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What makes Bell's original inequality unsuitable for experiments? Bell derives the inequality $$|E(\vec{a},\vec{b})-E(\vec{a},\vec{c})|\leq 1+E(\vec{b},\vec{c})$$ in his book Speakable and unspeakable in quantum mechanics. In this derivation he uses the assumption that when the axes $\vec{a}$, $\vec{b}$ are aligned, the outcomes of measurements $A,B$ of the spins of spacelike-seperated particles 1,2 along these axes respectively are anticorrelated so that the product $AB$=-1.
Other similar "Bell-type" inequalities such as the CHSH inequality do not use this assumption. The CHSH inequality has the form
$$-2\leq E(\vec{a},\vec{b}) + E(\vec{a},\vec{b}') - E(\vec{a}',\vec{b}) + E(\vec{a}',\vec{b}')\leq 2 $$
My question is, is there anything other than the above assumption made in the derivation of Bell's original inequality that makes it unsuitable to experimental testing? I.e. is it something to do with the form of Bell's inequality?
Most papers I've read just simply state that the original inequality isn't suitable for testing and gloss over some statement about perfect correlations.
| The original derivation assumed that every measurement would give a result, such as for example "particle with up-spin detected," or "particle with down-spin detected" But in real experiments some particles never affected the detectors; presumably they "leaked out" or somehow vanished, and no measurements occurred. Hence subsequent derivations provide for non-detections, usually scored as a zero result. E.g. the CHSH derivation considers three possible scores, +1, -1 and 0. This is why the S value that the inequality is solved for (if local hidden variables exist) calls for "S equal to or less than |2|," rather than just "S less than |2|." In other words the inequality is satisfied if S falls anywhere between -2 and +2. Of course non-detections are serious problems because they allow the fair-sampling loophole, but that's another matter: If 1000 entangled particles were launched and only 100 were measured, are these 100 representative of the entire population of entangled particles? This question could not even be considered by the original derivations.
| {
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Lagrange equation and a force derivable from a generalized potential I was reading the solution of this exercise and I have a doubt:
A point particle moves in space under the influence of a force derivable from a generalized potential of the form $$U(r,v) = V(r)+\sigma\cdot L $$
where $r$ is the radius vector from a fixed point, $L$ is the angular momentum about that point, and $\sigma$ is a fixed vector in space.
I need to find the components of the force on the particle in Cartesians coordinates, on the basis of Lagrange equations with a generalized potential.
This exercise is from the book "Goldstein - Classical Mechanics". I have the solution, but I really don't understand a step. If I convert $r$ to Cartesian coordenates, I have $r = \sqrt{x^2+y^2+z^2}$, and if I put the expression into the Lagrange equation: $$Q_{j} = \frac{d}{dt}\frac{\partial }{\partial \dot q_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right) - \frac{\partial }{\partial q_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right) $$
$$= \frac{d}{dt}\frac{\partial}{\partial \dot v_{j}}(\sigma\cdot[(x\hat{i}+y\hat{j}+z\hat{k})\times(p_{x}\hat{i}+p_{y}\hat{j}+p_{z}\hat{k})])- \frac{\partial }{\partial x_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right)$$
What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$? and why $\partial\dot q_{j} = \partial\dot v_{j}$?
If anyone can give me an explanation I would appreciate it :)
|
What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$?
You mean, why does V(r) disappear from the $\frac{\partial }{\partial \dot q_j}$ term, right? It's because V(r) is a function only of $q_j$ not $\dot q_j$. Those variables are treated as independent and so $\frac{\partial V}{\partial \dot q_j}=0$.
and why
$\partial\dot q_{j} = \partial\dot v_{j}$?
Looks like you have an extra dot... Should be:
$$
\dot q_{j} = v_{j}
$$
If I'm interpretting your notation correctly... I.e., the $q_j$ are the coordinates (which are just the cartesian coordinates. And the $\dot q_j$ are the velocities...
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Could I break the sound barrier using circular motion? (And potentially create a sonic boom?) Ok, Lets say I get out my household vaccum cleaner, the typical RPM for a dyson vaccum cleaner reachers 104K RPM, Or 1.733K RPS. In theory, this disc would be travelling with a time period of 0.00057692307 seconds, If we take the speed of sound to be 344.2 metres per second, a breach in the sound barrier is easily possible for an item on the edge of the disc.
One question remains: For an extremely strong disc , could an item stuck onto it break the sound barrier, and create a sonic boom?
| I'm surprised to see no mention of turbofan jet engines in the answers so far. In fact, the blade tips of most modern turbofan engines do reach supersonic speeds. As predicted by a comment above by tpg, this does produce shockwaves from each blade tip and what you hear is a 'buzzing' sound, which is commonly described as sounding like a buzzsaw. If you listen for it, you can usually hear this sound during the takeoff roll or during other flight phases where a high power setting is being used. In case of some engines, it can be quite loud.
You can hear this 'buzzsaw' sound distinctly on this video of a Rolls Royce-powered 777 takeoff as well as this video of a 757 takeoff.
It seems that Rolls Royce engines are much more prone to producing this sound than GE or P&W engines. For example, compare the above RR-powered 777 takeoff with this GE90-powered 777 takeoff. As a fun fact, the engine in the latter video is the most powerful jet engine ever created, the GE90-115b.
Rolls Royce filed a patent in 2010 for an intake duct liner seeking to attenuate this 'buzzsaw' sound.
| {
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Solitons and its infinite extension A soliton, for example the KdV equation solution, has the profile proportional to a hyperbolic secant squared ${\text{sech}}^{2}(x-ct)$. And since it is hyperbolic it has an exponential dependence, so it has an infinite span, it has tails that extend to infinite.
However the solitons after an interaction do reemerge like nothing have happened, except by an phase shift. What I don't understand is: how can we say that they don't interact after, or even before, the interaction has happened if they have infinite tails? Because I was wondering (to myself) that if the tails are infite so their tails (of two solitons) are always interacting.
[I hope it is not a stupid question]
| An exponentially decaying tail is almost like having no tail for all practical reasons. For example consider the yukawa potential for interaction through exchange of a massive particle, it is $\propto e^{-\mu r}/r$ which is even a stronger tail than the asymptotic behavior of the hyperbolic secant. There we say that the interaction has the the effective distance of $1/\mu$, and is practically zero otherwise.
Said differently, it is the same reason that two neutrons can be considered non interacting if the are father apart than the inverse mass of the pion (the particle exchanged in the effective yukawa description)
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Why temperature of liquid drops after spraying through a nozzle? We have tested in our lab as mentioned in the picture.
We connected hot water at $130^\circ F$ at $40 Psig$ to a nozzle (bottle sprayer). We measured the temperature differences inside tank and after spraying and found that there is a temperature drop of $50^\circ F$. We know that liquid water is sprayed and some heat is lost in the form of work done inside the nozzle. The surrounding temperature is kept at $75^\circ F$. This temperature difference is due to faster heat transfer happening due to increase in surface area of water as it is sprayed? Or does it have anything to to do with work done and surface tension?
| I think that the answer is more complicated than that. What holds liquid together when it is siphoned? People don't quite know however it would seem that there is some energy that holds it together. It is related to the vibration of each molecule in the system. When you force water through a nozzle you are in fact applying pressure on the water. This forces the water molecules together and reduces the temperature through reduced vibration for each molecule. You can not create or lose energy and therefore localized potential energy is built up in the system at the point of the nozzle. This is why you can fill a higher up tank of water when you put your finger over the end of a hose pipe. You have not created energy however because the water is cooler than what it otherwise would have been. The reason that you get the spray is because for the water to hold together you need energy. When you remove the vibration energy and instead create potential energy your water has less potential to hold together and has a greater propensity to spay. It is possible that the water will lose temperature to a colder environment when the spray forms, however I don't believe that this would be the key explanation.
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Understanding the Quantum Vacuum State In terms of the creation and annihilation operators $a_{j}$ and $a_{j}^{\dagger}$ (fermionic or bosonic, doesn't matter):
Is the vacuum state $\mid\mathrm{vacuum}\rangle$ exactly the zero vector on the Hilbert space $\mathcal{H}$ in question?
For a while I thought that the answer is yes, but if I think about a finite-dimensional Hilbert space I can never apply a matrix (a.k.a, some representation of $a_{j}^{\dagger}$) to the zero vector, and get out a vector that is not the zero vector. However, you can do this with the vacuum state.
| Don't be fooled by the zero inside the ket. That is just a label. For example in Scalar QFT, the vacuum state of interacting theories is usually denoted as $|\Omega\rangle$ rather than $|0\rangle$. So no, the vacuum state does not represent the null vector of the Hilbert space.
Rather, the vacuum state is defined to be the state with the lowest possible energy, so that an application of the annihilation operator gives zero, i.e,
$$ \hat{a} |0\rangle = 0$$
so that this state has the lowest possible non-zero energy,
$$ \hat{H}|0\rangle = \frac{1}{2}\hbar\omega|0\rangle $$
Edit: The zero vector is just a mathematical object with no physical interpretation. The zero vector is formally defined as the additive identity of the additive group so that,
$$\mathbf{u} + \mathbf{0} = \mathbf{u}$$
for all vectors $\mathbf{u}$.
| {
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Acceleration due to Gravity: Free Fall If we are in a free fall which implies we are accelerating at 9.8 m/sec every sec. And let's say that we are falling into a pit that has enormous depth. So isn't this be possible that we may accelerate and surpass the magnitude of speed of light?
| You make two wrong assumptions in your question, namely that if an object is accelerating the velocity would keep increasing ad infinitum without limit, and that the acceleration due to gravity on earth is always $9.8 m/s^2$ these are both not the case.
First of. The theory of relativity doesn't allow for objects that have mass to go faster than the speed of light. You can (informally) see this by the following equation.
\begin{equation}
m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}
\end{equation}
Where $m_0$ is the mass the object for an observer at rest with respect to the object, $v$ is the velocity of the object and $c$ is the speed of light in vacuum. When the object would reach the speed of light $c$ the numerator becomes zero and the mass would become infinitely big, and it would therefore require an infinite amount of force to accelerate it. It would also have an infinite amount of kinetic energy ($1/2mv^2$), which I hope you will understand is not possible.
Secondly the rate at which object accelerate toward the earth, or fall towards any object in general is not a constant. The classical formula for the gravitational acceleration $g$ is the following.
\begin{equation}
g = \frac{GM}{r^2}
\end{equation}
Where $M$ is the mass of the earth, $G$ is the gravitational constant and $r$ is the distance between the centre of the earth and the object. The reason people say the gravitational acceleration is $9.8m/s^2$ is that the change in $g$ is negligible if we are talking about object falling over distances on the order of several meters: for the derivation of this fact see: https://physics.stackexchange.com/a/35880/76430.
I hope this somewhat answers your question.
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Why is $F=ma$? Is there a straightforward reason? Why is force = mass $\times$ acceleration? I have searched in many sites but didn't actually get at it. Simply I want to know that if a mass in space moves (gains velocity thus further accelerates), how can I think, postulate and further believe that force = multiplication of mass and its acceleration ?
| The acceleration is the effect on the objects motion due to the cause(s), which is the vector sum/total of the forces acting on the object, i.e. the total force (composed of many forces) is responsible for the acceleration (which is just one vector).
The mass is just the constant of proportionality that links total force and acceleration.
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If the Earth is a good conductor of electricity, why don't people get electrocuted every time they touch the Earth? Since the Earth is a good conductor of electricity, is it safe to assume that any charge that flows down to the Earth must be redistributed into the Earth in and along all directions?
Does this also mean that if I release a million amperes of current into the Earth, every living entity walking barefooted should immediately die?
| There is a concept of "voltage of a step"* in energy industry - if a high voltage power line is leaking into the ground and isn't shut down, then near that point the ground voltage difference over a single human step (when one feet is closer than the other) can be enough to kill a person; that's why it may be dangerous to approach fallen wires after a storm or something like that. Nowadays it's less of an issue due to more automated detection and cutoff systems, but a few decades ago it was an important hazard.
The distance is meaningful for high-voltage lines, e.g. 30 kV - 330 kV range, but even for such amounts it's not long range - the voltage dissipates rather quickly; and even rather close you wouldn't get electrocuted if just standing there without making a long step.
[*] possibly a different term should be used in English, this comes from other languges.
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In $1$-dimensional space, how would the gravity generated by an electron affect a photon moving away from the electron if the photon can’t slow down? Suppose we had a universe obeying the same physical laws as our own. But it had only one spatial dimension (represented by the $x$ axis) and it was totally empty. There are just two point-like particles in this universe:
*
*An electron which is at rest.
*A photon which is moving away from the electron.
Yet we have two important rules that can’t be broken:
*
*A photon can’t slow down, its speed must always be equal to $c$.
*Gravity affects all form of matter, even photons.
.
So how would the gravity engendered by the electron affect the photon if it can’t slow down?
If this was in $3$-dimensional or $2$-dimensional space, there would be no problem since the photon could just be slightly deviated from its trajectory. But here the photon is moving away from the electron very precisely along the axis joining them, we’re in $1$-dimensional space, the photon can’t be deviated.
We've got a paradox over here!
| The energy of a photon is given by the equation E = hf where h is Planck's constant and f is frequency. The energy would decrease, making the frequency decrease (since h is constant). So, if the photon was blue light, then it would get redder and redder as time when on. There is a point, however, when your system eventually stops working. This is because the photon actually exerts a gravitational pull on the electron so eventually it would start moving. This doesn't change the answer, but it means your system cannot be maintained as stated, indefinitely. The electron will start moving . Photons exert a gravitational pull bacease of their contribution to the Stress Energy Tensor.
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Is it possible that every single isotope is radioactive, and isotopes which we call stable are actually unstable but have an extremely long half-life? I've read that tellurium-128 has an half-life of $2.2 \times 10^{24}$ years, much bigger than the age of the universe.
So I've thought that maybe every single isotope of every single atom are radioactive, and isotopes which we call "stable" are actually unstable but their half-life are immensely big (but not infinite), like $10^{100}$ years.
Is this a possible theory or are we truly $100$% sure that stable isotopes are really eternal?
| We are never 100% certain of anything. The scientific method falsifies wrong theories, but it does not verify those we colloquially call "correct" or "true"
If we tomorrow detect a normal oxygen atom decaying, we'll have to devise new theories to explain it.
But we don't expect the things we call stable to ever decay (that's why they're called stable). We have never seen them decay, and we - within the theories we currently accept as true - see no way how they could decay. Since those theories have done well by us in other cases, there is no reason to not trust them in this case (until evidence comes in that they are indeed false).
As a side note, Tellurium-128 is simply the nuclide with the longest half-life we have ever observed decaying. There are others, thought to be unstable with much longer half-lives, which are "observationally stable" in the sense that we've never observed them in sufficient quantity and duration to see them decay.
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Does the microgravity environment in highly elliptical orbits differ from circular orbits? I think everyone understands the microgravity environment broadcast from the ISS. But the ISS stays in a fairly circular orbit, the acceleration of gravity should be fairly uniform, the altitude and velocity changing very little.
But when a ship goes into an highly elliptical orbit (like a Geosynchronous Transfer Orbit), its velocity fluctuates alot. (Kerbals taught me that much, RIP) Does this generate any acceleration (g-force) effects that you would be able to note in the spacecraft?
Bonus points for any insights into a lunar trajectory.
| When you are in a synchronous circular orbit, the gravitational force equals the radial force, at all times. If the orbit is elliptical, the variations in the orbit will translate into force variations. Whether these variations would be perceived by humans, depends on how "elliptical" the orbit is, and how close to other celestial bodies the orbit takes the station (satellite, etc.). I suspect that the variations could go from imperceptible to very perceptible!
