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confusion on quantum field theory
Are field quanta infinite in extent as stated in Art Hobsons paper? What does this even mean? I've not seen any electrons or atoms that are infinite in size?
Am I getting something wrong somewhere?
My younger brother came home from school today and told us at the dinner table that when the sun burns out the Earth could be ejected from its orbit. Skeptical, I asked his source. He quoted his teacher and 'scientists'.
I've since googled if such a thing could happen without much luck - I haven't been able to locate or identify those scientists. Using my own rather rudimentary understanding of physics I don't believe this could happen. If we believe this source then the sun will have the majority of its mass when it burns out, leaving the gravitational effects largely unchanged (or surely not enough to eject our planet).
My question is - is there a way that the earth could conceivably be ejected from the solar system when the sun burns out?
Answer
There doesn't seem to be any large possibility of the earth wandering space unattached to (the remnants) of the Sun.
Scientific American covered Earth's long-term fate in 2008
Although scientists agree on the sun’s future, they disagree about what will happen to Earth. Since 1924, when British mathematician James Jeans first considered Earth’s fate during the sun’s red giant phase, a bevy of scientists have reached oscillating conclusions. In some scenarios, our planet escapes vaporization; in the latest analyses, however, it does not.
The answer is not straightforward, because although the sun will expand beyond Earth’s orbit, or one astronomical unit (AU), it will lose mass along the way. As a result, Earth should drift outward as the gravitational tug lessens over time. (At its maximum radius of 1.2 AU, the sun will have lost about one third of its mass, compared with its current heft.) In this way, Earth could escape solar envelopment.
That's escape envelopment, not escape orbit.
According to space.com
Collisions between planets might conceivably lead to the ejection of Earth from orbit
And as the sun ages, it is expected to swell and lose mass; previous studies have shown that could have significant effects on the planets in the next 7 billion years or so. Earth might be vaporized when this happens, or it might ? with a gravitational assist from a passing star ? be booted right out of the solar system. A study in 2001 by Laughlin, then at NASA, and Fred Adams of the University of Michigan put the odds of the Earth being ejected at one-in-100,000.
Fold this net into a cube and solve this 3D Masyu.
The original Masyu rules apply. In addition, question mark '?' should be replaced by either a black or white circle.
Side note: Yeah, it's almost a die... I'm adding '?' to make sure that the solution is unique. ><
Addendum: Move '?' from the center of side $4$ to the center of side $2$.
Answer
By adding a
white circle
we get
My question has a few parts concerning the speed of light in general relativity.
Firstly, time changes in response to gravity and speed. Therefore, as gravity effects time in an area of space, should the speed of light "change?" When I say "change" I understand that light's speed is finite but wouldn't the length of second change and then change light's speed in meters per second?
Secondly, does the speed of light actually change with gravity? Time stops at the speed of light but does that mean light cannot experience time? Can light experience slower time while travelling through warped space time?
Answer
The answer to your questions is very nuanced for the most part. I'll start with the easy answers: Light does not experience time, neither does it experience no time, the most accurate statement I could probably make off-hand is that it experiences null time. Null time does not mean time has stopped, that would be zero time; null time means null time, null is not a number (isNaN(null)==True). Photons have no frame of rest (inertial or otherwise), as such it is impossible to discuss the time experienced by them because in order to do so you have to imagine a frame where they are at rest (which is impossible). So I can't say that light experiences slower time when travelling through warped spacetime.
But as for your first question, that is more complicated. In the comments (now deleted) I said it might be best for a black hole person to answer this than a GR person. The reason I said that is because black hole people probably answer this question so much that they've got great wording. Now I stand by my opinions sometimes and this is one of those times. So I went and looked up a great answer by a black hole person that is much better than anything I could have come up with. But link-only answers aren't fun, so here's the TL;DR version:
Special relativity does not say the speed of light is always constant (even though, as I'll get to, you can cheat such that it is). It says the speed of light is always constant when measured locally from an inertial frame of reference. An inertial frame of reference is a non-accelerating frame. The equivalence principle says that an accelerating frame is locally indistinguishable from a gravitational field. So that gives some room for the speed of light to not be measured consistently.
If you are in free-fall, your acceleration due to gravity cancels out the gravitational field (equivalent to accelerating the other direction) such that you are now in an inertial frame. Thus, if you measure the speed of light where you are, I guarantee you will get $c$. However, as stated, this is only valid locally. Say I'm free.....free-falling (couldn't help myself, sorry) and you are somewhere else and I measure the speed of light where you are, I'll get a different value than $c$ because it's not local any more and the gravitational field still changes the passage of time around you.
If you are in a gravity field and not free-falling, then you are not in an inertial frame and so will (probably) not measure the speed of light as exactly $c$, not even locally (What is this? Sorry folks, that must have been a brain tumor on my part or something. Obviously it's constant locally, I must have been thinking about just beyond local. Like if you were measuring from $10km$ up the speed from a source on the ground, which is non-local. Wow, what magic sauce was I smoking when I wrote that? Why did nobody call this out until now?). On Earth, the gravity field is small so the effect of this makes the difference from $c$ extremely small. Near a black hole, this effect becomes pronounced and very noticeable (thus, black hole people are more accustomed to answering questions about it).
However, there is a way to cheat to make the speed of light always appear constant even non-locally. Your measurement of the speed of light changes because your measurements of length and time change separately. You can get around this by measuring the speed of light with reference to measurements of length and time that are also changing. In the link I provided, the author uses lunar orbits and Earth days as his length and time references. Even though you wouldn't measure these as the same number of kilometers or seconds in every frame, every observer in every frame will agree that they measure the speed of light as about $12000$ Lunar Orbits/Earth Day. Why is this cheating? It's pretty much like saying the speed of light is 1 lightsecond/second or 1 lightyear/year. But so long as you can measure the length of the Lunar orbit and the duration of an Earth day, you'll always have a constant number for the speed of light (assuming proper mathematical approximations are made).
If its true that energy isn't lost, its just transferred, where did the energy go from a falling object that hits the floor and stays there? It started with the most gravitational potential energy, a second before it lands it is converted to kinetic energy, and the kinetic is at the peak. But after it fell and stopped, where is all the energy?
Also, it is said you can't create energy. But when an object is at rest on the floor it has no energy, and after being picked up it has energy.
I know these are basic questions, but I can't get to the bottom of this.
Kohmoto from TKNN(Thouless-Kohmoto-Nightingale-deNijs) who described the topology of the integer quantum hall effect always stressed the importance of the 2D Brillouin zone being a donut due to periodic boundary conditions.
--> http://www.sciencedirect.com/science/article/pii/0003491685901484
Now I don't really see why this is relevant. Shouldn't the zeros of the wavefunction always lead to a Berry phase?. What would happen if we have a sphere instead of a donut? I think I am missing a major point here, because I can't see how the Gauss-Bonnet theorem that connects topology to geometry plays a role. For a sphere with no holes the Gaussian curvature gives us 4$\pi$, for a donut if gives us 0. In both cases Stokes' theorem should still give us a nonzero value?
Charles Kane then uses this argument to compare a donut with the quantum hall state and a sphere with an insulator. He then writes down the Gauss-Bonnet theorem and immediately talks about topological insulators, and again I don't see the connection or is it just an analogy and I shouldn't waste any time on this?
If there is a connection, I would like to know if there is a simple explanation for using the Gauss-Bonnet theorem in the context of topological insulators. I'm even more confused because Xiao-Gang-Wen said in a recent post here, that a topological insulator is NOT due to topology but due to symmetries...
We say that electrons are probability waves. Then if two or more elements form bonds by overlapping of orbitals then what happens to the wave function of the electrons? Is it possible that during bonding, these electrons still behave as waves and interfere?
Select the correct choice from the bottom 6 options to fill the final box above the horizontal line.
I stumbled upon this seemingly difficult question while doing a Mensa practice test. I have since been trying to figure it out but I can't. I am calling it seemingly difficult, because these questions always appear to be difficult, until you come up or are provided with, their simple and elegant answer. I hope someone can help me!
Answer
It is
the bottom right one.
Because
In each column, flip the top image over the horizontal axis. Overlay the second image; where both images have lines, erase the line. You end up with the third image.
