eventually matches

My first teapot can be from clean to dirty

Unfortunately!

My second teapot can be dirty

Most of the time, it is inky!

My third teapot can be dirty in sports

Sounds right, though I am not sure one does say it that way...

My first teapot reproduces everything

They do...

My second teapot reproduces data

It does....

My third teapot doesn't reproduce

Not in a way I understand...

mass - What would happen to a teaspoon of neutron star material if released on Earth?

I've read on NASA's page on neutron star that one teaspoonful of that star would weigh over 20 billion tonnes on Earth. If it was somehow possible to bring it to earth would it:

1. 
   

   Burn and disappear during Earth atmosphere entry?

   
   
   


2. 
   

   Assuming we have 20 billion tonnes of mass occupying the volume of a teaspoon here on Earth, would it fall through the ground under its own weight?

   
   

Answer

The reason that the density is so high is because the pressures are so immense. If we somehow teleported a teaspoonful of neutron star material to earth, it would very rapidly inflate because the pressures aren't high enough to crush it into its dense form. This would effectively be an enormous explosion.

It is difficult to describe what it would inflate out into - the neutron star material can be imagined as an incredibly dense soup of neutrons with some protons and leptons in small numbers. The protons and leptons would make neutron-rich elements like deuterium, but most of the matter would consist of free neutrons. These free neutrons would undergo beta decay to produce neutrinos, protons, and electrons, which would likely recombine to make very large amounts of hydrogen, some helium, and a few heavier atoms. In all of these cases, the atoms would be neutron-rich isotopes, though.

The behavior would look most like a very rapidly expanding gas. It would explode with such force that it wouldn't even need to "fall through the ground" - it would obliterate the floor entirely.

condensed matter - How does a phonon cause two electrons to attract each other and form a cooper pair?

We know that like charges repel each other. But my professor claimed that two electrons can attract each other as well. What he said was that due to screening an electron travelling at some speed won't repel another electron, but that they will, in some cases, attract each other due to weak phonon exchange. What does that mean? What is phonon exchange? Do two electrons really attract each other?

Answer

From the Wiki article Cooper pair:

In condensed matter physics, a Cooper pair or BCS pair is two electrons (or other fermions) that are bound together at low temperatures in a certain manner first described in 1956 by American physicist Leon Cooper. Cooper showed that an arbitrarily small attraction between electrons in a metal can cause a paired state of electrons to have a lower energy than the Fermi energy, which implies that the pair is bound. In conventional superconductors, this attraction is due to the electron–phonon interaction. The Cooper pair state is responsible for superconductivity, as described in the BCS theory developed by John Bardeen, Leon Cooper, and John Schrieffer for which they shared the 1972 Nobel Prize.

Although Cooper pairing is a quantum effect, the reason for the pairing can be seen from a simplified classical explanation. An electron in a metal normally behaves as a free particle. The electron is repelled from other electrons due to their negative charge, but it also attracts the positive ions that make up the rigid lattice of the metal. This attraction distorts the ion lattice, moving the ions slightly toward the electron, increasing the positive charge density of the lattice in the vicinity. This positive charge can attract other electrons. At long distances this attraction between electrons due to the displaced ions can overcome the electrons' repulsion due to their negative charge, and cause them to pair up. The rigorous quantum mechanical explanation shows that the effect is due to electron–phonon interactions.

mathematics - The barge ladder

You are the captain of a large barge and need to cross a canal in order to transport your barge from the sea to a lake above sea-level.

As you probably know, when you enter the lock, water is added in order to raise the water level until the intended above-sea level is reached.

As you know, this takes a long time.

Really bored, you start to do some math (as all bored people do):

Your barge is $9m\times3m$ long and $3m$ high, with $1m$ submerged in water.

A $1.75m$ long ladder is hanging at one side of the barge, and is used for climbing in the barge.

The lock is a $10m\times10m$ square and it's filled at a rate of $500$ liters per second.

How much time does the water take to reach the hanging ladder?

Answer

Infinity and beyond - the water will never reach the ladder.

Why?

Because your barge rises with the water level. Thus no matter how much the water rises the distance from the ladder to the water (the "height" of the ladder) will not change.

lateral thinking - Why didn't they arrest Bob?

Bob smashes the taillights of a stranger's car. But the cops arrest the car's owner, not Bob.

Why didn't they arrest Bob?

Answer

Because

He was being kidnapped

So

He smashes the lights while he is inside the boot to either alert someone or to try and make the cops pull the car over. They find Bob and the cops arrest the kidnapper.

reference frames - Instantaneous axis of rotation

If there is sphere rolling on the horizontal surface without slipping. Then we know the instantaneous axis of rotation is the point which is in contact with surface. Now my doubt is, why the angular velocity taken from centre of sphere and IAOR is same ?

Please someone help me.

definition - What is a conservative force?

Currently I have three different pictures to describe/understand conservative forces. For the moment I just want to get an electron from point A to point B. In the near surrounding is another electron and thus it gets complicated, they repel.

1. The first is like my "standard" imagination of forces. You make things move and as you move them, you might feel some sort of resistance, which pushes in the other direction. That means you have to apply constantly a force and constantly submit energy to keep the whole process alive. Well, it is a conservative force, so you are allowed to choose an arbitrary path as long as A and B stay.

However this confused me... because you could take paths that go through A several times before one reaches B, or you also just encircle B... In the end I have to imagine imaginary or negative energy to justify it. That seems wrong.

2. 
   

   I imagine the path like given mountain railway or a slide, the path for the electron is already predetermined and just push the electron and then it moves. And because there isn't really a different path, it is forced to move on that railway or whatever. Then I would imagine the force as an one time event, it also has the same flaw as the standard picture I have. I don't see how the given path cancel itself out. I might have to revise my imagination/picture.. but then, what is a force?

   
   


3. 
   

   My 3th imagination is just like a description of F(B) - F(A). Then I can explain it well, without problem. However then I don't understand for example the concept of force and force with the opposite direction..

   
   

Has someone an idea, how to help me with my problem? Maybe I just need more time.. I will check later this day for replies and maybe add to my thoughts, if needed.