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Black hole singularity in loop quantum gravity How is the singularity of a Black Hole treated in Loop Quantum gravity ? Does it go away ? And if it does, what's after the event horizon ?
| The theory of loop gravity would suggest that it would only become more dense in a singularity event. So It would not go away but become very very dense. So if anything the possibility arises That after the event horizon it would either become so dense it would simply just not be detected aside from its push of gravity. So if the theory dictates these events it would be small. Yet another theory suggests that it would be a repetitious event, such as it would go through and come out some where else coming out just as it was going in. So it wouldn't go away in either but become stronger and stronger in one theory and the other it would become stronger then return.
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Proton creation My question is simple: if proton antiproton pairs can be created in the laboratory (given certain energies) then they should also be able to be created in the universe at any time, not only during the Big Bang.
So, for example, a quasar could be a giant proton factory, or an accretion disc at the center of a galaxy could have energies much higher than any laboratory and the relativistic jets which are vertical to the disc could contain newly created protons, after the proton antiproton asymmetry is accounted for.
| Yes this is how proton cosmic rays are produced... quasars, supernovae, gamma rays bursts... also anti proton cosmic rays originate from proton anti proton creation in proton cosmic ray collision with nuclei in interstellar medium (otherwise anti protons cannot leave the vicinity of a matter dominated (as opposed to anti-matter) source)
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What exactly is the mass of a body? What determines it? The term "mass" is very common. But what does it depend on? How is it known?
| Since your tags are "newtonian-gravity" and "mass" I will attempt to answer this question in a classical framework. In classical mechanics, mass is essentially defined as a measure of an object's inertia.
Let me explain further. We have Newton's second law
$$F_\text{net}=ma$$
which is assumed to hold for all objects in classical mechanics. Suppose we apply a force of $1000\,\mathrm{N}$ to two objects. One is light, $m_1=10\,\mathrm{kg}$ and the other is heavy, $m_2=1000\,\mathrm{kg}$. Then the two objects have different accelerations ($a=F_\text{net}/m$)
$$a_1=100\,\mathrm{m}/\mathrm{s}^2,\quad a_2=1\,\mathrm{m}/\mathrm{s}^2$$
So the object that has more mass is harder to accelerate, i.e. has more inertia.
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Negative pressure How is negative pressure created in a fluid system?Isn't it counter intuitive that we are reducing the pressure of a system containing no molecules(zero pressure) to a lower value?
| 'Negative ' pressure is strictly a relative state; relative to what one may wish to define as zero pressure, and here on earth we chose to define that as one standard atmosphere of pressure which is about 760 mm Hg absolute pressure. If you are capable of removing all gas particles from a space, then you will achieve -760 mm Hg gauge pressure, but you cannot reduce pressure beyond that point
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Magnetic field and electric field induce one another forever A changing electric field produces magnetic field and vice versa. Does that mean that this process will carry on forever? Think of a circuit with a capacitor. The magnetic field due to the current at a point on the wire (by the Ampere-Maxwell law). But the current was changing with time, so it also meant that the magnetic field changed. And a changing magnetic field produces an electric field, so we have to go back again from the start by Ampere's law. It seems that this will go on forever. What is the final magnetic and electric field that I have to calculate?
| The idea in the comments above is a good one. The reason you don't need to worry about the order is that you're looking for an equilibrium solution. In terms of going on forever, it's broadly true. I mean electro magnetic radiation is exactly the kind of effect you're talking about. In a circuit there is normally a dissipative term, but in a steady state with a power supply you just need to solve the equations simultaneously
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Scalar and vector defined by transformation properties In Classical Mechanics, we are defining scalars as objects that are invariant under any coordinate transformation. Vectors are defined as objects that can be transformed by some transformation matrix $ \lambda $ .
Why is this important? I get that it leads to other properties such as the invariance of the dot product under coordinate rotations but how does this relate to physics? This is supposed to lead to another question but I will refrain from posting so that I may think a little about it.
I also have seen Noether's Theorem explaining that symmetries pop out conservation laws, such as the time independence of the Lagrangian gives you the Hamiltonian equating to the total energy of the system.
| Draw an arrow to represent a vector, with its length representing the vector magnitude. Draw a coordinate system and get the components of the vector. Now draw another coordinate basis, rotated with respect to the first, and get the components with respect to the new basis. The length of the arrow is the same in both systems - i.e length is invariant - and so there must be a relationship between the components in each system. That relationship is the vector transformation relation.
If it wasn't for invariants, mechanics would be impossible. The maximum displacement of a bridge, for example (and whether it falls down or not), would depend, quite literally, on how you look at it. Which is, of course, nonsense.
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Explanation for: A monopole antenna must contain a resistor (or equivalent) and therefore must have 2 terminals? Could someone explain why this sentence makes sense:
A monopole antenna transfers energy from electrical domain to the electromagnetic domain, hence must contain (equivalently) a resistor, hence must have 2 or more terminals.
Why must it contain a resistor and how can you conclude that you need 2 terminals because of the resistor?
Also,
A monopole is half of a dipole with groundplane in its symmetry plane and the drive is between antenna-feedpoint and ground(plane)
What is the symmetry plane? What does drive mean here? The transmitter?
|
Why must it contain a resistor
The author is most likely referring to the radiation resistance. A circuit delivers electrical energy to a resistor where it is converted to heat, i.e., the energy is not stored but is lost to the environment.
Analogously, a circuit delivers electrical energy to an antenna where it is converted to electromagnetic radiation which propagates away at the speed of light.
Thus, in this sense, the antenna is a 'radiation resistor'. It is a two-terminal system where one terminal is the antenna proper and the other terminal is ground.
What is the symmetry plane?
Essentially, there is an 'image' antenna on the other side of the ground plane. From the linked article:
The radio waves from an antenna element that reflect off a ground
plane appear to come from a mirror image of the antenna located on the
other side of the ground plane. In a monopole antenna, the radiation
pattern of the monopole plus the virtual "image antenna" make it
appear as a two element center-fed dipole antenna. So a monopole
mounted over an ideal ground plane has a radiation pattern identical
to a dipole antenna.
What does drive mean here? The transmitter?
Yes. Or, more precisely, the output terminals of the transmitter.
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Electron propagation
How can electrons travel in these beams if they repel?
|
How can electrons travel in these beams if they repel?
First of all, the picture you posted looks like lightning which is basically arcing, i.e., ionization of gas to create a conductive path. This is not what I would typically consider an "electron beam".
To answer you question: Creating and maintaining the integrity of an electron beam is not easy. It requires some engineering. But it is certainly possible, you just have to compensate for the repulsive force between electrons by some other forces, often referred to as electron "lenses" since they focus electron beams just like optical lenses focus light. This is done all the time with electron microscopes.
| {
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Will the electrostatic force between two charges change if we place a metal plate between them? If a thin metal plate is placed between two charges $+q$ and $+q$, will this cause a change in the electrostatic force acting on one charge due to another? What is the concept behind this? What will happen if the metal plate is thick?
| According to the formula of electric force, the electric force will be zero if a metal plate is inserted between the two charges. The charges will not experience any force because it uses the same concept as that of the hollow charged sphere.
| {
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Coriolis force: why is pole-to-equator air flow eastly? I have little knowledge in fluid dynamics, so this may be naive. But I have a question while reading a textbook about the Coriolis force, by which the rotation of the earth from west to east changes the air circulation pattern.
The text states that
As a ring of air about the earth atmosphere moves toward the poles,
its radius decreases. In order to maintain angular momentum, the
velocity of air increases with respect to the land surface, thus
producing a westerly air flow
The converse is true for a ring of air moving towards the equator --
it forms an easterly air flow.
I can't understand why an easterly flow should be generated in the "converse" part. To me, suppose we have a satellite staying put (w.r.t to earth coordinate system) above the earth. Since the earth is rotating west-to-east, the satellite should move east-to-west (westerly) relative to the earth surface, right?
How does the Earth rotation cause a easterly movement?
-- EDIT --
I was probably confused about the word "easterly", and thought it meant "to the east". As @rob pointed out, it means "from the east". In that case, the question still remains, how does the Earth rotation cause a "westerly" (i.e. west-to-east) movement as described in the first half of the quoted text?
(I really don't see the symmetry in the "conversely" part of the reasoning in terms of physics).
| $R = r(\cos A\cos B,\,\cos A\sin B,\,\sin A)$ where $A$=latitude, $B$=longitude and $r$=Earth radius
Let
\begin{align}\frac{dB}{dt}&=\frac{2\pi}{24\,{\rm hours}} \\
\frac{dA}{dt}&=v/r\end{align}
mass $m$ moving due N with speed $v$.
Differentiate the vector $R$ twice with respect to time to find the acceleration that is experienced by mass $m$. There are 3 component accelerations. The
$r\frac{dA}{dt}\frac{dB}{dt}$ acceleration component is the Coriolis's acceleration vector.
| {
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How can the Schwarzschild radius of the universe be 13.7 billion light years? So i was reading about Schwarzschild radius on Wiki and I found a interesting thing written there link.
*
*It says that the S. radius of the universe is as big as the size of the universe?
*How is this possible?
*Since most the universe is empty space shouldn't the S. radius of our universe be significantly smaller then 13.7 light years?
| Firstly we should note that the universe as a whole is not described by the Schwarzschild metric, so the Schwarzschild radius of the universe is a meaningless concept. However if you take the mass of the observable universe you could ask what the Schwarzschild radius of a black hole of this mass is.
For a mass $M$ the Schwarzschild radius is:
$$ r_s = \frac{2GM}{c^2} \tag{1} $$
If the radius of the observable universe is $R$, and the density is $\rho$, then the mass is:
$$ M = \tfrac{4}{3}\pi R^3 \rho $$
and we can substitute in equation (1) to get:
$$ r_s = \frac{8G}{3c^2} \pi R^3 \rho \tag{2} $$
Now we believe that the density of the universe is the critical density, and from the FLRW metric with some hair pulling we can obtain a value for the critical density:
$$ \rho_c = \frac{3H^2}{8\pi G} $$
And we can substitute for $\rho$ in equation (2) to get:
$$ r_s = \frac{H^2}{c^2} R^3 \tag{3} $$
Now, Hubble's law tells that the velocity of a distant object is related to its distance $r$ by:
$$ v \approx Hr $$
and since the edge of the universe, $r_e$, is where the recession velocity is $c$ we get:
$$ r_e \approx \frac{c}{H} $$
and substituting this in equation (3) gives;
$$ r_s = \frac{1}{r_e^2} R^3 \tag{4} $$
If $r_e = R$ then we'd be left with $r_s = R$ and we'd have shown that the Schwarzschild radius of the mass of the observable universe is equal to it's radius. Sadly it doesn't quite work. The dimension $R$ is the current size of the observable universe, which is around 46.6 billion light years, while the size used in Hubble's law, $r_e$, is the current apparent size 13.7 billion light years.
If I take equation (3) and put in $R$ = 46.6 billion light years and $H$ = 68 km/sec/megaParsec I get $r_s$ to be around 500 billion light years or a lot larger than the size of the observable universe.
| {
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Can gyroscope work in zero gravity? Most ships have two or more gyroscopes to balance on water, man made satellites uses gyroscope for orientation as they fall around earth. All these applications seems to be associated with gravity, therefore how can a gyroscope works in zero gravity?
|
man made satellites uses gyroscope for orientation as they fall around
earth. All these applications seem to be associated with gravity,
therefore how can a gyroscope work in zero gravity?
You are probably confusing or identifying the property of a gyroscope with the phenomenon of precession
The bicycle wheel (gyroscope) doesn't fall down and precesses because the force of gravity is deviated from the vertical axis to the horizontal one. What you see is gravity pushing the wheel around instead of pushing it downward.
But this happens because of the property of a gyroscope to oppose any change in the plane of rotation and shift it by 90°
This property is univerally valid and works in any conditions and also in zero gravity.
You can see here what happens in a spacecraft at zero gravity. The gyroscope does not rotate on its center of mass when is hit/pushed on one tip. You can see that, when it's not spinning, it rotates, exactly like on earth.
The only difference is that, if you hang it on one tip, it will stand still, it will not precess because no force is acting on it in any direction
| {
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Beginners Textbooks in physics Hello I am fifteen and I already know everything that my school has been teaching me so I have been going ahead. I have already been studying mathematics far past where I am at school, but I am very interested in physics. I want to learn everything up to advanced topics such as super-string theory. But to get there, I obviously have to start at the beginning. Any good textbooks out there for somebody like me? Preferably something with a lot of practice problems and that has many applications.
| The textbook my school uses for my AP Physics 2 course is the fifth revised edition Giancoli textbook. Like you, I am reading ahead and trying to absorb as much information as possible. I am sure many high school physics books are a great starting place to go further in studying physics at your level, but do not forget there are many resources online too. Just pick up a used textbook for cheap and get reading. Good luck brotha.
| {
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How do solar panels generate infinite electricity? We know that solar cells generate electricity utilizing the energy of the photon, but how can they generate electricity forever?
In a n-type terminal we have the bond of silicon and phosphorous so we have a free electron and when photons hit the panel they let the free electrons flow to the p-type terminal which is a bond of boron and silicon which has a free place for an electron.
But when all electrons have moved to the p-type from the n-type, then, how can they generate electricity forever when all electrons are at the p-type terminal? (because both the terminals are neutral)
| When photons from the Sun hit the crystallized silicon wafers in a solar panel, they energize electrons to become loose and make a complete trip around the closed circuit that include the solar collectors. So the solar panels do not lose electrons because they go out from one end and come back in from the other end
| {
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Spectral line formula Two years back my friend told me a simple formula for calculating the number of spectral lines. But, now I'm a bit confused about it
number of lines is =$ \frac{2(n-1)}{2}$ is this right or is there any error in it?
| You can derive it simply by noting that each level can have $n-1$ transitions,so we have
$n-1+n-2+...+1=n(n-1)/2$
| {
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Are there any scales other than temperature that have different zero points? For most physical measurements, zero is the same regardless of the units used for the measure:
$0 \mathrm{mi} = 0 \mathrm{km}$
$0 \mathrm{s} = 0 \mathrm{hr}$
but for absolute temperatures, different systems have different zeros:
$0 ^\circ\mathrm{C} \neq 0\,\mathrm{K}$
Are there any other physical, measurable quantities (other than temperature) that have different zero points?
I'm looking for measurable quantities that are applicable anywhere -- things like voltage or temperature, not local quantities like "distance from the Empire State Building".
| Sound can be measured in deciBels ($\mathrm{dB}$) but also as an intensity measured in $\mathrm{W/m^2}$.
$0\,\mathrm{dB}$ on this scale is equal to $1\times10^{-12}\,\mathrm{W/m^2}$.
| {
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Prove Christoffel Symbol Identity In a book I am reading, the following identity is claimed and then "left to the reader to prove." $g_{ij}$ is the metric tensor, and $\Gamma$ is the Christoffel symbol of the second kind with the appropriate indices.
$$\partial_k g_{ij} = g_{jl}\Gamma^{l}_{ki}+g_{il}\Gamma^{l}_{kj}$$
I have tried expanding the $g_{ij}$ term using its definition, $g_{ij}=\epsilon_{i}\cdot\epsilon_{j}$, but then I don't really know if a vector identity should be used. Moreover, I'm not even sure if that's even on the right track.
Could you possibly give me nudge in the right direction? Do I need to assume the covariant derivative of the metric tensor is zero?
| At the most basic level, you can just use the definition of the Christoffel symbols in terms of the metric:
$\Gamma^i_{jk} = \frac{1}{2}g^{is} (\partial_j g_{sk} + \partial_k g_{sj} - \partial_s g_{jk})$.
Plugging this into the right-hand side of your expression will yield the left-hand side.