I'm studying the moduli space of vacua for some supersymmetric gauge theory and I want to know specifically why it is important to know the geometry of this space. I know everything about the division between Higgs branch and Coulomb branch for a gauge theory, I know that the moduli space is a Kahler manifold and I know you can see the moduli space as a complex algebraic variety (or better affine variety). What I would like to better understand is what is the physical reason that brings us to think to the geometry of the moduli space? Why don't we just try to find out the properties of the moduli space of vacua just looking at a Lagrangian density? Is it then if a theory has no Lagrangian you can understand it anyway just looking at the geometry of the space?
Where can I find some good review or book where is explicitly written the importance of the geometry of the moduli space in order to understand the physics of a supersymmetric gauge theory?
Here I am again with a fifth See me once riddle. Have fun finding the answer!
see me once, I'm taped together as a twist.
see me twice, used by both chef and geologist.see me once, icky sticky but everywhere in use.
see me twice, my subjects I like to abuse.
Although the answer has already been found, here is a hint. (For the people reusing this riddle. Answers here do not have to include this hint - but feel free if you want to):
see me once, some people like to inhale
see me twice, eat me fresh or else I go stale
Here are the previous riddles in this series (the solutions there have nothing to do with this one, only the process of getting there).
#1, #2, #3, #4
Answer
I think you are:
Tar
see me once, I'm taped together as a twist.
"Tar Tape" used around the handle of a baseball bat.
see me twice, used by both chef and geologist.
Tartar sauce and Cream of Tartar
see me once, icky sticky but everywhere in use.
Used on roads everywhere
see me twice, my subjects I like to abuse.
tartar : a person of irritable or violent temper.
What's so not-quantum about tree-level diagrams?
Answer
The reasons were given here. Essentially, at tree level you recover classical results. Loop corrections are proportional to powers of $\hbar$ and these are quantum terms.
What is color?
“The property possessed by an object of producing different sensations on the eye as a result of the way the object reflects or emits light”…. You might say.
So take a white object and paint it “red”. What property of the paint now makes the object appear red?
“Pigmentation!” … you might say.
Good. Now that I’ve got you thinking in line with what I’m thinking, please help me understand what pigmentation actually is. I’m looking for an explanation that goes down to the microscopic level. What is it about the elemental structure of different pigmentation that alters the way that pigmentation (color) is seen by our eyes when applied over an object?
Answer
OK the answer is not microscopic level but beyond that. It is in atomic level.
These are the complex and detailed answers to the question
According to my teacher, an electron is point sized and it does not absorb or release energy. Moreover, my teacher says their orbital absorbs energy rather than the electron. If that is the case, then what about the photoelectric effect, in which electrons release energy after excitement?
How can it be explained using the laws of refraction that a light through optical centre of a lens passes undeviated?
If we assume the portion of the lens in the middle to be made of even number of alternately place up and down prisms, then it's clear, but why can the number of prisms not be odd?
Answer
Look at the lens as a slab of finite thickness. In a small vicinity around its center we may consider its opposing surfaces as parallel to each other. A ray incident on this area, at any angle, will be refracted twice at parallel interfaces. Therefore it will emerge on the other side of the lens on a direction slightly displaced, but parallel to the incident one. Since the parallel displacement is tiny as long as the lens is not too thick, it looks like the ray passes through the center undeviated.
Considering that light is in the 400-800 THz range, if you had an electrical oscillator that ran at that frequency connected to an aerial of some sort, would the antenna emit visible light, in the same manner that radio waves emit radio "light"?
We have specialized transistors that can run at many hundreds of gigahertz, and even some that work up to the few-terahertz regime. I can fairly easily see further advances perhaps producing oscillators at a hundred THz, which would be well into the infrared range, if not visible light.
Note: My grandma is a little odd. She only likes certain things. Solve this puzzle by providing phrases that are true. For example: Grandma likes A but not B
(Puzzle Credit: My friend Derek)
Answer
Grandma likes
Flowers but not plants
Fingers but not toes
Fridges but not magnets
In but not out
Pancakes but not sweets
Paper clips but not staples
Why?
Because grandma does not like Tea!
If I have two vectors $\vec{b}$ and $\vec{v}$, and I know that $$ \vec{b} \times \vec{v} = \vec{c} $$ and $$ \vec{b}\cdot\vec{v} = \lambda $$ can I find the $\vec{v}$ vector in terms of the $\vec{c}$ vector, $\vec{b}$ vector, and $\lambda$? I have been struggling with for quite sometime. And I hope it's time I asked for help.
Is it possible? How can I explain the randomness of spins in the paramagnetic phase with chaos theory? In this case, is the randomness apparent?
If yes, I think the temperature would be a reasonable parameter change that is mentioned in chaotic systems, which changing the temperature leads to phase transition and consequently making it disordered. I had a brief discussion with a professor of mine regarding this, and he mentioned that temperature can't be the parameter and the reason was something like "temperature is also random or too many possible states for a specific temperature." (He was speaking in another language that I'm not so good at.)
And can the Ising model be considered a nonlinear system? How do I find the time-evolution equation of the Ising model?
Are there any experts or physicists in this field who give some comments on these? I'm quite interested in phase transitions as well as chaos theory. I'm hoping to bridge and link these two fields in my mind.
I know that the difference between two musical notes is given by the sound frequency, and the difference in volume is given by the amplitude.
What I am wondering is why does the same note sound different on different musical instruments? What in the wave makes the difference between the sound of a harmonica and the sound of a violin singing the same note?
Answer
The different tonality of a note in different instruments stems from the different mixes of amplitudes in the harmonic frequencies that the instrument provides.
To be more concrete (and keeping to a slightly simplified view), you play the A note (440 Hz) and then you have the harmonic frequencies 880, 1320, 1760, ... ($440n$ where $n$ is the number of the harmonic). Each of the frequencies will have an amplitude or "volume" contributing to the sound produced by an instrument. Thus a particular set of amplitudes gives the instrument its tonality.
This is highly related to the concept of Fourier analysis used in many areas of physics.
I'm currently working on my thesis and I'm stuck on a question. I'm designing activity stations for disabled children to be used for equine therapy.
The stand is 9ft tall and I've calculated the the wind load at 0.94lbs, now I need to calculate the mass required to stop the station falling over (the concrete base is not a foundation sunk in the ground - it needs to remain portable)
The frontal area of the stand is evenly spread. Any help would be appreciated.
Answer
Assuming a square base of width $w$ with mass $M$, and a horizontal wind load $F_w$ at a height $h$, then the condition for static equilibrium is
$$\sum\tau = 0$$ $$\tau_{wind} + \tau_{base} = 0$$
Since if the sign tips, it would rotate around the edge of the base, that's a convenient axis about which to compute the torques: $$- F_w \frac{h}{2} + M g \frac{w}{2} = 0$$
Solving for M...
$$M = \frac{F_w h}{g w}$$
Plugging in your numbers (0.94 lbs-force = 4.18 N, 9 ft = 2.7 meters),
$$M = \frac{11.29 ~\rm Nm}{9.81 ~\rm{m/s^2} ~w} $$ $$M = \frac{1.15 ~\rm kg~m}{w} $$
If you prefer lbs and feet,
$$M = \frac{8.32 ~\rm lbs~ft}{w}$$
For example, if your square base has a width of 3 feet, then you need a minimum of $M = 2.77 ~\rm lbs$
Note 1: This analysis assumes the mass of the sign itself is negligible compared to the base. This is a conservative assumption, since extra mass on the sign will make it more stable for initial tipping.
Note 2: I'm very skeptical that a sign of any significant size would only experience a wind load of 0.94 lbs in any significant wind. I would double-check that figure.
EDIT: I revised my answer now that the OP made it clear that the sign is a rectangle that extends from the ground up to 9 feet.
As I understand it, the axion $a$ originates from the spontaenous symmetry breaking of $U(1)_{PQ}$. This symmetry being anomalous, and because of the QCD vacuum structure, a non vanishing term like $\frac{a}{f_a}Tr( G \tilde{G})$ is included in the Lagrangian, where $G$ is the gluon field strenght. This determines the axion couplings to gluons. Talking about a coupling to photons would mean to consider a term like $\frac{a}{f_a} F \tilde{F}$, where $F$ is the QED field strenght. I thought a term like $ F \tilde{F}$ could be expressed as a vanishing total derivative, unlike $Tr( G \tilde{G})$, so why are we talking about axion couplings to photons ?
Frequently I see the expression $$F = \frac{dp}{dt} = \frac{d}{dt}(mv) = \frac{dm}{dt}v + ma,$$ which can be applied to variable mass systems.
But I'm wondering if this derivation is correct, because the second law of Newton is formulated for point masses.
Furthermore if i change the inertial frame of reference, only $v$ on the right side of the formula $F = \frac{dm}{dt}v+ma$ will change, meaning that $F$ would be dependent of the frame of reference, which (according to me) can't be true.
I realize there exists a formula for varying mass systems, that looks quite familiar to this one, but isn't exactly the same, because the $v$ on the right side is there the relative velocity of mass expulsed/accreted. The derivation of that formula is also rather different from this one.
So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.
I am wondering when the missing dollar puzzle first appeared. I am sure I read it forty years ago or so, but maybe it is much older. Does anyone knows it?
Answer
It has been dated back as far as 1933, published in a book, Diversion and Pastimes by R.M. Abraham.
Source: http://www.psychologytoday.com/blog/brain-workout/200912/where-is-the-missing-dollar
Given the recent news about the discovery of an "Earthlike" planet orbiting Alpha Centauri (our nearest stellar neighbour) it got me wondering just how fast would spaceship have to travel to be able to reach Alpha Centauri within a person's lifetime (say 60 years)?
The reasoning is that even if we sent an unmanned probe - assuming that it would be a one way trip - the journey time would have to be sufficiently short to keep people's interest in the mission so that when it actually arrived and sent back data there would be somebody home to receive the data.
The added complication is that the craft would have to slow down sufficiently so that it could at least enter orbit around the star (I'm not going to suggest that it manages to orbit one of the planets!).
I suppose what I'm really interested in how this speed compares to the speeds we've managed to attain so far and thus get an idea of how much technology must advance for us to be able to even think of achieving this.
I know about relativistic effects due to travelling at a high percentage of the speed of light - but I'm not really interested in that here.
Answer
If you are not interested in relativistic effects, the answer to your question is easy to workout. According to Wikipedia, Alpha Centauri is 4.24 ly away (4.0114x$10^{16}\mathrm{m}$). So to get there in 60 years ($1892160000\mathrm{s}$).
So your non-relativistic answer is
$v = \frac{d}{t} = \frac{4.0114 \times 10^{16}}{1892160000} = 21200000 \mathrm{m}\,\mathrm{s}^{-1}$.
This is 21200 $\mathrm{km}\,\mathrm{s}^{−1}$. The fastest recored space flight was 24,791Mph which is around 11$\mathrm{km}\,\mathrm{s}^{−1}$ which is 0.05% of 21200$\mathrm{km}\,\mathrm{s}^{−1}$. This means we have to be able to get spaceships to travel 2,000 times faster than the fastest current spaceship.
Note, I believe satellites in geostationary orbits do $\approx 17\mathrm{km}\,\mathrm{s}^{−1}$.
Edit. The relativistic calculation can be found here.
Rutherford's experiments confirmed the existence of light-weight electron clouds in a mostly empty atom, and that they occupy some space around the nucleus. What made us conclude that they can move? Can't it be vice versa: nucleus moves around the electron cloud?
Can you find what corresponds to 5?
If: $$1=5$$ $$2=6$$ $$3=7$$ $$4=8$$
then, $$5=?$$
Note: the answer is not 9.
Answer
Is the answer simply
1, due to symmetry (thanks for the correction, @Peter!!) of the equals sign?
There are 3 gods, named Past, Present and Future, who all look identical. They sit in a throne room, with one in the middle, the others on the right and left. You are allowed to ask them a series of yes/no questions in order to discern their identities. Present will answer truthfully, Past will truthfully answer the previous question you asked, and Future will truthfully answer the next question you plan to ask.
In order to avoid paradoxes (which would destroy the universe), you must follow these rules:
What is the fewest number of questions you must ask to determine their identities, and how do you do this? (There is a solution, and it can proved optimal).
I got this puzzle from William Wu's puzzle website. You may also consider the "time is a flat circle" variant, where Past will answer your last question when your first question is directed at them, and Future will similarly wrap around to the first question you asked.
Answer
To solve the riddle in three questions it is essential to remember that the rules allow to choose the addressee of the question (but not the question) depending on the outcome of the question. For if the order of questions is fixed then the problem cannot be solved in three questions. To show this, assume we're asking Left first and Right last (any other order is equivalent). Then ? denotes a random answer:
Left/Middle/Right | L M R
--------------------+------
Past/Present/Future | ? ?
Past/Future/Present | ?
Present/Past/Future | ?
Present/Future/Past |
Future/Past/Present |
Future/Present/Past |
To distinguish the case Past/Present/Future from the other cases we would need an answer "Yes" in this one case and "No" in all other cases (or the reverse). If the answer is "No", we have only 2 questions remaining, which can only differentiate 4 cases, but 5 remain, making it impossible to solve in three questions if the question order is fixed.
So we need to ask questions in a manner that we will put the last question to Past or Present but not to Future. The second question needs to determine whether we put our first question to past: if we did then the answer to the first is useless and unneccessary, otherwise it serves to disambiugate the other cases:
Left/Middle/Right | L M
--------------------+----
Past/Present/Future | ? Y
Past/Future/Present | ? Y
Present/Past/Future | Y N
Present/Future/Past | Y N
Future/Past/Present | N N
Future/Present/Past | N N
Then we need to choose the addressee of the third question such that it disambiugates between the remaining cases. (That this is the optimal solution is obvious, since with three boolean questions one can disambiugate at most 8 cases - in this case indeed only 6 because Past answers at random if asked first. With two questions one could disambiugate only four cases, which would be insufficient - though with head-exploding questions nine cases could be disambiugated...)
That only helps to deal with 1. the possibility of hitting Past on the first question and 2. how to avoid hitting Future with the third question. However, they still might not answer the question they are supposed to answer: for Future, the next (2nd or 3rd) question must contain the question that must be answered on the previous (1st or 2nd) question; for Past, the previous (1st or 2nd) question must contain the question that must be answered for the next (2nd or 3rd) question:
| 1 2 3
--------+------
Past | - 1 2
Present | 1 2 3
Future | 2 3 -
Hence the questions must be patterned as follows:
( are you Present and [question 1] ) OR ( are you Past and [question 2] )( are you Present and [question 2] ) OR ( are you Future and [question 1] ) OR ( are you Past and [question 3] )( are you Present and [question 3] ) OR ( are you Future and [question 2] )Thus when e.g. Future is asked second, they will provide the answer to question 3, which will be exactly the answer that present would give of question 2.
Now we know that a solution is possible, but still must determine which questions to ask. Due to the pattern rule above we can start by considering the present case only. Consider again the response table given above, then we want the following pattern:
Left/Middle/Right | 1 2 3
Past/Present/Future | ? Y Y
Past/Future/Present | ? Y N
Present/Past/Future | Y N Y
Present/Future/Past | Y N N
Future/Past/Present | N N Y
Future/Present/Past | N N N
The first question, asked to L, can be "is L Present?" to give the desired pattern. The second question, asked to Middle, can be "Did I ask Past my first question?". Together these tell us whether L is Past, Present or Future. This has two consequences:
The following three questions shall be asked:
1. To Left:
are you Present?[*]
2. To Middle:( are you Present AND I asked Past my first question ) OR ( are you Future AND Left is Present ) OR ( are you Past AND ( (Left is Past AND Middle is Present) OR (Middle is Past) ) )?
3. To Middle if previous both answered with "No", to Left otherwise:( you are Present AND ( (Left is Past AND Middle is Present) OR (Middle is Past) ) ) OR ( are you Future AND I asked Past my first question )?[*] according to the pattern there should be
OR ( are you Past AND I asked Past my first question ), but that drops away since that applies only if question 2 is addressed to Past, but then the first question can't have been addressed to Past.
I came across this puzzle that involves a treasure hunt in a poem. The puzzle is as follows:
A legend of nine,
In the Andaman,
A treasure you may find,
But, certainly, the blood of men.Staring at the Sun,
At 30 to 6,
Al asked the captain,
“What is the fix?”
“The Sun” says the cap,
“And blood shall determine our fate”
And so he screams “Die, Al”
And so remain 8.Down the direction, so gracefully learnt, Atop a lonely tree, they find a nest.