However, one can obtain your expression directly from one of the properties of the Christoffel symbols; namely, that they are the connection coefficients of a metric-compatible affine connection (i.e. they can be used to construct a covariant derivative operator $\nabla_i$ which satisfies $\nabla_i g_{jk} = 0$). Expanding the equation $0 = \nabla_i g_{jk}$ out explicitly, we obtain
$0 = \nabla_i g_{jk} = \partial_i g_{jk} - \Gamma^s_{ij} g_{sk} - \Gamma^s_{ik} g_{js}$,
which gives
$\partial_i g_{jk} = \Gamma^s_{ij} g_{sk} + \Gamma^s_{ik} g_{js}$.
This is precisely the equation you're after.
| {
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Sun and planets orbit each other Do not the planets and the Sun revolve in orbits around each other and the shape of the orbit depends on where the center of gravity of the system is? The greater the mass of the Sun, the closer the orbit approximates a perfect circle.
| Yes, objects with mass all attract to each other and move each other of course, except that the star doesnt change it's theoretical orbital shape depending on it's mass, it probably just experiences small tidal forces that aren't much bigger than it's own centrifugal forces. The oscillation of the star position is a complex dynamic based on it's surrounding stars, activity, rotation, and fluidity that all work together. if the sun was twice it's current mass, the oscilaltions of its theoretical centre probably wouldnt be more curcular mathematically.
| {
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Path integral in quantum mechanics I am confused by the derivation in Srednicki QFT's chapter 6 from (6.8) to (6.9). In (6.8), we have
$$<q'',t''|q',t'>~=~\int DqDp \exp[i\int_{t'}^{t''}dt(p\dot{q}-H(p,q))],\tag{6.8}$$
and (6.9) we have
$$<q'',t''|q',t'>=\int Dq \exp[i\int_{t'}^{t''}dt L(\dot{q},q)].\tag{6.9}$$
It is clear for me that one can work out each infinitesimal integral $$<q_k|\exp[-i\delta t \frac{p^2}{2}]|q_{k-1}>~\sim~ \exp[\frac{i(q_k-q_{k-1})^2\delta t}{2}] $$ to derive the above formula. But I'm confused by the way that is presented in the book. It makes it sound like there is a more general way of computing path integral by finding the stationary point, i.e. given
$$\int Dp \exp[i\int f(p,t)dt]$$
is the result $$\exp[i\int F(t)dt]$$ where $F(t)=f(p(t),t)$ such that $p(t)$ is a stationary point of $f(p,t)$ with respect to $p$.
| I did not get my copy of Srednicki out but from what you have written...
Srednicki is referencing the method of steepest descent. Although these notes look to be better than wikipedia. Another page that is directly applicable to the quantum field theory case is here.
In short, exponential integrals may be estimated by the saddle points of the integrand. Using a Minkowski formalism the saddle points are related to how oscillatory the integral is.
| {
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Neutral $K$ and $B$ mesons decay to 2 photons? The neutral pion $\pi^0$ decays almost exclusively to 2 photons, $\pi^0 \rightarrow \gamma \gamma$,
which got me thinking:
Can we have $K^0 \rightarrow \gamma \gamma$ and $B^0 \rightarrow \gamma \gamma$ or are they forbidden for some reason?
| They are both allowed, but they are flavor-changing neutral currents that are loop-supressed. The $B^0\to\gamma\gamma$ decay occurs by the Feynman diagrams from arXiv:1010.2229 and I imagine that kaon decay proceeds similarly.
They are so suppressed that the $B^0$ decay hasn't been measured, but PDG report an upper bound on the branching ratio of $<3.2\times 10^{−7}$. The kaon decay has been measured as long ago as the 60s, remarkably. The branching ratios are much less than that of the pion, but, then again, the $\pi\to\gamma\gamma$ decay proceeds through the axial anomaly, so it's not surprising that it's atypical.
| {
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What does it mean that a magnetic field's flux vanishes through any closed surface? I'm reading the Britannica guide to Electricity and Magnetism, and I came across the following quote:
A fundamental property of a magnetic field is that its flux through
any closed surface vanishes.
Can someone explain this in simpler terms?
Source
| There are no magnetic monopoles. i.e. Unlike electric field flux, there are no sources or sinks of magnetic flux.
Therefore the amount of flux entering any closed volume must equal the amount exiting.
| {
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Ways for undergrad to understand physics research? I am a college freshman who is looking to research in a experimental physics lab this year. I am an engineering student and have taken basic Physics classes, but I lack the background knowledge to fully understand what is going on with the topics this lab is researching (Rydberg atoms, Rydberg atom collisions, heavy-Rydberg-ion-states, etc.) I've tried looking for review papers but have mostly found things from the 80s, (some of which were written by the professor I hope to research with). What resources are good for a relative novice to learn about these things? Sorry if this question is too broad!
| Wikipedia does a pretty good job at explaining the basics of this, in my opinion. Looking at review papers are very good for this and so are the papers(you can find the papers on sites like Pubmed and APS Journals). Just because the review papers are from the 80s, does not mean they are bad. Even if a tiny fraction of the information has changed it is still not bad. In fact, reading what the professor has written is probably one of the best things you can do, because you learn more about what he studies, more about the field and the quality of the papers he publishes. In addition to all of this, even though it may seem obvious, are books.
| {
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What is the capacitance and inductance of an ideal wire? I am reading a magazine for acoustics, one article is about the choice of speaker wire. The article said the ideal speaker wire should has no resistance, capacitance and inductance. I understand that for the low resistance in ideal case but doubts on the capacitance and inductance. We are driving in the speaker with AC signal, says ideal sinusoidal wave, in the text, it is given that the voltage drop on the capacitor which is connected to the AC power should be
$$v_c = I_{max}X_C\sin\omega t $$
where $X_C=1/(\omega C)$. My understanding is for ideal wire, the voltage potential on a short section on the wire should be zero, so we should have $v_c=0$ so does it mean $C\to \infty$ when $\omega\neq 0$?
Similar reasoning but on inductance, we have
$$v_L = -I_{max}X_L\sin\omega t $$
so for ideal wire, $X_L$ should be zero. Are those reasoning corret?
| This would actually be easier to answer over at the EE stackexchange site since there is a handy schematic editor built in.
First, note that, by speaker wire, we're actually referring to a speaker cable; in this case, a pair of wires.
For each wire, we can assign a series resistance and inductance (per foot), i.e., the $R$ and $L$ of each wire is in series with the load.
Since we do not want a voltage drop from one end of the wire to the other - we want all of the source voltage to appear across the load - the ideal case is that the resistance and inductance is zero, i.e., there is zero voltage drop from one end of the wire to the other.
Now, since we have a pair of wires, we can also define a mutual capacitance (per foot) between the wires. This capacitance appears in parallel with the load. If the capacitance is non-zero, higher frequency currents will be somewhat shunted around the load. So, ideally, the cable capacitance is zero, i.e., there is zero current shunted around the load.
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Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? I was asked by an undergrad student about this question. I think if we were to take away air molecules around the pencil and cool it to absolute zero, that pencil would theoretically balance.
Am I correct?
Veritasium/Minutephysics video on Youtube.
| No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's measured in GigaPascal)
| {
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Why do we use capacitors when batteries can very well store charges? Can batteries be used instead of capacitors? I am trying to figure out a basic, superficial and any obvious difference between the two.
| batteries are a much more efficient at storing electricity but in circuits, it makes much more sense to use capacitors in circuits as they are much more efficient for the short term storage of electricity. batteries are a lot more bulky and to work as a capacitor they would need to be rechargeable. it would not make sense to have two batteries in a single circuit anyway because then you would have two possibly conflicting power sources. furthermore batteries cannot be fitted as easily into microchips and cannot be used in all the same applications as capacitors. capacitors can also be used to smooth the irregularities in corrected DC current.
a capacitor acts as an dam for electricity that resists changes in potential difference (voltage).
a battery acts at a supply of electricity due to the potential difference across it.
| {
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General Relativity visualization software As I am approaching the study GR, I was wondering if there are softwares that allow a quick visualization of custom metrics, curvature, and particle motion even in the limited context of 2D space.
Playing with equations is fun, but it would be more fun if I could play with various parameters and see the outcome.
Obviously free would be better, but I am open to commercial programs.
| For particle/light motion in 2D space, my nomination would be GROrbits
It's free and requires a JVM to run, there is also a web start version for the brave ;)
Sorry but I've never found anything aimed at visualizing metrics or curvature (apart from plotting programs of course).
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Is there any physical meaning for the inverse metric? I've been wondering if we can attribute any physical meaning to the inverse metric. I mean when we talk about the metric itself, there are lots of insights we can have towards its role in spacetime, yet I cannot see any physical meaning for the inverse metric. For now, I just see it as tensor with the special property of giving the identity when joined with the metric. Rigorously speaking, I would say it is not even an "inverse" actually, as it doesn't map like one. But still, is there any physical way of interpreting this tensor?
| The tensor algebra is symmetric between one-forms and vectors. One could start with defining any of them first and then obtain the rest of the things.
The inverse metric tensor is a linear map that takes two one forms on a manifold and maps into $\mathbb{R}.$
$g^{\mu \nu}: A_\mu,B_\nu \rightarrow \mathbb{R}$
It of course tranforms like a vector with respect to both the indices.
So to answer your question, the inverse metric tensor is as physical as the metric tensor. Just as metric tensor provides a way to measure length of vector fields, inverse metric provides a way to measure length of one-form fields.
| {
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Why stars are white? According to Rayleigh Scattering According to Rayleigh Scattering, the red waves are capable of travelling a long distance, so that only we are seeing the Sun as reddish during Sunset and Sunrise. If this was true then all other stars must also appear red in colour, as the distance through the atmosphere travelled by those lights is further during sunset or sunrise.
Why do stars still appear white?
| The colour of stars as observed by an observer on Earth varies just like the colour of our own Sun, depending on where in the sky the source is relative to the observer.
However, the light of stars is generally too faint to notice this as clearly with the naked eye, because we cannot perceive colour for weak light sources.
| {
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Considering this hypothesis...is charge really quantized? [If anything goes against any mathematical or physical rules please let me know. I am a first year undergraduate student perusing a joint major in mathematics and physics so I do not have a complete background in those fields. I am just using my imagination and what I already know]
Consider an isolated electron in space. We know at all points in this space an electric field exists due to this electron (except perhaps at infinity).
Now, from Coulomb's Law we can find a vector for the electric field due to this electron at all points in this space.
Since this is possible, does this imply the charge of the electron constitutes an infinite number of very small charges 'dq' that each produce a linear field where, each oppositely positioned dq destroys the field within the electron (shown in attached picture)?
| The electron is an elementary particle in the underlying building blocks of matter organized in the elementary particles table of the standard model of particle physics. Elementary particles are point particles.
The standard model is a precis of a very large number of measurements (data) fitted by mathematical models of theoretical physics. A point particle has no charge distribution, its charge is given in the table in the link.
To understand elementary particle physics one needs quantum mechanics and this cannot be done except through serious study in the appropriate courses.
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"timestamp": "2023-03-29T00:00:00",
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Why is the singularity of the Big Bang not considered to be the center of the Universe? If the universe is expanding, then at some time in the past, it must have started from a single point but why this point is not the center of the Universe. Just like the singularity of black holes is its center?
| I'll answer your question with an analogy. Imagine a really small balloon, so small that it occupies a point. Now, imagine that the balloon is expanding uniformly outward from that point. Note that that central point is not part of the balloon. It's the same idea as to what happened with the BB. In this analogy, the universe is the surface of the balloon. The central point is where the universe started, where the BB happened, but that point is actually not part of the universe (it's not on the surface of the balloon). Incidentally, if you look anywhere on the surface of the balloon (anywhere in the universe), you'll notice that the surrounding area is expanding around the point you're looking at so it looks as though any point is the centre of expansion. That is also true of the universe. If you focus your attention on any point in the universe, it looks as though everything is expanding outward from that point so that it looks as though that point is the centre of expansion.
| {
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How does temperature relate to the kinetic energy of molecules? In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules. If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to distinguish the random vibration KE and KE in one direction?
Furthermore, if we accelerate a block of metal with ultrasonic vibrator so that the metal is vibrating in very high speed with cyclic motion, can we say the metal is hot when it is moving but suddenly become much cooler when the vibration stop?
|
In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules.
In the kinetic theory of gases random motion is assumed before deriving anything.
If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to distinguish the random vibration KE and KE in one direction?
The temperature is still defined by the random motion, subtracting the extra energy imposed . This is answered simply by the first part of @LDC3 's answer. Does your hot coffee boil in the cup in an airplane?
Furthermore, if we accelerate a block of metal with ultrasonic vibrator so that the metal is vibrating in very high speed with cyclic motion, can we say the metal is hot when it is moving but suddenly become much cooler when the vibration stop?
This is more complicated, because vibrations may excite internal degrees of freedom and raise the average kinetic energy for that degree of freedom. It would then take time to reach a thermal equilibrium with the surroundings after the vibrations stop. If one supposes that this does not happen, then the answer is the same as for the first part, it is the random motions of the degrees of freedom that define the kinetic energy which is connected to the definitions of temperature. So no heat will be induced by the vibrations.
| {
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How do we know that the cesium-beam frequency used in atomic clocks is always the same? Atomic clocks use cesium-beam frequency to determine the length of a second. This has shown that the period of orbit of the earth is decreasing.
But what experiment showed that cesium-beam's period was so terribly consistent?
Did they just run several atomic clocks and note that there was no drift? Couldn't many things that might change the frequency of one clock also impact the other so this would not work?
| The frequency is determined by the energy spacing between two configurations of the caesium atom. Caesium has a single electron in the outermost $6s$ orbital, and this electron can be aligned with or against the nuclear spin. These two configurations differ in energy by about 0.000038 eV, and transitions between them produce/absorb light with a frequency of 9,192,631,770 Hz. This is the frequency used to measure time.
The only way this frequency could change is if the energy spacing of the two configurations of the caesium atom changed. But these energies are dependent only on fundamental constants such as the electron charge, mass, coupling constant, etc., and these constants are, well, constant. That means the energy levels must be constant and hence the frequency of the light emitted must be constant as well.
It's not impossible that the fundamental constants actually aren't constant, but if they were changing the effects of the changes would be wide reaching and affect far more than just the caesium atom. We would certainly have noticed by now :-)
I'm not sure what experiments have been done to measure the constancy of the caesium energy levels, but note that international atomic time is derived from around 400 different atomic clocks in different parts of the world. If all the clocks were subject to the same change we wouldn't see it, but random changes in clocks would be immediately detected because the clocks would get out of sync.
| {
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Entropy - Gas Inside A Closed System Reaches Maximum Entropy Filling a box with a certain amount of gas with a specific total energy and allowing the gas to reach a maximum entropy state, what happens next?
Would the gas remain in a maximum entropy state indefinitely?
What would prevent the gas atoms/molecules to end up in a more orderly state at some point just out of coincidence? After all, the gas particles have some energy total in this closed system and will keep moving around.
Is it even possible that if we waited a long long time, those gas molecules could fall back in a state of minimum entropy at some point?
edit: Just how many maximum entropy states are there in a gas with N particles inside a closed system vs lesser entropy states? Would it be even more likely for a gas to decrease in entropy than remain in a maximum entropy state? If yes, then this would be a clear violation of the 2nd law of thermodynamics. It would have to be stated more precisely.
| The second law of thermodynamics says that the entropy of an isolated system can increase, but that entropy can not decrease without the addition of energy to the system, or the transfer of entropy to another system.
Increasing entropy has been associated with the arrow of time, as entropy seems to be the only quantity in physical processes that requires time to be directional.
I am unsure how one could determine if an isolated system had reached "maximum entropy", as that term seems to have several technical meanings, some of which are used in the social sciences:
http://en.wikipedia.org/wiki/Maximum_entropy
| {
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Why doesn't a magnet on a refrigerator slide down? Obviously gravity pulls objects towards the earth's surface. Now suppose I have a refrigerator and a magnet. The magnetic force is perpendicular to the gravitational force. So it is not counteracting it. So why does the magnet not get pulled to the ground since no other force is opposing it?