They hear the ill omens of the crows, And another life goes to waste.Expectedly,
Another turn of events,
Left the first mate forever lost,
Along with the tents.At 45 to 4, onward they went,
Towards the oasis in sight.
The most eager of all shattered their illusion,
And died due to the snake bite.Soon, fell upon them, the night,
The predators were approaching fast,
And with backs turned to the Polaris,
They headed towards the past.As the Sun lifted into the sky,
So did the spirits of the grey.
They were ready again to face the odds as before,
But with only mirrors leading the way.“R.I.P. comrade,” said the captain,
And engulfed himself in loneliness,
Only to find that he was back,
At that inn with his mistress;“The inn is the centre of everything,”
He admitted, amazed,
And it is there where I should have realized,
The curse of the maze.”
The above poem is famous in the islands of Andaman and Nicobar. The treasure is rumoured to be still there. Find the place where the treasure is buried.
Hints:
Use "breadcrumbs"
The Sun and time are great direction specifying tools
Read it aloud
The inn has many names.
More Hints:
(Para 1) At every turn there is a death
(Para 2) It points towards the north
(Para 3) The key is in the missing rhyme
(Para 7) Everything they did yesterday, with one minor change about the needle of the compass.
Even more Hints:
(Para 1) Every vertex is signified by a death.
(Para 2) The sundial can be used to signify an angle, given the latitude and the time of a place.
(Para 3) The direction is in the missing rhyme.
(Para 5) The Sun still guides the way.
(Para 6) Go back to the point where you started from.
(Para 7) Everything they did yesterday but with lateral inversion about the needle of a compass.
(Para 9) Properties of an equilateral triangle.
Let's assume we have a 1 kg cube on a flat uniform surface. The coefficients of friction between them are $\mu_k = 0.25$ and $\mu_s = 0.50$. This cube is moving at 1 m/s in the y direction, with whatever force is needed to overcome friction being applied. There is no force or motion in the x direction. Assume $g = 10\, {\rm m / \rm s^2}$ in the z direction.
Now, if a force is applied in the x direction which type of friction applies, static or kinetic? Edit: To clarify, if the cube is moving forward, and gravity is acting down, the new force is being applied to the right.
My rudimentary model of friction is that it acts like two pieces of sandpaper where the bumps can interlock when there is not motion, but when they are moving over top of each other they don't have time to settle in. In that model, I would think kinetic friction would apply, as the direction of movement shouldn't matter.
However, I realize that in reality friction is a result of molecular attractive forces, and probably a lot of other more complicated things I'm unaware of. Because of this, I suspect the answer may lie somewhere between static and kinetic friction.
Answer
I can´t fully come up with an explanation from more basic principles, but in the case you describe you will have kinetic friction. Or at least that is what all engineering books say...
There are a number of situations where this effect is clearly demonstrated:
While not entirely the same, a similar situation arises when a car rolling down a road with a strong side wind brakes and locks the wheels. While the wheels were rolling, there is no relative motion between the road and the tire, so there is static friction in effect, and unless the wind is really strong, as in a hurricane, the force will not overcome friction and the car will not skid sideways. Once the brakes are locked, the car starts skidding forward, because the force due to the inertia of the car overcomes friction. Once this happens, the car will also start skidding sideways, due to two factors:
the lesser important is that the coefficient of friction in effect once there is relative motion is the kinetic, not the static one.
the main effect is due to the fact that the direction of movement doesn't really matter at all: at the contact point you have a force due to inertia and a force due to the side wind, and once their combined magnitude exceeds friction, you will start having movement in the direction of that combined force, so forward but also to the side.
I am a haven for those with thick walls
I challenge convention with my transparent skin
My effect is quite big, though I can be small
I keep the bad stuff out, and the good stuff in.
What am I?
Answer
How about a
greenhouse
I am a haven for those with thick walls
like plants, whose thick walls refer to their stems Also, like BlueRaja-DannyPflughoeft suggested, plant cells normally have thick cell walls
I challenge convention with my transparent skin
greenhouses use transparent material (while plants are normally planted outside in the garden or in a house
My effect is quite big, though I can be small
greenhouses, even if they're small, can be effective at retaining heat
I keep the bad stuff out, and the good stuff in.
they can keep the bad stuff (chills, winds, bugs, etc) out, and the good stuff, like heat, in
EDIT: In regards to the title, with credit to user BlueRaja-DannyPflughoeft
the seven refers to the seven colours of the rainbow: ROYGBIV, where the middle is G -> green
This is an entry in the 14th Fortnightly Topic Challenge.
Ha, nice, a new fortnightly challenge. What's more, it is about history! It seems that people didn't like sports that much, huh? Fortunatly, there are (good) puzzlrs around here who seem to lke history. Like @Sleafr, who alredy postd two questins for ths challnge. Hey, what is that? I ment @Slefar, of corse. No, @Sleafar * ! God dmmt, what's happnng with the vowls? It's lke I am a @JnMarkPrr post No, t's @JonMarkPerry * ! Argh, cm on I can't evn pzzle prprly, th pnctutin s gong awy! Sht I hv to do it fst: the goal f this puzzl is to fnd the reltion betwn thse five famos quotsFindthir authrs, andyo'll fndth relatn! k, evnth spacsre vnishng... Rmmbrths s taggd hstry...
whndplmcndswrbgns
hstrwllbkndtmfrntndtwrtt
dsrfrmrpwrflthngnswdntltrpplhvgnswhshldwltthmhvds
thbttrgttknwmnthmrfndmslflvngdgs
wcnntlwsbldthftrfrrthbtwcnbldrthfrthftr
* : I'd like to thank them by the way, they inspired me to make this puzzle.
Answer
The quotes are:
1.
'When diplomacy ends, war begins' - Adolf Hitler
2.
'History will be kind to me, for I intend to write it' - Winston Churchill
3.
'Ideas are far more powerful than guns. We don't let our people have guns. Why should we let them have ideas?' - Joseph Stalin
4.
'The better I get to know men, the more I find myself loving dogs' - Charles de Gaulle
5.
'We cannot always build the future for our youth, but we can build our youth for the future' - Franklin D. Roosevelt
They are all quotes from:
The leaders of powers in WWII (1939-45):
Adolf Hitler - Leader of the Nazi Party and Chancellor of Germany (1933-45)
Winston Churchill - Prime Minister of Great Britain (1940-45/51-55)
Joseph Stalin - Leader of the Soviet Union (1920(s) -53)
Charles de Gaulle - Leader of France (1940-44)
Franklin D. Roosevelt - US President (1933-45)
Which is why it is part of the Fortnightly Topic Challenge
I don't understand how quantum mechanics (and therefore also quantum computers) can work given that while we work with quantum states, particles that this quantum state consist of cannot be observed, which is the most fundamental requirement.
If I am not mistaken, by "observed" we mean interaction with any other particle (photon, gluon, electron or whatever else). So my very important questions:
Aren't the particles this quantum state consists of interacting with each other? Why doesn't that cause the state to collapse?
Aren't all particles in the universe interacting with Higgs field and gravitons etc? Why doesn't that cause every quantum state to collapse?
I feel there is something very fundamental in quantum mechanics that I am not aware of, hence I would be very pleased to have these questions answered.
Answer
Aren't the particles this quantum state consists of interacting with each other? Why doesn't that cause the state to collapse?
We have a mathematical model for the observations we can make of any system in the micro world. This model is quantum mechanics and its predictions have been verified experimentally over and over again.
Observables are quantities we can measure about the particles and fields in the micro world. A main postulate is that to every observable there corresponds a quantum mechanical operator. These operators enter the quantum mechanical equations whose solutions given the boundary conditions describe a system in the micro world.
It is true that a quantum system is continually interacting within itself as described by the quantum model, and there can be continual interactions with the boundaries but interaction is not a synonym for a measurement. The continuous interactions are off mass shell, virtual, and within the bounds of the quantum mechanical solutions of specific energy levels and allowed states and conservation of quantum numbers. They are not measurements.
Aren't all particles in the universe interacting with Higgs field and gravitons etc? Why doesn't that cause every quantum state to collapse?
Collapse is fancy terminology for measurement . Nobody is measuring the higg's field continuous virtual exchanges that give mass to the elementary particles, nor the gravitons either. In fact gravitons are hypothetical particle because we have never measured one, in the way we have measured photons. Also nobody is measuring the virtual photons that keep the electrons in their energy levels around the nucleus.
The basic misconception is identifying "interaction" with measurement. A measurement necessarily means an interaction. An interaction is much more than a measurement.
Many years ago I helped to support an experiment conducted in Japan which investigated the effects of high frequency oscillation ventilation (HFOV) on the mixing and distribution of gas into the lungs.
But this experiment rather simplified the system by using a simple horizontal tube, perhaps 2 meters in length. At one end of the tube a plenum was placed, and at the other, an HFOV ventilator. Initially the system was filled with air. An oxygen cell was placed in the plenum to measure the $O_2$ concentration. At the start of each experiment, a small volume (maybe 500 mL) of pure oxygen was introduced slowly at the end where the HFOV ventilator was attached.
Although I can't recall all the details of the experiment, I do remember one result that's left me baffled all these years. The following two experiments:
(1) After introducing the oxygen the system was left unpowered so that natural diffusion would occur. (2) After introducing the oxygen, the HFOV ventilator was turned on and allowed to oscillate the gas column
Note that oscillating the gas column with the ventilator results in no net displacement of the gas. Perhaps only 100 mL back and forth displacement. And the volume of the full tube is significantly larger than the volume of the introduced oxygen.The HFOV machine was able to oscillate at frequencies between 4 and 20 Hz.
The expected result would be that the oscillations would promote the movement of the oxygen towards the plenum. But the opposite, confounding result occurred. The oscillations actually inhibited the movement of the oxygen by a significant time.
I'm guessing now but diffusion maybe took an hour without oscillation. But with oscillation it took maybe 30 minutes longer.
So my question: are there any known physical properties of oscillating flow that could explain such a result?
I tried researching to see if the work was ever published, but no luck. The physician I worked with was a Dr. Miao.
My question is about the article Swimming in Spacetime.
My gut reaction on first reading it was "this violates conservation of momentum, doesn't it?". I now realize, however, that this doesn't allow something to change its momentum; it only allows something to move (change position) without ever having a nonzero momentum. Since this is relativity, there is no simple relationship between momentum and velocity like p = mv, so this is all well and good. An object can move, with a constant momentum of zero, by changing its shape in a nontrivial cycle.
However, now I'm thinking about a different conservation law and I can't see how "swimming through spacetime" is possible without violating it. The conserved quantity I'm thinking of is the Noether charge associated with Lorentz boosts, which is basically x - (p/E)t, that is, the position of the center of mass projected back to time t=0. If p = 0, then the conserved quantity is simply x, the position of the center of mass. This obviously contradicts the whole swimming idea.
What's going on here? Is swimming through spacetime only possible if the spacetime is curved in some way that breaks the symmetry under Lorentz boosts? Or is there some error in my reasoning?
Answer
What's going on here? Is swimming through spacetime only possible if the spacetime is curved in some way that breaks the symmetry under Lorentz boosts? Or is there some error in my reasoning?
That is precisely the case. No error in your reasoning. In the case of a curved spacetime the "center of mass" of an extended body is no longer well-defined w.r.t external - i.e. located in an asymptotically flat region - observers.
In order to "swim" through spacetime one exploits the inhomogeneities of the gravitational field. The presence of these inhomogeneities breaks local Lorentz symmetry which is necessary for the mechanism to work.
In particular the scale of the swimmer and the inhomogeneities should be comparable. This is one reason why, at present, the construction of an actual swimmer is far beyond our technological means.
Edit: For those interested on extended body effects in GR there is are classic papers by Dixon. More recently Abraham Harte has done some amazing work along these lines Extended-body effects in cosmological spacetimes.
For analysis purpose, I am looking for raw data from real entanglement experiment with spin measured at two detectors, at various combinations of angles, but preferably at a single combination of angles.
To be precise, I am looking for individual pair detail data captured in the sequence it was measured as described below. The data can be for photons, but preferably for electrons.
Pair#----angle of detector1----spin outcome1----angle of detector2----spin outcome2
1---------30---------------------UP----------------120--------------------DOWN
2---------30---------------------UP----------------120--------------------DOWN
3---------30---------------------DOWN-----------120--------------------UP
and so on .. This is the example of information I need, format can be different as long as sequence is preserved. Angles 30/120 are also just for example, they can be any angles.
If someone could please point me to such data. It is fine with me even if it is very large amount of data, I will handle it for analysis purpose.
If there are multiple sources of data, more the merrier. It will be also helpful to have location/distance information about generation and measuring devices
I have read these questions:
Does a photon in vacuum have a rest frame?
Based on dmckee's answer, the answer is no to a photon's rest frame.
In the modern view each particle has one and only one mass defined by the square of it's energy--momentum four vector (which being a Lorentz invariant you can calculate in any inertial frame): $$ m^2 \equiv p^2 = (E, \vec{p})^2 = E^2 - \vec{p}^2 $$ For a photon this value is zero. In any frame, and that allows people to reasonably say that the photon has zero mass without needing to define a rest frame for it.
Now I understand that the graviton is hypothetical.
But it is meant to be massless like the photon. It is a gauge boson like the photon.
So we can write the same for a graviton: $$ m^2 \equiv p^2 = (E, \vec{p})^2 = E^2 - \vec{p}^2 $$
And this should be 0 in any inertial reference frame for the graviton too.
Now as per SR (second postulate):
Now for GWs (like EM waves), it should be true that:
In this case, gravity could be thought of as the same thing as light, and we could basically build SR onto GWs (instead of light). We could build the postulates of SR onto GWs, derive the laws of SR from the behavior of GWs (instead of light).
Question:
Does a graviton (I understand it is theoretical, but what is it meant to be) in vacuum have a rest frame?
Do GWs obey the laws of SR, so is the speed of GWs in vacuum the same for all observers regardless of the motion of the GW's source?
Answer
Special relativity is part of general relativity and gravitational waves should travel with the velocity of light.
Gravitons are hypothetical particles of a quantized general relativity. There exists a graviton that is massless and is like the photon, but also Kaluza Klein gravitons which have a mass and thus have variable velocities depending on the specific problem. These are research projects as can be seen by searching for "kaluza klein graviton"
Does a graviton (I understand it is theoretical, but what is it meant to be) in vacuum have a rest frame?
Some gravitons in some models may have a rest frame, but not the lowest order one, which is photon like massless and moving with velocity c.
Do GWs obey the laws of SR, so is the speed of GWs in vacuum the same for all observers regardless of the motion of the GW's source?
Classical gravitational waves also follow special relativity, in flat spaces . See here.
See also my answer here for kaluza klein models
From Jackson, problem 10.3:
A solid uniform sphere of radius $R$ and conductivity $\sigma$ acts as a scatterer of a plane-wave beam of unpolarized radiation of frequency $\omega$, with $\omega R /c \ll 1$. The conductivity is large enough that the skin depth $\delta$ is small compared to $R$. (a) Justify and use a magnetostatic scalar potential to determine the magnetic field around the sphere, assuming the conductivity is infinite. (Remember that $\omega \neq 0$.)
We'd like to show $\nabla \times {\bf B} = \nabla \cdot {\bf B}=0.$
I have two questions. First, what does it mean for a plane wave to have a definite frequency and be unpolarized? For example, are there many sources out of phase, all at a given frequency, radiating with varying amplitudes and in all directions? If this is true, then is it possible for the superposition of all these waves to vary faster than the original frequency due to interference?
Assuming that the above question is resolved and that we can make the long-wavelength approximation so the magnetic field is roughly constant over the sphere, why does
$$\nabla \times {\bf B} =0\neq \frac{1}{c^2}\frac{\partial {\bf E}}{\partial t}$$ (assuming that the electric field is just like an oscillating dipole and has a term free from powers of $\omega$)?
What happens microscopically when an electrical current starts to flow? I'd like to understand microscopically what happens in detail when electrons start moving (quasi-classically).
Electrons can have different velocity, they can produce electromagnetic fields, leads have free electrons and rigid atom cores and there exist electromagnetic fields. That's all the ingredients you should need?
Electrons only move due to EM fields, so basically this question boils down to what the EM fields look like and how they build up?! In steady state, what is the electric and magnetic field distribution in/around the lead? And what about the transient state?
What happens when you attack a battery to a lead? Are there EM fields between battery poles or why are electrons pushed? How do the EM field start to push electrons along an arbitraritly shaped long lead?