My instinct is that the magnetic force pulls the magnet closer to the refrigerators surface and this increases the friction. as a result the magnet does not fall to the floor. But I'm not sure this is correct.
| As you may know, the friction is proportional to the normal force of an object or in this case the force of attraction between magnet and refrigerator.
If your force is strong enough then the friction will be sufficient and the magnet will not slip (on earth the force of friction must exceed the mass of your magnet multiplied by 9.81 m/s).
If we assume your magnet weighs 0.010kg and is made of iron, while your refrigerator is steel then the coefficient of friction is 0.4 (Engineering Toolbox, Friction and Coefficients of Friction). The force exerted by the magnet on the steel is 5N.
Friction = 5 x 0.4 = 2N
Now we find that the Force down is equal to 0.010 x 9.81 ~ 0.1N
The magnet would stick
| {
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Since electron clouds of different atoms repel each other, does that mean that touch is the feeling of electromagnetic repulsion? Also when we rest our hand on an object does that mean we are effectively levitating because of the repulsion of the electron clouds?
| Yes ... but let's be careful to understand that the sensation of touch is a psychophysical phenomenon. The electrons at the surface of an object "push" against the electrons at the surface of your fingers. The electrons never touch each other. Your skin deforms a bit, and the nerves in your fingers detect this deformation, and send a signal to your brain. Your brain creates the sensation of touch.
I don't know if I'd use the word "levitate", but it is true that there is a tiny gap between the electrons.
| {
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What is the position as a function of time for a mass falling down a cycloid curve? In the brachistochrone problem and in the tautochrone problem it is easy to see that a cycloid is the curve that satisfies both problems.
If we consider $x$ the horizontal axis and $y$ the vertical axis, then the parametric equations for a cycloid with its cusp down is:
$$\begin{cases}
x=R(\theta-\sin\theta)\\
y=R(\cos\theta-1)
\end{cases}$$
A cycloid is the curve of fastest descent for an idealized point-like body, starting at rest from point A and moving along the curve, without friction, under constant gravity, to a given end point B in the shortest time.
A cycloid is also the curve for which the time taken by an object sliding without friction in uniform gravity to its lowest point is independent of its starting point.
We already know the parametric equations for the the geometry of the path of the mass, but I would like to know the position of the object over time.
I would like to know what is the solution to the equations of motion.
It would be like specifying the arc length as a function of time $s=s(t)$ in the figure:
To sum up, what I mean is:
Given certain initial conditions, what are the expressions $x(t)$ and $y(t)$ that give the position as a function of time $\mathbf{\bar{r}}(t)=\left ( x(t),y(t)\right )$
for a particle falling down a cycloid curve?
|
1. Brachistochrone
\begin{equation}
\boxed{\:
\begin{matrix}
x\left(\theta\right) = R\left(\theta-\sin \theta\right)\\
y\left(\theta\right) = R\left( 1-\cos \theta\right)
\end{matrix}\:}
\tag{b-01}
\end{equation}
\begin{equation}
\omega= \dfrac{\,\theta \,}{t}=\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\sqrt{\dfrac{\,g\,}{R}} =\text{constant}
\tag{b-02}
\end{equation}
\begin{equation}
\boxed{\:
\begin{matrix}
& x\left(t\right) = R\Biggl[ \sqrt{\dfrac{\,g\,}{R}}\,t-\sin \left(\sqrt{\dfrac{\,g\,}{R}}\,t\right)\Biggr]=R\Bigl[\omega\,t-\sin \left(\omega\,t\right)\Bigr]\\
& \\
& y \left(t\right)= R\Biggl[1-\cos \left(\sqrt{\dfrac{\,g\,}{R}}\,t\right)\Biggr]=R\Bigl[1-\cos \left(\omega\,t\right)\Bigr]
\end{matrix}\:}
\tag{b-03}
\end{equation}
\begin{equation}
s\left(t\right)=4R\Biggl[1-\cos\left(\dfrac{\theta}{2}\right)\Biggr]=4R\Biggl[1-\cos\left(\sqrt{\dfrac{g}{4R}}\,t\right)\Biggr]=4R\Biggl[1-\cos\left(\frac{\omega}{2}\,t\right)\Biggr]
\tag{b-04}
\end{equation}
Time of descent from point $\mathrm{A}(0,0)$ to lowest point $\mathrm{F}(\pi\,R,2R)$: from (b-02) with $\:\theta=\pi\:$
\begin{equation}
t\left[\mathrm{A}\rightarrow\mathrm{F} \right] = \pi\sqrt{\dfrac{\,R\,}{g}}
\tag{b-05}
\end{equation}
2. Tautochrone
\begin{equation}
t\left[\theta_{0}\rightarrow\theta\right]=\sqrt{\dfrac{\,R\,}{g}}\Biggl(\pi-2\arcsin\Biggl[\dfrac{\cos\left(\theta/2\right)}{\cos\left(\theta_{0}/2\right)} \Biggr] \Biggr)
\tag{t-01}
\end{equation}
Time of descent from $\:\theta_{0}\:$ to the lowest point $\:\theta=\pi\:$ : from (t-01) with $\:\theta=\pi\:$
\begin{equation}
t\left[\theta_{0}\rightarrow\pi\right]=\pi\sqrt{\dfrac{\,R\,}{g}}=\text{constant independent of $\theta_{0}$.}
\tag{t-02}
\end{equation}
Also from (t-01)
\begin{equation}
\cos\theta=\left( \dfrac{1+\cos\theta_{0}}{2}\right) \left[ 1+\cos\left(\sqrt{\dfrac{\,g\,}{R}}\,t \right) \right]-1
\tag{t-03}
\end{equation}
\begin{equation}
s\left(t\right)=4R\Biggl[\cos\left(\dfrac{\theta_{0}}{2}\right)-\cos\left(\dfrac{\theta}{2}\right)\Biggr]=4R\cos\left(\dfrac{\theta_{0}}{2}\right)\Biggl[1-\cos\left(\sqrt{\dfrac{g}{4R}}\,t\right)\Biggr]
\tag{t-04}
\end{equation}
\begin{equation}
\theta\left(t\right)=\arccos\Biggl[\biggl(\dfrac{1+\cos\theta_{0}}{2}\biggr) \left[ 1+\cos\left(\sqrt{\dfrac{\,g\,}{R}}\,t \right) \right]-1\Biggr]
\tag{t-05}
\end{equation}
\begin{equation}
x\left(t\right)=R\big[\theta\left(t\right)-\sin\theta\left(t\right)\bigr]
\tag{t-06}
\end{equation}
3. Cycloid(properties)
| {
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Why performing axial symmetry, results in the same masses for pion and sigma mesons? Under axial transformations, $\sigma$ and $\pi$ are rotated into each other:
$\vec{\pi} \rightarrow \vec{\pi}+ \vec{\theta} \sigma $,
$\sigma \rightarrow \sigma+ \vec{\theta}.\vec{\pi} $.
In arXiv:nucl-th/9706075, page 12, it is the written that if axial symmetry is a symmetry of QCD Hamiltonian, then $\sigma$ and $\pi$ should have the same eigenvalues, i.e., the same masses.
My question is that how this symmetry results in expecting the same masses for $\sigma$ and $\pi$? I know that these mesons have different masses and it is the result of spontaneous symmetry breaking, but if the symmetry was not broken, why should we expect the same masses for these states? How can we prove this?
| The mass terms for the $ \sigma $ and $ \vec{ \pi } $ fields are,
\begin{equation}
m _\sigma \sigma \sigma + m _\pi \vec{ \pi } \cdot \vec{ \pi }
\end{equation}
You have two terms that are going to turn into each other under a symmetry transformation. Thus they need to have the same coefficient in order to remain invariant under the symmetry (feel free to explicitly stick in the transformations if you are not comfortable with this). If we do have, $ m _\sigma = m _\pi \equiv m $ then the mass terms take the form,
\begin{equation}
m \Pi \cdot \Pi
\end{equation}
where $ \Pi \equiv \left( \vec{ \pi } , \sigma \right) $ and
\begin{equation}
\Pi \rightarrow \left( \begin{array}{cc}
1 & \vec{ \theta } \\
- \vec{ \theta } & 1
\end{array} \right) \Pi
\end{equation}
(you missed a minus sign above) under the axial symmetry. Since this is a unitary transform, it leaves the mass term invariant as expected. Thus we conclude that for the system to be invariant under axial transformation it requires the masses of the $\sigma $ and $\vec{\pi}$ be degenerate.
| {
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How can the speed of sound increase with an increase in temperature? I was reading a textbook. I found that it was mentioned the speed of sound increases with increase in temperature. But sound is a mechanical wave, and it travels faster when molecules are closer.
But an increase in temperature will draw molecules away from each other, and then accordingly the speed of sound should be slower. How is it possible that the speed of sound increases if temperature increases? What is the relation of speed of sound and temperature?
| Sound waves propagate through a medium as the result of collisions between molecules. At higher temperatures, molecules have greater kinetic energy, and as they move faster their collisions occur at greater frequency and they carry sound waves faster. Greater kinetic energy = less inertia = increased speed.
However, as sound waves are compressional waves traveling through a compressible medium, their speed depends not just on inertia of the medium, but also on its elasticity.
Generally, the closer together molecules are, the faster they will carry sound waves. Although distance between molecules tends to increase when a medium is heated, this is relatively less important to the speed of sound within a given medium than is the faster movement of the molecules.
| {
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Why can't we use a pair of polarizers (whose pass axis are oriented at 90 degrees) to protect from entry of harmful radiation? If a pair of polarizers can block EM waves passing through it, then can we use them for
*
*protecting against harmful radiation (in nuclear reactors and also in spaceships)
*in thermo flasks to prevent loss of heat by radiation
*attaining low temperature by blocking cosmic background radiation
I hope my question is clear.
| A polariser is quite good at blocking EM radiation, but not perfect.
No polariser is perfect, and does not block 100% of the radiation. This is typically specifed in terms of the extinction ratio. The best laboratory grade polarisers have extincition ratios of the order 100000:1 [1].
Such good polarisers are only possible over certain wavelengths, and for high-energy radiation (like x-rays, gamma rays), polarisers don't exist or have much lower efficiency. Additionally, for such types of radiation, the radiation itself could potentially damage the structure of the material itself.
Many types of harmful radiation are not electromagnetic, but rather high-energy charged particles, which are not blocked by polarisers.
For blocking thermal radiation from the outside to keep something, you also need to consider that the polariser emits its own thermal radiation, so you would need to cool it down. In this case, a metal heat shield in a cryostat is just as good. Thermos flasks that keep things warm need to reflect thermal radiation, whereas polarisers absorb it.
[1] http://www.thorlabs.de/newgrouppage9.cfm?objectgroup_id=752
| {
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Units of eigenvectors Consider for example a mass matrix $M$, $\lambda$ one eigenvalue and $X$ a corresponding eigenvector. Then $[M]=\text{mass}$ (the brackets indicate the "unit operator"), and $MX=\lambda X$ so $[M][X]=[\lambda][X]$, so $[\lambda]=\text{mass}$. That's why for example in oscillators, the pulsations $\omega$ are such that $[\omega^2]=[M^{-1}K]=\text{seconds}^{-2}$.
But what about the eigenvectors? I would tend to think that they are dimensionless, because during a change of basis $u=Pq$, $q$ and $u$ have the same units, while $P$ gathers the eigenvectors $X$; the vectors designate a change of basis but the vector space remains the same.
Is it so?
| Not just with eigenvectors, but with any vector.
I always wondered when you decompose any vector into magnitude and direction who gets the units. Is it $(1,0.1,0)\cdot3\mathrm{m}$ or $3\cdot(1\,\mathrm{m}, 10\,\mathrm{cm},0)$ ?
I think it is up to the user to interpret a decomposition any way they see fit.
With eigen vectors you have $$\hat{y}=A \hat{x} = \lambda \hat{x} $$
which is interpeted as $A$ and $\lambda$ carrying the units of $\hat{y}$, or $\hat{x}$ carrying the units of $\hat{y}$. Or it is both.
For mechanics to understand the units you need to phrase the problem in terms of forces
$$ K \hat{x} - \omega^2 M \hat{x} = 0$$
and then multiply by $M^{-1}$ giving $$M^{-1} K \hat{x} = A \hat{x} = \omega^2 \hat{x} $$
So in this case it would make sense to retain the units of distance for the eigenvectos $\hat{x}$. But it doesn't have to be like this. All it matters is that $\lambda \hat{x}$ is the same units as $A \hat{x}$.
| {
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How come the length of a wire does not affects on the circuit current? Today we started learning about the electromagnetic induction. Out teacher gave us the following explanation:
Suppose we have a conduction frame inside a magnetic field $\vec{B}$ going towards the page. That frame has a stationary edge (also a conductor), which we will call $AB$. $AB$ is pushed right with a starting velocity $\vec{v}$. Lets say $AB$ moves at a constant speed. As a result of the edge's movement, Lorentz force acts on it's inside charges. It's direction on the positive charges is up (towrads $A$), and on the negative charges, down (towards $B$). This seperation creates an electric potencial $V_{AB}$, which will cause an electric force $\vec{F_E}$ to act on the edge's charges as well. The situation will balance when:
$$F_B = F_E$$.
Thus
$$qvB=qE$$
We know that $E=\frac{V_{AB}}{d}$, where $d$ is the length of $AB$. Thus:
$$qvB=q\frac{V_{AB}}{d}$$
Thus
$$V_{AB}=vBd$$
This represents the voltage created between the two sides of the edge, which moves inside a magnetic filed. It will, obviously, create a current in the circuit, as follow:
$$I=\frac{V_{AB}}{R}=\frac{vBd}{R}\quad (1)$$
Where $R$ is the resistance of the circuit.
From the last equation, we can learn that a longer edge $d$, will produce a larger current $I$. So far so good.
But, if we dig down in the definition of the resistance of a wire, we will remember that:
$$R = \rho\frac{d}{A} \quad (2)$$
Where $d$ is the length of the conductor, $A$ is the cross-sectional area of the conductor and $\rho$ is the electrical resistivity. The last formula states that there is a direct relation between $R$ and $d$. If we combine $(1)$ and $(2)$, we produce the follow:
$$I=\frac{vBd}{\rho\frac{d}{A}}=\frac{vBA}{\rho}$$
Wait - the last formula shows us that the current $I$ has no relation to the length of the wire $d$ - but how is that possible? Moving a million metre wire and a one meter wire inside a magnetic field will produce the same current? My teacher said that isn't true, but we had no time to find the error, if such one trully exist.
Thanks in advance :)
P.S - i'm just a highschool student, so please use math as simple as possible to answer. Thanks.
| Right, but you are not getting something for nothing. What about the energy losses? These scale as $I^2 R$ and so does depend on the length (and area) of the wire.
$$ W = I^2 R = \frac{v^2 B^2 A^2}{\rho^2} \frac{\rho d}{A} = \frac{v^2 B^2 A d}{\rho}$$
| {
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How does electromagnetic radiation affect the velocity of a charged particle? I've heard that the acceleration of a charged particle releases electromagnetic waves. So let's say there is a charged electron moving forwards in a region with a downwards magnetic field. If the magnetic field is a certain strength, it should cause the particle to travel in a circular arc. This kind of motion is centripetal acceleration, and since the particle is accelerating and is charged, it should release radiation. Is this correct? And if so, how is the kinetic energy of the charged particle affected by this release of radiation?
| It will decelerate causing its speed to decrease, and because of $r=mv/qB$ the radius will decrease as well and you will get a spiral motion.
This deceleration due to radiation is known as the Abraham-Lorentz force of radiation backreation. Using these equations you can more precisely derive the spiral motion. This effect is also responsible for the failure of the classical picture of the atom, because of the collapse, which is saved by the uncertainty principle.
| {
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Can electricity flow through vacuum? People say yes, and give a wonderful example of vacuum tubes, CRTs. But can we really say that vacuum (..as in space) is a good conductor of electricity in a very basic sense?
| Summary: (?) A vacuum is the absence of atmosphere and is a neutral force--offering neither resistance nor conduciveness to proton/electron flow. The state of "vacuum" does not compare with the state of "space". Any/all space is a physical measure of distance and can be overcome by the optimal difference of potential.
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Making sense out of covariance and contravariance I just read about co- and contravariant vectors and I am not sure that I got it right:
If we imagine that we have a n-dimensional manifold $M$ then a tangent space is spanned by the vectors $\partial_1,...,\partial_n.$ These guys transform from one coordinate system to another by
$$ \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j}.$$
This transform is according to wikipedia called the covariant transform. Now, it is worth noticing that normally the covectors are the elements in the dual space. The basis vectors of the dual space are given by $dx_1,...,dx_n.$ They transform differently as
$$dx^i = \frac{\partial x^i}{\partial y^j} dy^j.$$
Despite, although we transform covectors this transform is called contravariant. So somehow it seems as if the kind of transform does not fit to the kind of vector we are considering here and I don't see why this happens.
If you have any questions, please leave me a comment.
| Basically, vectors are called contravariant because their components transform oppositely to the basis vectors: if our change of coordinates is such that
$$ \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j}$$
then if we have a vector $\mathbf{V}$, its components $V^i_x$ in the $x$ coordinates are related to its components $V^i_y$ by
$$V^i_x = \frac{\partial x^i}{\partial y^j} V^j_y.$$
By the same logic, 1-forms are called covectors or covariant vectors because their components transform like the basis vectors, while the basis covectors transform like the components of vectors.
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Why do spatial filters use microscope objectives, and not other types of lenses? A spatial filter is a device to 'clean up' a laser beam with an irregular intensity profile, and create a smooth Gaussian profile at the output.
It is usually said (e.g. here) that you need a microscope objective and a pinhole for this. The microscope objective creates the Fourier transform of the laser beam at its output. The pinhole acts as a low-pass filter in the Fourier plane of the lens, to remove unwanted high spatial frequency components of the beam.
Why is the microscope objective necessary, instead of any other lens with diffraction-limited performance (e.g. aspheric lenses)? What makes microscope objectives more suitable?
| Yes, that is a good answer. Microscope objectives are relatively cheap and well corrected on axis to provide a nice Airy pattern at the image. The pinhole usually sized so its diameter is the same as the first dark ring of the Airy pattern will provide a very clean beam when aligned properly. Also microscope objectives are easily held and small and you can handle them easily. Spatial filter hardware comes with mounts for microscope objectives. Can you imagine having to hold a tiny aspheric lens to do the same thing? Microscope objectives come in handy little packages and stuff.
| {
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Paramagnetic/ferromagnetic transition under a magnetic field The paramagnetic/ferromagnetic phase transition is an archetypal example of a continuous (or second-order) phase transition. When the temperature $T$ approaches the Curie temperature $T_c$, the magnetization $M(T)$, which is the order parameter of the transition, continuously goes to zero.
I heard (in an informal context) that under a constant nonzero magnetic field $H$, the transition becomes first order, but I was not able to find references which clearly confirm or infirm this statement, and I am not convinced of any of the possibilities.
So my question is : is the paramagnetic/ferromagnetic transition under magnetic field, $H\neq 0$, a continuous or a first order transition ? (It would be better with a reference, but I would also like an explanation.)
Thanks in advance.
| The Landau model for ferromagnetism has the following expression for the free energy density, as a function of temperature $T$ and magnetization $M$:
$$F(T,M)=F_0(T)+\dfrac{a}{2}(T-T_C)M^2+\dfrac{b}{4}M^4+\dfrac{c}{6}M^6+\mathcal{O}(M^6)$$
First order phase transition occurs when the first derivative of $F$ (namely, the entropy) is discontinuous as $T\to T_C$, which happens in the case of a non-vanishing external magnetic field, represented by $b<0$. The stable minimum of $F$, for $T<T_C$ (and $T\sim T_c$) is:
$$M_0^2=\dfrac{|b|}{2c}\left(1+\sqrt{1-\dfrac{4ac}{b^2}(T-T_C)}\right)\simeq\dfrac{|b|}{c}+\dfrac{a}{|b|}(T_C-T)$$
The first order phase transition, at $T=T_C$, is thus characterized by the discontinuity $\dfrac{|b|}{c}$ in the order parameter, which is reflected in the entropy:
$$S=-\left(\dfrac{\partial F}{\partial T}\right)_M=-\dfrac{dF_0}{dT}-\dfrac{a}{2}M^2$$
$$\lim_{T\to T_C}\left[S_{T<T_C}-S_{T>T_C}\right]=-\dfrac{a|b|}{2c}$$
| {
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Rotation in the x-t plane I am currently studying special relativity using tensors. My lecture notes (which happen to be publicly accessible, see top of page 99) say that the standard configuration can be viewed as a rotation in the x-t plane. Can anyone explain this a bit? Is there a good way to visualize it?
The Standard configuration;
I would prefer an intuitive (not rigorous) explanation over a maths one, but if you have a maths one then for completeness it would be good to post it.
All other notes from these lectures are available here.
| Bernhard Schutz discusses this reasonably well in his book A First Course in General Relativity.
Consider sending a light beam horizontally along an $x$ axis and then receiving it back again. A space time plot of this would look like
Here's an example of the rotation you are describing
And this is how everything becomes distorted when you create such an $xt$ plane rotation:
Note some major properties
*
*Simultaneity broken
*It shrinks moving objects (length contraction)
*Slows down moving clocks (ie time dilation)
But it works and allows $$v_{light} = c$$
See this video for a visual (it's really cool!):
http://youtu.be/C2VMO7pcWhg
And buy/rent/borrow Schutz's book. He nicely walks you through this part. The SR section is actually my favorite part in the whole book.
In matrix form,
$$\begin{bmatrix}
ct'\\
x'\\
\end{bmatrix}= \begin{bmatrix}
\cosh \theta & -\sinh \theta\\
-\sinh \theta & \cosh \theta \\
\end{bmatrix}\begin{bmatrix}
ct\\
x\\
\end{bmatrix}$$
Video Explaining Rotation Matrix:
http://youtu.be/lRnGpBJQluQ
Note:
The OP mentioned tensors. Someone might like to add a bit about the role of tensors in Lorentz transformations.
| {
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Do the same equations of motion imply the same Lagrangians? If two Lagrangian (densities) $\mathcal{L}$ give the same equations of motion, are they equivalent?
| If by 'equivalent' you mean equal, then no. They can clearly differ by a constant, but they moreover can differ by a total time derivative. So if two lagrangians $L_{1}$ and $L_{2}$ are such that $L_{1} - L_{2} = \frac{\mathrm{d} \Phi}{\mathrm{d}t}$ for some function $\Phi$, then they lead to the same equations of motion. You can find a proof of this in Jose and Saletan's Classical Dynamics: A Contemporary Approach.
EDIT: Although if I recall correctly, they only deal with Lagrangians of two freedoms (a generalized position and velocity), leaving the more general case as an exercise.
| {
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"Hollow Earth" like Theories for pedagogical porpuses I recently encountered the Hollow Earth theory, and I realize that This kind of theories can be used to give a really interesting discussion of a physical law.
For example, If we take the hypothesis that the earth is hollow and if we ignore the border effects of the pole holes and approximate it as a hollow sphere, we know that the gravitational field inside a hollow sphere is:
$$\Psi_{earth}=\begin{cases} 0 &\mbox{if } R < r_{earth} \\
\frac{mG}{R^2} & \mbox{if } R > r_{earth} \end{cases} $$
Using this we can conclude that the people living inside wouldn't feel any kind of gravitational attraction from the earth, so they would be floating.
We also can conclude that if we put a sun inside the earth the things inside would get attracted to it and there is just no way of nothing being there to start with.
Now that I've said this:
Do you know any other theories that can be used with this approach ?
*
*I think this is a really good way to teach students today, the students that are really interested in video games and in science fiction. There are two really interesting videos related to this :
Can the earth be a hollow sphere/ ¿Podría estar hueca la Tierra? (Spanish) :
The Birth Of Magic
| Due to the rotation of the Earth, "inner-Earth-dwellers" would feel a fictitious centrifugal force pointing away from the axis of rotation. Ask your students how strong that force would appear to be.
Once they realize it's a very weak force indeed, ask them to determine how fast the Earth would be spinning to give inner-Earth-dwellers the appearance of normal Earth gravity.
| {
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Angular Momentum Expectation Values in Spherical Coordinates I have a homework problem that asks:
Using the spherical harmonics calculate $\langle J_x \rangle$, $\langle J_y \rangle$, $\langle J_z \rangle$ in the state $|l,m\rangle$. Use the derivative forms of the $J_{i}$ in spherical coordinates.
I'm not looking for a direct answer, but I am trying to understand how to approach this problem. I have the representations of the $J_i$ in spherical coordinates, but how do I use these to calculate the requested expectations values?
I've asked for help for a specific part of this derivation in Mathematics stackexchange at: https://math.stackexchange.com/questions/1265825/integrating-associated-legendre-polynomials
| You have to compute $\int d\Omega\ Y_{lm}^*(\theta,\phi)\hat{J}_iY_{lm}(\theta,\phi)$ where $d\Omega=\sin\theta d\theta d\phi$, $J_i$ are the angular momenta operators represented in position space and $Y_{lm}$ are the wavefunctions for the state $|lm\rangle$, i.e. spherical harmonics.
| {
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Derivation of the energy-momentum tensor for an imperfect fluid In chapter 7 of the "Physical Foundations of Cosmology" Mukhanov uses this energy-momentum tensor for an imperfect fluid:
$$T^\mu_\nu = (\rho + p)u^\mu u_\nu - p\delta^\mu_\nu - \eta(P^\mu_\gamma u^{;\gamma}_\nu+P^\gamma_\nu u^\mu_{;\gamma}-\frac{2}{3}P^\mu_\nu u^{\gamma}_{;\gamma}) $$
where $\eta$ is the shear viscosity coefficient and $P \equiv \delta^\mu_\nu - u^\mu u_\nu$ is the projection operator.
Where does this relation come from? can you introduce some references for derivation of this energy-momentum tensor?
| Here is a sketch of where it comes from. First just consider the perfect fluid terms and note the thermodynamic relation
$$
\rho + p = \mu n + T s,
$$
where $T$ and $s$ are temperature and entropy, $\mu$ and $n$ are a chemical potential and number density.
We also have a relation for derivatives of $p$
$$
dp = n d\mu + s dT.
$$
Now if you take the divergence and dot with $u$
$$0=u^\nu \nabla_\mu T^\mu_\nu = \nabla_\mu (\rho+p)u^\mu - u^\nu\nabla_\nu p$$
Now use the thermodynamic relations and the fact that the number density is conserved and you get
$$ 0= T \nabla_\mu s u^\mu.$$
so this expresses that the entropy current is conserved in the perfect fluid.
The new viscosity terms will modify this expression but they are chosen in such a way that the divergence of the entropy current will be strictly positive in order to satisfy the second law.
You can work it out yourself if you rewrite the viscosity terms as a four index symmetric traceless tensor contracted with $\nabla_\gamma u_\delta$. After you take the divergence and dot with $u$ as above, you end up with both $\nabla_\mu u_\nu$ and $\nabla_\gamma u_\delta$ contracted into this tensor.
In general there can be more terms, for instance the bulk viscosity. The details can be found in this review paper
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Normalizing a wave function in a mixed well So I got this potential and want to solve for the even wavefunctions
Since it's symmetric around the origin I need only to look at the interval $[0,b]$ and solve for the wavefunction there. The energy is lower than $V_0$ so I will get exponentials in $[a,b]$ and sine and cosine in $[0,a]$.
\begin{cases}
A\cos(kx) + B\sin(kx), & \text{for }0 < x < a \\
Ce^{Kx} + De^{-Kx}, & \text{for }a < x < b
\end{cases}
Now I use the requirement that psi needs to be continuous at $a$, derivative contionuous at $a$, zero at the infinite wall and since I only look at half of the potential I need to add the condition that the derivative should be zero at $x=0$ for even functions of $\psi$. If I do this I get
\begin{cases}
A\cos(ka) + B\sin(ka) = Ce^{Ka} + De^{-Ka}, & \text{(Continuity at $x=a$) [1]} \\
-Ak\sin(ka) + Bk\cos(ka) = K(Ce^{Ka} - De^{-Ka}), & \text{(Derivative continuous at $x=a$) [2]} \\
Ce^{Kb} + De^{-Kb} = 0, & \text{(Wavefunction should be zero at the wall) [3]}\\
-Ak\sin(0) + Bk\cos(0) = 0, & \text{(derivative at $x=0$ should be zero) [4]}
\end{cases}
From [4] one can see that B have to be zero and from [3] I can express $C$ in terms of $D$ but here is where I get stuck. I got two equations that I can use to solve for $A$ now, but I get two different answers depending on if I use [1] or [2]
\begin{cases}
A = \frac{C(e^{Ka} - e^{-Ka+2Kb})}{\cos(ka)}, & \text{If I use [1]} \\
A = \frac{-KC*(e^{Ka} + e^{-Ka+2Kb})}{k\sin(ka)} & \text{If I use [2]}
\end{cases}
Which one am I supposed to use in the normalization? Or are they equal if you just rewrite them in some way?
| Assuming you've done the algebra correctly, these equations can be solved for a relationship between $k$ and $K$, which should lead to the quantization of energy levels in terms of $a$, $b$, and $V_o$. Then you solve for $C$ in terms of $A$ from either equation (you MUST get the same result with either) and then normalize.
| {
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Conceptual explanation of the Single particle partition function The Single particle partition function is defined mathematically as
$$\text{Z=$\sum $}g_ie^{\left(\frac{-E_i}{K_BT}\right)}$$
But what is the physical interpretation of the partition function and it's significance to Thermodynamics? I'm seeking a simple yet understandable intuition.
|
But what is the physical interpretation of the partition function and
it's significance to Thermodynamics? I'm seeking a simple yet
understandable intuition.
The partition function has one simple physical interpretation in terms of Thermodynamic functions: Its natural log is proportional to the Free Energy (the proportionality constant is the negative inverse temperature).
The Free Energy, which is well-known from Thermodynamics, is given by
$$
F=E-TS\;,
$$
where E is the Thermodynamics Energy, T is the temperature, and S is the Entropy.
From a Statistical Mechanics perspective, remember that the probability to be in a state $n$ is given by
$$
p_n\equiv \frac{e^{-E_n/T}}{Z}\;,
$$
where $Z$ is your partition function, T is the temperature, and $E_n$ is the energy of state $n$.
The statistical definition of Entropy is
$$
S=-\sum_n p_n \log(p_n)
$$
$$
=-\sum_n p_n (-E_n/T-\log(Z))
$$
$$
=E/T+\log(Z)
$$
I.e.,
$$
S-E/T=\log(Z)=-F/T
$$
| {
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What's the differences between time in Physics and time in everyday use? OK. This question might sound as not a good question, but the word 'time' is so confusing to me. I mean thermodynamics says time is the product of entropy. Relativity says time is relative. Quantum Mechanics says time doesn't exist, and that we can derive any given equation without involving time like Kepler's law. And I thought time measurement such as hour or second is just what we invented for conveniences on daily lives.
So are definition of time in Physics difference from time we associated in everyday life?
| In everyday (or casual) use "time" is often (mis-)taken to mean
*
*duration, or
*a coordinate assignment to indications or entire events ("coordinate time").