[EDIT: Ideally an explanation with the Drude model (which partly derives from Fermi model) or an explanation why that model isn't sufficient. Also stating the EM fields consistent with the electron density distribution would be important (i.e. $\vec{E}(r,\theta,z)$ and $\vec{B}(r,\theta,z)$) because otherwise it's hand-wavy arguments.]
(Please consider all remarks in this question. I know common arguments for parts of the question, but I've never seen a full microscopic in detail explanation.)
My understanding so far:
A moving charge produces a magnetic field, B, in an analogous way to a current produces one.
A magnetic field has an energy density which is proportional to B squared.
My question is: Does a charged particle in motion have an additional energy associated with it's motion due to this magnetic field?
I suspect the answer is no, for a number of reasons, but want to check.
Additional work would then have to be done to accelerate a charged particle. This would be work against a force, and I can see no mechanism which could produce that force (other than just 'the increase in field energy').
If the charge is a point charge then the energy density would diverge close to that charge. (The volume integral would also diverge). So moving a charge would require infinite energy. I'm guessing this is similar to why you can't define the electric field energy of a single point charge, it doesn't interact with anything as it dwindles into infinity.
There is this idea of relativity in Classical Mechanics:
The laws of mechanics valid in an inertial frame must also be valid in any frame moving uniformly with respect to it.
I was just trying to apply these to the case of the law of conservation of momentum and the law of conservation of angular momentum.
Let there be an inertial frame S and another frame S' moving with velocity $\mathbf{\vec{v}}$ w.r.t to S with:
$$\mathbf{\vec{r}}'_i = \mathbf{\vec{r}}_i - \mathbf{\vec{v}}t$$
$$\mathbf{\vec{v}}'_i = \mathbf{\vec{v}}_i - \mathbf{\vec{v}}$$
For momentum conservation: In frame S', putting $\dfrac{d}{dt} \sum_i \mathbf{\vec{p}}'_i = \mathbf{0}$ and substituting $\dfrac{d}{dt} \sum_i \mathbf{\vec{p}}_i = \mathbf{0}$ of frame S in it:
$$\dfrac{d}{dt} \sum_i \mathbf{\vec{p}}'_i = \dfrac{d}{dt} \sum_i \mathbf{\vec{p}}_i - \dfrac{d}{dt} \sum_i m_i \mathbf{\vec{v}} = \mathbf{0} - \mathbf{\vec{v}} \dfrac{d}{dt} \sum_i m_i$$
If this has to be $\mathbf{0}$, then $\sum_i m_i = 0$
Now, on to angular momentum. In frame S:
$$\dfrac{d}{dt} \sum_i \mathbf{\vec{L}}_i = \dfrac{d}{dt} \sum_i (\mathbf{\vec{r}}_i \times m_i\mathbf{\vec{v}}_i) = \mathbf{0}$$
Am trying to prove the law in frame S' from the law in S:
$$\dfrac{d}{dt} \sum_i \mathbf{\vec{L}}'_i = \dfrac{d}{dt} \sum_i \mathbf{\vec{L}}_i - \dfrac{d}{dt} \sum_i (\mathbf{\vec{r}}_i \times m_i\mathbf{\vec{v}}) - \dfrac{d}{dt} \sum_i (\mathbf{\vec{v}}t \times m_i \mathbf{\vec{v}}_i)$$
$$= \mathbf{0} - \dfrac{d}{dt} \sum_i (\mathbf{\vec{r}}_i \times m_i\mathbf{\vec{v}}) - \dfrac{d}{dt} \sum_i (\mathbf{\vec{v}}t \times m_i \mathbf{\vec{v}}_i)$$
$$= - \sum_i m_i (\mathbf{\vec{v}}_i \times \mathbf{\vec{v}}) - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{r}}_i \times \mathbf{\vec{v}}) + \sum_i m_i (\mathbf{\vec{v}}_i \times \mathbf{\vec{v}}) - \sum_i m_i (\mathbf{\vec{v}}t \times \mathbf{\vec{a}}_i) - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{v}}t \times \mathbf{\vec{v}}_i)$$
$$= - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{r}}_i \times \mathbf{\vec{v}}) - \sum_i m_i (\mathbf{\vec{v}}t \times \mathbf{\vec{a}}_i) - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{v}}t \times \mathbf{\vec{v}}_i)$$
$$= \mathbf{\vec{v}} \times \sum_i \dfrac{dm_i}{dt} \mathbf{\vec{r}}_i - \mathbf{\vec{v}}t \times \sum_i \mathbf{\vec{F}}_i$$
But this is what I wanted to prove to be $\mathbf{0}$. I stil have to prove the following:
For a system of particles at $\mathbf{\vec{r}}_i$ with mass $m_i$, which have forces $\mathbf{\vec{F}}_i$ acting on them such that $\sum_i \mathbf{\vec{r}}_i \times \mathbf{\vec{F}}_i = \mathbf{0}$, given $\sum_i \dfrac{dm_i}{dt} = 0$; how do I prove:
$$\mathbf{\vec{v}} \times \sum_i \dfrac{dm_i}{dt} \mathbf{\vec{r}}_i = \mathbf{\vec{v}}t \times \sum_i \mathbf{\vec{F}}_i$$
for any arbitrary $\mathbf{\vec{v}}$ and for all time $t$.
Answer
Using your notation of
$$ \boldsymbol{r}_{i}' =\boldsymbol{r}_{i}-\boldsymbol{v}\,t \\ \boldsymbol{v}_{i}' =\boldsymbol{v}_{i}-\boldsymbol{v} $$
and with the assumption that $\dot{\boldsymbol{v}}=0$ (uniform motion of frame S') form the linear and angular momentum expressions on the S frame.
$$ \boldsymbol{p} =\sum_{i}m_{i}\boldsymbol{v}_{i} \\ \boldsymbol{L} =\sum_{i}\left(\boldsymbol{r}_{i}\times m_{i}\boldsymbol{v}_{i}\right) $$
Now look at linear and angular momentum in the S' frame and relate them to the ones from S.
$$\require{cancel} \begin{aligned} \boldsymbol{p}'&=\sum_{i}m_{i}\boldsymbol{v}_{i}'=\sum_{i}m_{i}\left(\boldsymbol{v}_{i}-\boldsymbol{v}\right)=\boldsymbol{p}-\left(\sum_{i}m_{i}\right)\boldsymbol{v}=\boldsymbol{p}-m\,\boldsymbol{v}\\\boldsymbol{L}'&=\sum_{i}\left(\boldsymbol{r}_{i}'\times m_{i}\boldsymbol{v}_{i}'\right)=\sum_{i}\left(\boldsymbol{r}_{i}-\boldsymbol{v}\,t\right)\times m_{i}\left(\boldsymbol{v}_{i}-\boldsymbol{v}\right)\\&=\sum_{i}\left(\boldsymbol{r}_{i}\times m_{i}\boldsymbol{v}_{i}\right)-\sum_{i}(\boldsymbol{r}_{i}\times m_{i}\boldsymbol{v})-\boldsymbol{v}\,t\times\left(\sum_{i}m_{i}\boldsymbol{v}_{i}\right)+\cancel{\boldsymbol{v}\,t\times\left(\sum_{i}m_{i}\right)\boldsymbol{v}}\\&=\boldsymbol{L}+\boldsymbol{v}\times\sum_{i}\left(m_{i}\boldsymbol{r}_{i}\right)+\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)\times\boldsymbol{v}\,t \end{aligned}$$
To show that these quantities are conserved, take the derivative (assuming that $\frac{{\rm d}\boldsymbol{p}}{{\rm d}t}=0$ and that $\frac{{\rm d}\boldsymbol{L}}{{\rm d}t}=0$)
$$\begin{aligned} \frac{{\rm d}}{{\rm d}t}\boldsymbol{p}'&=\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{p}}-m\,\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{v}}=0\\\frac{{\rm d}}{{\rm d}t}\boldsymbol{L}'&=\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{L}}+\frac{{\rm d}}{{\rm d}t}\left[\boldsymbol{v}\times\sum_{i}\left(m_{i}\boldsymbol{r}_{i}\right)\right]+\frac{{\rm d}}{{\rm d}t}\left[\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)\times\boldsymbol{v}\,t\right]\\&=\boldsymbol{v}\times\sum_{i}\left(m_{i}\frac{{\rm d}}{{\rm d}t}\boldsymbol{r}_{i}\right)+\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{p}}\times\boldsymbol{v}\,t+\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)\times\boldsymbol{v}\\&=\boldsymbol{v}\times\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)+\boldsymbol{p}\times\boldsymbol{v}\\&=\boldsymbol{v}\times\boldsymbol{p}+\boldsymbol{p}\times\boldsymbol{v}=0 \end{aligned}$$
Reflection and transmission (Fresnel equation) of polarized light are treated in many optics or electromagnetism books.