In contrast, the current, correct and careful meaning of "time" in Physics is based on Einstein's definition of "time" as "the position of the little hand of my watch"; i.e. generally as any one particular indication of any particular participant; or (also) as the entire ordered set of indications of any particular participant.
| {
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Is there a rotational equivalent to newtons laws? Newtons three laws of motion appears to apply only for linear motion:
*
*An object remains at rest or moves in a straight line at uniform velocity unless a force is applied.
*Force is mass times acceleration.
*Every action causes an equal and opposite reaction.
Is there a rotational equivalence? For example:
1'. Every body rotates around a fixed axis at uniform angular velocity unless a torque is applied
2'. Torque is Moment of Inertia times angular acceleration
3'. When one body exerts a torque on another; there is an equal and opposite torque applied on the first body by the second.
First are these actually correct; if not, what are the correct equivalence; and who formulated them?
| There is a rotational equivalence, but it is not what you stated. The problem, as pointed out by @curiousOne, is that conservation of angular momentum does NOT imply rotation about the same (fixed) axis. But I think a simple restatement like this could work:
*
*if no torque acts on a body, its angular momentum will remain unchanged
*rate of change of angular momentum is proportional to applied net torque
*when two bodies interact, the torque that A applies to B is equal and opposite to the torque that B applies to A, so that the angular momentum of the combined system (A+B) is preserved.
I believe that addresses the objections raised to your earlier version. Note that "axis of rotation is unchanged" is fundamentally different from "angular momentum is unchanged".
| {
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Correct vector space of eigenkets of angular momentum When we say an particle is in the state:
\begin{equation}
|l,m\rangle,
\end{equation}
what is the underlying state space, as a vector space? Is it a tensor product vector space, of dimension:
\begin{equation}
l\times(2l+1)\ ?
\end{equation}
How can I find the matrix representation of the angular momentum operators that act on the $2l+1$ vector space in that tensor product? I am used to angular momentum operators taking the form of a cross-product:
\begin{equation}
x_ip_j - p_ix_j,
\end{equation}
but can we still do that for the $2l+1$ dimensional space corresponding to $m$?
| My understanding of this limited, but this might help (too long for a comment):
The state space is spanned by the set of simultaneous eigenstates of the Hamiltonian, $ \hat L^2$, and $ L_z $. In fact, they form an orthonormal basis of a Hilbert space $ H $ which is the state space.
Out of convenience, we denote the eigenstates by the quantum numbers, indexing them with $ n, \ell$ and $m $ which correspond (though are not equal to) their respective eigenvalues for each of the operators.
I suppose that a state with only $\ell$ and $ m $ specified lies in the subspace of $ H $ with an orthonormal basis equal to the set of simultaneous eigenstates with those quantum numbers.
| {
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How to measure temperature of a laser cooled sample at picoKelvin temperatures? I'm reading about laser cooling.. my question is: how can the temperature of the sample be measured? (using laser cooling we can reach $10^{-12}K...)$
| I hope someone with more knowledge will pop into thread, but here is my education. There might be number of ways to measure such low temperatures. One I find fascinating is starting with material, namely Bose-Einstein condensate. Reference is this one: Cooling Bose-Einstein Condensates Below 500 Picokelvin, Leanhardt et al. Science, 12 September 2003.
Shortly, you start with notion that Bose-Einstein condensation (BEC) phase transition temperature is directly related to number of particles:
$\omega$ here is frequency of field that is used for trapping condensate. 2500 atoms in condensate carry temperature of 450 pK. Number of particles was measured using optical absorption.
| {
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Feynman rules for gauge bosons and Goldstone bosons Does anyone know where I can find:
*
*gauge boson propagators (in an unfixed gauge) for the unphysical
Electroweak gauge bosons $A^1_\mu$, $A^2_\mu$, $A^3_\mu$ and
$A^4_\mu$ whose combinations give the physical $W^+_\mu$, $W^-_\mu$,
$A_\mu$ and $Z_\mu$? (With the form of the Higgs doublet clearly
stated).
*vertex couplings for $A^1_\mu$, $A^2_\mu$, $A^3_\mu$ and $A^4_\mu$
with the Higgs
*propagator for the Higgs boson field
*propagator of the Golstone boson arising from the Higgs mechanism in
EW
*couplings of the Goldstone boson to the fields
I want to compute radiative corrections to some of the propagators of $A^1_\mu$, $A^2_\mu$, $A^3_\mu$ and $A^4_\mu$ due to the higgs boson, but I would like to do that (analytically) without deriving all feynman rules by myself.
| For anyone interested, I found a good document with all Feynman rules and different notations:
http://arxiv.org/abs/1209.6213
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Does friction act on a wheel rolling at a constant speed One of the things I've seemed to have taken for granted is that its the friction the floor exerts on a rolling wheel that prevents slip from occurring.
However, I ran into something that challenges that assumption: the contact point of the wheel on the floor has zero velocity relative to the floor, and no other lateral forces appear to exist. This implies that there is no friction. Otherwise the wheel would accelerate/decelerate, which there seems to be no mechanism for (ignoring drag, assuming no slip).
So, is it true to say no friction occurs on a wheel rolling at a constant speed? It is difficult to test as drag is difficult to isolate from the issue. No friction would imply that if the wheel rolled from a rough surface to a perfectly smooth surface, the wheel would not slip.
Is this effect analogous to a spring, such that if the wheel is overspinning, friction acts to slow the spinning down (spring extension, tensile force), if it underspins, friction acts to speed the spinning up (spring compression, compressive force), and a wheel spinning without slip experiences no friction (spring natural length, no force)?
| There might be normal friction acting on the rolling wheel, namely static friction. The static friction force, $F_f$, is often written as,
$$
F_f \leq \mu_s F_n,
$$
where $\mu_s$ is the coefficient of static friction and $F_n$ the normal force, in this case the weight of the wheel.
Note the less-than-or-equal-to sign. The magnitude of this friction force can change depending on the other forces applied to the wheel. The friction force namely prevents the wheel from slipping. If however this force exceeds the right hand side of the equation, then the wheel will start to slip. In that case you need to use the coefficient of kinetic friction, which is typically lower. The magnitude of this force does not vary from zero to an upper limit.
When a wheel is rolling without slipping there will be another kind of friction. Namely due to the deformation of the wheel and the surface at the point (or rather the area) of contact.
| {
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Sub-structure of hyperfine levels In studying introductory atomic physics I have come across fine structure splittings in energy levels due to spin-orbit coupling. Which has a sub-structure called hyperfine structure which comes from a coupling of the nuclear spin and the total angular momentum.
My question is, is there any theoretical reason (as I assume there's no experimental evidence) that there's a sub-structure to the hyperfine levels in some atoms, in certain conditions?
| As a matter of fact the list of the corrections to the hydrogen atom goes on and on. This is a list of corrections to the hydrogen atom and their order of magnitude for comparison.
*
*Bohr energy, which is very rough version of the hydrogen atom $\sim\alpha^2m_ec^2$
*Spin orbit coupling (AKA Fine structure of hydrogen) $\sim \alpha^4m_ec^2$
*Hyperfine splitting $\sim \alpha^4m_ec^2 \left( \frac{m_e}{m_p}\right)$
*Lamb shift because of the vacuum fluctuations of the electromagnetic field, $\sim \alpha^5m_ec^2$
where $\alpha$ is the fine structure constant, $m_e$ and $m_p$ mass of electron and proton respectively and $c$ the speed of light. Notice that $\alpha \approx 1/137$ and $\frac{m_e}{m_p}\approx1/2000$. This should give an idea how the corrections get smaller and smaller. However with current technology we are able to experiment with much better precession so they are also experimentally confirmed. Furthermore I also believe that there are also further corrections but I personally don't know any of it.
At the very least you can do relativistic corrections „although the mean speed of the electron in hydrogen is only 1/137th of the speed of light” (Wikipedia) and with the development of quantum gravity I believe there will be other corrections.
| {
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If the electrostatic potential is zero, why doesn't the electric field have to be zero? I thought the relation between the electrostatic field $\vec E$ and the electrostatic potential $V$ is as follows:
$$\vec E = - \nabla V$$
Thus, when $V$ is zero, $\vec E$ is also zero.
| It depends on what you mean when you say $V=0$. In the context of the equation:
$$\vec{E}=-\nabla V$$
which holds specifically in electrostatics $V$ is a scalar field, meaning that it is actually a function which assigns every point in space a scalar value. $\vec{E}$ is a vector field, which assigns a vector to every point in space. Thus, both the electric field and the potential are dependent upon position. This can be shown more explicitly as:
$$\vec{E}(\textbf{r})=-\nabla V(\textbf{r})$$
where $\textbf{r}$ is a position vector. Now, if $V(\textbf{r})=0$ for all $\textbf{r}$ then certainly the gradient is also zero everywhere, and thus, the electric field is zero everywhere.
On the other hand $V(\textbf{r})$ may equal zero for only some $\textbf{r}$. For example, at the point $P$ midway between two point charges, one with charge $+q$ and the other with charge $-q$ the potential is zero, assuming infinity as the reference point. However, if you move even slightly away from this point, the potential is non zero. The fact that the potential is changing at point $P$ indicates that the gradient at this point is non zero. Thus, the electric field at $P$ is non zero, even though the potential itself is zero at $P$.
Note that this is true for electrostatics, but, as Sebastian mentions in a comment below, it is incomplete in the context of electrodynamics. This is simply because the expression you point to relating the electric field to the potential only holds for electrostatics. For treatment of the more general case, please see Sebastian's and Alexander's fine answers.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Weight distributions If a man is standing on two weighing machines (scales), with one foot on each, Will both machines show equal weight or his weight will be distributed in two machines?
|
$N_1$ the reading of force on the first weighing machine
$N_2$ the reading of force on the second weighing machine
$X_1$ the horizontal displacement of first leg from COM
$X_2$ the horizontal displacement of second leg from COM
we know that the man is in equilibrium. so
the weights shown on both meters will have their sum equal to the force of gravity on the man.
$$N_1 + N_2 = W$$
then the forces must satisfy that the total torque on the man is zero, that is
$$ N_1X_1 = N_2X_2 $$
by solving $$ N_1 = WX_2/(X_1+X_2) $$ $$ N_2 = WX_1/(X_1+X_2)$$
the thing is : it depends. it depends on the distances that your legs spread horizontally with the center of mass.if both distances are same then the weights are equal.
| {
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Is Interpretation of state vectors and density matrices according to Frequentist or Bayesian interpretation of probability? I asked a question on math stack exchange what does probability mean. I did not know about Frequentist and Bayesian interpretation of probability previously. So according to which interpretation are the density operators and amplitude squares of state vector defined ? I am reading the introduction to quantum information from Nielsen and Chuang. For measurement operators $\{M_m \}$ such that $\sum_m M_m^{\dagger}M_m=I$ the book defines probability that the outcome is $m$ as $\langle \psi|M^{\dagger}M_m|\psi\rangle$ when measuring state $|\psi\rangle$. They also define similarly in terms of density operator.
So is all this according to Frequentist or Bayesian interpretation of probability or are both interpretations used whenever one seems more suitable ?
| It's pretty hard to be Bayesian about quantum mechanics without believing in some sort of underlying hidden-variable theory. Such theories are highly unpopular in modern culture (not to mention experimentally falsified in the majority of cases) and so the overwhelming interpretation amongst physicists is an operationalist/frequentist one.
| {
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Took a picture of my laptop screen with my iPhone. The yellowish pattern in the image look like magnetic lines. How is this possible? The pattern seems consistent with the magnetic force lines of a bar magnet.
| That looks like a Moire pattern to me. You have a camera with a grid of pixels on the imaging element and a screen with a grid of (colored) pixels. These elements don't line up exactly, so you get the odd patterns.
Try taking another image with the camera slightly twisted along the lens axis or slightly angle the lens axis to the laptop. If it's a Moire, then the image will look quite different in both cases.
And if I am understanding it right, these type of Moire patterns are
made as a result of overlapping of curved grid lines
No, they're similar to aliasing effects. Imagine taking a picture of a grid. The optics will put a picture of the grid onto the imaging sensor (CCD). At some distance/zoom, the lines of the grid will be almost exactly the same distance apart as the CCD elements. If the lines fall between the elements, they won't be seen easily. If they fall exactly on the elements, they are seen easily.
But your laptop screen isn't all the same distance from the lens. It's flat, so the edges are farther than the center is. This means the angular separation of the LCD pixels changes from the center to the edge. This apparent change in the grid separation from one part of the image to another makes the bright/dark/bright/dark areas appear. They're curved because the apparent grid size changes with distance (so you get sort of circular patterns).
Your original image looks to me to be a very lucky shot. The grid in the middle is free of much distortion over a large area. Pretty neat.
| {
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4 dimensional interpretation Has it ever been hypothetized that, in a 4 dimensional space, being time the 4th D, one body could travel through the dimensions at the combined speed of $c$?
If a body is at rest in the classical 3 dimension, it would travel through time at $c$, but if traveling at $c$ in space, it would be resting in the "time" dimension...
| As it happens, you are absolutely correct.
The velocities we encounter in everyday life are 3D velocities that are vectors defined as:
$$ \vec{v} = \left(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right) $$
In special relativity we use a 4D velocity called the four-velocity, and this is a four-vector defined as:
$$ \vec{v} = \left(c\frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau}\right) $$
where the quantity $\tau$ is called the proper time. The proper time is the time shown on a clock carried by the moving object.
But there's something funny about this four-velocity. Suppose we choose coordinates $(t, x, y, z)$ in which I am not moving. Then $dx/d\tau = dy/d\tau = dz/d\tau = 0$. But I am moving in time, at one second per second, so $dt/d\tau = 1$. In that case my four-velocity is:
$$ \vec{v} = (c, 0, 0, 0) $$
And the magnitude of my four velocity is $c$. In other words I am moving at the speed of light even when I am stationary.
In fact you can easily prove that the magnitude of the four-velocity is always $c$. I won't do that here because I suspect the maths is a bit more in depth than you want (shout if you do want the proof and I'll edit it in). But basically when you're moving the $dx/d\tau$ etc are not zero but time dilation changes $dt/d\tau$ to compensate, so the magnitude always remains $c$.
| {
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Question about ohmic conductors I'm having some trouble understanding Ohm's law. My trouble is with the different ways it is described when referring to ohmic and non-ohmic conductors. If someone can answer this question I think it will clear up my doubts. (I made up this question myself -- it's not homework!)
Which one of A and B is wrong, and why?:
A:
A non-ohmic conductor is one whose resistance changes with increasing
temperature, while an ohmic conductor is one whose resistance doesn't change
with increasing temperature.
B:
A non-ohmic conductor is one whose temperature changes with increasing
voltage, while an ohmic conductor is one whose temperature doesn't change
with increasing voltage.
| Ohm's law assumes the temperature remains constant.
An Ohmic conductor is one in which the current flowing through it is proportional to the voltage applied across it.
A non-ohmic conductor is one in which the voltage and current are not linear.
A) The resistance of most conductors increases as the temperature increases, however being ohmic and not ohmic is not the reason.
B)What causes heating in a conductor is the current flowing through a conductor. The Power = current^2 * Resistance = voltage^2 / resistance
This power is converted into heat which increases the temperature as time goes on. The actual voltage does not matter as much as the how long it has been on the conductor.
| {
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What is the magnetic flux density "outside" the solenoid when AC current is passing through it? I know that there are well defined equations explaining the magnetic flux density in the solenoid.
However what about magnetic field outside the solenoid?
How is the magnetic flux density related with the current?
UPDATE : Sorry may be the original post is misleading. I have updated some of the terms. I was wanted to ask about the magnetic flux density
To be more precise:
Consider the case where we have one solenoid placed at coordinate $(0,0,0)$ and AC current $I$ is passing through it.
At the same time at point $(x,y,z)$ we observed the magnetic flux density $B$
How can we write $B$ in terms of $d=\sqrt{(x^2+y^2+z^2)}$ and $I$?
| Approximately zero for a solenoid of infinite length. As far as the magnetic field goes, nothing changes from the situation of a direct current passing through the solenoid. The magnetic flux is homogenous inside, and the magnetic flux outside is approximately zero (it's the same magnetic field as inside the solenoid but spread out in all the space around it (to infinity), so you have nearly zero magnetic flux).
What changes inside AND outside is that the changing current causes a changing magnetic flux inside the solenoid (also outside but they are negligible for the reasons stated before). That causes an induced azimuthal electric field (its direction with respect to the current has to do with the rate of change of $I$ which is $dI/dt$).
That electric field is induced inside and outside of the solenoid.
| {
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Two objects with mass m with charge q connected to spring I am taking Physics 201 on Yale Open Courses. I having trouble with the 5th problem in the problem set 1, which says as follows:
Two sphere of mass m and negligible size are connected to two identical springs of force constant k as shown in Figure 1. The separation is a. When charged to q Coulombs each, the separation doubles. (i) What is k in terms of q,a,and ε0? (ii) Find k if the separation goes to a/2 when the charges are ±q. (iii) In case (i) suppose the charge on the right is held fixed while that on the left is displaced by a tiny amount x and released. Find the resorting force F = −kex and the (angular) frequency ω of small oscillations. (I call the effective force constant for oscillations as ke to distinguish it from the k for the springs.)
I am having trouble with question iii; I understand the how to approach the problem but after the Taylor Expansion it seems like they are dropping terms without reason ( specifically: ka/2 and the first term in the taylor expansion(1)). I tried many ways but none of them seem to work out.
Here is their solution:
\begin{align}
F_{\text{tot}} =& −k(−\frac{a}{2} + x) − \frac{1}{4\pi\varepsilon}\frac{q^2}{(2a − x)^2} \\
=& \frac{ka}{2} − kx − \frac{1}{4\pi\varepsilon}\frac{q^2}{4a^2}\frac{1}{(1-\frac{x}{2a})^2} \\
=& \frac{ka}{2}−kx− \frac{q^2}{16\pi\varepsilon a^2}(1 + 2\frac{x}{2a} +\dots) \\
=& −\left(k + \frac{q^2}{16\pi\epsilon_0 a^3} \right)x \\
≡& −k_{\text{eff}}x \\
\\
&\omega = \sqrt{(k + \frac{q^2}{16\pi\epsilon_0 a^3}) / m} \\
\end{align}
| For part (i), the separation doubles when the charges are +q each.
The force at that moment is
$$F = \frac{q^2}{4\pi \epsilon_0 (2a)^2}$$
Now they are connected to two identical springs (not shown, I imagine these go to "opposite walls") meaning that each spring will be compressed by $\frac{a}{2}$.
It follows that
$$k\frac{a}{2} = \frac{q^2}{16\pi \epsilon_0 a^2}$$
And that is why those two terms cancel...
| {
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Feasible way to cool water (ice) to temperatures around -20 degree celsius Is there an easy way to cool water to temperatures around -20 or -30 degree celsius that is economically and technologically feasible for a high school student. Can it be done using typical household equipments? Thanks :)
| Some domestic, commercially marketed deep freezers are able to reach that range of temperature. Here is one for example that advertises -20deg F, so almost -30 deg C.
It really depends also on the amount of mass you want to lower the temperature to. Whatever device to be considered must provide outward heat flow to compete with whatever heat influx the mass would be exposed to from the environment. Here's an idea. For a small amount of mass you can use endothermic chemical reactions which absorb heat. For example barium hydroxide octahydrate and ammonium thiocyanate. If you mix these two chemicals in the right stoichiometric ratio you can reach cold enough temperatures to freeze water in seconds. See the Demo Here.
But careful, barium compounds are toxic. Also not your everyday household item.
| {
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Bending moment of a cantilever beam The following procedure is here.
Consider a cantilever fixed at one end and loaded at the other one. In cartesian coordinates (if $y$ is horizontal and $x$ vertical, meaning that the load acts parallel to the $x$ axis) the equation of the curvature is:
$$\frac{\frac{d^2x}{dy^2}}{\left[1+\left(\frac{dx}{dy}\right)^2\right]^{3/2}}=\frac{M(y)}{EI},$$
where $M(y)$ is the moment of bending, $I$ the moment of inertia of the cross-sectional area of the beam, and $E$ is the Young's modulus of the beam's material.
Considering the variable $z=dx/dy$ and the lenght of the beam:
$$s(l)=\int_0^l\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy,$$
where $l$ is the projection of the beam onto the $y$ axis. Then we can write:
$$\frac{z}{\sqrt{1+z^2}}=\int_0^y\frac{M(y)}{EI}dy = G(y).$$
The bending moment is $M(y)=P(l-y)$, where $P$ is the applied force parallel to the $x$ axis. Therefore
$$\frac{ds}{dy}=\left(1-\frac{P^2}{E^2I^2\left[ly-\frac{y^2}{2}\right]^2}\right)^{-1/2}.$$
After this introduction I present my question:
I carried out an experiment of bending spaghettis by placing them horizontally with a fixed end, and loaded at the other end. The aim was to find the spaghettis' Young's modulus.
Unfortunately the theory developed for these deflections is mainly broad for small deflections (they take some approximations in the first equation I wrote). However I measured $l,P,I$ to be $0.17, 0.1, 5.75\times 10^{-12}$, in mks.
My idea is to integrate numerically the last equation so that $s(l,E)=L$, where $L$ is the length of the spaghetti: $0.2$m. This means that the integration is made over $y$, from $0$ to $l$. Then, since $l$ is known, $s$ really depends on $E$. So there must be some $E$ for which the integral is $L$. However I need a really large $E$ (~$1\times 10^9$ Pa) so that the expression in the square root is positive, and this doesn't make sense because the spaghettis shouldn't have such a big $E$.
What could be wrong with this approach?
| Generally speaking, Young's modulus is measured in terms of Gigapascals, or $10^9$ Pascals, or at least Megapascals, or $10^6$ Pascals. Hence, your answer should be fine. Here is a table with Young's modulus values for day to day materials. Notice how $E$ is measured in Gigapascals.
| {
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Is it possible to write the fermionic quantum harmonic oscillator using $P$ and $X$? The Hamiltonian of the quantum harmonic oscillator is
$$\mathcal{H}=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2$$
and we can define creation and annihilation operators
$$b=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}b^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(X-\frac{i}{\omega}P)$$
where the following commutation relations are fulfilled
$$[X,P]=i\hbar\qquad{}[b,b^{\dagger}]=1$$
and the Hamoltonian can be written
$$\cal{H}=\hbar\omega\left(b^{\dagger}b+\frac{1}{2}\right).$$
Now, it is also known that we can define a fermionic quantum harmonic oscillator with the Hamiltonian
$$\cal{H}=\hbar\omega\left(f^{\dagger}f-\frac{1}{2}\right)$$
where $f$ and $f^{\dagger}$ satisty the following anticommutation relation
$$\{f,f^{\dagger}\}=1.$$
What I am trying to get is a Hamiltonian for the fermionic harmonic oscillator using $P$ and $X$. I have tried defining
$$f=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}f^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(-X-\frac{i}{\omega}P)$$
because after imposing the anticommutation relation $\{X,P\}=i\hbar$ for $X$ and $P$ (as I guess would suit a fermionic system) these definitions of $f$ and $f^{\dagger}$ imply $\{f,f^{\dagger}\}=1$. Nonetheless, for the Hamiltonian I get
$$\mathcal{H}=\frac{P^2}{2m}-\frac{1}{2}m\omega^2X^2$$
where I get an undesired minus sign. My question is then the following: is it possible (with an appropriate definition of $f$ and $f^{\dagger}$ in terms of $X$ and $P$) to obtain the first hamiltonian I have written from the fermionic oscillator Hamiltonian written in terms of $f$ and $f^{\dagger}$?
| Fermions are strange beasts in many ways. The first problem you will encounter, and which will make it impossible to write an harmonic oscillator for fermions is the following:
The fermion ladder operators $f$ and $f^\dagger$ require that $\{f,f^\dagger\}=1$. Translated to $X$ and $P$ this means that $\{X,P\}=i\hbar$. But is also means that $\{X,X\}=0$ and $\{P,P\}=0$ since they are now fermionic operators. As a result the Hamiltonian can at most have bilinear terms in $X$ and $P$.
Especially the terms $X^2$ and $P^2$ are forbidden, so no "Harmonic oscillator"-style Hamiltonian exists.
| {
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A general relativity question about the Einstein equations? Assuming a Robertson-Walker metric to describe homogeneous and isotropic cosmological models, Einstein equations with cosmological constant reduce to these 3 non-linear ordinary differential equations for a perfect fluid:
\begin{align}
\dot{\rho} &= -3H(\rho + P) \tag{1} \\
\dot{H} &= -H^2-4\pi G(\rho + 3P)/3 + \lambda/3 \tag{2} \\
H^2 &= 8\pi G\rho/3-K/a^2+\lambda/3 \tag{3}
\end{align}
Here, dot represent the derivative with respect to time and the function $H=H(t)= \dot{a}/a$ where $a = a(t)$ is the scale factor.
I know that (1) is the conservation of energy equation but i can't understand what (2) and (3) represent?
| If what you want a gut-feeling of what these equations mean, I can share mine.
Equation (2) is a consequence of the two others. Take the time-derivative of (3), remembering that K and $\lambda$ are constants and $H=\dot a/a$, combine this derivative with (1) and (3) and you get (2). So no gut-feeling for that one, just trivial algebra.
Now gut-feeling of (1):
The mass volumic density can change in two ways: either by changing the volume for the same amount of mass (term $-3H\rho$ ) or creating or losing energy (which is the same as mass, of course) because of positive or negative work of the pressure (term $-3HP$). OK, you might not be really convinced, but think it over. If you ignore pressure, then the mass in a comoving volume is strictly constant, and (1) for zero pressure just tells you that the mass volumic density behaves like the inverse cube of $a$.
Finally, equation (3).
Multiply by $a^2$. Then the right-hand-side is $\dot a^2$, kinetic energy per unit mass. Second term, if we had $\rho a^3$ that would be total mass. But we just have $\rho a^2$, so this is mass $M$ divided by radius $a$. $(8\pi GM/3)/a$ opposite of the gravitational potential energy per unit mass. Now $K$ is a constant, total energy. For $\lambda=0$ you have kinetic energy as the total energy minus (negative) potential energy, An intuitive feeling for $\lambda a^2$.... that's another story.
I hope that helps... This is just gut-feeling, it would need a lot of work to make is precise. But his is the way I look at them.
| {
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Does Time change over temperature? I am not a physicist, I am just an engineer.
But I dared to ask whether the temperature changes the perception of time.
Let's consider a particle that "stops" at absolute zero. I was thinking as a hypothesis, that our perception of time changes and the particle actually does not stop at all.
| If everything were at zero temperature you probably* would not be able to distinguish between the past and future. Mathematically time would still exist, in the same way that spatial directions still exist on a completely featureless plane, but since it would not be measurable (even if there were something around that could do a measurement!), it is debatable whether it could be considered physically meaningful.
However, this has nothing to do with being at zero temperature itself. The world would also be time-symmetric if it were in equilibrium at a finite temperature. It is the fact that you are at equilibrium that makes time become unimportant, not at which temperature this equilibrium occurs. There is a fascinating history of theories about the relation between being out of equilibrium and time's arrow, summarized up for example in the popular account From Eternity to Here by Sean Carroll.
*It's not completely clear that this is true, actually. Google around for 'time crystals,' a proposal by Nobel Laureate Frank Wilczek. I'm not sure of the current status of this proposal.
| {
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Phase added on reflection at a beam splitter? If we have light of a particular phase that is incident on a beam splitter, I assume the transmitted beam undergoes no phase change. But I thought that the reflected beam would undergo a phase change of $\pi$. I have, however, read that it undergoes a phase change of $\pi/2$.
Which is it, and why?
| https://arxiv.org/abs/1509.00393
The answer is $\pi/2$. Interestingly, it doesn't matter if it is plus or minus $\pi/2$ or if the phase change is in transmission or reflection in analysis of quantum or classical interferometry.
From the above referenced article:
"Quantum optics essentially provides black-box models of the beamsplitter. They all agree on the existence of a π/2 phase-shift, even if its sign and precise location (on the transmitted or reflected beams) are uncertain. Such inconsistencies, however, are not critical for what concerns the respect of energy conservation principle."
| {
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How can tangential acceleration from a radial force be explained? A mass is attached to a rope, and put into a circular motion. If I pull the string from the center, the tangential speed of the mass will increase (by conservation of angular momentum).
I am applying a force only in the radial direction, so how can the tangential velocity increase if there is no tangential force?
| No. The easiest way to see this without invoking rotating reference frames is to write out Newton's Law's in polar coordinates, which work out to be:
\begin{align*}
F_r &= m \ddot{r} - m r \dot{\phi}^2 \\
F_\phi &= m r \ddot{\phi} + 2 m \dot{r} \dot{\phi}
\end{align*}
From these, it's pretty easy to see that if we have $F_\phi = 0$ and $\dot{r} \neq 0$ (i.e., no tangential forces but a decreasing radius), we will still get $\ddot{\phi} = \dot{\omega} \neq 0$.
The question you might ask, then, is why do Newton's Laws look so strange in polar coordinates? Why are those extra terms present on the right-hand side of the equations? The answer is that the "real" equation is that $\vec{F} = m \ddot{\vec{r}}$. If we express $\vec{r}$ in, say, Cartesian components ($\vec{r} = x \hat{x} + y \hat{y}$), then the vectors $\hat{x}$ and $\hat{y}$ are constant with respect to time; thus, $\ddot{r} = \ddot{x} \hat{x} + \ddot{y} \hat{y}$. However, if express the particle's position in polar coordinates ($\vec{r} = r \hat{r}$), then $\hat{r}$ is not constant with respect to time; so when we take the derivative of $\vec{r}$ with respect to time, we get unexpected terms. See the above link for the full derivation.
| {
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What is the curve described by the water of a wet spinning tennis ball? I'm looking at this picture
from this site and I'm curious about what is the curve described by the water. The involutes
$$x=r\left[\cos(\theta+n)+\theta\sin(\theta+n)\right]\\
y=r\left[\sin(\theta+n)-\theta\cos(\theta+n)\right]$$ for different $n$ seem to make it right at least near the ball:
But I'd like to know precisely the idea of how the curve develops. What I'm thinking is that the water leaves off the ball about the normal at each point because of the adhesion to it and then it should just spread 'freely'. Can the evolution of the water be described easily or has it been done accurately before?
| I think you are right. (Involute spiral: Wikipedia)
If you take a string wound around a stationary ball and unwind it, its end traces an involute spiral.
Similarly, if the ball is rotating and the string runs out in one direction, it is the same curve with respect to the ball.
A drop of water leaving the surface of the ball should travel in a straight line out from the center of the ball, at a velocity equal to its tangential velocity when it was on the surface.