If $E_s$ and $E_p$ is incident electric field with s-polarization and p-polarization, respectively, reflected electric field would be $E'_s = r_s E_s$ or $E'_p = r_p E_p$, where $r$ is reflection coefficient and $'$ means reflected field.
Then electric field of any polarized light can be calculated by decomposition.
For instance, since $E_{45^{\circ}} = \frac 1 {\sqrt{2}} ({ E_p + E_s })$, $E'_{45}$ would be $\frac{1}{\sqrt 2} (E'_p + E'_s )$
However, how can I express electric field of reflected unpolarized light?
Is Einsteins Equivalence theorem in General Relativity correct? It seems to me that it neglects the fact that gravitational acceleration depends upon separation distance squared, thus neglecting the effect of tidal forces.
For example, as I sit on earth, I experience the affect of earth's gravity; Although the acceleration on my head is slightly less than the acceleration acting on my feet. If I make the claim that my frame is equivalent to me being in a space ship traveling at g, doesn't that mean my whole body is accelerating at g uniformly? This is contrary to the previous statement though.
Answer
The equivalence principle, as stated correctly by Einstein, says that these two situations are equivalent:
So, as you noted, this does not apply to the gravitational field of the Earth. Imagine you are in an elevator, free falling towards the Earth. You could let go of two pens - what you would see is that the two pens would come closer to each other, since each of them would be falling towards the centre of the Earth. You, as an observer, could then say with certainty that you are in a gravitational field.
In a hypothetical uniform gravitational field no experiment could reveal, whether you are in a gravitational field or not. Einstein took this thought experiment as a motivation for the development of General Relativity. In the mathematical construction of General Relativity, the equivalence principle does not play an important role.
If you are standing on earth's surface, in what direction (and at what speed) should you throw a 1Kg uniform sphere of radius 0.1 meters in order to put it into lower earth orbit? Assume that there is air, but it is not moving relative to the earth
Answer
If we ignore air resistance for a moment, then all orbits in an inverse square force like gravity are closed. this means that if you throw something hard enough it will complete one orbit then return to its starting point i.e. your hand.
So if you throw the object downwards it obviously hits the Earth, and if you throw it straight upwards it goes up then down and hits the Earth. For all the angles in between the object will go into orbit, though it will be an exceedingly low orbit since once an orbit it will pass through the point where you released it.
If you want to put the object into a roughly circular orbit you can't do this just by throwing it. Satellites are launched using a rocket trajectory like this:
And this requires a continuous boost from the rocket. You can't replicate this with a throw.
Adding air now means that if you throw the object hard enough to get it out of the atmosphere it will simply burn up, just as satllites burn up when the reenter the atmosphere.
My understanding is that an observer can measure the precise location of a particle so long as the corresponding uncertainty in momentum measurement is not an issue and vice-versa.
Say there is such an observer, interested in the precise position of a particular particle. Now, consider a second, independent observer, unbeknownst to the first, who is trying to measure the exact momentum of the same particle without caring about the position. As a thought experiment, we assume that the two observers are somehow able to access the same particle at the same time in some way without being aware of each other.
Can both observers get their desired results?
Answer
They could not - don't think of the observers as people, think of them as experiments. If two individual experiments were taking place on the same particle simultaneously, there's no reason why they couldn't be combined into a singular experiment, and you'd then have an experiment that is able to determine both a precise position and momentum, which is impossible.
Here is something I see :
Let's say the ideal fluid(water here) of density $\rho$ is drawn from a source by a motor and thrown upwards with a velocity $v$. Now we take the power of motor be constant and fluid to flow in streamlines (same velocity all over because of same cross section area) ,then applying Bernoulli's Equation for two points shown we get the $$P_{down}-P_{up}=\rho g h$$
$P_{down}=P_{atm}$ because the down point is open to air. So, we get $$P_{up}=P_{atm}-\rho g h$$
which will go negative if $h>10m$. How is it possible?
Answer
which will go negative if $h>10m$. How is it possible?
It is not possible that $P_{up}$ will go negative. It will experience cavitation. This is one of those cases where you have to realize that the equation you're using is a special case of a larger theory, step back, and employ the larger theory.
If you step back from the concept of water as an absolute liquid, then it's clear that it has a state diagram. Going from some liquid place on the state diagram to $P=0$ will cross a phase transition line. Probably just once, and probably the liquid to vapor line. That means the water boils.
As you hit this threshold of $10 m$ between atmosphere and the lowest pressure point in the system, that point will turn to gas. There will possibly be some significant change in temperature, but we can imagine it happens slowly so everything is held at room temperature. So draw a line on the PT diagram of water of constant T, decreasing pressure.
I have to make some adjustments to your diagram now. Firstly, you imagined an open tube pointing down. I'll have to revise this to a tube that snakes around so that it has a defined level. Take a pipe filled with water and turn the open end down - the water is likely to fall out. We can often ignore this in small (common aquarium) pipes because of surface tension effects.
In addition to that, let's imagine the cavitated area that is now filled with water vapor. I'll, of course, assume your pipe is strong enough. Not to worry, plenty of pipes are.
There's some detail that I didn't include. But let's start making observations:
As the pump pushes more liquid through the system, that liquid will overflow through the vapor region and fall down, joining the water in the outlet pipe.
But somehow, despite all of the reasons it shouldn’t work, it does.
Scientists at NASA just confirmed it.
Now in this question - it is strongly suggested that this doesn't work.
So at best, Shawyer has invented a very inefficient and expensive fan.
Even further - this answer states:
Shawyer's "analysis" is a mess, incoherent and deeply confused about fundamental aspects of relativity.
My question is: Has NASA confirmed that Roger Shawyer's EmDrive thruster works?
Edit:
There may be an answer here. I just don't know what it means.
Edit #2
I don't believe this is a duplicate of the other question - as it is not clear how much to trust NASA's engineers - or what their comment actually means.
Answer
No, NASA has not confirmed that. What NASA has confirmed, again, is that it has some rather nutty folks working for it.
I am reading Steven Weinberg's book Gravitation and Cosmology. He makes a big deal out of the equivalence principle and showed a bunch of deductions you can make based on it. This surprised me since other books I have read haven't emphasized it as much.
My Question:
Is there an accepted set of axioms or principles that constitute the core premises of GR from which many, most or all relevant properties can be deduced?
Answer
General relativity can be constructed from the following principles:
The Principle of Equivalence
Vanishing torsion assumption ($\nabla_XY-\nabla_YX=[X,Y]$)
The Poisson equation (or any other equivalent Newtonian mechanics equation)
Explanations:
The Equivalence Principle can be used to show that spacetime is locally Minkowskian, i.e. the laws of special relativity hold in an infinitesimal region around a freely-falling observer. This is equivalent to the mathematical idea that a manifold of dimension $n$ is locally homeomorphic to $\mathbb{R}^n$. This allows to do two things (that I can think of at the moment). We conclude that spacetime is a manifold. We also can make the substitutions $\eta\rightarrow g$ and $\partial\rightarrow\nabla$, which yields the correct (there are exceptions) GR equations.
This is required for the geodesic equation to be obtainable from a variational principle because it implies the Christoffel symbols are symmetric. This condition is relaxed in certain theories such as Einstein-Cartan theory or string theory.
Simply put, we need this equation to fix the constants in Einstein's equation.
All treatments of GR use the Principle of Equivalence. Weinberg's treatment especially so. The reason for this has to do with his background as a physicist. Weinberg was (and is) one of the greatest particle physicists alive. His dream was to write a coherent quantum field theory for gravity. In his mind, $g_{\mu\nu}$ being called the metric tensor is an "antiquated" term left over from when Einstein learned differential geometry from his friend Grossmann and Riemann & co.'s old papers$^1$. In Weinberg's mind, $g_{\mu\nu}$ is just the graviton field, and any connection to geometry is purely formal$^2$. In texts such as Carroll, Straumann or Wald, they use the EP to make the connection $$\tag{1}\text{Equivalence Principle}\implies\text{Spacetime is a manifold}$$ From that point on, spacetime being a manifold is assumed. Weinberg, however, was of the opinion that gravity had nothing to do with geometry and manifolds and this mathematical description was a pure formality. He has to stress the EP because philosophically he didn't accept (1).
$^1$ See the first paragraph of section 6.9.
$^2$ See, for instance, page 77, where he calls the geodesic equation a mere formal analogy to geometry.
Gluons bond quarks into baryons (i.e., protons and neutrons). For example, two up quarks and one down quark form a proton while one up quark and two down quarks form a neutron. Is there one gluon per one baryon or two gluons per one baryon or what is the ratio of gluons to baryons?
Answer
I asked a question very much like this several years ago (in person, not online, but someone else asked it here): "what's the baryon asymmetry of the proton?" Thinking, of course, about the three valence quarks and the so-called "sea" of quark-antiquark pairs.
After talking with several people in my department and at conferences and at other places I visited, I finally came to an unsatisfactory conclusion: the answer to my question depends on the momentum scale at which you examine the proton. For a proton at rest, you don't really have evidence of sea quarks. For a proton at modest energy, evidence of excitations in the strong field begin to appear — but "nucleons and mesons" are a more succinct set of degrees of freedom at low energy than "quarks, antiquarks, and gluons." It's only if you look at relatively high energy — or equivalently, at very short distances — that quarks and gluons become the most parsimonious way to describe the strong interaction.
I think you're going to find the same sort of frustrating non-answer to your question about gluons. Gluons are the force-carrying particle in QCD, like the photon is the force-carrying particle in electromagnetism. If you have a hydrogen atom in its ground state, which is pretty well-described by the one-photon-exchange Coulomb potential, the equivalent question would be something like "how many virtual photons are exchanged as the electron orbits the proton once?" But of course as you study quantum mechanics you learn that "orbit" isn't really a good description of the electron-proton interaction for low-energy hydrogen states, and likewise the virtual photons are not really things you can count.
In multipole expansion, we use monopole, dipole, quadrupole or octupole to describe an electromagnetic field. But I saw someone use sextupole to describe transition states.
If we expand an electromagnetic field by spherical harmonics, $\ell=0,1,2,3$ represent monopole, dipole, quadrupole and octupole. Do we sextupole in electric field? For different $m$ in the expansion, are they also different modes?
It is common knowledge among the educated that the Earth is not exactly spherical, and some of this comes from tidal forces and inhomogeneities but some of it comes from the rotation of the planet itself. The deformation from the rotational effect makes it more oblate spheroid-like, or as I would prefer, "like a pancake". Here is one site illustrating the behavior, and image:
Literature exists detailing the mathematical expectations for a rotating planet using just hydrostatic forces, for example, see Hydrostatic theory of the earth and its mechanical implications. I like to imagine a ball of water in space held together by its own gravity. I also don't want to deviate from consideration of only hydrostatic (and gravitational) forces because I think it is sufficient for this discussion.
It would seem that the solution of the described problem is in terms of a small change in radius as a function of the azimuth angle, or z-coordinate if you take the axis of rotation to be the z-axis. This is using rotational symmetry. In other words, Earth's deformation due to rotation does not depend on longitude.
I want to ask about the extreme case. Imagine a planet rotating so fast that it is a very thin pancake. What will occur in this case? I am curious:
It seems to me that it would be logical for the high-rotation case to break up into 2 or more separate bodies. The reason is that a 2 body system is stable an can host a very large angular momentum. But would it be an instability that leads to this case? When would such an instability occur and could a rotating planetary body deform in a different kind of shape from the beginning, such as a dumbbell-like shape, which would transition into a 2-body system more logically than the pancake shape?
To sum up, how would a pancake shape transition into a dumbbell shape? Or would it? What are the possibilities for the described system?
See title: Does one age quicker at higher altitudes?
A few years ago I heard that you would age slightly faster (or slower) at higher altitudes.
Is this true? What is the theory or evidence for this?
Or to put it a little differently:
Does time move faster or slower at higher altitudes?
Answer
The term you are looking for is Gravitational Time Dilation.
Gravity has effects on the surrounding space-time, directly dependent on the mass of the bodies. This effect is not noticeable to us on Earth unless considering a large timescale, since the Earth is nowhere near massive enough for it, as opposed to extremely massive celestial objects (e.g: black holes and neutron stars).
To give an idea to how little it matters on Earth (same Wikipedia article as first link):
Clocks that are far from massive bodies (or at higher gravitational > potentials) run more quickly, and clocks close to massive bodies (or at lower gravitational potentials) run more slowly. For example, considered over the total time-span of Earth (4.6 billion years), a clock set at the peak of Mount Everest would be about 39 hours ahead of a clock set at sea level. This is because gravitational time dilation is manifested in accelerated frames of reference or, by virtue of the equivalence principle, in the gravitational field of massive objects.[4]
[Emphasis mine]
Using those numbers as reference, we can calculate that if an observer at sea level stayed there for 100 years, someone who would have stayed on the Everest would be older by roughly 0.003 seconds.
Technically yes, relative to an observer on Earth, a person at higher altitudes will age faster.
Why is that in absence of a manually fixed point, a body shows its turning effect on application of a torque with its axis of rotation through its centre of mass? my attempt: is there a definition of the centre of mass that im mising?
This is Part 5 of the Murder of the President brainteaser/riddle series. If you have not already, check out the answers for Murder of the President - Part 1 posted by Nit, Murder of the President - Part 2 and Murder of the President - Part 3 both posted by Joe Z., and Murder of the President - Part 4 posted by el jefe. Each part will give you a clue and you must solve it. This part will have several clues. Use all knowledge you have of cryptography, ciphers, past puzzles, etc. You should also use Google. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are who they are now. Please post your answers in spoiler tags.
Here's the riddle:
You arrive at the London NYC Hotel at 2:26 a.m. You've already called the staff at the hotel to let them know you were coming. On the way up, you were beginning to hear reports that the President was murdered. You head up to room 118. You look around for a bit until you find a box under the covers. On top of the box is a note.
It says
nvnmjoonhenuan
m2m
121
0+1
ttfflhesi
You check out the padlock. It is a one 5 digit code. Under the box there is another note.
Padlock:
ZUTYS
After you figure out this code, you open up the box and find, of course, 6 coded notes. Some notes are typed, others are handwritten. (Italics will indiacte handwritten and bold will indicate typed)
Here are the notes
#1
ERP BTRCBIV PTUKU GTI CUBL
#2
WAPHFS RPL MU UMK SQOF GWAWFGX UCTME UPCG
#3
M DT AJF MMUPIS - YQ
#4
SAQEHY WQG NT XLF VMOIA DUWI
#5
BVB YPPX LEXDK NJ
#6
VVHC ASL FNOXIS JETM ZVVOIU ZWTJPK
LLIFVFXZIDN
What is the code to get into the box? What do all the notes mean?
Good Luck.
Hint:
The messages are coded using the method and seed answered by el jefe. The sequence of numbers is continuous (it doesn't restart for each message). Also, you must add one to every zero in the sequence
Note: I will be posting Part 6 (the final part) in one to two days. I will select the correct answer for Part 5 before, but I will post the answer for Part 5 if no one gets it
Answer
Here are the decoded notes. Italics for handwritten, bold for typed.
#1: ALL WRITTEN NOTES ARE TRUE
#2: NUMBER ONE IS THE ONLY CORRECT TYPED NOTE
#3: I AM THE KILLER - VP
#4: NUMBER SIX IS THE RIGHT CLUE
#5: YOU WONT CATCH ME
#6: PREZ TOD WINTER FALL SUMMER SPRING CHEESESTEAK
So I suppose:
Our next stop is Hotel Four Seasons in Philadelphia (as in Philadelphia cheesesteak). Also, the VP is off our list of suspects.
"PREZ TOD" must refer to the President's time of death, which was November 5, 2014 at 1:09 p.m. That's most likely our clue for the room number (109 maybe?)
About decrypting the messages:
The notes are encrypted with a vigenére cipher, using the sequence of pseudorandom numbers obtained by using a variation of John von Neumann's middle square method using 121 as a seed, taking three numbers from the end (offset by one) and replacing all zeroes with ones. For those interested, here is the ugly but working Python implementation I ended up with.