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Density of states of 3D harmonic oscillator Consider the following passage, via this image:
5.3.1 Density of states
Almost all of the spin-polarized fermionic atoms that have been cooled to ultralow temperatures have been trapped by magnetic fields or focused laser beams. The confining potentials are generally 3D harmonic traps. So let's consider this case in more detail. You might be interested to note that Fermi's original paper on fermionic particles considered this case, not the 3D box case above. As we saw previously, ignoring the zero-point energy in each dimension the eigenvalues (accessible energy states) are given by $\epsilon(n_x, n_y, n_z)=n_x\hbar\omega_x + n_y\hbar\omega_y + n_z\hbar\omega_z$. In order to evaluate the various integrals, we first need to obtain the density of states per unit energy. A rough way to do this is to simply set $k_i=n_i$, so that
$$\epsilon^2 = k_x^2(\hbar\omega_x)^2 + k_y^2(\hbar\omega_y)^2 + k_z(\hbar\omega_z)^2 \equiv k^2(\hbar \overline\omega)^2,$$
where $\overline \omega = (\omega_x\omega_y\omega_z)^{1/3}$ is the mean frequency, and $dk_i/\epsilon_i=1/\hbar\overline \omega$. Because $k_i=n_i$ now rather than $k_i=\pi n_i/L$, th 3D density of states is given by
$$g(\epsilon) = \frac{k^2}{2} \frac{dk}{d\epsilon} = \frac{\epsilon^2}{2(\hbar\overline\omega)^3}.$$
for the first displayed equation,
shouldn't be $\epsilon^2 =\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 + 2\epsilon_{n_x}\epsilon_{n_y} + 2 \epsilon_{n_x}\epsilon_{n_z} + 2\epsilon_{n_y}\epsilon_{n_z}$..?
if I assume $\omega_i=\omega$ for $i=x,y,z$
by
$\epsilon_{n_x}=\hbar \omega n_x $
$\epsilon_{n_y}=\hbar \omega n_y $
$\epsilon_{n_z}=\hbar \omega n_z $
$\epsilon_{n_x,n_y,n_z}=\hbar \omega(n_x +n_y +n_z)$
let $\vec{k}=(k_x,k_y,k_z)$ where $k_i=n_i$
$$\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 = \hbar^2 \omega^2 (k_x^2 + k_y^2 +k_z^2 ) = \hbar^2 \omega^2 k^2 \not=\epsilon^2~?$$
And for second displayed equation, why it's not $$\frac{\pi k^2}{2} = \frac{1}{8}4\pi k^2~?$$
| we only need to separate the energies
$$\epsilon_i = \hbar \omega_i n_i$$
then
$$\epsilon^2 =\epsilon_x^2 + \epsilon_y^2 + \epsilon_z^2 = \left(\hbar \omega_x n_x\right)^2 + \left(\hbar \omega_y n_y\right)^2 + \left(\hbar \omega_z n_z\right)^2$$.
we can make it better into
$$\epsilon^2 = \hbar^2 \left(\omega_x^2 n_x^2 + \omega_y^2n_y^2+\omega_z^2n_z^2\right)$$
yeph. same with the answer
| {
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"url": "https://physics.stackexchange.com/questions/185501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does the water bottle not rotate when it is half full? Consider this water bottle:
When it is full and thrown up in the air, it rotates at a constant velocity.
When it is less than 1/8th full, the water bottle rotates even faster than when it was full.
When it is half full, however, the water bottle rotates for one half-spin, and then it stops rotating.
Why is this?
By the way, I tested it with a 16.9oz bottle, but the bottles are mathematically similar.
| i dont know if anyone cares yet, but i had this topic more or less at a physicist tournament:
The thing is: once u start rotating the bottle (for the somersault) the center of mass lays a few cm (depending on the bottle) under the waterlevel...therefore the water above (we know it cuz rotations happen aroung the center of mass), it will start sloshing/climbing up the bottle because of centrifugalforces. So then the perpendicular radius increases to the pivot and so does the moment of inertia (I = m*r^2)
so if L (angular momentum stays constant because of newtons law of conservation of momentum) L = I * omega
if L stays constant and I increases, our omega which is the angular velocity, decreases... Thats the same when ice skaters do their rotations and spread out their arms to go slower or pull their arms in to go faster
| {
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"url": "https://physics.stackexchange.com/questions/185590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Electron-Positron Annihilation: How is charge conserved at the verticies? How is reaction possible? The verticies do not conserve charge. Also, why is the arrow for the positron pointing downwards when as time increases, the positron should move towards its vertex?
Sorry, I just a bit all confused about this diagram and how it's formed
| It is the way one reads/writes Feynman diagrams, a particle going backwards in time is the antiparticle. The electron radiates a gamma, and continues to meet the positron , annihilating charge with another photon. Two real particles are needed for momentum conservation in the center of mass, and two photon vertices are the simplest case within the standard model.
| {
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The central density of a nucleus remains roughly constant? Let A be the atomic number
The density of a nucleus is computed using the ratio of the number of protons and neutrons, A, to the volume of the nucleus (which at my current level, is assumed to be a sphere)
The math is trivial.
Conceptually, the argument for the central density of a nucleus remaining roughly constant stems from the suggestion (from my text) that each nucleons feels a force only from it's nearest neighbour. As more nucleons are added to the nucleus, the central density increases but still lies within a range sufficient to be classified as being 'constant'.
This does not convince me. Is it necessarily true that a nucleon experiences constant force only from it's nearest neighbour?
Suppose there exists a finite number number of nucleons, n, in the nucleus. Each nucleon then experiences a force from (n-1) number of nucleons.
The other nucleon experiences also a force from (n-1) number of nucleons. By induction, and since these particles are indistinguishable-protons and neutrons are fermions- is it also not true then that every $$nucleon_i$$ experiences a constant force?
Could someone fill me in conceptually?
| If you were to imagine each nucleon as a hard sphere, then the packed volume of such a sphere will be proportional to the number of nucleons and the sphere will have the same mean density, regardless of the number of nucleons.
Think about the forces experienced by one of these "hard spheres". It feels a strong force from each of its nearest neighbours, but nucleons that are not in "contact" with it exert no influence at all.
This analogy is reasonably appropriate for nucleons bound together, because the strong nuclear force has a short range and becomes very repulsive at small separations. In other words, if we double the equilibrium separation between a pair of nucleons, the strong force is still attractive but so small it can be neglected, whereas if we halve that distance the force become very strongly repulsive.
| {
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"timestamp": "2023-03-29T00:00:00",
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Physical reason behind having greater amplitude when driving frequency$ < $ natural frequency than that when driving frequency $>$ natural frequency This is quoted from A.P.French's Vibrations & Waves:
If the driving force is of low frequency relative to the natural frequency, we would expect the particle to move essentially with the driving force. This is equivalent in saying that $m\dfrac{d^2x}{dt^2}$ in $$m\dfrac{d^2x}{dt^2} + kx = F_0\cos\omega t$$ plays small role compared to the term $kx$. The amplitude is controlled by spring constant.
On the other hand, at frequencies very large compared to the natural frequency, the opposite situation holds. The $kx$ becomes small compared to $m\dfrac{d^2x}{dt^2}$. In this case, we expect a relatively smaller amplitude of oscillation than the above case.
$\bullet$ Why does $kx$ become prominent in the first case & not in the second case?
$\bullet$ What is the physical reason behind that the first case has greater amplitude & the second one smaller?
| This is just a footnote to Name's answer (which you should accept because it's correct) to give a slightly more intuition based argument.
If the driving force changes slowly compared to the natural frequency of the system then the system can move fast enough to stay in phase with the driving force. So most of the time the system is already moving in the direction the driving force is pushing it, and the force will accelerate the motion so the resulting amplitude of the oscillation will be big.
If the frequency of the driving force is a lot higher than the natural frequency of the system then the system cannot move fast enough to stay in phase with the driving force. This means some of the time the driving force is acting opposite to direction the system is moving, so it's slowing the motion not accelerating it. This means the amplitude of the motion will be reduced.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why do electric sparks appear blue/purple? Electric sparks tend to appear blue or purple or white in color. Why?
| The answer is that electrical excitation of air molecules is able to produce lots of excited singly ionised nitrogen ions.
The electronic structure of singly ionised nitrogen has a number of allowed radiative transitions, where the outer excited valence electrons can rearrange themselves into lower energy configurations.
The most prominent turn out to be those transitions corresponding to emitted photons at 443, 445 and 463 nm, and it is these that are responsible for the blue airglow.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
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How much is if my internet provider says "we have increased your WiFi connection by 2 decibels"? I understand the definition of decibel as a logarithmic unit for intensity of anything. I just cannot imagine its "actual size" when comes to the wireless connection.
Can you provide something illustrative which would help me to really understand how much it is if my internet provider says
"We have increased your WiFi connection by 2 decibels"?
Is it a significant improvement, or is it just a trifle?
Something like some kind of a table of values (in decibels) saying that the WiFi connection is poor/fair/good/excellent withing these intervals, or anything else practical.
I mean the question like a request for explanation of the practical perception of a particular physical unit. Like if I asked "How do we perceive increasing light wavelength by 100 nm, is it a completely different color or is it almost the same?"
ADDED:
Some asked in comments what exactly they did. It is a small square-like antenna. They physically came to my house and turned the antenna several degrees north :) to better aim their access point transmitter. And I live in a rural area.
| Let me assume your internet speed is the power of the router, divided by the square of the distance from the router (as the signal is spread over the surface of a sphere with area $4\pi r^2$).
$$
\text{speed} \propto \frac{P}{r^2}
$$
Your power increased by $2\,\text{db}$ - this means a fractional increase of $(10^{1/10})^2 \approx 1.6$. With your new router you can achieve the same speed at distance $r^\prime$ as you could with the old router at a distance $r$. To find the relation between $r$ and $r^\prime$, note that
$$
\frac{P^\prime}{P} = (10^{1/10})^2 = \frac{r^{\prime 2}}{r^2}
$$
which implies
$$
r^\prime = 10^{1/10} r \approx 1.26 r
$$
so you might get a $25\%$ increase in range, which isn't bad.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quality Factor in a Parallel LC Circuit I was wondering if there is a notion of a quality factor in a parallel LC circuit, since there is no resistance.
One can show that this circuit has a resonance frequency as follows:
Impedance:
\begin{equation}
\frac{1}{Z_{parallel}}=\frac{1}{Z_{C}}+\frac{1}{Z_{L}}=\frac{1}{\frac{1}{i\omega C}}+\frac{1}{i\omega L}=i\omega C+\frac{1}{i\omega L}
\end{equation}
\begin{equation}
Z_{parallel}=\frac{1}{i\omega C+\frac{1}{i\omega L}}=\frac{i\omega L}{1-\omega^{2}LC}
\end{equation}
Resonance frequency:
\begin{equation}
\omega_{0}=\frac{1}{\sqrt{LC}}
\end{equation}
How about the quality factor? For a RLC circuit it would be:
\begin{equation}
Q=\omega_{0}RC=\frac{RC}{\sqrt{LC}}=\frac{R}{\sqrt{\frac{L}{C}}}
\end{equation}
\begin{equation}
Z_{0}=\sqrt{\frac{L}{C}}\Rightarrow Q=\frac{R}{Z_{0}}
\end{equation}
But since there is no resistance, how does one think about the quality factor in this circuit?
|
since there is no resistance.
That's not quite correct; in fact, there is infinite parallel resistance or, better, zero parallel conductance.
Recall that, for a parallel RLC circuit, the circuit elements are parallel connected. If the parallel resistance were zero, the Q would be zero since the resistance is effectively an ideal wire shunt across the L and C.
For the parallel RLC, the greater the (parallel) resistance, the greater the Q. In the limit as the resistance goes to infinity, there is simply a parallel LC circuit for which the Q is 'infinite'.
In other words, there is no dissipation and, at the resonance frequency, the parallel LC appears as an 'infinite' impedance (open circuit).
(The above assumes ideal circuit elements - any physical LC circuit has finite Q).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Bifurcate Killing Horizon of the Ads-Schwarzschild solution I need your help.
I'm considering the AdS-Schwarzschild solution:
$ds^2=V(r)dt^2 + \frac{1}{V(r)}dr^2+r^2dΩ^2$
where $V(r)=1-\frac{2m}{r}+\frac{r^2}{l^2}$ with $m$ the mass of the black hole and $l$ the radius of the AdS spacetime.
Now, given $r=r_+$ one of the roots of the function $V(r)$, I've already shown that $r=r_+$ is a Killing horizon for the Killing vector field $\xi=\partial_t$. What I want to prove also is that $r=r_+$ is a bifurcate Killing horizon.
Is there any simple way to do that?
Thanks :)
| A bifurcate Killing horizon is a killing horizon in which the past and future horizons intersect at a point in the Penrose diagram, which is the bifurcation two-sphere. At this point, the generator of the horizon vanishes, which is to say $\xi = 0$ as opposed to being simply null.
In order to show that this is the case in any particular spacetime, you need coordinates which are regular at this intersection point. Here $t$ is ill-defined as a coordinate on the future horizon, so these coordinates are no good. You'll need Kruskal coordinates in order to evaluate this directly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Two people are holding either end of a couch, is one person exerting more force than the other? I was carrying a couch with my flatmate yesterday, and I started thinking about this.
Often when carrying objects like this, one person will be taller and has thus lifted his end of the couch higher than the other person's. Additionally, one can intentionally lift his end of the couch higher, for comfort.
My question is, in situation is one person carrying more weight than the other?
| I believe the following diagram tells you everything you need to know: when the center of mass is above the support point, then the lower point will carry more of the weight since it is acting closer to the center of mass ($x_1 < x_2$) and torque balance requires that $F_1 x_1 = F_2 x_2$. Conversely, if the center of mass is below the support, the higher point will carry more of the weight when the object is tilted.
Understanding this diagram really helps when you are moving large objects up or down the stairs: the lower person should grab the object higher up (say the arms of the couch) if you want to achieve more equal weight distribution.
ADDENDUM
In the second case, if the person on the right pushes against the couch, it is possible for the person on the left to end up carrying the entire weight of the couch. In the case of carrying things up stairs that can easily happen... Which is why movers often use webbing straps to permit/ensure a more equal distribution of effort (and to allow lifting with shoulders/back/knees and not just arms).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/186869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
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What is the physical interpretation of second quantization? One way that second quantization is motivated in an introductory text (QFT, Schwartz) is:
*
*The general solution to a Lorentz-invariant field equation is an integral over plane waves (Fourier decomposition of the field).
*Each term of the plane wave satisfies the harmonic oscillator equation.
*Therefore, each Fourier component is interpreted as a harmonic oscillator in ordinary QM
*The $n$'th energy level of each Fourier component is now interpreted as $n$ particles.
Everything in 1-3 looks like a sensible application of ordinary QM to a field. But how did 4 come about? What is the justification?
| The term is practical, but in reality, there is only one quantization.
This was to stretch it a little, as I then consider the Klein-Gordon equation as classical. That is motivated by that I consider Maxwell's equations as classical. Both equations are on the same level and so I consider Klein-Gordon to be classical. The classical normal mode vibrations of Maxwell or Klein-Gordon fields are standing waves. The individual vibrations are quantized as one-dimensional harmonic oscillators.
The second quantization of fermions is only a way to manage the antisymmetry condition. Here the first quantization (as given by Heisenberg, Schrödinger, and Dirac) is the real quantization.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 3,
"answer_id": 2
} |
Heating suprecritical water in a closed water tank How does the pressure of water in a closed tank evolve in the following setting:
- closed tank of 2 liters (filled up with water)
- water initially at 25°C and pressurized to 3 bars
The water is now heated up to 130°C, thus remaining a fluid (based on water property tables). The specific volume of water increases roughly by 6% due to the temperature increase. Since the water is in a closed tank, it cannot expand, thus the pressure needs to increase.
Which equation will allow me to calculate the resulting pressure in the water tank?
Thanks in advance for any hints and help
| Fluid reserved in a fixed volume and with no inlet and outlet of mass variations are represented well by a parameter called internal energy, although, it does not variates expressive by the pressure, but it does from the temperature. It is an important parameter to systems like yours.
I usualy use a software that does the calculations for me, so i have good precision and i dont need to keep consulting the termodinamical tables. https://www.irc.wisc.edu/properties/
As if you know the variation of the specific volume, you can enter the 25C and 3bar for pressure and get the spec. volume. Then, use your variation of 6% to increment it... now you can enter the 130C as temperature and the specific volume to calculate the pressure you want :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/187201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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